# Gauge pressure

1. Jun 13, 2005

### MAPgirl23

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure 1.01x10^5 Pa and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa.

What then is the gauge pressure at the bottom of the water?
** I got 4000 Pa

By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

** Here I was thinking of 1500 Pa / 1000 = 1.5 --> 1.5/9.8 ???

2. Jun 13, 2005

### OlderDan

That looks like the right idea. You could be a little more careful with units, and writing actual equations.

3. Jun 14, 2005

### quark

When you maintain a gauge pressure of 1500Pa above water, to get a final gauge pressure of 2500Pa at the bottom of the vessel, the water column should exert a pressure of 1000Pa. If 'x' units of water column creates a pressure of 2500Pa, what fraction of 'x' creates 1000Pa? You have to drain the rest of fluid.

What does "Here I was thinking of 1500 Pa / 1000 = 1.5 --> 1.5/9.8 ???" mean?

4. Jun 14, 2005

### OlderDan

The problem asks how much the water level needs to be reduced. That can be found directly from the pressure change without calculating the initial and final heights, assuming of course there is at least enough water to drain. It is certainly worth checking that condition, but since the inital water column pressure was 2500Pa you will be able to reduce it by 1500Pa without running out of water.

The final statement surely needs better presentation, but it does awkwardly reflect the relationship between pressure, gravitational constant, height and density needed to find the change in height.

$$\Delta P = \frac{\Delta{W}}{A} = \frac{ \rho \Delta{V} g}{A} = \frac{ \rho A \Delta{h} g}{A} = \rho \Delta{h} g}$$

$$\Delta{h} = \frac{ \Delta P}{\rho g}$$

5. Jun 14, 2005

### MAPgirl23

so is the height = 1500 Pa/(1000 * 9.81) ?

6. Jun 14, 2005

### MAPgirl23

If the air pressure is increased by 1500 Pa the pressure at the bottom of the container is 4000 pa

If the original head of water gives a pressure of 2500 Pa and the water is to be drained to reduce the pressure from the head of water to 1000 pa
Then the level is to be reduced by 1500/2500 = 3/5ths

7. Jun 14, 2005

### MAPgirl23

how do I express that in cm?

8. Jun 14, 2005

### OlderDan

That would be the change in height as the water is drained. If you keep track of your units in your calculations instead of just writing numbers, your answer will be dimensionally correct. If you need to express the answer in cm, you might have to convert an answer in meters to cm. I think you know how to do that.

9. Jun 15, 2005

### quark

If we have to go by p = hdg then the initial data is redundant. That is why I thought that the liquid column height should be expressed in terms of percentage.

10. Jun 15, 2005

### OlderDan

I don't see any redundancy. The data says the initial pressure is one atmosphere, and the guage pressure at the bottom of the water column is 2500 Pa. From that you certainly can deduce the initial height of water using the pressure and density, but there is no redundant specification of that height. Raising the gas pressure increases the guage pressure by the same amount (1500 Pa) to 4000 Pa as calculated by the OP. Lowering the water level while maintaining gas pressure to lower the pressure by 1500 Pa leads to a determination of the change in height of water. The actual height of the water has no bearing on how much has to be drained, as long as there is enough water to do it. The initial guage pressure and water height could have been anything, and the guage pressure changes and the change in water height would have been the same. The percent change is not material to the problem.