9B2 :  Steam Power Plant Operating on the Rankine Cycle  9 pts 

Consider the Rankine Power Cycle shown below. Steam is the working fluid. The hot and cold thermal reservoirs are at 500^{o}C and 10^{o}C, respectively. The boiler operates at 12 MPa and the condenser operates at 100 kPa. The  


pump is isentropic and the turbine has an isentropic efficiency of 84%. The pump and turbine are adiabatic. The temperature of the surroundings is T_{surr} = 300 K. a.) Construct a neat, fully labelled TS Diagram of the cycle. 

Calculate… b.) Q & W for each unit in the cycle, in kJ/kg. c.) The thermal efficiency of the power cycle. d.) The total entropy generationin kJ/kgK and the total lost work in kJ/kg for the cycle. 

e.) The net work that would be produced if this cycle were completely reversible and the state of all four streams remained the same as in the actual cycle.  
Read :  The TS Diagram is pretty standard. The pump is isentropic, but the turbine is not. In order to determine the Q's and W_{S}'s we will need to determine all the H's. H_{2} and H_{4} come straight from the Steam Tables. Then use entropy to determine H_{1} and H_{3}. Plug the H's into the 1st Law for each unit to determine all the Q's and W's. Once we have all the Q's and W's in part (b), we can calculate the thermal efficiency from its definition. The key to part (d) is that, for a cycle, DS = 0. So, S_{gen} = DS_{univ} is just DS_{furn} + DS_{cw}. Because the furnace and cooling water behave as thermal reservoirs, we can evaluate the change in their entropy from the definition of entropy. Lost work is just T_{surr} S_{gen}. The easiest way to evaluate W_{S,rev} is to use the definition of W_{S,lost} and the values of W_{S,act} and W_{S,lost} that were determined in parts (b) and (d), repectively. W_{S,rev} = W_{S,act} + W_{S,lost}.  
Given:  P_{2}  12000  kPa  Q_{41} = Q_{23} =  0  kJ/kg  
T_{2}  450  ^{o}C  T_{furn}  500  ^{o}C  
h_{s,turb}  84%  773.15  K  
P_{3}  100  kPa  T_{cw}  10  ^{o}C  
x_{4}  1  kg vap / kg  283.15  K  
T_{surr}  300  K  
Diagram: 


Let's organize the data that we need to collect into a table. This will make it easier to keep track of the values we have looked up and the values we have calculated.  
Stream  State  T (^{o}C)  P (kPa)  X  H (kJ/kg)  S (kJ/kgK)  
1  Sub Liq  100.42  12000  N/A  429.9  1.3028  
2  Super Vap  450  12000  N/A  3209.8  6.3028  
3S  Sat Mix  99.61  100  0.8256  2281.3  6.3028  
3  Sat Mix  99.61  100  0.8914  2429.9  6.7013  
4  Sat Liq  99.61  100  1  417.50  1.30276  
Additional data that may be useful.  
State  T (^{o}C)  P (kPa)  X  H (kJ/kg)  S (kJ/kgK)  
Sat Vap  324.68  12000  1  2685.4  5.4939  
Sat Liquid  324.68  12000  0  1491.46  3.49671  
Sat Vap  99.63  100  1  2674.9  7.3588  
Sat Liquid  99.63  100  0  417.5  1.3028  
Part b.)  In order to evaluate Q and W_{S} for each process in the cycle, we will apply the 1st Law to each process.  
All the devices operate at steadystate and any changes in kinetic and potential energies are negligible.  
The boiler and condenser have no shaft work interaction and the pump and turbine are both adiabatic.  
Therefore, the relevant forms of the 1st Law for the four devices are:  
Boiler : 

Eqn 1  
Turbine : 

Eqn 2  
Condenser : 

Eqn 3  
Pump : 

Eqn 4  
In order to evaluate all the W's and Q's in Eqns 1  4, we must first determine H for all four streams.  
We can immediately evaluate H_{2} and H_{4} because we know both P_{2} and T_{2} and we know P_{4} and we know that it is a saturated liquid, x_{4} = 0. So, we can lookup H_{2} and H_{4} in the NIST Webbook.  
H_{2}  3209.8  kJ/kg  H_{4}  417.50  kJ/kg  
In order to determine H_{1}, we must make use of the fact that the pump is adiabatic and internally reversible and that changes in kinetic and potential energies are negligible. Under these conditions, the shaft work done at the pump can be determined from the Mechanical Energy Balance Equation:  

Eqn 5  
Cancelling terms yields : 

Eqn 6  
Because the liquid water flowing through the pump is incompressible, the specific volume is constant, and Eqn 6 simplifies to: 


Eqn 7  
Now, we can lookup V_{4} and use it in Eqn 7 to evaluate W_{S,pump}:  V_{4}  0.0010432  m^{3}/kg  
W_{S,pump}  12.414  kJ/kg  
Now, we can use W_{S,pump} and Eqn 4 to determine H_{1}. 

Eqn 8  
H_{1}  429.9  kJ/kg  
Next, we need to use
the isentropic efficiency of the turbine to determine H_{3}. 

Eqn 9  
Solve Eqn 9 for H_{3} : 

Eqn 10  
Now, we must evaluate H_{3S} before we can use Eqn 10 to determine H_{3}. H_{3S} is the enthalpy of the turbine effluent IF the turbine were isentropic. Therefore, S_{3S} = S_{2}.  
S_{2}  6.3028  kJ/kgK  S_{3S}  6.3028  kJ/kgK  
Now, we know the values of two intensive properties at state 3S: S_{3S} and P_{3S}, so we can fix this state and determine H_{3S} by interpolation in the NIST Webbook. First we must determine the phase of state 3S.  
At 100 kPa :  T_{sat}  99.63  ^{o}C  
^{}  
H (kJ/kg)  S (kJ/kgK)  Since S_{sat liq} < S_{3S} < S_{sat vap}, state 3S is a saturated mixture.  
Sat Liq:  417.50  1.30276  
Sat Vap :  2674.9  7.3588  T_{3S}  99.63  ^{o}C  
Determine x_{3S} from the specific entropy, using: 

Eqn 11  
x_{3S}  0.8256  kg vap/kg  
Then, we can use the quality to determine H_{3S}, using: 

Eqn 12  
H_{3S}  2281.3  kJ/kg  
Now, plug H_{3S} back into Eqn 10 to determine H_{3} :  H_{3}  2429.86  kJ/kg  
Now that we know the specific enthalpy of all four streams, we can use Eqns 1  3 to evaluate the W_{S,turb} and the two remaining Q's.  
Q_{12}  2779.9  kJ/kg  
W_{S,turb}  779.97  kJ/kg  Q_{34}  2012.4  kJ/kg  
Part c.)  The thermal efficiency of a power cycle is defined by: 

Eqn 13  
Where : 

Eqn 14  
Now, we can plug values into Eqns 13 & 14 to complete this part of the problem.  
W_{cycle}  767.55  kJ/kg  
h  27.61  %  
Part d.)  The total entropy generation for the cycle is equal to the entropy change of the universe caused by the cycle.  

Eqn 15  
Because the furnace and the cooling water are isothermal (they behave as thermal reservoirs) we can evaluate the change in their entropy directly from the definition of entropy.  

Eqn 16  

Eqn 17  
Now, we can put values into Eqns 16, 17 & 15 :  DS_{furn}  3.5956  kJ/kgK  
DS_{cw}  7.1070  kJ/kgK  
S_{gen}  3.5115  kJ/kgK  
We can calculate lost work using : 

Eqn 18  
W_{S,lost}  1053.44  kJ/kg  
Alternatively, we can compute the lost work using :  

Eqn 19  
W_{S,lost}  1053.44  kJ/kg  
Part e.)  The reversible work can be determined from the definition of lost work :  

Eqn 19  
Solving for W_{S,rev} yields : 

Eqn 20  
W_{S,rev}  1821.0  kJ/kg  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  See the diagram at the beginning of this solution.  
b.)  Device  Step  Q (kJ/kg)  W_{S} (kJ/kg)  c.)  h  27.6  %  
Boiler  1  2  2780  0  
Turbine  2  3  0  780  d.)  S_{gen}  3.51  kJ/kgK  
Condenser  3  4  2010  0  W_{S,lost}  1053  kJ/kg  
Pump  4  1  0  12.4  
e.)  W_{S,rev}  1820  kJ/kg  
HEY !  Why is it that : 

???  W_{S,rev} / Q_{H}  65.5%  
1T_{C}/T_{H}  63.4%  
Because the reversible cycle also exchanges heat at the turbine !  
In fact, the reversible turbine absorbs heat reversibly from a third thermal reservoir. This reservoir is a very special thermal reservoir because it is always at the same temperature as the working fluid in the turbine. This is pretty tricky because the temperature of the working fluid changes as it passes through the turbine !  
Consider the definition of entropy generation: 

Eqn 21  
But, T_{sys} is NOT constant ! 

Eqn 22  
The value of Q_{23} is equal to the area under the path for Step 12 on the TS Diagram, but we cannot evaluate it because we do not know the equation of the path, T = fxn(S).  
We could evaluate Q_{23} by applying the 1st Law to the entire reversible cycle, because Q_{12}, Q_{34} and Q_{41} are the same for the reversible cycle as they were for the actual cycle. So, we can use the values calculated in part (c).  
1st Law, Reversible Cycle : 

Eqn 23  

Eqn 24  
Q_{23,rev}  1053.44  kJ/kg  
Let's double check that this cycle is indeed reversible when it GAINS heat from the surroundings at the turbine.  

Eqn 25  
We already calculated DS_{furn} and DS_{cw}, above. So, now we need to calculate DS_{surr}.  

Eqn 26  
Plug in values:  DS_{surr}  3.5115  kJ/kgK  
Now, plug values into Eqn 26 :  DS_{univ}  0.0000  kJ/kgK  
There is one impossible aspect about our imaginary reversible cycle. The surroundings (at 300 K) must reversibly supply heat to the steam inside the turbine (which is always at a higher temperature than 300 K). That would require a heat pump and that would be a very complicated turbine, indeed ! This is ok because this is just a hypothetical reversible turbine anyway. It just seems strange. 