Solutions Introduction to the Second Law
A geothermal power plant cycle uses 10 kg/s of isopentane. The cycle consists of a
turbine, a condenser, a pump, and a heater. The heat addition in the heater, which has no
useful work, is 5 MW. The pump, which has negligible heat transfer, has a power input of
15 kW. The turbine has an inlet temperature and pressure of 120oC and 1 MPa, negligible
heat transfer, an output pressure of 100 kPa, and an outlet entropy that is equal to the inlet
entropy. All heat rejection takes place in the condenser which has no useful work.
1. What is the power output of the turbine.
The turbine has one inlet and one outlet. For a steady open system with negligible kinetic and
potential energies the first law for negligible heat transfer becomes:
outinuhhmW
.
From the isopentane tables we have hin = h(120oC, 1 MPa) = 696.38 kJ/kg and sin = 2.1514
kJ/kg·K. At the outlet with Pout = 100 kPa, sout = sin = 2.1514 kJ/kg·K lies between 60oC and
70oC. Interpolation gives the outlet enthalpy and the turbine power as follows,
kg
kJ
Kkg
kJ
Kkg
kJ
Kkg
kJ
Kkg
kJ
kg
kJ
kg
kJ
kg
kJ
hout
97.6081428.21514.2
1428.21980.2
09.60676.624
09.606
kJ
skW
kg
kJ
kg
kJ
s
kg
hhmW outinu1
97.60838.69610
= 874.1 kW
2. What is the cycle efficiency?
The cycle efficiency is given by the equation h = |W|/|QH|, where |W| is the net work output and
|QH| is the heat input. Here the net work is the turbine work output minus the pump work input.
Thus |W| = 874.1 kW – 15 kW = 859.1 kW. The heat input is given as 5 MW = 5000 kW. This
gives the efficiency as follows:
kW
kW
Q
W
m
Q
m
W
Q
W
HH
H5000
1.859
= 17.18%
3. What is the heat rejection in the condenser?
We can apply the general law for a cycle that |QH| = |QL|+ |W| to solve for the heat rejected, |QL| =
|QH| - |W|. As in problem 2, we can multiply this equation by the mass flow rate and apply the
data from problem 2 to give.
kWkWWQQWmQmQm HLLH 1.8595000
|QL| = 4141 kW
docsity.com