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gphsuvat
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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:
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}
##
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} ##
## \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 ## ?
Homework Statement
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}
##
Homework 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} ##
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 ## ?