Thermodynamics - Calculating Work/Heat, Piston/Cylinder

1. Apr 24, 2015

gphsuvat

First, I did find a previous discussion from a few years ago related to this problem. However, my specific question is further along in the problem than the other person got:

1. The problem statement, all variables and given/known data
A piston-and-cylinder device contains 5 kg of water initially at 150 ◦C and 0.20 MPa. The frictionless piston is then pushed slowly in an isothermal process until the volume of water becomes 10% of its initial value. Calculate the heat and work exchanged between the device and the surroundings during this process. You must start from the appropriate form of the balance equations and simplify them.

Following is data that I looked up from Steam Tables for (1= Superheated vapor) and (2=Saturated Vapor)
1. Superheated
$\hat V = 0.9596 m^3/kg$
$\hat S = 7.2795 \frac{ \text{kJ}\ }{ \text{kg K}\ }$
$\hat U = 2576.9 \frac{ \text{kJ}\ }{ \text{kg}\ }$

2. Saturated Vapor
$\begin{matrix} \hat V ^v = 0.3928 m^3/kg & \hat V ^L = 0.001091 m^3/kg \\ \hat S ^v = 6.8379 \frac{ \text{kJ}\ }{ \text{kg K}\ } & \hat S ^L = 1.8418 \frac{ \text{kJ}\ }{ \text{kg K}\ }\\ \hat U ^v = 2559.5 \frac{ \text{kJ}\ }{ \text{kg}\ } & \hat U ^L = 631.68 \frac{ \text{kJ}\ }{ \text{kg}\ } \\ \Delta \hat U _{evap} = 1927.9 \frac{ \text{kJ}\ }{ \text{kg}\ } \end{matrix}$

2. Relevant equations
The system is initially a superheated vapor (all of it is vapor phase). It is isothermally compressed to a saturated mixture of liquid and vapor. Process is completely reversible.
In the other thread, Chestermiller mentioned that there were two things to consider:
1) isothermal compression to saturation point
2) isothermal and isobaric compression to equilibrium vapor pressure/temperature.
I am unsure how to calculate the two of these together...

$\hat V = w^v \hat V ^v + (1-w^v ) \hat V ^L$
$w^v = {\frac{ \hat V - \hat V^L }{ \hat V ^v - \hat V ^L}}$
$dU = Q + W \rightarrow dU = Q - W \text{ as W here is only PV work.}\$
$dS = \frac{Q}{T} \rightarrow Q = TdS \rightarrow Q = T \int_1^2 dS \rightarrow Q = MT( \hat S _2 - \hat S _1 )$
$\hat V = \frac{V}{M}$

3. The attempt at a solution
$\hat V _1 = \frac{V_1}{M} \rightarrow V _1 = \hat V _1 M \rightarrow V _1 = (0.9596 m^3/kg)( 5 kg) = 4.798 m^3$
$V _2 = 0.1 V_1 = 0.4798 m^3$
$\hat V _2 = \frac{V_2}{M} \rightarrow \frac{0.4798 m^3}{5 kg} = 0.09596 m^3/kg$

$w^v = \frac{ \hat V - \hat V^L }{ \hat V ^v - \hat V ^L} \rightarrow \frac{0.09596 m^3/kg - 0.001091 m^3/kg}{0.3928 m^3/kg - 0.001091 m^3/kg} = 0.242$
$w^v=0.242 and w^L = (1-0.242) = 0.758$

Need to find the second entropy..
$\hat S _2 = w^v \hat S ^v + w^L \hat S ^L = (0.242)(6.8379 \frac{ \text{kJ}\ }{ \text{kg K}\ } ) + (0.758)(1.8418 \frac{ \text{kJ}\ }{ \text{kg K}\ }) = 3.051 \frac{ \text{kJ}\ }{ \text{kg K}\ }$

Thus the heat for the system

$Q = MT( \hat S _2 - \hat S _1 ) = (5 kg)(423 K)(3.051 \frac{ \text{kJ}\ }{ \text{kg K}\ } - 7.2795 \frac{ \text{kJ}\ }{ \text{kg K}\ }) = -8945 \frac{ \text{kJ}\ }{ \text{kg}\ }$

This is where I am confused on how to proceed, to solve for the work.
I know for an ideal gas, $\Delta U = 0$ but, I do not think that $\Delta U = 0$ in this case? I do not know if I need to use $\Delta U_{evap}$ ?

$\Delta U = Q + W \rightarrow M( \hat U _2 - \hat U _1 ) = Q - W$
$\hat U _2 = w^v \hat U ^v + w^L \hat U ^L = (0.242)(2559.5 \frac{ \text{kJ}\ }{ \text{kg}\ }) + (0.758)(631.68 \frac{ \text{kJ}\ }{ \text{kg}\ }) = 1098.21 \frac{ \text{kJ}\ }{ \text{kg}\ }$
$W = Q - M( \hat U _2 - \hat U _1 ) = (-8945 \frac{ \text{kJ}\ }{ \text{kg}\ } ) - (5 kg) (1098.21 \frac{ \text{kJ}\ }{ \text{kg}\ } - 2576.9 \frac{ \text{kJ}\ }{ \text{kg}\ } ) = -1551.55 kJ$

Therefore the W and Q for system $W = - 1551.55 kJ$ and $Q = -8945 kJ$

Do I need to account for the $\Delta U_{evap}$ ? Is the thought process correct? Or does $U=0$ ?

2. Apr 24, 2015

Staff: Mentor

I don't know what my thought processes were a few years ago regarding this problem, but what you've done here certainly looks correct to me now. You've automatically taken into account the heat of vaporization when you calculated $\hat{U}_2$ . ΔU is not equal to zero because of the change of phase. Note however, that the $\hat{U}$'s of the vapors in the two states are close to being equal. The difference is due to non-ideal gas behavior.

Chet

3. Apr 24, 2015

gphsuvat

Thank you for the reply. To make sure I understand the concept, when calculating $\Delta U$ the heat of vaporization does not need to be added, because the calculation already accounted for it?
I am also somewhat confused with this outcome for Work because I thought that work done ON a system in compression is positive (As in, W > 0).

I started to become confused when I saw another classmate's work, they had done this:
$\Delta U _{evap} = 1927.9 \frac{ \text{kJ}\ }{ \text{kg}\ } = (5 kg)(1927.9 \frac{ \text{kJ}\ }{ \text{kg}\ }) = 9639.5 \text{kJ}\ = \Delta U$
$\Delta U = Q + W$
$W = \Delta U - Q \rightarrow W = 9639.5 \text{kJ} \ - (- 8974.3 \text{kJ} \ ) = 18613.8 \text{kJ} \$

So...yeah =? is work negative when isothermal compression of a real gas goes through a phase change, and work positive in a isothermal compression of an ideal gas??

4. Apr 25, 2015

Staff: Mentor

In your calculation, you employed the internal energy of the vapor, which is equal to the internal energy of the liquid plus the heat of vaporization. (add 'em up to check). So, as I said, you did actually automatically include the heat of vaporization.
Your equations for the first law are a little schizophrenic. You have written both $ΔU=Q+W$ and $ΔU=Q-W$ on the same line in two different places. The first form of the equation is used when you are referring to the work done ON the system, and the second form is used when when you are referring to the work done BY the system. In your calculations, you used the second form, so the calculations determined the work done BY the system, which, in compression, is negative.

Your friend's calculation uses the first form of the first law equation, which assumes that you are referring to the work done ON the system. However, his calculation is totally wrong, whereas, your calculation is correct. (With these gaps in your understanding, I find it amazing that you were able to do the calculations correctly).
In compression, if W refers to the work done BY the system, it is always negative, irrespective of whether the gas is real or ideal, and irrespective of whether a phase change occurs.

Chet