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Solutions to Homework Problems
1. A 500 MW power plant, operating at full capacity, has a heat rejection of 3.5x10^9 Btu/hr.
What is its Efficiency?
In this problem, we are given W = 500 MW and Q L = 3.5x10^9 Btu/hr. Using the equation Q H
= Q L + W , we find Q H = 3.5x10^9 Btu/hr + (500x10^6 W)(3.412 Btu/kWh) = 3.5x10^9 Btu/hr +
1.705x10^9 Btu/h = 5.206x10^9 Btu/h. We can thus compute the efficiency as follows.
x Btu h
x Btu h
Q
W
H
9
9
2. It takes 2.2 million tons of coal per year to fuel a 1000-MW power plant that operates at a
capacity factor of 70%. If the heating value of the coal is 12,000 Btu/lb, calculate the plant’s efficiency and the heat rate.
In this problem, we are given the following quantities, where W without a subscript denotes the annual average power:
m x tons year Q Btu lb
W
W
W MW
fuel^2.^210 / c^12 ,^000 /
6
max
max
We want to find:
(dimensionl ess) ( Btu / kWh ) W
Q
and HeatRate Q
W
Equations and basic calculations:
( 1000 MW)( 70 %) 700 MW 700 , 000 kW W
W
W W
max
(^) max (^)
- 03274 x 10 Btu/hr 24 hours
day
365 days
year
ton
2000 lb
lb
12 , 000 Btu
year
- 2 x 10 tons Q m Q
9
6 fuel c
With these values for power output and heat input we can calculate the desired final results.
kWh
Btu 8 , 611 700 , 000 kW
hr
- 03274 x 10 Btu
W
Q
HeatRate
9
3. Problem 4 of chapter 5 in the
text.
GIVEN: Gas turbine cycle with data as shown. The net power output is 5,000 kW and the heating value of the fuel is 47.1 MJ/kg
FIND: (a) the mass flow rate of air and (b) the thermal efficiency of the cycle.
APPROACH: this solution uses the equations in Hodge, which are slightly different in form from those in a typical engineering thermodynamics class.
The equations in Hodge use the equations for an ideal gas with constant heat capacity. Hodge accounts for the temperature of the heat capacity by using different values of heat capacity different components in the system. The Hodge equations assume that each component in the gas turbine cycle is steady flow with negligible kinetic and potential energy changes. The turbine and the compressor each have one inlet and one outlet so that the general first law for these components is q = wu + hout – hin. We assume all devices are adiabatic. In the combustor, all the chemical energy of the fuel is converted to thermal energy in the exhaust stream, with no external heat loss.
We can use equation (5-14) in Hodge to find the outlet temperature of the compressor.
K K
P
P
T T
k
k
c
1 1.^4
- 41
1
1
2 2 1
Similarly, equation (5-20) in Hodge gives the outlet temperature of the turbine. The inlet pressure to the turbine is 3% less than the outlet pressure of the combustor giving P 3 = (1 – 3%)P 2 where P 2 = 5.5P 1 5.5(97 kPa) = 544.5 kPa. Thus P 3 = 0.97(544.5 kPa) = 517.5 kPa. Here we use Hodge’s value of k = 1.334 for the turbine.
K
kPa
kPa K P
P
T T
k
k
t^1895.^1
- 5
- 334
1 1. 3341
3
4 4 3
The fuel air ratio can be found from equation (5-22) in Hodge for an adiabatic combustor (using Hodge’s value for the mean heat capacity in the combustor).
, 3 2
MJ
kJ
kg
MJ
K K
kg K
kJ
H
c T T
m
m
v
pcombustor
air
fuel
Compressor P 2 /P 1 = 5. c = 84%
Turbine t = 88%
Combustor (P 3 – P 2 )/P 2 = 3%
1
2 3
T 1 = 30oC = 4 303.15 K P 1 = 97 kPa
T 3 = 1000oC = 1273.15 K
P 4 = 100 kPa
