Name: ME 166 Energy Systems Engineering Design California State University at Fr
ID: 2086244 • Letter: N
Question
Name: ME 166 Energy Systems Engineering Design California State University at Fresno April 4th 2018 Mid Term Test You must answer all the questions. It is assumed that you have agreed to produce your answers without the consultations of any other person or academic services. Show your calculations. Please answer all the questions. Problem 1 Time allowed: 1 hr 10 mins A power plant operating on a Rankine cycle has a maximum temperature of 700°C, minimum temperature of 46°C, heat addition of 3523.5 kJ/kg, turbine work out 1668.1 kJ/kg and pump work in 30.3 kJ/kg. (1) What is the efficiency of this power cycle? (30 points) (2) What is the second law efficiency of the power plant? (10 points) (3) If the stcam flow rate is 15 kg's in the stated power plant, and the electricity generator has an efficiency of97%, what is the maximum rate of electricity conversion? (20 points) (4) Assuming the power plant has a capacity factor of 0.8, how many electricity units can be produced annually? (10 points)Explanation / Answer
Given maximum temperature is T1 = 700 deg C = 973 K
Minimum Temperature T2 = 46 deg C = 319 K
assuming ambient temperature = T0 = 300 K
Net work output = turbine work -pump work = 1668.1-30.3 = 1637.8 kJ/kg
(1) cycle efficiency = Net work output/ heat input = 1637.8/3523.5 = 0.4648 = 46.48%
(2) second law efficiency:
net available energy = heat input * (1- T2/T1) = 3523.5 x (1-319/973) = 2368.3
second law efficiency = Net work output/net available energy = 1637.8/2368.3 = 0.6916 = 69.16%
(3) total work output for the given steam input rate of 15 kg/s = 15*1637.8 = 24567 kW
maximum electricity generation rate = 24567 x generator efficiency = 24567 x .97 =23830 kW = 23.83 MW
(4) since the given capacity factor = 0.8
from the definition, capacity factor = actual production in a time period/maximum production in that time period
no. of hours in 1 year = 24 x 365 = 8760
1 electricity unit = 1 kWh
actual production units = 23830 x 8760 x 0.8 kWh = 167000640 kWh = 167000640 units of electricity
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