Ratio of Pressures Involving Piston Strokes

[eoneil]

Homework Statement

If the cross-section of the barrel of a pump is 5 square centimeters, the stroke of the piston is 20 centimeters, and the volume of the receiver, in which the air is drawn from, is 1 liter, calculate the ratio of the pressure of the air in the receiver to the original pressure after three strokes of the pump.

Homework Equations

p=F/a
p=hDg
p1V1 = p2V2

The Attempt at a Solution

Trying to visualise the problem. Compare the pressure before to the pressure after. The piston could be going downward, drawing air from below. The piston sits on top of the gas, adding its weight to the pressure of the atmosphere pushing down on the gas. That gas, 1L (0.001m^3), would then exert hydrostatic pressure upward on the piston, at 10^5Pa.

p=F/a = (9.81m/s^2)/(0.01m^2)=981
Using p1V1=p2V2, solving for either pressure.

The volume of space the piston compresses= (0.05m^2)(0.20m)= 0.01m^3 (is it correct to assume that the cross-section of the pump, 0.05m^2, is incorporated into the volume calculation?)

The weight of the piston is unknown and the temperature doesn't factor in either. So no equation including those two variables is necessary.
The ratio involves only the volume of air displaced by the piston, and the volume of air in the receiver.

Using p1V1=p2V2, we can solve for p2.
p2=p1V1/V2=(10^5Pa)(0.01m^3)/(0.001m^3)= 14841Pa, this is the pressure the 1L of air in the receiver exerts on the piston.

Then the ratio of pressure would be 10^5Pa/14841Pa= 1/10th

Is this right? How do I incorporate 3 strokes of the piston?

 

[gneill]

Presumably there's a one-way valve between the piston end and the receiver tank. It will open when the pressure in the piston barrel equals that in the receiver.

So, the piston compresses the volume of air in the barrel until the valve opens (doing some amount of work on that air), then after the valve opens, the piston is pushing the air into the receiver and doing more work while moving the air and compressing the larger volume. When the piston stroke is complete, the receiver will have a new amount of gas at a higher temperature and pressure. Repeat three times.

Could you do it all in one go if you pretend that the piston is three times as long?

 

[eoneil]

I thought the receiver was the component responsible for pressure onto the piston, not the inverse. I guess I'm having trouble conceptualizing it. I even looked for a couple piston animations online.

Do you mean to adjust the calculation to make the piston 3x its original volume?

 

[gneill]

I thought the receiver was the component responsible for pressure onto the piston, not the inverse. I guess I'm having trouble conceptualizing it. I even looked for a couple piston animations online.

Do you mean to adjust the calculation to make the piston 3x its original volume?

I was just wondering if the amount of work done would be the same in both cases. I can't think why not. Never mind, carry on.

 

[rwisz]

Your approach and calculation seem correct. To incorporate 3 strokes of the piston, you would need to consider the volume of air displaced by each stroke. Since the piston stroke is 20cm, that means each stroke would displace 0.01m^3 of air (as you calculated). Therefore, after 3 strokes, the total volume of air displaced would be 0.03m^3. Using p1V1=p2V2, you can calculate the new pressure after 3 strokes as p2=p1V1/V2=(10^5Pa)(0.03m^3)/(0.001m^3)= 3x10^6Pa. Then the ratio of pressure would be 3x10^6Pa/14841Pa= 202.3, which means the pressure has increased by a factor of 202.3 after 3 strokes.