Example Problem with Complete Solution

9B-2 : 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 500oC and 10oC, 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 Tsurr = 300 K.
a.) Construct a neat, fully labelled TS Diagram of the cycle.
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/kg-K 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 WS's we will need to determine all the H's.  H2 and H4 come straight from the Steam Tables.  Then use entropy to determine H1 and H3.  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, Sgen = DSuniv is just DSfurn + DScw.  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 Tsurr Sgen.  The easiest way to evaluate WS,rev is to use the definition of WS,lost and the values of WS,act and WS,lost that were determined in parts (b) and (d), repectively.  WS,rev = WS,act + WS,lost.
Given: P2 12000 kPa Q41 = Q23 = 0 kJ/kg
T2 450 oC Tfurn 500 oC
hs,turb 84% 773.15 K
P3 100 kPa Tcw 10 oC
x4 1 kg vap / kg 283.15 K
Tsurr 300 K
Find: a.) See TS-Diagram, below. c.) h ??? %
b.) Wstep ??? kJ/kg d.) Sgen ??? kJ/kg-K
Qstep ??? kJ/kg WS,lost ??? kJ/kg
For steps 1-2, 2-3, 3-4 & 4-1 e.) WS,rev ??? kJ/kg
Assumptions: 1 - Each component in the cycle is analyzed as an open system operating at steady-state.
2 - All of the processes except the turbine are internally reversible.
3 - The turbine is adiabatic and the pump is isentropic.
4 - Condensate leaves the condenser as saturated liquid.
5 - No shaft work in the boiler or condenser.
6 - Changes in kinetic and potential energies are negligible.
Equations / Data / Solve:
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.
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 (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
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 (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
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 WS for each process in the cycle, we will apply the 1st Law to each process.
All the devices operate at steady-state 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 H2 and H4 because we know both P2 and T2 and we know P4 and we know that it is a saturated liquid, x4 = 0.  So, we can lookup H2 and H4 in the NIST Webbook.
H2 3209.8 kJ/kg H4 417.50 kJ/kg
In order to determine H1, 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,
Eqn 6 simplifies to:
Eqn 7
Now, we can look-up V4 and use it in Eqn 7 to evaluate WS,pump: V4 0.0010432 m3/kg
WS,pump -12.414 kJ/kg
Now, we can use WS,pump and Eqn 4 to determine H1.
Eqn 8
H1 429.9 kJ/kg
Next, we need to use the isentropic
of the turbine to determine H3.
Eqn 9
Solve Eqn 9 for H3 :
Eqn 10
Now, we must evaluate H3S before we can use Eqn 10 to determine H3.  H3S is the enthalpy of the turbine effluent IF the turbine were isentropic.  Therefore, S3S = S2.
S2 6.3028 kJ/kg-K S3S 6.3028 kJ/kg-K
Now, we know the values of two intensive properties at state 3S: S3S and P3S, so we can fix this state and determine H3S by interpolation in the NIST Webbook.  First we must determine the phase of state 3S.
At 100 kPa : Tsat 99.63 oC
H (kJ/kg) S (kJ/kg-K) Since Ssat liq < S3S < Ssat vap, state 3S is a saturated mixture.
Sat Liq: 417.50 1.30276
Sat Vap : 2674.9 7.3588 T3S 99.63 oC
Determine x3S from the specific entropy, using:
Eqn 11
x3S 0.8256 kg vap/kg
Then, we can use the quality
to determine
H3S, using:
Eqn 12
H3S 2281.3 kJ/kg
Now, plug H3S back into Eqn 10 to determine H3 : H3 2429.86 kJ/kg
Now that we know the specific enthalpy of all four streams, we can use Eqns 1 - 3 to evaluate the WS,turb and the two remaining Q's.
Q12 2779.9 kJ/kg
WS,turb 779.97 kJ/kg Q34 -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.
Wcycle 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 : DSfurn -3.5956 kJ/kg-K
DScw 7.1070 kJ/kg-K
Sgen 3.5115 kJ/kg-K
We can calculate lost work using :
Eqn 18
WS,lost 1053.44 kJ/kg
Alternatively, we can compute the lost work using :
Eqn 19
WS,lost 1053.44 kJ/kg
Part e.) The reversible work can be determined from the definition of lost work :
Eqn 19
Solving for WS,rev yields :
Eqn 20
WS,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) WS (kJ/kg) c.) h 27.6 %
Boiler 1 - 2 2780 0
Turbine 2 - 3 0 780 d.) Sgen 3.51 kJ/kg-K
Condenser 3 - 4 -2010 0 WS,lost 1053 kJ/kg
Pump 4 - 1 0 -12.4
e.) WS,rev 1820 kJ/kg
HEY ! Why is it that :
??? WS,rev / QH 65.5%
1-TC/TH 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, Tsys is NOT constant !
Eqn 22
The value of Q23 is equal to the area under the path for Step 1-2 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 Q23 by applying the 1st Law to the entire reversible cycle, because Q12, Q34 and Q41 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
Q23,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 DSfurn and DScw, above. So, now we need to calculate DSsurr.
Eqn 26
Plug in values: DSsurr -3.5115 kJ/kg-K
Now, plug values into Eqn 26 : DSuniv 0.0000 kJ/kg-K
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.