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The Artful Bodger
12-23-2010, 12:28 AM
.... I have a need to transform air pressure by about 10:1. For example if the pressure is 2.5psi and want to produce a pressure of .25psi.

OK, I guess the obvious way is to use interconnected pistons of differing surface area but I am wondering at the alternatives. This is for a form of novelty instrumentation.

dp
12-23-2010, 12:45 AM
It's called a lever! :)

macona
12-23-2010, 12:51 AM
Just build a t-shaped piston-cylinder. Big side pressure in, small side has the gauge. 10x the surface area on the big side. Shouldnt need a return spring if it is going to a pressure gauge.

If you need a constant flow at 1/10th the high pressure you could use an analog pressure transducer through an op-amp to a electrically actuated air regulator.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=190462266147#ht_500wt_1156

darryl
12-23-2010, 01:29 AM
I can see a constant flow using a venturi effect- or perhaps the opposite of it. A stream of air slightly impinges on the opening of a tube, and some mechanical adjustment sets the ratio between the pressures. If a constant flow isn't needed, a pair of bellows of some kind would do the job. A lever connects the two, and the location of the pivot point sets the ratio.

If the lower pressure air is going to be draining away through leakage or other, then an altered regulator of some sort could be used- the higher pressure controls the regulator, which is initially set to an output of 5 psi with a certain input psi. As the input pressure changes, the regulator valve sees the change and allows more output pressure- hopefully in a linear fashion.

Regulators work by having the output pressure close off a valve to the input pressure. A Diaphram moves the valve, responding to the lower pressure, and the large area of the diaphram coupled with the relatively small area of the valve makes it so the high pressure can be sealed off with little change in the pressure on the diaphram. If the diaphram is lots smaller, and the valve area bigger, the high pressure will overcome the valve and pressure will increase on the low side.

Basically what you're asking for is a very poor regulator, but good at what it's doing.

10 to 1- let's see- imagine an input tube with an area of say 1 sq inch. That tube is connected to another tube that's 10 sq inches in area, and protrudes into it for a short way. A piston in the larger tube has a section of smaller tubing in it that's a close fit over the input tube. This slides over the OD of the input tube, hopefully with minimal air leakage. The cavity on this side of the piston is open to atmosphere. Inside this second smaller tube is a gasket that can close off the input tube when it comes up against the end of it. When it's spaced away, air can flow around it and occupy the space in the larger tube, on the other side of the piston. This cavity may have an output tube, or maybe just a pressure indicator fed from it.

In any event, pressure coming in pushes the valve open, which moves the piston with it. When pressure is built up on the other side of the piston to 1/10 of the input pressure, the piston closes off the valve. A constant 1/10 pressure is maintained in the output tube. You may need to bleed off air on the low pressure side in some small amount so any leakage past the valve won't simply raise the secondary pressure slowly to the full input pressure.

Evan
12-23-2010, 01:33 AM
It's air so if a very small leak is ok then an orifice followed by a bleed needle valve will do the job. Needle valves are very linear over a wide range of pressure which is why they are used in carburetters. Admit air to the reservoir which then passes through a fixed orifice to a second reservoir and adjust the bleed for the correct pressure in the second reservoir. The ratio will be maintained unless the second reservoir is also feeding some other device.

The Artful Bodger
12-23-2010, 02:12 AM
All good information thank you folks.

Evan, I think yours might be the choice as it is nice and uncomplicated and I assume easy to adjust.

Now the project it to make a simple glass tube manometer mounted on the the den wall to indicate wind speed from a roof mounted pitot tube. I will be using water in the tube hence the need to reduce the pressure.

The Artful Bodger
12-23-2010, 03:01 AM
If I understand you Evan this is how it will look, only the upward tube would be visible the rest behind a panel. I am aiming for about 1 metre high.

http://farm6.static.flickr.com/5009/5284550647_0cb88bcdfa.jpg (http://www.flickr.com/photos/25239206@N06/5284550647/)
manotube (http://www.flickr.com/photos/25239206@N06/5284550647/) by aardvark_akubra (http://www.flickr.com/people/25239206@N06/), on Flickr

Air pressure at 30 mph is about 2.4 psi (I stand to be corrected) which would be about 5 feet of water obviously too much for a practical size instrument hence the need to reduce the pressure.

Now the next question is, will I get a linear scale for wind speed? I think it will be a square law scale?

darryl
12-23-2010, 03:16 AM
Wow! If I stood in a 30 mph wind, I would be facing an estimated pressure against my body of 1000 psi- assuming about 3 sq ft of frontal area. I don't think that's right. I think for what you're doing you might have to experiment a bit when some wind blows. I doubt you'd find the water column would change by so much that you'd have to reduce pressure-

The Artful Bodger
12-23-2010, 03:19 AM
Wow! If I stood in a 30 mph wind, I would be facing an estimated pressure against my body of 1000 psi- assuming about 3 sq ft of frontal area. I don't think that's right. I think for what you're doing you might have to experiment a bit when some wind blows. I doubt you'd find the water column would change by so much that you'd have to reduce pressure-

You could well be right as I am by no means overly confident on that bit of information.:)

Evan
12-23-2010, 07:09 AM
I think for what you are doing all you need is a bleed valve on the inlet of the manometre. That should serve for calibration purposes.

I think it is linear since the scale on a pitot powered airspeed indicator is linear.

rp designs
12-23-2010, 09:25 AM
If all I were trying to do was to scale a measurement I would build a twin manometer. The fluid you measure exerts pressure on two different types of fluid sending the columns up two different heights then it is just a matter of selecting fluid densities such that the ratio of heights works out to be your desired scaling.

Evan
12-23-2010, 10:27 AM
If all you want is an anemometer the simplest design is a light bulb, a few volts and a milliamp meter. Use a panel light bulb and knock the glass off it carefully so as not to damage the filament. Power it with just enough current so that the filament barely glows as seen in the dark. Measure the current in a wheatstone bridge so you can adjust the balance and therefor the calibration. The faster the wind the cooler the filament and the higher the current. It is called (wait for it...) a hot wire anemometer.

Peter.
12-23-2010, 12:45 PM
Isn't that how some mass airflow meters work in cars?

Black_Moons
12-23-2010, 01:28 PM
30mph != 2.4psi, Or I would of allready stuck a ram air scoop on my moped.

I seem to recall around 30mph it was closer to 0.1psi, It was however exponential.

http://www.homebuiltairplanes.com/forums/aircraft-design-aerodynamics-new-technology/8799-pressurizing-ram-air-scoop-2.html#post83919

Mentions that its only 0.95psi at 200knots. (With math to back it up)

I have read more then once that ram air is basicly 'usless' to improving engines unless you exceed 100kph, ie, Much less then 10% (or 1.4psi) improvement. So I highly doubt you are getting 2.4psi from 30mph.

The Artful Bodger
12-23-2010, 02:48 PM
Hmmmm.... obviously my figure of 2.4psi at 30mph is wildly off! Maybe 2.4lbs per sq ft! But I dont know, yet.:)

Evan, there are about a dozen varieties of anemometer that I have fiddled with over the years but this particular project is for a den decoration where the manometer will fit right in! Regarding your comments concerning the position of the restrictions I do think two restrictions are required. In my diagram, using the 'electric analogy', the two needle valves are to provide resistance and to get a voltage (i.e. pressure) change across a resistance there must be current flow. The first restriction reduces from pitot pressure to whatever is required by the manometer, the second restriciton reduces that manometer pressure to ambient.

P.S. For a really cool, almost "steam punk", example of a real anemometer using pitot pressure Google "Dines Pressure Tube Anemometer"

http://wwwdelivery.superstock.com/WI/223/1895/PreviewComp/SuperStock_1895-15669.jpg

Evan
12-23-2010, 06:02 PM
The first restriction reduces from pitot pressure to whatever is required by the manometer, the second restriciton reduces that manometer pressure to ambient.

Yes, of course. But, what I was thinking was bleeding air at the input of the manometer. The tube running from the pitot acts as the first restriction. The manometer itself is the second reservoir.

The Artful Bodger
12-23-2010, 06:19 PM
Yes, of course. But, what I was thinking was bleeding air at the input of the manometer. The tube running from the pitot acts as the first restriction. The manometer itself is the second reservoir.

That is true.