5C6 :  Analysis of a Steam Power Cycle  8 pts 

The steam power plant, shown below, operates at steadystate with negligible heat losses to the surroundings and negligible pressure drops due to friction in the boiler and condenser.  


If the mass flow rate of the steam is 11.3 kg/s, determine… a.) The power of the turbine and the pump b.) The velocity at the outlet of the pump c.) The heat transfer rates in the boiler and in the condenser 

d.) The mass flow rate of cooling water required in the condenser e.) The thermal efficiency of the power cycle 

Data: P_{1} = 120 kPa, T_{1} = 55^{o}C, P_{2} = 10,000 kPa, T_{2} = 50^{o}C, D_{2} = 0.05 m, P_{3} = 10,000 kPa, T_{3} = 700^{o}C, P_{4} = 120 kPa, x_{4} = 0.95 kg vap/kg, T_{cw,in} = 20^{o}C, T_{cw,out} = 45^{o}C  
Read :  Cycle problems of this type usually require you to work your way around the cycle, process by process until you have determined the values of all of the unknowns. This is a good approach here because the problem statement asks us to determine the values of unknowns in every process in the cycle. The only decision is where to begin. We can begin with the turbine because that is the 1st question and also because we have enough information to answer part (a). We know T_{3} and P_{3}, so we can determine H_{3}. Stream 4 is saturated mixture with known P_{4} and x_{4}, so we can also determine H_{4}. With the usual assumtions about kinetic and potential energy, we can determine W_{turb}. In fact, because we know the T and P of streams 1 and 2 as well, we can analyze the processes in this cycle in any convenient order. So, we will let the questions posed in the problem determine the order in which we analyze the processes. We will apply the 1st Law to the pump, the boiler and the condenser, in that order. Use the Steam Tables in the NIST Webbook.  
Diagram: 


Given:  m  11.3  kg/s  P_{3}  10000  kPa  
P_{1}  120  kPa  T_{3}  700  ^{o}C  
T_{1}  55  ^{o}C  P_{4}  120  kPa  
P_{2}  10000  kPa  x_{4}  0.95  kg vap/kg  
T_{2}  55  ^{o}C  T_{cw,in}  20  ^{o}C  
D_{2}  0.05  m  T_{cw,out}  45  ^{o}C  
Find:  W_{turb}  ???  MW  Q_{boil}  ???  MW  
W_{pump}  ???  kW  Q_{cond}  ???  MW  
v_{2}  ???  m/s  m_{cw}  ???  kg/s  
h_{th}  ???  
Assumptions:  1   Changes in kinetic and potential energy are negliqible in all the processes in the cycle.  
2   The pump and turbine are adiabatic.  
3   All of the heat that leaves the working fluid in the condenser is transferred to the cooling water. No heat is lost to the surroundings.  
Equations / Data / Solve:  
Part a.)  Begin by writing the 1st Law for the turbine, assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation, velocity or pipe diameter information to use.  

Eqn 1  
If we assume that the turbine is adiabatic, we can solve Eqn 1 for the shaft work of the turbine :  

Eqn 2  
Now, we must use the Steam Tables to determine H_{3} and H_{4}. Let's begin with stream 3.  
At a pressure of 10,000 kPa, the saturation temperature is :  T_{sat}  311.00  ^{o}C  
Because T_{3} > T_{sat}, we conclude that stream 3 is superheated steam and we must consult the Superheated Steam Tables. Fortunately, there is an entry in the table for 10,000 kPa and 700^{o}C, so no interpolation is necessary.  
H_{3}  3870.0  kJ/kg  
Stream 4 is a saturated mixture at 120 kPa, so we need to use the properties of saturated liquid and saturated vapor at 120 kPa in the following equation to determine H_{4} :  
At 120 kPa : 

Eqn 3  
H_{sat liq}  439.36  kJ/kg  
H_{sat vap}  2683.1  kJ/kg  H_{4}  2570.9  kJ/kg  
Now, we can plug H_{3} and H_{4} back into Eqn 2 to answer part (a) :  W_{turb}  14.680  MW  
Part b.)  Write the 1st Law for the pump, assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation or velocity data and we are given only the outlet pipe diameter. Also, assume the pump is adiabatic, Q_{pump} = 0.  

Eqn 4 

Eqn 5  
Now, we must determine H_{1} and H_{2}. We know the T and P for both of these streams, so we should have no difficulty determining the H values.  
T_{sat}(P_{1})  104.78  ^{o}C  T_{1} < T_{sat}, therefore we must consult the Subcooled Water Tables.  
T_{sat}(P_{2})  311.00  ^{o}C  T_{2} < T_{sat}, therefore we must consult the Subcooled Water Tables.  
The NIST Webbook provides these enthalpy values without interpolation.  
H_{1}  230.34  kJ/kg  H_{2}  238.74  kJ/kg  
Now, we can plug H_{1} and H_{2} back into Eqn 5 to answer part (b) :  W_{pump}  94.857  kW  
Part c.)  Here, we need to consider the relationship between velocity, specific volume and crosssectional area.  

Eqn 6  
where : 

Eqn 7  
A_{2}  0.0019635  m^{2}  
From the NIST Webbook :  V_{2}  0.0010101  m^{3}/kg  
Now, we can plug values into Eqn 6 to answer part (c) :  v_{2}  5.813  m/s  
Part d.)  Write the 1st Law for the boiler, assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation, velocity or pipe diameter data. There is no shaft work in a boiler.  

Eqn 8 

Eqn 9  
We determined H_{2} in part (b) and H_{3} in part (a), so all we need to do is plug numbers into Eqn 9.  
Q_{boil}  41.033  MW  
Part e.)  Write the 1st Law for the condenser assuming that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation, velocity or pipe diameter data. Use the working fluid as the system so that Q_{cond} is the amount of heat transferred to the cooling water. There is no shaft work in a condenser.  

Eqn 10 

Eqn 11  
We determined H_{1} in part (b) and H_{4} in part (a), so all we need to do is plug numbers into Eqn 11.  
Q_{cond}  26.448  MW  
Part f.)  In order to determine the mass flow rate of the cooling water, we must write the 1st Law using the cooling water as our system. For this system, Q_{cw} =  Q_{cond} because heat leaving the working fluid for the cycle enters the cooling water.  
Q_{cw}  26.448  MW  
Assume that changes in kinetic and potential energy are negligible. This makes sense because we have no elevation, velocity or pipe diameter data. There is no shaft work for the cooling water system.  

Eqn 12  
We cannot use the Steam Tables to determine the enthalpy of the cooling water because we do not know the pressure in either stream. The next best thing we can do is to use the specific heat of the cooling water to determine DH_{cw} using:  

Eqn 13  
If we further assume that the specific heat of liquid water is constant over the temperature range 20^{o}C  45^{o}C, than Eqn 13 simplifies to:  

Eqn 14  
We can then combine Eqn 14 with Eqn 12 to obtain :  

Eqn 15  
Finally, we can solve Eqn 15 for m_{cw} : 

Eqn 16  
All we need to do is look up the average heat capacity of water between 20^{o}C and 45^{o}C.  
NIST Webbook :  C_{P,cw}(50^{o}C)  4.1813  kJ/kgK  
C_{P,cw}(20^{o}C)  4.1841  kJ/kgK  C_{P,cw}  4.1827  kJ/kgK  
Let's use :  C_{P,cw}  4.182  kJ/kgK  
Then :  m_{cw}  252.97  kg/s  
Part g.)  The thermal efficiency of this power cycle can be determined directly from its definition.  

Eqn 17  h_{th}  0.3555  
Verify:  None of the assumptions made in this problem solution can be verified.  
Answers :  a.)  W_{turb}  14.68  MW  e.)  Q_{cond}  26.4  MW  
b.)  W_{pump}  94.9  kW  f.)  m_{cw}  253  kg/s  
c.)  v_{2}  5.81  m/s  g.)  h_{th}  35.5  %  
d.)  Q_{boil}  41.0  MW 