From Robert Balmer\'s Modern Engineering Thermodynamics: Chapter 8, Problem 6 –

Question

From Robert Balmer's Modern Engineering Thermodynamics:


Chapter 8, Problem 6 – An engineer claims to be able to compress 0.100 kg of water vapor at 200°C and 0.100 MPa in a piston-cylinder arrangement in an isothermal and adiabatic process. The engineer claims that the final volume is 6.10% of the initial volume. Determine:
a. The final temperature and pressure
b. The work required
c. Show whether the process is thermodynamically possible

I'm looking for an explanation as much as an answer. A detailed explanation will help me solve similar problems on my own. Thank you for any help.

Explanation / Answer

here initial stae = P1 = 0.1 MPa and T1 = 200 C

Now, at 0.100 Mpa and 200 C, stam is in superheated state, and from steam table specific volume of the steam = v = 2.172 Kg/m^3

V initial = V1 = m x v = 0.1 x 2.172 = 0.2172 m3

and specific internal energy = u1 = 2658.225 KjKg

Now, V final = V2 = 6.10 % of V1 = 0.061 x 0.2172 = 0.0133 m3

a) Now, as the process is isothermal, T = 0

T2 = T1 = 200 C

To calculate pressure after compression, we look into steam table for T2 = 200 C and V2 = 0.0133 m3 (V2 we calculated above)

And hence P2 at 200 C and 0.0133 m^3 is P2 = 1.5548 MPa

and also specific internal energy = u2 = 850.59 Kj/Kg

b) As ther is only expansion work involved , W done = Pdv

Now, By the relation:

Tds = du + Pdv

as adiabatic process is done ,ds = 0

PdV = - dU = - m x (u2 - u1)

             = 0.1 x (2658.225 - 850.59)

            = 180.7635 KJ

Work done on the system = 180.76 KJ

c) Now, here we have assumed , entopy chage of the system = 0 , Ssystem = 0

and since, tno heat flows to surrounding,Ssurrounding= 0

Ssystem + Ssurrounding = 0 = S universe.

but S universe. of universe is always greater than 0, hence the process is not thermodynamically possible, as we have got S universe = 0.

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