pistion cylinder 1

[nulax23]

I will be taking my PE exam soon so figured I would get PE review book at the library. Unfortunalty the answers were all ripped out of the back. I manged to answer almost all but this question has stumped me:

A perfectly insulated piston-cylinder assembly initially contains masses of water at
different conditions separated by an adiabatic membrane as shown in the figure.
The lower section of the piston contains 0.04 kg of saturated liquid at a pressure of
0.3 MPa. The upper section of the piston contains 8.0 kg of vapor at 200 ºC and 0.3
MPa. The diameter of the piston is 20 cm.
Now, the membrane is broken and the total volume of the system is observed to
change, A) Does the piston move up or down? What
distance in cm? B) In either case, since the piston moves, the
water will either gain or loose energy. And how much energy
does the water gain or loose during the process?
Thanx for the help

 

[iainuts]

Ouch! Great question. Answer will take too long to actually calculate though, not enough time right now. Very quick, here's how it's done:

(Note: the drawing referenced may indicate how pressure varies with piston movement, but since there's no picture, I'll assume the piston has a constant force on it, such as by a given weight. This is very important, since if the force on the piston is constant, than the final pressure will be equal to the initial pressure, 0.3 MPa.)

1. Draw a control volume around the first mass (saturated liquid) of 0.4 kg. When the diaphragm breaks, let's assume there's no mixing, only heat moves across this control boundry. So the control boundry may increase in volume, but not change total mass. Now apply the first law to this control volume, and you get:
U1i - U1f = Q + W1

2. Draw a control volume around the second mass (superheated vapor) of 8.0 kg. Again, let's assume there's no mixing, only heat moves across the control boundry. Applying first law to this:
U2i - U2f = Q + W2

3. Note that Q in eq. 1 and eq. 2 above are equal in magnitude, but opposite in sign (one is positive, the other negative). Whatever heat is lost by CV 2, is gained by CV 1.
- The result is 2 equations with 4 unknowns.
Unknowns are: Q, W1, W2, Tf (where Tf = final temperature)
- Note also T1f = T2f (ie: the final temperatures of CV 1 and CV 2 are equal, otherwise there will be continued heat flux untill the temperatures equalize.

4. We need 2 more equations, which are defined by the work.
W1 = P A L
where P = Pressure (0.3 MPa)
A = area of piston
L = distance piston moves
The change in density of the gas in each CV is what we need to determine the work. Note that as density decreases in the CV with saturated liquid, work is done by CV 1. But the density is also defined by pressure and temperature (or alternatively by pressure and internal energy).

Note that the final volume minus the initial volume gives the volume change, which is a change of volume at constant pressure. Thus you can derive the work done for each control volume using the same variables of pressure, temperature and/or internal energy.

- In the end, you should have 4 equations and 4 unknowns. The equations are 2 equations for the control volumes with boundries around the 0.4 kg and 8.0 kg mass respectively, and 2 equations relating final density to the work done by each control volume.
answer A: The piston moves up or down depending on the final calculations for work done by CV 1 and CV 2
answer B: The total work done is the sum of work by CV 1 and CV 2

Hope that helps. Thanks for the great question.

Dave.