# Example Problem with Complete Solution

5C-6 : Analysis of a Steam Power Cycle 8 pts
The steam power plant, shown below, operates at steady-state 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: P1 = 120 kPa, T1 = 55oC, P2 = 10,000 kPa, T2 = 50oC, D2 = 0.05 m, P3 = 10,000 kPa, T3 = 700oC, P4 = 120 kPa, x4 = 0.95 kg vap/kg, Tcw,in = 20oC, Tcw,out = 45oC

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 T3 and P3, so we can determine H3.  Stream 4 is saturated mixture with known P4 and x4, so we can also determine H4.  With the usual assumtions about kinetic and potential energy, we can determine Wturb.  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 P3 10000 kPa
P1 120 kPa T3 700 oC
T1 55 oC P4 120 kPa
P2 10000 kPa x4 0.95 kg vap/kg
T2 55 oC Tcw,in 20 oC
D2 0.05 m Tcw,out 45 oC
Find: Wturb ??? MW Qboil ??? MW
Wpump ??? kW Qcond ??? MW
v2 ??? m/s mcw ??? kg/s
hth ???
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 H3 and H4.  Let's begin with stream 3.
At a pressure of 10,000 kPa, the saturation temperature is : Tsat 311.00 oC
Because T3 > Tsat, 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 700oC, so no interpolation is necessary.
H3 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 H4 :
At 120 kPa : Eqn 3
Hsat liq 439.36 kJ/kg
Hsat vap 2683.1 kJ/kg H4 2570.9 kJ/kg
Now, we can plug H3 and H4 back into Eqn 2 to answer part (a) : Wturb 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, Qpump = 0. Eqn 4 Eqn 5
Now, we must determine H1 and H2.  We know the T and P for both of these streams, so we should have no difficulty determining the H values.
Tsat(P1) 104.78 oC T1 < Tsat, therefore we must consult the Subcooled Water Tables.
Tsat(P2) 311.00 oC T2 < Tsat, therefore we must consult the Subcooled Water Tables.
The NIST Webbook provides these enthalpy values without interpolation.
H1 230.34 kJ/kg H2 238.74 kJ/kg
Now, we can plug H1 and H2 back into Eqn 5 to answer part (b) : Wpump -94.857 kW
Part c.) Here, we need to consider the relationship between velocity, specific volume and cross-sectional area. Eqn 6
where : Eqn 7
A2 0.0019635 m2
From the NIST Webbook : V2 0.0010101 m3/kg
Now, we can plug values into Eqn 6 to answer part (c) : v2 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 H2 in part (b) and H3 in part (a), so all we need to do is plug numbers into Eqn 9.
Qboil 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 Qcond is the amount of heat transferred to the cooling water.  There is no shaft work in a condenser. Eqn 10 Eqn 11
We determined H1 in part (b) and H4 in part (a), so all we need to do is plug numbers into Eqn 11.
Qcond -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, Qcw = - Qcond because heat leaving the working fluid for the cycle enters the cooling water.
Qcw 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 DHcw using: Eqn 13
If we further assume that the specific heat of liquid water is constant over the temperature range 20oC - 45oC, 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 mcw : Eqn 16
All we need to do is look up the average heat capacity of water between 20oC and 45oC.
NIST Webbook : CP,cw(50oC) 4.1813 kJ/kg-K
CP,cw(20oC) 4.1841 kJ/kg-K CP,cw 4.1827 kJ/kg-K
Let's use : CP,cw 4.182 kJ/kg-K
Then : mcw 252.97 kg/s
Part g.) The thermal efficiency of this power cycle can be determined directly from its definition. Eqn 17 hth 0.3555
Verify: None of the assumptions made in this problem solution can be verified.
Answers : a.) Wturb 14.68 MW   e.) Qcond -26.4 MW
b.) Wpump -94.9 kW   f.) mcw 253 kg/s
c.) v2 5.81 m/s   g.) hth 35.5 %
d.) Qboil 41.0 MW 