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jontwo13
07-27-2006, 08:39 PM
We have been trying to find an answer can we run a--- (single phase)230 volt submersible pump on (single phase) 208 volt. Every person I ask says you can't run it on 208 3 phase. That is not what I asked them.

QSIMDO
07-27-2006, 08:44 PM
I asked a similar question once and was told it's all the same.

....?

If it's all the same then why not pick one and be done?

Anyway, result is the pump works just fine.

Arcane
07-27-2006, 09:42 PM
Short answer is yes, it will run. If the nametag does not specifically state it is a dual voltage motor(208/230 V), chances are it will burn out when you try to actually pump water with it.

Evan
07-27-2006, 10:03 PM
Every person I ask says you can't run it on 208 3 phase. That is not what I asked them.

Actually, that is what you asked. Single phase 208 is what you get from line to line on a three phase service. To run properly the motor must be 200 volt rated. If it isn't it will draw excessive current and overheat.

Wareagle
07-28-2006, 12:00 AM
Evan is on the mark. Ohm's law states that for a given load (wattage) if the voltage goes down, the amperage goes up, and vise versa. The motor you have (240 volt) would be running with 208 volts at the best case. Factor in the footage that the circuit is, and there will be some voltage drop in addition to using 208V. The fact of the voltage being low would mean the windings would carry more current than it is designed for to perform the work, and would shorten the life of the motor considerably; if not burn it up in short order.

Your options in this case would be to add a transformer to bring your voltage up to the 240V that you need, find a 208V rated motor, or hook it up and see how long it lasts (with the understanding that it may not live very long).

J Tiers
07-28-2006, 12:08 AM
Then again, if you don't ask for max gpm at max lift, it may last a long time.

Just idling at 208 won't hurt it at all, it is the extra current under power that kills it by heating.

Cutting current in half reduces heating by 4X, so even a relatively small amount below full power would have significantly less heating and might make up the 10% difference in voltage.

At 70% power, heating is reduced to half. At 90% power, heating is reduced 20%.

CCWKen
07-28-2006, 12:17 AM
LOL... When does a water pump idle? :D http://img.photobucket.com/albums/0903/CCWKen/Personal/icon_lol.gif

RobDee
07-28-2006, 10:14 AM
It idles in a shallow application like a spring as opposed to a deep well. It runs at reduced load when it is designed to run to a specific depth, say 200' and it is only 100' down the well.

Evan
07-28-2006, 11:06 AM
Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.

Example: block the intake of your vacuum. The motor speeds up.

JCHannum
07-28-2006, 11:56 AM
The power requirements of a pump, whether loaded or unloaded will depend on the design of the pump.

Positive displacement pumps aside, rotary pumps can have several different designs, a turbine pump will behave quite differently than a centrifugal. A centrifugal with an open face impeller will behave differently than one with a closed face. Impellers can be designed to be overloading or non overloading, and can be trimmed to use only a certain amount of power. There are many more variations that will influence the performance of a pump.

Depending on the application and pump design, the motor may or may not hold up. The best source of information would be to contact the manufacturer or supplier. More information about the performance characteristics of the pump are needed to give good advice.

Generally speaking, running a motor that far below the nameplate voltage will result in it's destruction unless load is reduced. If a clamp on ammmeter is available, the load can be evaluated, and if actual amperage is less than the nameplate by a factor approaching or exceeding the difference in voltage, it will possibly survive.

Evan
07-28-2006, 12:08 PM
In this case we are talking about a submersible well pump. They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.

RobDee
07-28-2006, 12:51 PM
Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.

The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.

Example: block the intake of your vacuum. The motor speeds up.

You’re confusing velocity with current load. A motor with no load will turn faster then one with a load. The loaded motor, spinning slower will have greater current required to turn the shaft regardless of how fast it spins. This is torque (newton/ meters). The greater the torque the higher the current or work required to do it.

The current load increases with respect to the amount of work it is doing. Not the speed it attains.

Joule = 1 watt/second.

Velocity does not enter into the equation.

The vacuum intake blocked creates a greater load and therefore a higher current draw. It does not speed up, it slows down regardless of the increased volume of the motor.

As for the well pump, I agree with JC.

Evan
07-28-2006, 01:09 PM
The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.


I believe you are the person confused. The lower the output pressure the greater the flow and therefore the greater the mass moved.

The pressure of the outlet of a pump is determined by the degree of resistance to flow. As the resistance goes up the flow rate goes down as does the amount of mass moved while pressure goes up.



You’re confusing velocity with current load. A motor with no load will turn faster then one with a load.

Precisely. No confusion there either. The fact that the motor speeds up is an indication of reduced load which proves my point.


The vacuum intake blocked creates a greater load and therefore a higher current draw. It does not speed up, it slows down regardless of the increased volume of the motor.
No it doesn't. Try it.

BTW, that last quote of what you said doesn't really make sense.

RobDee
07-28-2006, 01:12 PM
In this case we are talking about a submersible well pump. They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.

Anything that does work must adhere to laws of physics. The greater the force the greater the work. The less the force the less work being done. Period.

A motor that is moving a specific volume of water does more work then a motor that moves less water. Period.

The more work a motor does the more current it requires to do that work. Period.

We can’t get something for nothing as much as we would like to!

Evan
07-28-2006, 01:23 PM
Anything that does work must adhere to laws of physics. The greater the force the greater the work. The less the force the less work being done. Period.

A motor that is moving a specific volume of water does more work then a motor that moves less water. Period.

The more work a motor does the more current it requires to do that work. Period.

We can’t get something for nothing as much as we would like to!

So? Pressure isn't work.

You don't know how pumps work. If a deep well pump is set deep enough that it is working to it's maximum pressure, say 200 psi, then no flow at all happens. All that happens is that it stirs the water. The only work it does is to heat the water it is stirring. The lower the head the more work it does and the higher the current.

JCHannum
07-28-2006, 01:39 PM
A submersible well pump merely indicates a pump that is totally submerged in the well. They can take on as many configurations as any other pump, and can be centrifugal or turbine, high head, low head, high flow, low flow or a combination depending on the requirements of the application.

The work performed by a rotary pump is a product of the volume moved and the pressure developed. The design characteristics of the pump will determine it's performance against a closed discharge.

In some cases as flow and cooling effect decreases, dangerous pressures can be generated in the pump head from the energy input. I had a 6" X 6" high volume high pressure irrigation pump ultimately rupture from the steam pressure generated under these conditions. It was inputting 100 HP to a volume of something less than 20 gallons. It did not take long for problems to arise.

Again, unless more information is given in the form of an actual pump curve, and pumping conditions, an acurate answer cannot be provided.

A vacuum cleaner is not a well pump, or even a water pump, but a high volume air handler. The vacuum cleaner works not on vacuum but volume and velocity, and entrains the material in the airstream and drops it out in the tank where the velocity drops. Shutting off the suction removes the medium being moved and the power requirement drops. Shut off the discharge and the motor will slow. If the suction of a true vacuum pump is shut off, power requirements will rise and the motor will slow.

J Tiers
07-28-2006, 01:47 PM
If you block the vacuum cleaner input, you take OFF the load, it is just stirring air, and it speeds up.

If you block the OUTPUT, it may or may not speed up, depending on how it is set up.

Generally when you reduce the product of mass flow and back pressure power intake goes down. That is pretty much the definition of power.......

A blocked vacuum has zero mass flow.........

In the case of a pump, there are TWO key specs covering this.....

One is zero pressure max flow. All pumps have one, it is just a matter of what it is. Zero pressure is really "minimum head".

The other is the max back pressure at which a certain flow is maintained.

At the max back pressure, where the flow is back pressure limited, you have a max power situation. Max head, max available flow.

At zero pressure max flow, the power is usually less than full power, unless the outlet is restricted inside somewhere, instead of simply being a matter of the displacement vs rpm, inlet opening, etc, etc. If the inlet can handle any flow at all, then the minimum head situation may be closer to max power.

In between, you may have various flow rates and power draws.

After that, you have the question : "is it a demand pump, or is it a continuous flow?".

A demand pump has a duty cycle, and will heat less if not run as much.

A continuous flow pump has to be designed to operate all the time and cooling etc designed accordingly.

I see JCH typed faster than I did........

RobDee
07-28-2006, 01:50 PM
Rob
The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.

I believe you are the person confused. The lower the output pressure the greater the flow and therefore the greater the mass moved.

Think about this: What creates the pressure? It’s the volume of water! The greater the volume the greater the pressure. Less water being moved = lower pressure = less work = lower current.

The pressure of the outlet of a pump is determined by the degree of resistance to flow. As the resistance goes up the flow rate goes down as does the amount of mass moved while pressure goes up.

As the resistance to movement goes up the torque goes up! How much water movement we have at this point is moot.
We have a motor spinning at 1000 rpms. We take our hand and slow down the motor. What happens to the current? It goes up proportionally to the torque (work) required to keep the motor turning.


You’re confusing velocity with current load. A motor with no load will turn faster then one with a load.


Precisely. No confusion there either. The fact that the motor speeds up is an indication of reduced load which proves my point.

You’re contradicting yourself.
Here’s a quote from you:

“Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

How can this happen? It can’t !
Is pressure equivalent to load? Yes, absolutely. If we have a pump with a one inch pipe and a one inch head does it take more force to move it then a one inch pipe with six inches of head on the same pump.
No!
Which one does more work? The 6 inch long pipe. Which one has a greater pressure? The six inch long pipe.

The vacuum intake blocked creates a greater load and therefore a higher current draw. It does not speed up, it slows down regardless of the increased volume of the motor.

No it doesn't. Try it.

Ok, Take a vacuum, turn it on with no obstructions. Now stick it in a bucket of water. Is more work being done? Yes. Is the torque greater? Yes. Joules = torque, as the torque goes up so does the amount of work.

Joules = watt/seconds = power. Power = EI. As the power goes up the current (I) goes up.

Evan
07-28-2006, 01:53 PM
A vacuum cleaner is not a well pump, or even a water pump, but a high volume air handler. The vacuum cleaner works not on vacuum but volume and velocity, and entrains the material in the airstream and drops it out in the tank where the velocity drops. Shutting off the suction removes the medium being moved and the power requirement drops. Shut off the discharge and the motor will slow.
No it doesn't. Try it. It speeds up.

A vacuum is a fluid handler. At subsonic flow velocities air acts as incompressible fluid and all the fluid flow equations apply. The centrifugal pump in a vacuum works in exactly the same manner as the centrifugal impeller in a water pump and all the same math applies.

The same math is used to model aircraft wings as is used to model boat hulls, the only difference is in viscosity and reynolds number.


A submersible well pump merely indicates a pump that is totally submerged in the well. They can take on as many configurations as any other pump, and can be centrifugal or turbine, high head, low head, high flow, low flow or a combination depending on the requirements of the application.
Want to bet that this isn't a multi stage centrifugal submersible pump we are discussing?


The work performed by a rotary pump is a product of the volume moved and the pressure developed.
No it isn't the product. The curve is far from linear. To increase pressure means to decrease the outlet diameter for a given pump (simplification). As that is done flow friction goes up dramatically and non-linearly so that the amount of work done decreases with pressure.

jontwo13
07-28-2006, 02:06 PM
I can't check the motor plate unless I pull the well pump out 400 feet. I saw a 1/2 horsepower 10 stage submercible pump raise the water about 500 feet, the more stages you have the deeper it will lift from. This pump had 00000 pressure
dropped it about a 1 foot and it would let the water seep over the end of the pipe took it about an hour. Poor fellow had to get a larger hp motor and pump. I get into a lot of strange things helping people for free.

Evan
07-28-2006, 02:14 PM
Ok. Then it is a multi stage centrifugal submersible pump. For that depth of well 1/2 hp isn't close to good enough. I run a one hp Grundfoss 22 stage pump 350 feet down and even it starts to lose pressure when the well is drawn down.

BTW, dropping the pump one foot reduced the head by one foot so it was able to pump a tiny amount.

JCHannum
07-28-2006, 03:04 PM
Jerry is correct, the vacuum cleaner may or may not slow down if the outlet is blocked depending on the type.

I will not bet on the type of pump it is, as I have no idea of how it is being used. The only other information given is that it is at 400' depth. Without more information, an informed answer cannot be given. If it is a domestic water system, the chances are that it is a multi stage centrifugal pump. But that still gives insufficient information to answer the question properly.

The output is not a direct product, and I did not state that is linear. It is a combination of both. Different impeller configurations or pump designs will produce very different end results.

You can design a centrifugal pump that will go to a given shut off head or one that will continue to increase pressure until the motor horsepower is exceeded. Similarly, you can design a pump or impeller that will pump a given volume at wide open discharge, or one that will continue to increase volume pumped until it burns up the motor or becomes suction limited.

The effects of flow on aircraft wings and boat hulls deals with the flow past those surfaces and are not related to pump design as being discussed

Boucher
07-28-2006, 03:13 PM
Red Jacket Pump motors were built with a dual voltage capacity. The windings were designed to sustain the heavier current. This was done as an inventory reduction verus cost of mfg. consideration. You degraded the output volume by 10% for 208 volt operation. Franklyn Electric through litigation forced them out of the motor business. Franklyn has 208 volt motors avaliable. At least they did when I retired 3 years ago. One way to partially compensate for this would be to downsize the pump. ie one hp motor on 3/4 hp pump. In my trade area our voltage ran at the max. and 230 volt motors ran ok on 208 volts as long as they were not sized to max performance.

RobDee
07-28-2006, 03:17 PM
So? Pressure isn't work.

Is a spring holding a detent pin, work? Is your kitchen table working because its legs are exerting pressure against the floor? Of course not.

Pressure isn't work. It might be kenetic energy but it isn't work. Pressure acting against a moving force means that the force has to do work to overcome that pressure. The work is being done by the object overcoming the pressure.

You don't know how pumps work. If a deep well pump is set deep enough that it is working to it's maximum pressure, say 200 psi, then no flow at all happens. All that happens is that it stirs the water. The only work it does is to heat the water it is stirring. The lower the head the more work it does and
the higher the current.

When did this happen? When did 'for every action there is an equal and opposite reaction' (Newton’s Third Law) stop working here?

The motor is doing the same amount of work! The difference is the heat displaced.

Please you’re telling me “Nope!” and “You don’t know pumps” when you’re thinking is failing the simplest laws of physics.

Again your quote:

“Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

This was your opening statement and it is wrong. Pumps, motors and porch swings all work within the laws of Physics. Let’s try to remember that before we tell people what they don’t know.

Evan
07-28-2006, 03:23 PM
Odd. The other posts didn't show up for a while.


Think about this: What creates the pressure? It’s the volume of water! The greater the volume the greater the pressure. Less water being moved = lower pressure = less work = lower current.

No, not so. It doesn't work that way. The flow rate in a water system does not increase linearly with pressure because friction effects dominate the flow rate. Pressure is not a measure of work any more than voltage is. The amount of power is dependent on the total flow. Flow increases with pressure if all else remains equal but not in a linear fashion.

In the case of the submersible pump the pressure is determined by the depth of the pump (assuming the same pump) so all does NOT remain equal. As it goes deeper not only does it have to lift the water further but the friction increases as the pipe grows longer to the surface. It does less work as it moves less water until eventually, when deep enough, it moves none at all. At that point the pump is operating at maximum pressure and zero flow rate.


As the resistance to movement goes up the torque goes up! How much water movement we have at this point is moot.

Not, it is not moot. If we are not moving water out of the hole against the force of gravity then we are doing less work. Even if we ignore the lifting if all we do is stir the water then we are doing less work than if we pump it. Since if we don't move the water then we are not accelerating mass. Instead, the majority of the water in the impeller simply rotates along with the impeller subject to the friction effects in the housing.


“Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

How can this happen? It can’t !
Is pressure equivalent to load? Yes, absolutely.

Pressure is NOT equivalent to load. Load is equivalent to work done. If you look at the ratings for any well pump you will find that as the pressure the pump must deliver goes up the amount of water delivered goes down. The greater the volume of water delivered in a given amount of time the more work that is done. This applies to all pumps, not just well pumps.


If we have a pump with a one inch pipe and a one inch head does it take more force to move it then a one inch pipe with six inches of head on the same pump.
No!
Which one does more work? The 6 inch long pipe. Which one has a greater pressure? The six inch long pipe.

Which one moves the most water in a given time?

RobDee
07-28-2006, 04:30 PM
Think about this: What creates the pressure? It’s the volume of water! The greater the volume the greater the pressure. Less water being moved = lower pressure = less work = lower current.

No, not so. It doesn't work that way. The flow rate in a water system does not increase linearly with pressure because friction effects dominate the flow rate. Pressure is not a measure of work any more than voltage is. The amount of power is dependent on the total flow. Flow increases with pressure if all else remains equal but not in a linear fashion.

What are you talking about? Where did I say the flow rate increased linearly? Go back and read why I made the above statement. Within our one inch pipe the greater head the more pressure. Period.
Friction? Who cares! We have a load on a pump, the pump is drawing 5 amps. I don’t care what it’s moving, how much friction reduces the efficiency. It doesn’t matter. Now we’re back to Newton’s Third Law. We can not destroy or create matter. Period. You’re introducing efficiency, it’s not a factor.

In the case of the submersible pump the pressure is determined by the depth of the pump (assuming the same pump) so all does NOT remain equal. As it goes deeper not only does it have to lift the water further but the friction increases as the pipe grows longer to the surface. It does less work as it moves less water until eventually, when deep enough, it moves none at all. At that point the pump is operating at maximum pressure and zero flow rate.

ibid

As the resistance to movement goes up the torque goes up! How much water movement we have at this point is moot.

Not, it is not moot. If we are not moving water out of the hole against the force of gravity then we are doing less work. Even if we ignore the lifting if all we do is stir the water then we are doing less work than if we pump it. Since if we don't move the water then we are not accelerating mass. Instead, the majority of the water in the impeller simply rotates along with the impeller subject to the friction effects in the housing.

No, No, No!
You’re completely ignoring Newton’s Third Law again.

Since if we don't move the water then we are not accelerating mass.

This is a given, what's your point?

“Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

How can this happen? It can’t !
Is pressure equivalent to load? Yes, absolutely.

Pressure is NOT equivalent to load. Load is equivalent to work done. If you look at the ratings for any well pump you will find that as the pressure the pump must deliver goes up the amount of water delivered goes down.

First, see my last post where you attempted to establish pressure as work.
Pressure is not work but the greater the pressure the greater work needed to overcome it. So is pressure equivalent to the load? The less pressure, the less load on the pump. The more pressure the greater load on the pump.
Secondly, you’re sighting the limitations of a pump, any pump. All equipment has a maximum power handling ability. You’re mixing power handling ability and Physics.

The greater the volume of water delivered in a given amount of time the more work that is done. This applies to all pumps, not just well pumps.

No, this falls within the laws of Physics, not just pumps and I agree with it.

If we have a pump with a one inch pipe and a one inch head does it take more force to move it then a one inch pipe with six inches of head on the same pump. No!
Which one does more work? The 6 inch long pipe. Which one has a greater pressure? The six inch long pipe.

Which one moves the most water in a given time?

One more time. This was in response to your statement:

“Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

And it shows your statement is incorrect.

JCHannum
07-28-2006, 04:41 PM
Pressure is absolutely a measure of work. Otherwise why does it take more horsepower to increase the pressure in the holding tank of an air compressor from 50 to 150 psi? Very little flow involved there. Similarly, increasing or maintaining the pressure in a pumping system will require some amount of power, regardless of the flow in the system. The power consumed will possibly drop since the work of maintaining the volume of flow has been removed.

When sizing a pump drive, many factors must be taken into consideration. Product, temperature, volume pumped, output pressure and suction supply are the basic minimums. Other factors can be pipe type and smoothness, system losses, laminar or turbulent flow, viscosity, suspended solids and many more.

With a submersible pump, pressure developed is not related to the depth of the pump, but the depth of the water in the well. Discounting friction losses of course, the work done is a result of the height the water is raised above the well level plus the system pressure plus the volume pumped. It will not be a linear relationship, but change any one variable and power consumed will change.

Since most deepwell pumps are centrifugal, at some point they will reach shutoff head, and flow will stop. Power may drop or not. This is not true of all pumps, PD pumps will continue to pump until something breaks. The motor or the system.

Evan
07-28-2006, 04:54 PM
It might be kenetic energy but it isn't work.
Pressure isn't kinetic energy either, it's potential energy.

My statement “Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.” is correct.


From Engineers Edge:


Centrifugal pumps generally obey what are known as the pump laws. These laws state that the flow rate or capacity is directly proportional to the pump speed; the discharge head is directly proportional to the square of the pump speed; and the power required by the pump motor is directly proportional to the cube of the pump speed.
If the pump's speed does not change and the flow rate does because of changes in head then the inverse of the power law holds true. The power required decreases as the cube of the reduction in flow rate of the pump.

http://www.engineersedge.com/fluid_flow/pump_laws.htm

I do not appreciate comments of this nature:


This was your opening statement and it is wrong. Pumps, motors and porch swings all work within the laws of Physics. Let’s try to remember that before we tell people what they don’t know.
I suggest you check the facts before attempting to argue them.

Evan
07-28-2006, 05:14 PM
Pressure is absolutely a measure of work.
No it isn't. Pressure is a scalar value and does not indicate work done. That is why air compressors are rated in cubic feet per minute at a certain pressure. I don't understand why this isn't clear. You wouldn't buy an air compressor because it was rated at 150 psi only. The volume of air moved is what determines how much work it does.

It's like saying that a 9 volt battery contains more power than a 6 volt battery.


With a submersible pump, pressure developed is not related to the depth of the pump, but the depth of the water in the well.
It isn't related only to either one of those factors. It is related to the lift required above the surface of the water in the well regardless of the depth of the pump or the water.

RobDee
07-28-2006, 05:58 PM
Quote:
It might be kenetic energy but it isn't work.

Pressure isn't kinetic energy either, it's potential energy.

Of course, you’re right it is potential energy.

My statement “Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.” is correct.


From Engineers Edge:
Quote:
"Centrifugal pumps generally obey what are known as the pump laws. These laws state that the flow rate or capacity is directly proportional to the pump speed; the discharge head is directly proportional to the square of the pump speed; and the power required by the pump motor is directly proportional to the cube of the pump speed."


Yikes!
Do you understand the quote? Where does it say that the pump “runs at maximum load when the outlet is at minimum pressure”? Nowhere.
Now what controls the pump speed? It’s the power and in order to keep the pump going at a specific speed under changing load requires different amounts of power. The greater the load the more power required to maintain the speed. Minimum pressure has nothing to do with it!

If the pump's speed does not change and the flow rate does because of changes in head then the inverse of the power law holds true. The power required decreases as the cube of the reduction in flow rate of the pump.

What?
Let’s see the pump speed doesn’t change but the flow rate does. So what’s reducing the flow rate? Friction? Does the power required change? No, the efficiency changes. The power required is still the same as if the flow rate didn’t change. The load is the same AND your statement is still wrong and doesn’t fit the quote you gave.

I do not appreciate comments of this nature:
Quote:
This was your opening statement and it is wrong. Pumps, motors and porch swings all work within the laws of Physics. Let’s try to remember that before we tell people what they don’t know.

Then don't make statments you can't support, stay within the boundaries of basic Physics, don't tell people what they don't know when it is you who doesn't know and I won't make them.

I suggest you check the facts before attempting to argue them.

I did and showed several times your misunderstanding of the basic laws of Physics. Would you like me to go back and enumerate them for you?

Again, your original statement is wrong and you have established nothing in its defense.

Evan
07-28-2006, 07:17 PM
I did and showed several times your misunderstanding of the basic laws of Physics. Would you like me to go back and enumerate them for you?

Please do.

One more time. I shall try to make this as simple as possible so you can understand it.

Given a pump running at constant speed, in this case a well pump, we have a certain maximum flow rate possible for that pump. That flow rate is determined by the resistance to flow on the outlet of the pump and the work required to overcome the force of gravity.

To simplify let's eliminate the lift and consider only the flow rate against resistance to flow.

The flow rate is dependent on the resistance to flow. The power expended to create the flow is dependent on the flow rate and varies by the cube of the flow rate.

Note that it doesn't matter if the head of the pump is created by lifting against gravity or working against resistance (friction or a restriction). They are equivalent and are denoted in the term "dynamic head". The dynamic head of the pump running at constant speed determines the flow rate.

So, the lower the dynamic head of the pump the greater the flow rate. The greater the flow rate the greater the power used by the pump.

The greater the dynamic head of the pump the lower the flow rate. The lower the flow rate the less power the pump consumes.

So, "A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load." is correct

It is correct because with minimum head we have minimum pressure developed and maximum flow rate.

Note that I said when the outlet is at minimum pressure. That means the outlet of the pump.

A pump down a well so deep that it is able to produce no pressure at the top of the well is running at maximum pressure at the outlet of the pump. Since it cannot go higher it produces zero flow rate.

Theoretically the power consumption should drop to a very low level but it does not because the pump is then operating at zero efficiency, cavitation is taking place in the impellers and energy is wasted. Still, it uses less power in that instance than if the outlet of the pump is unrestricted and minimum pressure is developed along with maximum flow.

Evan
07-28-2006, 08:17 PM
Here is some more on this.



Pressure and Head

If the discharge of a centrifugal pump is pointed straight up into the air the fluid will pumped to a certain height - or head - called the shut off head. This maximum head is mainly determined by the outside diameter of the pump's impeller and the speed of the rotating shaft. The head will change as the capacity of the pump is altered.
The kinetic energy of a liquid coming out of an impeller is obstructed by creating a resistance in the flow. The first resistance is created by the pump casing which catches the liquid and slows it down. When the liquid slows down the kinetic energy is converted to pressure energy.

It is the resistance to the pump's flow that is read on a pressure gauge attached to the discharge line. A pump does not create pressure, it only creates flow. Pressure is a measurement of the resistance to flow.







Centrifugal Pumps (http://www.engineeringtoolbox.com/centrifugal-pumps-d_54.html) are "constant head machines" Note that the latter is not a constant pressure machine, since pressure is a function of head and density. The head is constant, even if the density (and therefore pressure) changes.

http://vts.bc.ca/pics/waterhp.gif

http://www.engineeringtoolbox.com/pumping-water-horsepower-d_753.html

Boucher
07-28-2006, 08:22 PM
Something is not working on this site today. will try to repost one that disappeared. Electrical power is the product of the volts X Current. Pump power is related to the product of the pressure X the volume. If you have specific questions I will try to reply.

Evan
07-28-2006, 08:43 PM
Pump power is related to the product of the pressure X the volume.

No, pump power consumption is not a straight product. It is non-linear. Resistance to flow creates pressure. Resistance to flow is equal to the square of the flow rate. Pump power consumption is equal to the cube of the flow rate.

RobDee
07-28-2006, 09:36 PM
Rob:
I did and showed several times your misunderstanding of the basic laws of Physics. Would you like me to go back and enumerate them for you?

Please do.

Again.

Evan:
You don't know how pumps work. If a deep well pump is set deep enough that it is working to it's maximum pressure, say 200 psi, then no flow at all happens. All that happens is that it stirs the water. The only work it does is to heat the water it is stirring. The lower the head the more work it does and
the higher the current.


Rob:
When did this happen? When did 'for every action there is an equal and opposite reaction' (Newton’s Third Law) stop working here?

The motor is doing the same amount of work! The difference is the heat displaced.
-----------------------

Evan
Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.

The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.

One more time. I shall try to make this as simple as possible so you can understand it.

Spare me this. We’re not talking Plank’s Ultraviolet Catastrophe here or Fast Fourier Transforms. This is the ad hominem tact. ‘I know what I’m doing and I’ve tried to explain it but you’re to daft to get it.’

It’s the, ‘if I say it enough times ,no matter how wrong my thinking and physics are it will become truth.’

Given a pump running at constant speed, in this case a well pump, we have a certain maximum flow rate possible for that pump. That flow rate is determined by the resistance to flow on the outlet of the pump and the work required to overcome the force of gravity.

To simplify let's eliminate the lift and consider only the flow rate against resistance to flow.

The flow rate is dependent on the resistance to flow. The power expended to create the flow is dependent on the flow rate and varies by the cube of the flow rate.

Note that it doesn't matter if the head of the pump is created by lifting against gravity or working against resistance (friction or a restriction). They are equivalent and are denoted in the term "dynamic head". The dynamic head of the pump running at constant speed determines the flow rate.

So, the lower the dynamic head of the pump the greater the flow rate. The greater the flow rate the greater the power used by the pump.

The greater the dynamic head of the pump the lower the flow rate. The lower the flow rate the less power the pump consumes.

So, "A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load." is correct.

Yikes, again!!

We’re talking about CURRENT LOAD on the motor not load with respect to quantity of water!
You keep confusing issues. The more water moved the more energy consumed and the more power required. Period!!

The man’s question was with respect to motor current load. My response was with respect to motor current load. Yours wasn’t.

Get it??

How many times do you want to go down this road? I’m as bored as if I was watching a mud fence in the rain all day!

Geeze! I have to go build something or I’ll fall asleep and dent my head on the keyboard!

Evan
07-28-2006, 10:11 PM
The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load

No, the lower the pressure (pressure = resistance) the greater the mass moved.


Rob:
When did this happen? When did 'for every action there is an equal and opposite reaction' (Newton’s Third Law) stop working here?

The motor is doing the same amount of work! The difference is the heat displaced.It hasn't stopped working. The lower the pressure the greater the mass moved and the more work the motor does, per Newton's law.


Yikes, again!!

We’re talking about CURRENT LOAD on the motor not load with respect to quantity of water!
You keep confusing issues. The more water moved the more energy consumed and the more power required. Period!!
Hey, I think you are beginning to get it. That is finally correct.

I'll post the quote from the engineer's toolkit site again, in case you missed it.



A pump does not create pressure, it only creates flow. Pressure is a measurement of the resistance to flow.

RobDee
07-28-2006, 10:30 PM
Rob
The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load .


No, the lower the pressure (pressure = resistance) the greater the mass moved.

I lean against a 3,000 lb. car. It resists my forward movement. In order to move it I must exert more force which = energy, joules, power.

I lean against a 20 lb. baby carriage it poses less resistance to move. Less power. Period.

You’re saying as the mass increases the force decreases. That’s not what Newton and I say. We say F=MA (force = mass x acceleration.)

Congratulations, you’re still failing physics 101. Tell me one more time how you want to make it simple for me. I like that patronization.

Rob:
When did this happen? When did 'for every action there is an equal and opposite reaction' (Newton’s Third Law) stop working here?

The motor is doing the same amount of work! The difference is the heat displaced.

It hasn't stopped working. The lower the pressure the greater the mass moved and the more work the motor does, per Newton's law.

Wrong again. Ibid.

Yikes, again!!

We’re talking about CURRENT LOAD on the motor not load with respect to quantity of water!
You keep confusing issues. The more water moved the more energy consumed and the more power required. Period!!

Hey, I think you are beginning to get it. That is finally correct.

What, that you’re still confusing issues?

It can’t be the rest of my statement because that is in conflict with the above statements by you.

But hopefully soon you’ll get it.

Evan
07-28-2006, 10:45 PM
Go read up on how pumps work. Then talk.

RobDee
07-28-2006, 11:08 PM
Go read up on how pumps work. Then talk.

I’ll try to find some books that don’t defy the laws of physics and common reason, although I imagine you have the last of those well ensconced in your own library.

J Tiers
07-29-2006, 12:23 AM
A few more things to think about.....

Pumps DO create pressure, because pressure is what causes flow..... if the pump could not create pressure, it could not push any flow against a head....it would not be a pump! Something might have been taken out of context on that quote....

Pressure (or height) is POTENTIAL energy. FLOW is KINETIC energy.

These two things balance....and interchange. If you impart a certain energy to water, it may be expressed EITHER way, or as a combination.

As potential energy stored, such as water having been lifted up a certain duistance.

As kinetic energy, water in motion.

You can convert one to the other. If you use the hose to "blast" pieces of debris off your driveway, you are using the kinetic energy in the water flow, and converting some to pressure that causes stuff to move.

If you shoot the hose into the end of a pipe that bends upwards, you can (probably inefficiently) convert kinetic energy into potential emergy.... a certain height of water will stand in the pipe, held by the pressure created by teh water into the open end.

If you put a tank up there at just below that height, you can "pump" water up to that tank with the hose. I have dewatered trenches with a hose and pipe that way when water-drilling pipe under driveways etc.

A "ram" pump converts kinetic energy into potential energy, usually with a large loss.

OK, so much for that....

When water is flowing THE PRESSURE IS LOW. When water is prevented from flowing, THE PRESSURE IS HIGH. it is a conservation of energy issue.

A venturi flow meter uses that fact. So does (usually in air) a pitot tube.

Hence, a pump ADDS ENERGY to water. The expression of that energy may be in terms of potential energy, or kinetic energy, at your choice.

So a pump will usually pump as much as possible within its HP against back pressure, and will tend to be a constant HP device, like a fan.

BUT..... a high head pump has limited flow capability, usually, So you only get max energy into the water at near max head. At lower heads, the energy imparted to the water, (and hence the power required) is often less than max, due to other restrictions on flow, such as the pipe coming to the pump, etc.

THAT is why I suggest that a high head pump may require less power when used at a significantly lower head than max.

And why the 208 does not have to be "instant motor death".

That and the duty cycle if it is a demand pump, because the usual limit on motor HP is really heating. It can stand over-current if not prolonged enough to overheat the motor.

BTW, that 400 feet of cable has to lower motor voltage somewhat......which they have already taken into account, we hope.

JCHannum
07-29-2006, 12:30 AM
Whether you call it pressure, head or resistance to flow, it ultimately is the same animal and it has a direct effect on the power consumed by a pump. Volume also has a direct effect on power consumed. Change the value of one or the other, and power consumption will change.

Neither is linear, and I am not going to dig out the references to go into the calculations as there are far too many variables in pump design and application to make a simple, one size fits all statement. Suffice it to say that the same horsepower is capable of moving a large volume at a low pressure or a small volume at a high pressure. Much can be done between those two extremes to tailor the pump's performance to give the desired characteristics.

Please note the quote from Engineer's Edge, whoever they may be, contains the word generally. That covers a wide range of differences in pump design.

The characteristics of a well pump in a home water system are merely one small example of the possibilities, and those characteristics cannot be applied across the entire range of rotary pumps. PD pumps are another entirely different animal, and the statement "pumps is pumps" cannot be made.

Air compressors do follow much the same principles, as they are basically nothing more than pumps themselves. They are described as producing a given CFM at a given pressure for the sake of convenience. For any given horsepower, if the output pressure is increased, the CFM numbers drop.

Picked at random from their 2005 catalog, a Grainger 1WD21 compressor with 5HP motor will produce 20.4 CFM @ 100 PSI, and 16.5 CFM @ 150 PSI. The exact same compressor with a 7-1/2HP motor will produce 25.2CFM @ 100 PSI, and 23.5 CFM @ 150 PSI.

Comparing a 6 volt battery to a nine volt battery does only tell a part of the story as to the power available. You cannot tell the power the batteries will put out without knowing the amp/hour rating. Similarly, you cannot tell the power required to drive a pump with out knowing the operating parameters which consist of pressure and volume and a lot of other variables dealing with the installation and characteristics of the fluid being pumped.

BTW, when replying to one of my quotes, please include my entire quote rather than just a portion and then rephrasing what I have said to prove your point. ie;

I said:
"With a submersible pump, pressure developed is not related to the depth of the pump, but the depth of the water in the well. Discounting friction losses of course, the work done is a result of the height the water is raised above the well level plus the system pressure plus the volume pumped. It will not be a linear relationship, but change any one variable and power consumed will change."


You replied;
"Quote:
(With a submersible pump, pressure developed is not related to the depth of the pump, but the depth of the water in the well.)

It isn't related only to either one of those factors. It is related to the lift required above the surface of the water in the well regardless of the depth of the pump or the water."

RobDee
07-29-2006, 12:33 AM
BUT..... a high head pump has limited flow capability, usually, So you only get max energy into the water at near max head. At lower heads, the energy imparted to the water, (and hence the power required) is often less than max, due to other restrictions on flow, such as the pipe coming to the pump, etc.

THAT is why I suggest that a high head pump may require less power when used at a significantly lower head than max.


Exactly!

The more head the more energy required to reach it. Thank you, someone who hasn’t completely trashed Newtonian Physics!

RobDee
07-29-2006, 12:40 AM
Whether you call it pressure, head or resistance to flow, it ultimately is the same animal and it has a direct effect on the power consumed by a pump. Volume also has a direct effect on power consumed. Change the value of one or the other, and power consumption will change.

Yes! Thank you.

J Tiers
07-29-2006, 01:12 AM
Exactly!

The more head the more energy required to reach it. Thank you, someone who hasn’t completely trashed Newtonian Physics!

You better say "the more energy required to reach it PER UNIT MASS raised".

Just creating a shutoff pressure head without flow uses little energy aside from internal frictional losses. In the limit, a centifugal pump stirs the fluid, and a piston pump simply compresses the fluid (air, we hope) and then expands it, without flow (with water it would stall, or break something).

When in doubt about an odd sounding claim....it is often a good idea to do an estimate of the energy balance..... If it looks like energy is being created from nothing, or seems to have no "destination", you have a problem..... It even works for economics, if you assume that money is power....... (insert icon for rim-shot here)

Fasttrack
07-29-2006, 01:33 AM
Evan is right in theory - and for an instant when flow stops it would be true but, also as Evan pointed out, other things have to be taken into consideration such as cavitation. Also, with no water flow we must consider what is going on at the bottom of the well. Consider, for instance, a torque converter in a car. If it were not for the stator, when you put your car in gear at idle the engine would die because the fluid is slung off of the converter pump (aka the impeller) and thrown to the turbine. If the turbine can't move, or moves at a substantially slower speed than the impeller (such as stop and go situations) then the fluid bounces off of the impeller maintaining a considerable amount of energy - because of the fins on the turbine and the nature of the flow, this fluid being bounced back is in the opposite direction of the impeller and would kill the engine. Thus, auto manufacturers designed a stator with a one-way clutch. This allows that fluid to be redirected in such a maner that it helps movement of the impeller. As both halves reach near equall speeds, the fluid circulates and moves in the same direction as the impeller - the one way clutch in the stator allow it to turn in the correct direction. That is why TC increase your torque (up to as much as 2.5:1) at very low rpm (rarely higher than 2000 rpm). Incidently that is why the TC is the hottest part of the transmission. At idle or acceleration, alot of the energy of the fluid bouncing back is turned into heat as it hits the stationary stator (which is held in place by the stator support shaft). This is why the cooler lines run directly from the TC and then back to the pan in most auto trannies.

Alright, now that i've gone sufficiently far astray :D - my point is all of you are right to some extent or the other. Evan is right in theory, and he makes this clear when he says "Theoretically the power consumption should drop to a very low level but it does not because the pump is then operating at zero efficiency, cavitation is taking place in the impellers and energy is wasted."


p.s. pressure can't be created with out resistance - theoretically pumps only create flow, but then they must, neccessarily, produce pressure because there is always some resistance.

maybe this calls for an experiment...

Fasttrack
07-29-2006, 01:35 AM
"(with water it would stall, or break something)." :D

I burst an air hose when i was younger doing something very stupid with, shall we just say, a positive displacement pump, a garden hose, some buckets, and a desire to soak all the little kids around... ;) I soon learned that it worked much better by compressing air on top of the fluid so i could store the energy to be released when i saw a suitable target :D

Evan
07-29-2006, 02:11 AM
Well, accomplished a few things in the shop.

---------------------------

Yes you can say "pumps is pumps". Pumps move mass. It takes energy to move mass. One measure of that energy is horsepower.

33,000 lbs lifted one foot in one minute= one horsepower.

The less mass you lift in a unit of time the less energy you use. Pumping mass against friction is equivalent to lifting and the exact amount of work is measured the same way. The amount of mass moved per unit time is the measure of the energy required to move it.

As the friction increases the amount of mass moved per unit of time decreases. The pressure increases. The horsepower required to move less mass per unit time goes down.

The comparison to an air compressor breaks down because the air is compressible in that situation. Freely flowing air at less than about 1/2 the speed of sound acts incompressible but this is not the situation in a compressor. In a compressor the air is not free to flow.

Water however is essentially incompressible as it acts as a fluid that has already been compressed to 210,000 bar (and stays that way). To compress it to 50% of free volume requires another 210,000 bar. Common amounts of pressure applied to water do not represent any significant stored energy. That is why water is used for hydrostatic testing.

So, pressure does not represent significant stored energy ( with water). Mass flow is the measure of the power required by a pump. Resistance to flow results in higher pressure and less flow. Less flow represents less energy used to produce that lesser flow.

Evan
07-29-2006, 02:19 AM
One other thing I forgot to mention:

When water flows through a pipe at constant velocity the forces acting on it are exactly balanced. The water experiences no net force and flows by momentum alone. This is a direct example of Newton's laws of motion.

JCHannum
07-29-2006, 10:03 AM
The formula for determing motor horsepower for pumps is:

GPM X Head in Feet X Specific Gravity/3960 X Efficiency of Pump

For fans & blowers it is;

CFM X Pressure (PSI)/33000 X Efficiency

I have not located one for compressors, but it undoubtably will follow the same format with a different constant and include density of the gas being compressed.

The difference encountered in various types and configurations of pump will have a marked effect on the efficiency number, and design can skew the performance of the pump to favor either it's pressure or volume output. You acnnot have both for the same power input.

If one horsepower = 33,000lb lifted one foot in one minute, that proves the point. Relating that to pumping, 33,000lb equates to mass or volume, one foot equates to pressure one minute is the time. You cannot separate the three.

RobDee
07-29-2006, 10:12 AM
Here is my original statement and Evan's response:


It idles in a shallow application like a spring as opposed to a deep well. It runs at reduced load when it is designed to run to a specific depth, say 200' and it is only 100' down the well.


Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.

Example: block the intake of your vacuum. The motor speeds up.

Friction, stall, etc. never entered into the equation.

I made the basic and true statement shown above.

If we take a pump and stick it in a small bucket with a foot of water above it (1 foot head) it will require less energy to evacuate it then it would if the pump was in an underground aquifer down 100 feet. Period!

Evan said, “Nope.” OK nope to what? That it takes less energy to move the water with a hundred foot of head then one foot? Wrong.
He went on to make more erroneous statements that I pointed out in our debate and when he asked for an example of his failure to adhere to basic principles of physics I showed him again.

Blocking the intake of a vacuum has nothing to do with it and it is a poor analogy.

We’re moving water the more you move the more energy required to move it. Period! And that is what I said in my initial post. Again, friction stall, etc. has nothing to do with it. That’s not the question the person asked. He asked about voltage which is affected by the current draw and the less current the better chance the pump will have to survive at the lower voltage.

My answer directly addressed that problem. Evan’s answer, not only didn’t but it had no bearing on my statement, no matter how much ‘fluff’ he surrounded it with.

So what was the "Nope" Evan. Nope F=MA is wrong? Do you want to trip over that same stone again?

Evan
07-29-2006, 10:54 AM
Block the outlet of the vacuum and it also speeds up and for the same reason as blocking the inlet. The analogy is perfect.

In all of this it appears you have forgotten or overlooked one very important factor. The pump we are dealing with has one important characteristic. It is constant RPM. The higher the friction the less water it is able to move. This equates to lower power consumption.



My answer directly addressed that problem. Evan’s answer, not only didn’t but it had no bearing on my statement, no matter how much ‘fluff’ he surrounded it with.

So what was the "Nope" Evan. Nope F=MA is wrong? Do you want to trip over that same stone again?
My answer bears directly on the problem. The motor is not loaded less by having less head to work against. It is loaded more. I have explained the reasons why. Whether you like it or not they are correct.

BTW, dismissing what you don't understand as "fluff" is not a valid debating tactic.

hitnmiss
07-29-2006, 11:02 AM
Evan is correct.

If a pump motor is selected that is just able to lift water say 100 feet at x flow and the pipe on the outlet breaks (pump free flows with almost no back pressure) the motor will draw MORE amps and burn up.

J Tiers
07-29-2006, 11:08 AM
So, pressure does not represent significant stored energy ( with water). Mass flow is the measure of the power required by a pump. Resistance to flow results in higher pressure and less flow. Less flow represents less energy used to produce that lesser flow.

Unfortunately, that is HALF of the issue.....

What you are saying is that "horsepower" has no meaning......

Pumping water out of a well is equivalent to lifting an equal weight in teh form of a block of lead up the same distance. We'll let pulley friction take the place of flow resistance if you want to be silly-fussy here.

If you lift that lead up slowly, and have little friction, a cricket could pull the rope. Hardly any power (time rate of energy expenditure) is used. BUT the energy eventually goes into the lead, since the potential energy represented by the lead at a height must be supplied.

If you lift more mass of lead, or lift the same mass faster, OR LIFT IT FARTHER IN THE SAME TIME, it takes MORE power.

In the case of the pump, clearly if you pump MORE water the same lift, it takes more power, despite the fact that the pressure is the same.

If you pump half the water per second (less flow) 3 x higher, it STILL takes MORE power since the total energy per unit time imparted to the water is greater.

Therefore the statement in bold above is false without more qualifiers.

BTW, in general, with water, pressure is an indicator of the potential energy. When combined with knowlege of the mass of water available at that pressure, it can be a direct energy measure, since it would normally represent a measure of height, as in a penstock at a high head power station.. I.e. I have X amount of acre feet of water available with Y pressure in the penstock outlet, so I have Z amount of potential energy available.

Thios is the whole deal behind pumped-storage peaking power plants. You seem to be implying, whether you wanted to or not, that they do not work.


When water flows through a pipe at constant velocity the forces acting on it are exactly balanced. The water experiences no net force and flows by momentum alone. This is a direct example of Newton's laws of motion.

In a "snapshot" that can be said to be true......

It is more accurate to say that the flowing water has a certain energy content. That energy is being lost as heat of friction constantly, and replacement energy must be added if the water is not to slow down.

That replacement may be expenditure of potential energy, as in downhill flow, or it may be replaced by a pump at the inlet.


***********
Lets not get confused about the deepwell pump.....

The "shutoff head" of a particular TYPE of pump at a particular power input is NOT related directly to the power needed to lift water. It is a function of the pump design.

If you had a frictionless positive displacement pump, THEN you could relate power directly to mass flow and head. Assuming you could adjust the pump to produce any desired flow, you could accept ANY power (energy per unit time) input, and produce a corresponding flow against ANY head.

Our cricket could pump water to the top of mount Everest. But very, very little.......

As far as the open-flow burn-up, depends on the pump..... If the motor speeds up because it cannot get enough water input, it will not burn up, because it will be headed towards an idling power input, and the higher motor back EMF will reduce current flow. If the pump can get all teh water it wants, it might indeed draw more power. If you put a fan impeller (designed to blow air) into water, it will burn up the motor because the volume flow it is designed to produce represents far more mass, and takes more power, so motor slows and burns up.

On teh other hand, if you take a water pump impeller and run it in air, its electric motor will idle at near synchronous speed, because it cannot get enough mass flow to load it. (don't bother mentioning racing boat props...different item entirely)

Now, the pump that the O.P. has, evidently WILL pump water from the well. I assume so, anyway. If it is an induction motor, it is semi-synchronous, its impeller speed is more-or-less fixed, and it has a certain shut-off head. The amount pumped will vary with height due to pump considerations. What its open flow rate is, I do not know.

If it is rated to pump X gpm from 400ft, it may or may not pump enough more from 250ft to absorb the same power.

My answer to the 208V is still that it may use enough less power to be fine at 208. They are water-cooled, although that is taken into account in the design.

If the pump & motor is rated for continuous flow, but is here used for a demand system, then if it runs less than about (208/240)% of the time, it will NOT overheat.

RobDee
07-29-2006, 11:36 AM
Block the outlet of the vacuum and it also speeds up and for the same reason as blocking the inlet. The analogy is perfect.

First, not all vacuums exhibit this. Nor do all pumps.

In all of this it appears you have forgotten or overlooked one very important factor. The pump we are dealing with has one important characteristic. It is constant RPM. The higher the friction the less water it is able to move. This equates to lower power consumption.

My car goes down the road. I step on the brakes and the same time step on the gas. Does it take more energy to keep my car moving? Yes! To overcome friction requires more torque. More torque from the motor means that it requires more energy to keep the motor moving at the same rpms.
More friction can never equal less power consumption! F=MA. Poor Newton must be turning in his grave.

My answer directly addressed that problem. Evan’s answer, not only didn’t but it had no bearing on my statement, no matter how much ‘fluff’ he surrounded it with.

So what was the "Nope" Evan. Nope F=MA is wrong? Do you want to trip over that same stone again?

My answer bears directly on the problem. The motor is not loaded less by having less head to work against. It is loaded more. I have explained the reasons why. Whether you like it or not they are correct.

You just made this statement. “The pump we are dealing with has one important characteristic. It is constant RPM.” So now we have this motor with less head which means less friction spinning at the same rpm (your criteria) as the same motor at the same rpm with greater head and thus greater friction and you’re saying the less friction, lower head, needs more current to maintain the same rpms.

Wrong!

Fasttrack
07-29-2006, 11:48 AM
"It is more accurate to say that the flowing water has a certain energy content. That energy is being lost as heat of friction constantly, and replacement energy must be added if the water is not to slow down."

The forces are balanced, and, provided the water is moving the same distance, the energy will be the same. You seem to be suggesting that there is a continually increasing expenditure of energy. If this were true, the pump would trip the circuit breaker in a matter of a few moments. To some extent it will draw more current as it runs - the components may heat up causing increased electrical resistance, increased mechanical friction, etc - these would cause the load to grow slightly but not very much.

At start up the pump must provide enough force to accelerate the water in the opposite direction as the force of friction and the force of gravity acting upon the water column. Once it has provided enough force to accelerate the water upward, the load decreases because, as Evan said, it is moving by momentum. The heat lost is related to the force the pump is constantly exerting to equal the force of friction. There is no such thing as "friction" energy - friction is just an opposistion to flow that usually converts kinetic energy to heat energy.

Evan
07-29-2006, 11:48 AM
JT,

To lift water against gravity is the same as pumping it against resistance. That is why the term "dynamic working head" is used. It represents the resistance to flow regardless of the reason. To increase the amount of power consumed when the head is greater requires that the motor either spin faster or the impeller be larger. In that case then what you say is true. If the impeller size and rpm stay constant the flow rate decreases as the inverse square of the head and the motor power as the inverse cube.

I am not ignoring the distance the water is lifted. It is because the relation is exponentional that the power decreases with increased head since the amount of water moved reduces by the inverse square of the lift.

RobDee
07-29-2006, 11:58 AM
Evan is correct.

If a pump motor is selected that is just able to lift water say 100 feet at x flow and the pipe on the outlet breaks (pump free flows with almost no back pressure) the motor will draw MORE amps and burn up.

Is that what we are dealing with? A pump that can lift water to just 100 feet? Or an open outlet?

Is that what my initial statement said? Where did I say anything about a pump running open or clogged?

We are talking about different depths of a pump and the power required to raise that water.

Now does it take more power to lift water to a greater level? Yes.

RobDee
07-29-2006, 12:05 PM
JT,

To lift water against gravity is the same as pumping it against resistance. That is why the term "dynamic working head" is used. It represents the resistance to flow regardless of the reason. To increase the amount of power consumed when the head is greater requires that the motor either spin faster or the impeller be larger.

Not true. The motor can maintain the same rpms but require more energy to do so regardless of the size of the impeller. JT is correct with no caveats.

J Tiers
07-29-2006, 12:12 PM
The forces are balanced, and, provided the water is moving the same distance, the energy will be the same. You seem to be suggesting that there is a continually increasing expenditure of energy.


NOPE....... But there is a constant energy loss. The only reason the forces are balanced is that there IS constant energy input at the inlet. The system is leaking energy like a sieve.



At start up the pump must provide enough force to accelerate the water in the opposite direction as the force of friction and the force of gravity acting upon the water column. Once it has provided enough force to accelerate the water upward, the load decreases because, as Evan said, it is moving by momentum. The heat lost is related to the force the pump is constantly exerting to equal the force of friction.

That is misleading.... sure, you have to accelerate the first mass of water to that flow rate....

BUT you have to accelerate EVERY PART of the total mass of water to that flow rate...... it isn;t automatic, it has to be done to every bit of water that enters the pipe as long as you are pumping.

PLUS, you have to supply the frictional losses.

PLUS, you have to supply the energy necessary to lift the water to whatever height you are pumping it..... (I am separating the lift from acceleration, since you could just be pumping thru a level pipe)



There is no such thing as "friction" energy - friction is just an opposistion to flow that usually converts kinetic energy to heat energy.

exactly. Friction converts *work* to heat.


I am not ignoring the distance the water is lifted. It is because the relation is exponentional that the power decreases with increased head since the amount of water moved reduces by the inverse square of the lift.

Eh?

Potential energy is E(p) = m x g x h ......( mass, gravitational acceleration, height)

http://jersey.uoregon.edu/vlab/PotentialEnergy/

If you lift 1 kG twice as far it is twice the energy input to do it....

Try lifting full buckets from a well sometime, and you will quickly see that. It gets no easier nor harder as the bucket nears the top.....

The increased back pressure at the higher head is simply representative of the extra energy required to lift each mass increment higher.

As you push a kG into the bottom, one comes out the top, and all the kG of water in the pipe is lifted at once.

Therefore, you exert a certain pressure and move a mass a certain distance. That immediately gives you the "work" done.

If extra energy is used, where does it go?

If less energy is used, where does the other portion of the needed energy come from?

Evan
07-29-2006, 12:34 PM
Not true. The motor can maintain the same rpms but require more energy to do so regardless of the size of the impeller. JT is correct with no caveats.
But it doesn't do that. If you like I can arrange a practical demonstration for you.

I have a fire fighting water tank.

http://vts.bc.ca/pics/fire1.jpg

The engine is a 3hp gasoline motor. It operates a centrifugal pump. Because of the intended use the governor has been disabled since the pump normally runs at constant load. The load is 100 feet of 3/4" industrial hose and a nozzle. Under that load it pumps about 10 gpm.

I have a valve directly on the output of the pump to shut off the water flow. When pumping against the load the engine runs at a certain RPM, easy to hear. If I shut off the valve while pumping against the load the rpms don't change noticeably. However, if I disconnect the load (hose) so that the pump is free to pump against nearly zero resistance and open the valve the rpms slow noticeably and the volume of water pumped is greatly increased. Close the valve and the motor speeds up. Open it, it slows down.

If you like I can make a sound video of the operation.

Millman
07-29-2006, 12:55 PM
{{ I can make a sound video of the operation.}} Evan, now that would be a cool video, you shooting down a bear attack!

RobDee
07-29-2006, 02:02 PM
But it doesn't do that. If you like I can arrange a practical demonstration for you.

Any motor can do that! Motor characteristics are the same and dependent on the type of motor. An AC motored table saw running and not cutting wood will run at say 3450 rpms. If we start cutting a piece of wood the saw will try to maintain the rpms and draw more current to do so.

A pump with the outlet blocked with function with respect to the impeller BUT the motor functions as all motors of that specific type function AND any motor with greater friction will draw greater current. PERIOD!

I have a fire fighting water tank.

The engine is a 3hp gasoline motor. It operates a centrifugal pump. Because of the intended use the governor has been disabled since the pump normally runs at constant load. The load is 100 feet of 3/4" industrial hose and a nozzle. Under that load it pumps about 10 gpm.

First, there is no reason to disable the governor. It has a function, a cold engine for instance.

I have a valve directly on the output of the pump to shut off the water flow. When pumping against the load the engine runs at a certain RPM, easy to hear. If I shut off the valve while pumping against the load the rpms don't change noticeably.

Because the resistance is near the maximum capable output of the pump. In effect the pump sees the load and a blocked outlet relatively equally.

However, if I disconnect the load (hose) so that the pump is free to pump against nearly zero resistance and open the valve the rpms slow noticeably and the volume of water pumped is greatly increased. Close the valve and the motor speeds up. Open it, it slows down.

Are you using a deep well pump? I doubt it. Because deep well pumps come in different HP ratings. As the well gets shallower the HP goes down, not up!

Did you run the experiment with a line one half or one quarter as long? Did you run it with the line straight up in the air?

So in your pump what is happening? What is changing the motor load? I blow through a straw unencumbered then I blow through the same straw with it blocked. Which takes more energy? Obvious isn’t it?

Now, back to my statement that as the head goes down the power required goes down. You don’t get something for nothing.

A 1 hp pump at 100 feet down uses less energy then the same situation 200 feet down. We’re not blocking outlets and we’re not running open. An open impeller and a blocked impeller exhibit different characteristics then a well pump at different levels. That’s why different depths run different HP pumps.

You’re giving me apples and oranges.

The basic principles of physics didn’t change in your pump. It’s not an enigma. The impeller and the motor all work based on physics.

Evan
07-29-2006, 07:38 PM
Once again, I suggest you read up on the affinity laws for pumps (and fans). You will find the answers there.

Here is the basic relationship: Note what it says in the red box above the performance graph. I will be happy to discuss this further once you have informed yourself. Also note that according to the affinity laws the reason for lower flow is not relevant.

The information you need may be found here:

http://www.pumpworld.com/Centrifugal%20Pump%20Performance%20Curve.htm

http://vts.bc.ca/pics/pumpcurve.gif

RobDee
07-29-2006, 09:31 PM
Geeeeeze!!

Did you look at the graph?

What does it say?

It says the higher the flow the lower the head FOR THE SAME HORSEPOWER.

The lower the flow the higher the head FOR THE SAME HP.

Get it??

That means with a 1 hp motor the flow must go down as the head goes up to keep a 1 hp motor working at different heads.

It doesn’t mean that higher head = lower hp!!!!!

Here’s THEIR formula:

HP = (flow in gpm x head in feet x specific gravity) / 3906 x effiency %

OK What does this say????

It says that if the head goes up the HP must go UP.

If the flow goes up the HP must go UP.

If the efficiency goes down the HP must go UP.

Yikes 1000 times!!

Please stop this I have nothing against you personally! Really!!!

And I’ve enquired into pumps much more then I ever wanted to.

I’m done I don’t want to talk about pumps anymore. I’m pulling my well pump out of the ground and crushing it my vice. I’m getting my drinking water from the pond using buckets from now on!

Pax vobiscum,

Rob Dee

Evan
07-29-2006, 10:36 PM
That means with a 1 hp motor the flow must go down as the head goes up to keep a 1 hp motor working at different heads.
Correct. The part you don't get is that the laws are reciprocal. If the flow goes down at the same head (close the valve for instance) then so does the horsepower (requirement). That is what dynamic head is about. The dynamic head increases. The statements in the red box hold in all cases for centrifugal pumps.

edit: Read it again. It says "higher head= lower flow, lower flow=lower horsepower"


It says that if the head goes up the HP must go UP.
Yes, to maintain the same flow.


If the efficiency goes down the HP must go UP.
Yes, to maintain the same flow at the same head.



If the flow goes up the HP must go UP.


Yes, at the same dynamic head.

So what?

If the dynamic head goes up and the flow goes down the horsepower required goes down. It isn't linear. Reducing the flow by half reduces the horsepower requirement by eight times.






Watch out for giardia.

JCHannum
07-29-2006, 10:49 PM
Go back to my earlier post today which includes the industry standard equation for determining the horsepower for driving a centrigugal pump:
HP=GPM X Head in Feet X specific Gravity/3960 X Efficiency of Pump.

This is a basic, simplistic starting place. The attached link is from Goulds ITT, a major pump manufacturer, and addresses the affinity laws, which express the mathematical relationships between the several variables involved in pump performance:

http://www.gouldspumps.com/cpf_0010.html

These variables are; head, capacity, brake HP, pump speed and impeller diameter. An understanding of the information, and working through a few different applications will clearly show that these variables are interrelated, and changing one will very definitely change the others.

This link is a guide to the Gould product line, they have one of the wider ranges of pump varieties available:

http://www.gouldspumps.com/Model_Numeric.html

Note that most of the perfomance curves plot flow on one axis and head on the other. Horsepower is determined from these parameters, and is usually plotted in individual curves for differing RPM & impeller diameter.

It will give you a pretty good idea of why a blanket statement that might apply to one minor facet of the performance of one individual pump type cannot be applied across the entire range of pumps that fall into the broad classification of centrifugal pump.

Evan
07-29-2006, 10:54 PM
[
It will give you a pretty good idea of why a blanket statement that might apply to one minor facet of the performance of one individual pump type cannot be applied across the entire range of pumps that fall into the broad classification of centrifugal pump.
The "blanket statement" is based on the affinity laws. They are based on the Navier-Stokes flow equations. They are how all fluid flow is calculated. It is not a "minor facet".

JCHannum
07-29-2006, 11:51 PM
The blanket statement is the assumption that one spot on one curve of one pump can be applied across the board to all pumps.

As the laws are applied across the spectrum, a predictable pressure/volume outcome can be calculated for a given pump casing design with a given impeller design and diameter, operating at a given horsepower and RPM pumping a given fluid.

Both the Goulds information and the Engineering Toolbox explain this quite clearly.

It is my only contention that power is required to produce pressure, and if no presure exists in a system it will not function any more than it will function without flow. Pressure creates flow as flow creates pressure, and they are interdependant. You will not be able to move a fluid through a system without both being present. The squares, cubes or fiducaries per hectacre have no meaning other than to obfuscate the issue and explain it in mathematical terms.

When a centrifugal pump goes to shutoff head, the power consumption will be reduced. It will not become zero, but power will continue to be consumed to maintain that pressure, and some pressure will be maintained. The actual pressure and power consumption will vary greatly with the pump design.

J Tiers
07-30-2006, 12:21 AM
Sheesh..................

Lets get real...... Listen to JCH and Rob if you will not listen to me.....

There is a basic energy requirement concerning raising water, or ANY weight, to a certain height at a certain rate.

YOU CANNOT AVOID THIS ENERGY EXPENDITURE.... it is required.

After that, there are other factors......

MAYBE the pipe is such that the TDH includes a lot of "flow head", which is equivalent head due to flow resistance AT THAT FLOW RATE.

But usually, the pump user doesn't want to pump so much volume that flow rate head is a huge part of the energy requirement. And, it is easily cured with a larger pipe at relatively low cost, even for a deep well, although the TDH cannot be ignored at 400 foot heads, since the pipe is necessarily small.

But, that isn't a pump problem, its a pipe problem.

Then also, every sort of pump has a characteristic curve due to its design. At higher heads, it does not pump as efficiently, because of back leakage, etc. at some point the flow just stops....

A particular pump may have a certain efficiency etc, etc, at any head. But that is that pump, not applicable to every possible type of pump equally.

If you look at the equations at the site YOU recommended, there is no sign of the inverse square law or whatever non-linear thing you cited.....

The HP is linear with both TDH and flow rate (assuming that flow does not affect TDH much).

Naturally, if the TDH includes a lot of pipe frictional head, it will become non-linear because flow will interact with TDH significantly. You can see the effect of pipe size with the THD calculator.

Please......THINK..... Formulas must be used correctly and sensibly. A simple thought experiment will often debug an incorrect conclusion.

You first must account for the work done in lifting the weight. Everything else additional in terms of required power is due to the exact mechanism used to DO that lifting.....

If the extra stuff dominates, the efficiency is low, and the mechanism can be improved.

I think your ideas of non-linearity etc, are derived from assumptions made about the pump and the piping etc. You need to state those, or the result will be interpreted any way the reader sees fit.

I might mention...... the curves for the pump showing flow vs TDH... they curve over..... where they intersect the GPM axis (not shown) is the open flow GPM.... and the point where they intersect the TDH axis is the shutoff head. So both exist and are finite points.

And, looking at the HP curves, as the head approaches zero, it is noticeable that the curve will pass through lower and lower HP points. The 6" curve appears to be heading for a point at about 3/4 HP and maybe 180 gpm for zero head.

I reckon his pump will be OK. But it would be simple enough to boost the voltage if there is a concern.

Fasttrack
07-30-2006, 12:49 AM
Quote:
Originally Posted by Fasttrack
The forces are balanced, and, provided the water is moving the same distance, the energy will be the same. You seem to be suggesting that there is a continually increasing expenditure of energy.



NOPE....... But there is a constant energy loss. The only reason the forces are balanced is that there IS constant energy input at the inlet. The system is leaking energy like a sieve.


Quote:
At start up the pump must provide enough force to accelerate the water in the opposite direction as the force of friction and the force of gravity acting upon the water column. Once it has provided enough force to accelerate the water upward, the load decreases because, as Evan said, it is moving by momentum. The heat lost is related to the force the pump is constantly exerting to equal the force of friction.


That is misleading.... sure, you have to accelerate the first mass of water to that flow rate....

BUT you have to accelerate EVERY PART of the total mass of water to that flow rate...... it isn;t automatic, it has to be done to every bit of water that enters the pipe as long as you are pumping.

PLUS, you have to supply the frictional losses.

PLUS, you have to supply the energy necessary to lift the water to whatever height you are pumping it..... (I am separating the lift from acceleration, since you could just be pumping thru a level pipe)



I think you may have misunderstood me - though you are right my post was misleading. I should have said that the energy per unit time, i.e. work will be the same. In your earlier post you seemed to suggest that for every unit time the pump operated the energy consumption increased. I merely meant to point out that as the pump runs, provided head stays the same, the work done remains the same because the forces are balanced. You can't add the frictional losses and energy required to lift it in such a manner. Unit time and distance will remain the same for the pump and the pipe, so they can be cleared from the equation and you will see that forces are balanced - also energy is balanced - hence it maintains the law of conservation of energy. The force required to lift the water and overcome friction (and once the water is flowing) is equal to the force supplied by the pump. If it were not equal the water would not make it to the top. The only time the pump must exert a greater force than the force required to raise the water and overcome friction is when the pump is accelerating the water from the bottom of the well. Thus plugging the intake will allivieate the load (again neglecting cavitation and the effect of the water bouncing against the internals of the pump). By this logic, we can extrapolate this to the output by considering flow. By increasing the head, we decrease the flow and thus the pump has to accelerate less water. On the other hand we are moving the water a greater distance (requiring more energy) and with greater frictional losses (also more energy)...so will pump head really matter very much at all?

Fasttrack
07-30-2006, 12:54 AM
"When a centrifugal pump goes to shutoff head, the power consumption will be reduced. It will not become zero, but power will continue to be consumed to maintain that pressure, and some pressure will be maintained. The actual pressure and power consumption will vary greatly with the pump design."


I'm sold on that point :)

J Tiers
07-30-2006, 12:57 AM
Yep, I did misunderstand you.... looks like we agree.

RobDee
07-30-2006, 01:21 AM
When a centrifugal pump goes to shutoff head, the power consumption will be reduced. It will not become zero, but power will continue to be consumed to maintain that pressure, and some pressure will be maintained. The actual pressure and power consumption will vary greatly with the pump design.

Yes, exactly and this is why Evan's own pump motor rpm doesn’t change significantly when he shuts down the outlet. Because flow stops does not mean pressure stops.

Evan:

You’ve proven it yourself.

Now go back to my statement that as head rises so does HP requirement all else being equal which you have been trying to tell everyone is not true.

Now, here’s your statement:

”You don't know how pumps work. If a deep well pump is set deep enough that it is working to it's maximum pressure, say 200 psi, then no flow at all happens. All that happens is that it stirs the water. The only work it does is to heat the water it is stirring. [b]The lower the head the more work it does and the higher the current.”

This is just wrong. You’re making a false premise and that is because flow stops pressure stops. This is incorrect again as your own pump seems to show. (I’m not there and you say you have no governor on the motor). You can’t say because pressure stops in the line that pressure stops in the pump. The higher the head the more power required to achieve it all else being equal.

Hopefully you understand this now.

RobDee
07-30-2006, 01:24 AM
"When a centrifugal pump goes to shutoff head, the power consumption will be reduced. It will not become zero, but power will continue to be consumed to maintain that pressure, and some pressure will be maintained. The actual pressure and power consumption will vary greatly with the pump design."


I'm sold on that point :)

Yep!


.......................

Evan
07-30-2006, 12:04 PM
Bleah. Been offline due to satellite failure since yesterday. Still not fixed and I am not going to waste the time needed to use the dial-up connection. I'll be back.

Evan
07-30-2006, 07:43 PM
RobDee posted:



Quote:
Originally Posted by Fasttrack
"When a centrifugal pump goes to shutoff head, the power consumption will be reduced. It will not become zero, but power will continue to be consumed to maintain that pressure, and some pressure will be maintained. The actual pressure and power consumption will vary greatly with the pump design."

I'm sold on that point

RobDee said: "Yep!"
So, now you starting to agree with me as well. However, that statement is only partly correct. Power consumption will be reduced but power is not consumed to maintain the head of water. Maintaining a non-moving head of water requires no energy. See further below.

JT said



There is a basic energy requirement concerning raising water, or ANY weight, to a certain height at a certain rate.

YOU CANNOT AVOID THIS ENERGY EXPENDITURE.... it is required.
Of course.


The HP is linear with both TDH and flow rate (assuming that flow does not affect TDH much).
No. The assumption that flow doesn't affect total dynamic head is entirely invalid. Flow rate is the overwhelming component of total dynamic head. Friction varies by the square of flow rate for any given size pipe. Also, the change in horsepower to produce a given flow rate isn't linear.




SUBJECT : The pump affinity laws 2-1

There are occasions when you might want to permanently change the amount of fluid you are pumping, or change the discharge head of a centrifugal pump. There are four ways you could do this:

* Regulate the discharge of the pump.
* Change the speed of the pump.
* Change the diameter of the impeller.
* Buy a new pump

Of the four methods the middle two are the only sensible ones. In the following paragraphs we will learn what happens when we change either the pump speed or impeller diameter and as you would guess other characteristics of the pump are going to change along with speed or diameter.

To determine what is going to happen you begin by taking the new speed or impeller diameter and divide it by the old speed or impeller diameter. Since changing either one will have approximately the same affect I will be referring to only the speed in this part of the discussion.

As an example:

The capacity, or amount of fluid you are pumping, varies directly with this number.

* Example: 100 Gallons per minute x 2 = 200 Gallons per minute
* Or in metric, 50 Cubic meters per hour x 0,5 = 25 Cubic meters per hour

The head varies by the square of the number. (my note: same as friction+ lift=total dynamic head)

* Example : a 50 foot head x 4 (22) = 200 foot head
* Or in metric, a 20 meter head x 0,25 ( 0,52) = 5 meter head

The horsepower required changes by the cube of the number.

* Example : a 9 Horsepower motor was required to drive the pump at 1750 rpm.. How much is required now that you are going to 3500 rpm?
* We would get: 9 x 8 (23) = 72 Horse power is now required.
* Likewise if a 12 kilowatt motor were required at 3000 rpm. and you decreased the speed to 1500 the new kilowatts required would be: 12 x 0,125 (0.53) = 1,5 kilowatts required for the lower rpm.

http://www.mcnallyinstitute.com/02-html/2-01.html

These laws are commutative, as they must be. Inverse relationships must hold. If doubling the flow rate requires 8 times the power then halving the flow rate must require 1/8 the power (ignoring small changes in efficiency losses).

Changing the impeller size changes the flow rate and the power required. Conversely, changing the flow rate changes necessary impeller size and the power required. If impeller size remains the same but the flow rate is changed the power required changes as the mass being moved changes, the same as it does with changes in impeller size. Note that this is one of the four options listed in the above quote, " Regulate the discharge of the pump".

Definitions:

Kinetic energy: Energy of motion relative to the frame of reference. Stasis in the frame of reference = zero kinetic energy. Kinetic energy is a vector value.

Potential energy: Energy that mass holds because it is in a higher energy state than the ground state in it's frame of reference. Does not include motion in the frame of reference. Potential energy is a scalar value.

Scalar: A quantity that has a magnitude but no direction. Speed is an example of a scalar quantity. Velocity is not scalar as it has direction.

Vector: A physical quantity characterized by measurement of both magnitude and direction.

Matter may have both potential and kinetic energy at the same time.

A centrifugal pump does not create pressure. Pressure is the scalar measure of resistance to the motion of the fluid as kinetic energy is converted to potential energy. If the water accelerated by the pump encounters no resistance then the pressure is zero and the mass flow is maximum, requiring the maximum amount of power. If the mass flow is minimum then the power required is minimum. The only power required at zero mass flow is due to losses that occur within the pump at all mass flows. In a real pump these losses increase as the flow rate decreases because the efficiency curve is not flat. They do not however increase as fast as the power requirement decreases with lower flow rate.

It does not require power to maintain a static, non-moving head of water any more than it does to support a weight on a shelf. The energy to create the head of water has already been expended and converted to potential energy. "Holding" that non-moving water above the pump is not responsible for energy consumption by the pump. If you place a check valve at the pump outlet to prevent the water from back flowing the situation won't change the slightest bit. The head of water would remain and so would the the pressure even with the pump turned off. No energy would be expended. Turn the pump back on and still nothing changes as the kinetic energy of the water from the pump will overcome the pressure of the static head of water and open the valve if it is able. This method is used although the valve is placed at the inlet to prevent the pump running dry. The pressure does not come from the pump, it it is a property of the water that has left the pump. The sole job of the pump is to move water by giving it kinetic energy.

Pressure is a measure of potential energy, such as the hydrostatic pressure of a non-moving head of water. It does not measure kinetic energy as pressure is a scalar value with no vector attribute. The pump imparts kinetic energy to the water, not potential energy. The pressure measured results from the kinetic energy being converted to potential energy by doing work against resistance. Pressure measured at the pump outlet does not represent pressure produced by the pump, it represents water that has been moved by the pump against some form of resistance and now has potential energy because it has been raised from a lower energy state to a higher one by the kinetic energy imparted by the pump.

If a pressure gauge is place in a free moving stream of water flowing from the outlet of the pump as an open jet no pressure will be measured, yet the pump is obviously doing work. The reason no pressure is measured is because the water has only the kinetic energy imparted by the pump. This makes it apparent that pressure is not the measure of work. All of the work the pump does is imparted in the form of kinetic energy. Pressure is not a measure of kinetic energy.

Note: If you place your hand in the stream you will feel force. This is often called pressure but that is an incorrect use of the term. What you feel is not pressure but the kinetic energy of the water acting against your hand in accordance with Newton's laws of motion. This may seem a trivial distinction but it is far from trivial. Pressure has no vector component and cannot be used to describe matter in motion.


You first must account for the work done in lifting the weight. Everything else additional in terms of required power is due to the exact mechanism used to DO that lifting.....
All resistance to flow is included in the total dynamic head calculation. It makes no difference if the resistance to flow is because of gravity, pipe friction or both. They both have the same net effect.

The proportion of work done to lift versus work done to overcome pipe friction depends entirely on flow rate. The work done to lift mass is directly proportional to the amount of mass lifted and how far lifted, a linear relationship. Pipe friction increases as the square of the flow rate regardless of pipe size. So, the ratio between the two is overwhelmingly dependent on flow rate, not the amount of lift, although the amount of lift does change the length of the pipe. For common sizes of pipes and flow rates pipe friction limits the flow rate unless flow rate is very low.

J Tiers
07-30-2006, 08:11 PM
No. The assumption that flow doesn't affect total dynamic head is entirely invalid. Flow rate is the overwhelming component of total dynamic head. Friction varies by the square of flow rate for any given size pipe. Also, the change in horsepower to produce a given flow rate isn't linear.

It is PERFECTLY valid for many flow rate and static head situations. You simply HAVE to consider the applicable formulas here....

The calculator on the site gave me a 25 foot increase on a 350 foot static head at a few GPM flow using 3/4" plastic pipe.

Well whoopeeeeee DING! That isn't exactly "overwhelming".

So what if it is some power of flow rate? It isn't squat for an adder. not even 10%.

So the problem isn't that your FORMULAS are invalid, it is applying them inappropriately.

Now, if you have a large static head, and tiny pipe, with a large flow, then the drop in flow rate at a higher static head might reduce the TDH versus more flow at a lower static head.

But the problem would be that THE SYSTEM WOULD BE A REALLY POOR DESIGN for the lower head.

Basically that is "shooting a straw man".

RobDee
07-30-2006, 10:07 PM
Evan:
Everytime someone gets you in a corner and you have no way out you do the switch.

Here’s an example:

Quote Evan:

The lower the head the more work it does and the higher the current.

Ok this is clearly wrong. The lower the head = less water moved and the thus the less current required.

Again, here’s the formula:

HP= (flow x head x specific gravity)/ 3906 x eff.

So if the head is lower as you said above then what happens to the HP?

It goes down. But you have a bad habit of changing your statements when you’re confronted with your errors.

As I show below here:

Quote Rob:
We’re talking about CURRENT LOAD on the motor not load with respect to quantity of water!
You keep confusing issues. The more water moved the more energy consumed and the more power required. Period!!

Evan:
Hey, I think you are beginning to get it. That is finally correct.

Now you’re agreeing with me. This clearly contradicts the statement you made above and which I have showed is clearly wrong.

Not only is it wrong but my original statement which fits the above formula you disagreed with.

Again:
Rob:
It runs at reduced load when it is designed to run to a specific depth, say 200' and it is only 100' down the well.

Evan:
‘Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.”

My statement above was with respect to the LOAD CURRENT of the motor.

Your statement does not work, according to the formula above. If we lower the output pressure (lower head for instance) then the HP required to achieve that head also goes down.

Once again, you’re wrong.

As for the your ‘fire hose’. We have two problems here. First we are not there to measure the results and secondly you have disabled the governor so we don’t know what the engine is trying to do. One note when I was a kid we built lots of go carts out of old lawn mower engines. The kids who couldn’t get the carbs right used to disconnect the governor to keep the engine from wandering up and down.

Frankly there is no reason to disconnect your governor it keeps the engine stable under different load and temperature conditions. But then I’m not there to observe it or for that matter, your pump.

One more time. Pumps are as dry a subject to me as popcorn farts.

You’ve been caught in your own double talk by several people here, maybe it’s time to let it go and stop trying to subterfuge it with definitions of vector and scalar quantities. Most of us are way above that but if you disagree I would be more then happy to discuss Laplace Transforms or the Miller effect and capacitance loading of ‘H’ bridge high side mosfets with you here if you like.

JCHannum
07-30-2006, 10:19 PM
When measuring the performance of a pump, the only meaningful measurements are those taken at the discharge of the pump. Any other location adds system effects to the performance, and does not give an accurate picture of what the pump is doing.

During this discussion, and in many other discussions of pump performance, pressure and head are used interchangeably, and while there are differences, it is acceptable practice. this is confirmed by Engineering Toolbox.

Now to point out a couple of errors in reasoning. If a pump goes to shut off head, flow will be zero, and power consumption will be reduced. Pressure at the pump will reflect the shutoff head value, but the only way the pump will be able to maintain that pressure is to use power. If the pump is shut off, the head pressure will cause the column of water or the system pressure to return to suction pressure.

Holding a weight static does require work. If it did not, you would be able to hold a 500# weight at arm's length for hours on end as long as someone else placed it there.

Installing a check valve above the pump discharge will permit removal of the pressure head from the pump, and allow the pump to be stopped without the water flowing back. As soon as the check valve closes and the pump stops, the pressure at the pump goes to zero.

The purpose of the check valve being placed on the suction side of a pump is to maintain a prime in the pump head, so it will be primed when restarting.

(Please God do not let us get into a discussion of priming and self priming vs. non self priming pumps and the effects of NPSH since the lifting of the water imposes more and different power demands.)

A direct quote of the Engineering Toolbox definition of Affinity Laws:

"Affinity Laws
Turbo machines affinity laws are commonly used within pumps and fans to calculate volume capacity, head or power consumption when speed - rpm, or wheel diameters are changed."

Please note that they are used to describe the CHANGES in pump performance due to changes in impeller speed or diameter. They do not necessarily apply to the initial design of the pump, and misapplying them will obviously lead you astray.

Again, and simply, centrifugal pumps can be designed to produce a wide variety of end results of flow and pressure, but as long as either flow or pressure is being produced, power will be consumed.

Evan
07-30-2006, 10:27 PM
(I said)
The lower the head the more work it does and the higher the current.
(RobDee said)
Ok this is clearly wrong. The lower the head = less water moved and the thus the less current required.
What!? My statement is correct. Read it again. Lower head = higher flow rate.


Every thing I have said so far is correct.

A centrifugal pump is a variable displacement pump. It performs work by accelerating water.

At zero head it accelerates the maximum amount of water to the maximum possible velocity and does the maximum possible work. There is no other situation where it can accelerate more water to a higher velocity and do more work. Therefore the motor is drawing the highest current when pumping against zero head.

As the head increases the effective displacement decreases as the flow rate decreases. The amount of water accelerated decreases and the pump does less work. The current requirement decreases.

When the pump is operating at maximum head the effective displacement of the pump is zero. The pump accelerates no water, no water enters the pump and no water leaves the pump. The pump does no work accelerating water. The only work done is due to friction losses in the pump.

[edit]

It occurs to me that perhaps you have the term head confused. Low head means less resistance to the pump, less lift, shorter pipe.

JCHannum
07-30-2006, 10:42 PM
The only work done is due to friction losses in the pump.

Well, you are finally getting closer to the correct answer and you will admit that power is being consumed.

The work being done is not entirely due to the friction losses, but due to a combination of friction losses, the relative inefficiency of the pump and it's basic design parameters. Of these, the design parameters will be the ultimate determination of the amount of power consumed.

Evan
07-30-2006, 10:50 PM
Well, you are finally getting closer to the correct answer and you will admit that power is being consumed.
I never said otherwise.

Quote myself, the first response to you:


In this case we are talking about a submersible well pump. They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.
As the effective displacement of the pump nears zero the pump begins to operate in a very inefficient manner due to cavitation and stirring losses, as I pointed out before. Still, the current is lower the greater the head.

RobDee
07-30-2006, 10:52 PM
Evan:
The lower the head the more work it does and the higher the current.

Rob
"Ok this is clearly wrong. The lower the head = less water moved and the thus the less current required. "

Evan
What!? My statement is correct. Read it again. Lower head = higher flow rate.

Lower head = less work done!! Geeeeeze!!!!!

Look at the formula:

HP = (flow x head x sg)/ 3906 x eff.

What happens if the head goes down? The HP required goes down!! The current goes down. THE FLOW RATE ONLY GOES UP IF THE HP STAYS THE SAME.

My god man this is surreal. If every thing stays the same then the HP has to go DOWN.

HP isn’t a constant here!! It’s requirement is the result of the criteria established by the formula.

Every thing I have said so far is correct.

Oh boy! This is sad you believe this.

It occurs to me that perhaps you have the term head confused. Low head means less resistance to the pump, less lift, shorter pipe.

Yikes!! I'm pinching myself here. First you wash physics down the drain now it's math!

CCWKen
07-30-2006, 10:57 PM
I'm glad we FINALLY got all this cleared up.

Don't you? :D

Evan
07-30-2006, 11:04 PM
Look at the formula:

HP = (flow x head x sg)/ 3906 x eff.

What happens if the head goes down? The HP required goes down!! The current goes down. THE FLOW RATE ONLY GOES UP IF THE HP STAYS THE SAME.
I see where you are confused by that. That formula assumes a constant flow rate. With the same pump at different heads the flow rate isn't constant and everything I have said is correct.

[edit]

Actually, the formula makes no assumptions, you do. Of course the HP requirement goes down with less head for the same flow rate. Normal submersible pumps are not hp/rpm adjustable so that isn't applicable.

J Tiers
07-30-2006, 11:09 PM
Holding a weight static does require work. If it did not, you would be able to hold a 500# weight at arm's length for hours on end as long as someone else placed it there.

Installing a check valve above the pump discharge will permit removal of the pressure head from the pump, and allow the pump to be stopped without the water flowing back. As soon as the check valve closes and the pump stops, the pressure at the pump goes to zero.


To DYNAMICALLY hold the pressure may require work.

To simply HOLD it as the check valve does, requires NO WORK, because there is pressure but no movement.

.
.

Evan is correct that he has not said anything that is *never* true.

BUT, those things are not necessarily true ABOUT THE POINT BEING DISCUSSED.

The dynamic head is ALL the head for pumping horizontally, where there is no static head. Obviously "dominant".

But for a well pump, the pipe and flow rate are not likely to be sized such that the dynamic head is dominant. So the things said were NOT very true about that case, assuming a decent design.

It is necessary to correctly choose the applicable formulas to suit the details of the problem at hand. It is a lesson that is useful to learn for any technical field, as I am sure Evan is quite aware, since he is a very smart person.

As for the orignal pump, I don't think it will burn up if cycled, and probably not if run more continuously. But it is easy enough to boost the voltage if there is a question, since we won't be seeing the motor any time soon!

Are we done?

RobDee
07-30-2006, 11:12 PM
Still, the current is lower the greater the head.

The current can’t be lower the greater the head. That would mean that the deeper the well the less HP required.

Go back and look at the formula. When the head goes up the HP goes up. EVERYTHING ELSE STAYS THE SAME. We don't manipulate the formula to fit our thinking.

P=EI, P= watts, HP = watts, If watts goes down and 'E' is a constant the 'I' goes down.

If 'P' goes down then HP goes down.

J Tiers
07-30-2006, 11:18 PM
The current can’t be lower the greater the head. That would mean that the deeper the well the less HP required.

Go back and look at the formula. When the head goes up the HP goes up. EVERYTHING ELSE STAYS THE SAME. We don't manipulate the formula to fit our thinking.

P=EI P= watts 1 HP = 746 watts

If 'P' goes down then HP goes down.

Evan is assuming a system with flow losses much much greater than the static head......

In that case, it could be possible...... IF flow decreases enough due to head. But, if the flow head is so much greater a part of the TDH, the static head increase might not make much difference, and his example may not actually work.

Possibly there is a combination of values where it will.

Evan
07-30-2006, 11:18 PM
That would mean that the deeper the well the less HP required.

Depends on what flow rate you are wanting.


When the head goes up the HP goes up. EVERYTHING ELSE STAYS THE SAME. We don't manipulate the formula to fit our thinking.

No we don't. When the head goes up the HP goes up to maintain the same flow rate.

Evan
07-30-2006, 11:24 PM
Evan is assuming a system with flow losses much much greater than the static head......

In that case, it could be possible...... IF flow decreases enough due to head. But, if the flow head is so much greater a part of the TDH, the static head increase might not make much difference, and his example may not actually work.

Possibly there is a combination of values where it will.

It works in all cases for a centrifugal pump. At some point as the head increases the flow rate drops so that the increasing head due to gravity exceeds the increase in head due to pipe friction due to increased depth making the pipe longer. Even then the only thing the motor sees is the TDH.

RobDee
07-30-2006, 11:26 PM
I see where you are confused by that. That formula assumes a constant flow rate. With the same pump at different heads the flow rate isn't constant and everything I have said is correct.

[edit]

Actually, the formula makes no assumptions, you do. Of course the HP requirement goes down with less head for the same flow rate. Normal submersible pumps are not hp/rpm adjustable so that isn't applicable.

The formula determines how much HP is required for a specific depth, flow, etc.

If we design a pump for 100 feet and 10 gpm all other factors being equal does it require the same HP as if we design for 200 feet? NO. The HP requirement goes up.

You’re assuming a constant flow rate. The formula doesn’t. You’re making the assumption. The formula allows us to determine how much HP we need at a specific depth.

Now we’re getting down to basic math skills. You have to be kidding. This is 9th grade Algebra.

J Tiers
07-30-2006, 11:35 PM
If you are not referring to flow head, as I understood you to mean..........

I think a glance at the pump curves would show that there is an area at the top limit of head for the pump in question where that may be true....

After the "inflection point" of the pump curve, the flow drops off so that it crosses the HP lines going lower, and shows decreasing flow.

But below the inflection point, the flow drops off slower than teh head increases, so that power requirement increases with head.

In any case, to get the SAME flow at a larger head, requires more power......

The other biz is a special case of a certain pump type, and a certain portion of the pump capability.

Evan
07-30-2006, 11:39 PM
If we design a pump for 100 feet and 10 gpm all other factors being equal does it require the same HP as if we design for 200 feet? NO. The HP requirement goes up.

Yes it does, for the same flow rate.

You’re assuming a constant flow rate. The formula doesn’t.

No, as I said the formula doesn't assume anything.

You’re making the assumption. The formula allows us to determine how much HP we need at a specific depth.

Yes it does. Same flow rate at greater depth requires more horsepower, as per the formula. That does not contradict anything I have said.


How about your basic reading skills?

RobDee
07-30-2006, 11:42 PM
Rob:
“That would mean that the deeper the well the less HP required. “

Evan:
Depends on what flow rate you are wanting.

Everything else stays the same. If we want to find out how one aspect of a fomula affects the output then we only vary that aspect. Let's try to do that from now on.


When the head goes up the HP goes up. EVERYTHING ELSE STAYS THE SAME. We don't manipulate the formula to fit our thinking.

No we don't. When the head goes up the HP goes up to maintain the same flow rate.


It’s your statement , you’re contradicting yourself this is your statement:

Still, the current is lower the greater the head.

When we make a statement like this it is technically understood that all other factors remain the same. Your statement is false. It fails the formula.

You’re tripping over your own words again.

Evan
07-30-2006, 11:44 PM
But below the inflection point, the flow drops off slower than teh head increases, so that power requirement increases with head.

That reflects the reality of the flow rate becoming low enough that friction no longer dominates the flow rate. It also reflects the fact that the pump losses dominate the current consumption because it is operating in a very inefficient manner. As I said very early:



They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.

JCHannum
07-30-2006, 11:44 PM
.
Are we done?

I am almost.

Evan said;

"It does not require power to maintain a static, non-moving head of water any more than it does to support a weight on a shelf. The energy to create the head of water has already been expended and converted to potential energy. "Holding" that non-moving water above the pump is not responsible for energy consumption by the pump. If you place a check valve at the pump outlet to prevent the water from back flowing the situation won't change the slightest bit. The head of water would remain and so would the the pressure even with the pump turned off. No energy would be expended. Turn the pump back on and still nothing changes as the kinetic energy of the water from the pump will overcome the pressure of the static head of water and open the valve if it is able. This method is used although the valve is placed at the inlet to prevent the pump running dry. The pressure does not come from the pump, it it is a property of the water that has left the pump. The sole job of the pump is to move water by giving it kinetic energy."

I refer specifically refer to; "It does not require power to maintain a static, non moving head of water...." and; "'Holding' that non moving water above the pump is not responsible for energy consumption by the pump."

These statements seem to indicate that the water will remain at that head or pressure by some magical means other than power being inputted to the pump. He is likening this to the weight on the shelf, when it is actually more comparable to the weight at the end of your arm.

Once a mechanical means is installed to hold it, either a shelf for the weight or a check valve for the head pressure, the power requirement will go away.

This is part of my initial comment, and I will stick by it;

"Depending on the application and pump design, the motor may or may not hold up. The best source of information would be to contact the manufacturer or supplier. More information about the performance characteristics of the pump are needed to give good advice."

Recommending a course of action based only on assumptions and minimal information is not my idea of doing someone a favor. In this case misinformation might not create serious problems, but it is best to warn of any potential hazards.

Evan
07-30-2006, 11:47 PM
RobDee said


It’s your statement , you’re contradicting yourself this is your statement:

Still, the current is lower the greater the head.

There is no contradiction.
Implicit in my statement is that we don't change the type of pump or it's horsepower as we change the head. Everything I have said has clearly been related to the same pump at different depths. You are attempting to confuse the issue by posting a strawman argument and taking my words out of context.

Evan
07-30-2006, 11:55 PM
JC said

I refer specifically refer to; "It does not require power to maintain a static, non moving head of water...." and; "'Holding' that non moving water above the pump is not responsible for energy consumption by the pump."

These statements seem to indicate that the water will remain at that head or pressure by some magical means other than power being inputted to the pump. He is likening this to the weight on the shelf, when it is actually more comparable to the weight at the end of your arm.

Once a mechanical means is installed to hold it, either a shelf for the weight or a check valve for the head pressure, the power requirement will go away.
Whether it seems correct or not, it is correct. There is nothing magical about it. When the pump has reached the maximum head that it can pump the water above the pump is not moving. No further energy is added to it. The forces are balanced and the net force between the pump and the head are zero. It makes no difference if we add a valve or not. The water from the head cannot enter the pump because of the kinetic energy of the water in the pump. The water in the pump cannot leave the pump because of the potential energy in the head. They balance precisely so no exchange of energy occurs. (ignoring small amounts of heat transfer and turbulence at the interface)

darryl
07-31-2006, 12:12 AM
Heck, all the guy wanted was a drink of water from 500 feet down:)

Nice discussion, guys. Got my brain going. I hope I don't dream of water flow tonite, I might pee the bed:)

JCHannum
07-31-2006, 12:14 AM
JC said
The water from the head cannot enter the pump because of the kinetic energy of the water in the pump.

That head is the total dynamic head, which is the total head when the pump is running. The energy to maintain it has to come from somewhere.

RobDee
07-31-2006, 12:30 AM
Evan:
The lower the head the more work it does and the higher the current.

Rob
"Ok this is clearly wrong. The lower the head = less water moved and the thus the less current required. "

Evan:
What!? My statement is correct. Read it again. Lower head = higher flow rate.

The lower the head the more work it does and the higher the current.

OK, here is where you manipulate the formula and increase the flow rate to fit you thinking because in order for the statement you made to be true the flow rate must increase. However, below when confronted about this you say,”
Same flow rate at greater depth requires more horsepower, as per the formula. That does not contradict anything I have said.

but it clearly does because in order for your statement:

The lower the head the more work it does and the higher the current.

to be valid the flow rate must go up and not just to balance the equation. It must go above its previous value for the current to go up.

Again:
HP = flow x head x sg / 3906 x eff.

Now remember that we are talking about how much power is required at different heads and current is a direct aspect of that power. This is why we have different HP pumps and HP is a product of the formula. We use the formula to establish the HP required.

RobDee
07-31-2006, 01:01 AM
It’s your statement , you’re contradicting yourself this is your statement:


Evan,
Still, the current is lower the greater the head.

There is no contradiction.
Implicit in my statement is that we don't change the type of pump or it's horsepower as we change the head. Everything I have said has clearly been related to the same pump at different depths. You are attempting to confuse the issue by posting a strawman argument and taking my words out of context.

Still wrong.

OK let’s keep the same HP pump.

According to the formula if the HP doesn’t change and we change the flow or the head proportionately the current is equal at all depths. Not less at greater heads as you propose. A pump at 200 feet should require less HP then the same pump at 100 feet if you are correct.

Maybe as we approach the two limits there might be some change but the pump will exhibit the same characteristics as the graph through the majority of its designed range. That means the current will be relatively constant and not less at greater depths within the designed range.

This I can accept.

Evan
07-31-2006, 01:37 AM
According to the formula if the HP doesn’t change and we change the flow or the head proportionately the current is equal at all depths. Not less at greater heads as you propose. A pump at 200 feet should require less HP then the same pump at 100 feet if you are correct.

That is what I maintain. The reason it is so is because of the non linear relationship between horsepower, head and flow rate. As flow rate goes down the horsepower required to maintain that flow rate goes down even faster, by the third power. Third power curves are very steep. The pump doesn't "know" that it is deeper. It only sees less water to move. It imparts the same amount of kinetic energy to the water regardless of the head. The head determines how much of that water leaves the pump. The less that leaves the pump the less that must be accelerated.

Naturally, inefficiency begins to play a large part as the flow rate drops well below what the pump is capable of with no head. But, the power expended to actually pump water goes down as the head goes up until it reaches zero when no water is pumped.

Do not take this to mean the pump does not consume power at that state. As I have repeatedly said, it does due to losses. The pump does not expend power maintaining the head at zero flow. It cannot, no energy is added to the head.

When the head is static with the pump running the head is in a steady state. A steady state (the head of water) neither gives up energy or absorbs it. All energy consumed by the pump in that steady state is due to loss in the pump since no water is pumped.

Once again, it is no different than a weight on a shelf. It took energy to lift the weight to the shelf but once it is there no further energy is required.

I reiterate, the pump does not produce pressure, it gives the water kinetic energy. That kinetic energy is converted to potential energy as the water moves through the pressure gradient in the head. The loss of kinetic energy is exactly equal to the gain in potential energy, minus friction. If the water cannot leave the pump no kinetic energy is transferred.

J Tiers
07-31-2006, 09:11 AM
That reflects the reality of the flow rate becoming low enough that friction no longer dominates the flow rate. It also reflects the fact that the pump losses dominate the current consumption because it is operating in a very inefficient manner. As I said very early:

A) that is for a specific pump type, not a general rule.

B) the friction does NOT dominate the flow rate. Use teh calculator at the site you posted.... quite reasonable pipe sizes give very low friction heads, FAR LESS THAN THE STATIC HEAD. Pump inefficiency...... now THERE you are getting to the meat of the matter....

C) But, even there, it only matters where on the pump curve you are.... Which is why pumps may have multiple stages.... so each stage operates in an area where it is more efficient. You can pick a case where power decreases with head, because flow is approaching zero.

D), That does NOT mean the flow external to the pump is influencing flow head that much, it may simply mean that pump with those stages is stalled out and just flailing water around in the casing....

e) IF it moved the same water at a higher head, it would take more power.

JCHannum
07-31-2006, 11:32 AM
The fallacy of Evan's reasoning is applying pump theory to the performance of an individual pump.

A centrifugal pump will consume maximum power at maximum flow, and minimum power at maximum head. What those maximum and minimum figures are depends solely on the pump design parameters. Without detailed knowledge of those parameters, as well as the system's conditions, any prediction of the pump's performance is merely a guess.

While the various laws are indeed true, it is the design of the pump that will ultimately determine it's performance under a given set of conditions.

Although it is possible for a pump to have zero flow at it's shutoff head, maintaining that head consumes power. When the pump is in that condition, the head is a dynamic head and requires power to maintain it. The loss of flow is due to the design and efficiency of the impeller and casing.

Leaving theory and returning to reality, the only way to determine how a given pump will perform is to examine and thoroughly understand the performance curve of that particular pump. Almost all performance curves are closed, and the operation of the pump outside of the curve is unpredictable. You cannot apply the performance of one type of pump to any other pump except in very general terms, and using general terms to predict an outcome is never a good idea.

The statement "The pump gives the water kinetic energy" is only partially true. The pump sitting there with the motor turned off is capable of nothing at all. It is only when the impeller is turning that anything as accomplished. Turning that impeller requires energy, which is then converted into the pump's output which is some combination of head and volume.

Interestingly, there are many high presure centrifugal pumps that will not pump unless a head is present. If started a with an open discharge, they will not move any water. They must be started against a closed discharge, and the discharge valve is opened only after a significant pressure is attained. Again, it is all in the design.

SherpaDoug
07-31-2006, 01:01 PM
It is easy to convert 208VAC to 220VAC. What you need is a 220V to 12V bell transformer whose SECONDARY is rated 12V at whatever amperage you want to use at 220V. Wire the transformer primary to the 208V input and wire the secondary in SERIES with the load (pump). Depending on which way you wire the secondary you will either get 208V + 12V = 220V, or 208V - 12V = 196V. If you get it wrong on the first try just swap the secondary wires.
To be percise the 220V transformer running on 208V will only give 11.3V for a net of 219.3V, but the pump will never notice the difference. If you really want closer to 240VAC find a transformer with a higher secondary voltage (240V - 208V = 32V). In the field this is called a "Boost Transformer".

SherpaDoug

Evan
07-31-2006, 01:31 PM
The fallacy of Evan's reasoning is applying pump theory to the performance of an individual pump.
Fallacy? What fallacy?


A centrifugal pump will consume maximum power at maximum flow, and minimum power at maximum head.
That is what I have been saying all along.


Additional confirmation of this may be seen here (http://www.irrigation.org.au/download/irrig2006/presentations/W2/W2%20Martin%20Newey.pdf) It shows that reducing the output flow by "throttling" increases the pressure and reduces the power consumption.


Further:


The power consumed by the pump (at the pump shaft) is:
http://vts.bc.ca/pics/pumppow.gif
where Ppump: power consumed at the pump shaft
SG: specific gravity of the fluid
delta HP: Total Head
Q: flow through pump
eta pump : pump efficiency


http://www.fluidedesign.com/solved_pumping_problems.htm#q414



Also, more on power curves here:
http://www.lawrencepumps.com/newsletter/news_v02_i11_nov05.html





JT,

The calculation of pipe loss is far from simple and the pump curve doesn't show the story.

The equations for friction loss are complex and iterative so tables are normally used. Here (http://www.mcnallyinstitute.com/Charts/friction_1.2_1.5.html) is one. Assuming the pump can supply the flow rate with 1 1/4" sched 40 pipe at 85 gpm the loss of head due to friction equals the gravity head. At 150 gpm it is triple the gravity head. However at 50 gpm it is only 1/3 the gravity head. It's that non-linear exponential curve at work.

As for power expended to maintain the shut off head, there isn't any. Water can't enter the pump because it is in "churn". If you "magically" place a cap on the outlet nothing changes at all. The pump still churns and the head stays the same. Power consumption does not change.

J Tiers
07-31-2006, 01:42 PM
JT,

The calculation of pipe loss is far from simple and the pump curve doesn't show the story.

The equations for friction loss are complex and iterative so tables are normally used. Here (http://www.mcnallyinstitute.com/Charts/friction_1.2_1.5.html) is one. Assuming the pump can supply the flow rate with 1 1/4" sched 40 pipe at 85 gpm the loss of head due to friction equals the gravity head. At 150 gpm it is triple the gravity head. However at 50 gpm it is only 1/3 the gravity head. It's that non-linear exponential curve at work.

As for power expended to maintain the shut off head, there isn't any. Water can't enter the pump because it is in "churn". If you "magically" place a cap on the outlet nothing changes at all. The pump still churns and the head stays the same. Power consumption does not change.

I am well aware of flow head and its character.....

The pump curve wasn't what I referred to..... But at reasonable home well pump flows, and quite reasonable sized pipes, the head calculator on your favorite site shows the flow head quite a bit smaller than the static.....

What you say above just indicates that it is possible to get a reasonable system design that avoids really stupid problems like a flow head higher than the static head...... And the pumps for that use I have dealt with (and we are presumably discussing) don't produce 85 GPM at 350 foot heads...... so they don't need nor use big pipes.

.
.
It SURELY DOES require power to hold the head IF YOU USE A CENTRIFUGAL PUMP TO DO IT. If you don't think so, shut off the switch.

Of course, a check valve will do the job too, and requires zero power.

So a pump simply isn't efficient at holding a head of water..... but it may do fine for actual pumping... the check valve won't do that very well without help.

Evan
07-31-2006, 03:08 PM
It SURELY DOES require power to hold the head IF YOU USE A CENTRIFUGAL PUMP TO DO IT. If you don't think so, shut off the switch.
Not according to physics. The power consumed by the pump is NOT going into the head of water. This isn't a matter of semantics either. The pump enters a condition where it is the same as a closed valve on the outlet. In fact, the pump does normally have a foot valve. With no flow the foot valve will be closed. So, what is holding the head? The pump or the foot valve?

As for flow rates, common sizes of deep well pumps will provide from 20 to 90 gpm at 100 feet. The split between gravity head and friction will be close to equal for many installations. It isn't just the lift that must be considered. The water has to get to the pressure tank. In some cases that adds a lot more pipe. My well is about 75 feet from the house plus 350 feet deep so I have over 400 feet of pipe. A neighbour has his well at the bottom of the hill so he has 100 feet of lift to get out of the well, another 200 feet lift to get up the hill and a total run of about 500 feet of pipe. Deep wells are the norm here and one friend of mine has a well 450 feet deep.

[edit]
Note: In my well as in most the water is much higher than the pump. The amount of pipe is much longer than the lift.

JCHannum
07-31-2006, 03:10 PM
"A centrifugal pump will consume maximum power at maximum flow, and minimum power at maximum head.
That is what I have been saying all along."

No, at one point, you said:

" "Holding" that non-moving water above the pump is not responsible for energy consumption by the pump."

My statement uses the words maximum and minimum, never zero. It also includes two more sentences, which you conveniently omit again.

As I stated the fallacy of your reasoning is that you are applying theory to actual. In a perfect world, that might work, but the world of centrifugal pumps is far from perfect.

If you will take the time to look at a few different styles of the Goulds, or any other major, full line manufacturer of pumps, and study the various curves, it will become quite obvious that a centrifugal pump can be designed to handle any condition of flow & head imaginable. No one pump will handle every situation, and if a pump is operated out of its design parameters, no one can accurately predict how it will behave.

With the limited amount of information given in the initial question, it is impossible to accurately predict how that pump will react to the lower voltage. The best guess is that it might, but that is still nothing more valid than a guess. There are several basic questions that need to be asked, and no one has asked them as yet.

Evan
07-31-2006, 03:17 PM
Did you actually visit the sites I posted? If no water is being pumped then no energy is consumed to pump water. The gpm term "Q" goes to zero in the formula I posted for power consumption.

The operation of centrifugal pumps is similar enough that generalizations are valid. They all act as variable displacement pumps. That is the primary governing factor that determines the main characteristics. Many of the sites that discuss these characterisitics use generic performance curves to illustrate the points. That is because a generic curve is close enough for the job of showing the general operation of centrifugal pumps.

JCHannum
07-31-2006, 03:20 PM
Not according to physics. The power consumed by the pump is NOT going into the head of water. This isn't a matter of semantics either. The pump enters a condition where it is the same as a closed valve on the outlet. In fact, the pump does normally have a foot valve. With no flow the foot valve will be closed. So, what is holding the head? The pump or the foot valve?

In the case of a deep well pump, the foot valve serves merely to hold the water column after the pump stops. This to eliminate the need to refill the system or to maintain the pump prime. With the motor running, the valve clapper or ball, depending on valve type, will merely drop due to gravity, and has no effect. The water column will remain at the dynamic discharge head as long as the pump continues to run.

Evan
07-31-2006, 03:28 PM
My point is that there is no distinction between the pump "holding" the head and the valve. The fact remains that no energy is introduced to the column of water above the pump when that water is not in motion. The fact also remains that the pump does not of itself produce pressure. That is shown by measuring the pressure of the flow when at zero head. The flow is maximum, the pressure is zero.

JCHannum
07-31-2006, 03:42 PM
Evan, I have visited those sites, and many, many others, as well as having worked closely with most major pump manufacturers and suppliers during 40 plus years in plant, facilities and maintenance engineering in the petrochemical, consumer chemical and food processing industries. I have specified, purchased, installed and maintained hundreds of pumps of all types in all applications from micro dispensing metering pumps to several million gallon a day waste handling pumps.

Limited as my experience may be, the one simple fact I have learned is that you cannot expect a generalized statement to apply across the spectrum of centrifugal pumps.

The question of where the power is being applied is moot. Power is being expended to maintain the dynamic discharge head imposed on the pump.

In the equation posted, the power required will also go to zero if any other number on the top is zero. Basic math there, and why the equation does not hold true in all conditions.

jontwo13
07-31-2006, 03:48 PM
I think I will leave it like it is it works fine. If I change the wireing it may not last long. Let the person who is renting the house pay for the electric for the pump.

Evan
07-31-2006, 03:51 PM
In the equation posted, the power required will also go to zero if any other number on the top is zero.

Of course, but the flow rate is the only number that can go to zero while running in normal operation. This happens on my well if I draw it down too far which is why the pressure switch I have cuts off the power if the tank pressure drops below 20 psi.

JCHannum
07-31-2006, 03:55 PM
That is shown by measuring the pressure of the flow when at zero head. The flow is maximum, the pressure is zero.
Er, didn't you say you couldn't measure pressure of flow?

The pressure of the water column above the pump is meaningless. Refresh your memory on the difference between static and dynamic discharge heads.

Pump performance is measured at the pump discharge. If the pump is running and producing a dynamic discharge head, there will be pressure at the discharge. The only way this pressure is maintained is by turning the impeller and this will consume power. If the pump is stopped, and a check valve is holding the water column, this is static discharge head, no power requirement and a totally different situation.

Evan
07-31-2006, 04:08 PM
Er, didn't you say you couldn't measure pressure of flow?

You can't measure any pressure in a free flow because there is no pressure gradient. No kinetic energy has been converted to potential energy.


Pump performance is measured at the pump discharge. If the pump is running and producing a dynamic discharge head, there will be pressure at the discharge. The only way this pressure is maintained is by turning the impeller and this will consume power. If the pump is stopped, and a check valve is holding the water column, this is static discharge head, no power requirement and a totally different situation.

The head is also static at shut off when the pump is running. The pump is considered to be "churning" or idling.

JCHannum
07-31-2006, 04:24 PM
Of course, but the flow rate is the only number that can go to zero while running in normal operation. This happens on my well if I draw it down too far which is why the pressure switch I have cuts off the power if the tank pressure drops below 20 psi.

The key words that confirm my argument are "while running in normal operation". This means within it's design parameters. Once a pump is running off it's particular curve, the theories can not be applied.

JCHannum
07-31-2006, 04:34 PM
The head is also static at shut off when the pump is running. The pump is considered to be "churning" or idling.

Not true. The pump is producing dynamic discharge head as long as it is running. If you reduce the power input, the dynamic discharge head, with no check or foot valve in the system, will decrease as power is reduced.

J Tiers
07-31-2006, 05:46 PM
Not according to physics. The power consumed by the pump is NOT going into the head of water. This isn't a matter of semantics either. The pump enters a condition where it is the same as a closed valve on the outlet. In fact, the pump does normally have a foot valve. With no flow the foot valve will be closed. So, what is holding the head? The pump or the foot valve?

With no valve, the only thing holding the water column against gravity is the pump. I will guarantee that with no check valve, if you shut off the pump the water column will drop... what the power is doing isn't extremely important to this point........ ;)

However, you are doing no work on the water in the pipe, although you may be doing plenty on the water in the pump casing as it swirls. Hence a check valve will hold as well or better, and consume no energy doing it.

Actually, the whole statement was meant to be a bit of a reducto ad absurdum.... any energized pump is drawing power, whether or not it is doing anything useful.... Obviously you wouldn't use a pump, which draws SOME power if running, to hold the water column without some really good reason.

And equally obviously, the check valve uses no power, yet it holds the column...... A far better solution, and proof that no energy (in our reference frame) is required to actually hold the water column motionless.

RobDee
08-13-2006, 07:22 AM
I’ve been away for awhile but I did want to finish my views on this subject.


According to the formula if the HP doesn’t change and we change the flow or the head proportionately the current is equal at all depths. Not less at greater heads as you propose. A pump at 200 feet should require less HP then the same pump at 100 feet if you are correct.


That is what I maintain. The reason it is so is because of the non linear relationship between horsepower, head and flow rate. As flow rate goes down the horsepower required to maintain that flow rate goes down even faster, by the third power. Third power curves are very steep. The pump doesn't "know" that it is deeper. It only sees less water to move. It imparts the same amount of kinetic energy to the water regardless of the head. The head determines how much of that water leaves the pump. The less that leaves the pump the less that must be accelerated.


First, you’re contradicting yourself. As flow rate goes down so does the HP is correct and fits the formula. Of course the pump doesn’t know it’s deeper but it still must adhere to the formula and simple physics! Go back and look at the formula again. I don’t care how steep any curves are your answer defies the formula and my statement above it.

And my answer is not a straw man. The original question was whether a pump could be run at less voltage. Less voltage equals less HP. (HP = watts, watts = EI, E = Volts, p goes down then HP goes down)

I said basically the power required depends on the head. You said “nope”.

The formula clearly shows us the answer. As the head goes down, with all other factors being equal, the HP required to maintain that head also goes down. Period!

You simply can not keep the HP a constant in the formula. That is basically his question. Will a 240 volt well motor run at 208 volts? Yes, absolutely. What changes? The HP the motor is capable of producing changes. So if we take a 240 volt well pump and run it at 208 volts what will change in our well? According to the formula the head, the flow rate or both must also go down. Period.

Here is what Shurflo shows:

http://www.mrsolar.com/page/MSOS/CTGY/shurflo