Steam is
contained in a piston and cylinder device with a freefloating piston. Initially, the steam occupies a volume of 0.18 m^{3} at a pressure of 500 kPa. 

The steam is slowly heated until the temperature is 300^{o}C, while the pressure remains constant. If the cylinder contains 0.65 kg of steam, determine the heat transfer and the work in kJ for this process. 























Read : 
We know the values of two intensive variables
for state 2: T and P, so we can determine the values of all other properties in this state. 













Therefore we can
calculate DU directly. 













We can also use the
definition of work for an isobaric process to evaluate W_{12}. 













Once we know W_{12} and DU, we can use the 1st Law to evaluate Q_{12}. 












Given: 
V_{1} 
0.18 
m^{3} 


Find: 
Q_{12} 
??? 
kJ 


m 
0.65 
kg 



W_{12} 
??? 
kJ 


P_{1} 
500 
kPa 








P_{2} 
500 
kPa 








T_{2} 
300 
^{o}C 


















Diagram: 





Assumptions: 
1  
Changes in kinetic and potential energies are negligible. 



2  
The process is a quasiequilibrium process. 












Equations
/ Data / Solve: 




















Choose the water inside the cylinder as the system. 













Apply the integral form of the 1st Law to the process: 


















Eqn 1 













If we assume that changes in kinetic and potential energies are negligible, then Eqn 1 simplifies to : 

















Eqn 2 













We can evaluate W_{12} from the definiton of work applied to an isobaric process. 


















Eqn 3 













Let's combine Eqns 2 and 3: 























Eqn 4 













We still need to
lookup the same amount of data in the Steam Tables, V and H, but the calculations are just a
little bit simpler and faster using H than using U. 













Before we can look up H, we need to determine the state
of the water in the cylinder. 













Calculate V_{1} from : 




V_{1} 
0.2769 
m^{3}/kg 













At 500
kPa : 




V_{sat liq} 
0.0010925 
m^{3}/kg 








V_{sat vap} 
0.37481 
m^{3}/kg 













Since V_{sat liq} < V_{1} < V_{sat vap}, we conclude that a saturated
mixture exists in the cylinder at state 1. 













So, we must next
evaluate the quality of the steam. 















Eqn 5 


x 
0.7381 
kg vap/kg 












Then, we can use the quality to evaluate the specific enthalpy : 






















Eqn 6 













H_{sat liq} 
640.09 
kJ/kg 








H_{sat vap} 
2748.1 
kJ/kg 



H_{1} 
2196.0 
kJ/kg 













Next, we need to
determine the phases present
in State 2. We can do this by comparing T_{2} to T_{sat}(P_{2}). 













In the saturation pressure table of the Steam Tables we find: 

T_{sat}(P_{2}) 
151.8 
^{o}C 













Because T_{2} > T_{sat}(P_{2}), state
2 is a superheated
vapor. 

















From the NIST Webbook or the Superheated Tables of the Steam Tables we obtain the
following data: 













V_{2} 
0.52261 
m^{3}/kg 



H_{2} 
3064.6 
kJ/kg 













Now, we can plug
values back into Eqns 3 and 4 to evaluate Q_{12} and W_{12}: 













W_{12} 
79.8 
kJ 



Q_{12} 
564.6 
kJ 












Verify: 
The assumptions made
in this problem solution cannot be verified. 












Answers : 
W_{12} 
80 
kJ 

Q_{12} 
565 
kJ 













