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Calculate the thermal efficiency of a Carnot cycle heat
Chapter 5, Problem 44HP(choose chapter or problem)
Problem 44HP
Calculate the thermal efficiency of a Carnot cycle heat engine operating between reservoirs at 300°C and 45°C. Compare the result to that of Example 4.7.
Example 4.7
Consider the simple steam power plant, as shown in Fig. 4.11. The following data are for such a power plant where the states are numbered and there is specific pump work as 4 kJ/kg.
State |
Pressure |
Temperature or Quality |
1 |
2.0 MPa |
300°C |
2 |
1.9 MPa |
290°C |
3 |
15 kPa |
90% |
4 |
14 kPa |
45°C |
Determine the following quantities per kilogram flowing through the unit:
a. Heat transfer in the line between the boiler and turbine.
b. Turbine work.
c. Heat transfer in the condenser.
d. Heat transfer in the boiler.
Since there are several control volumes to be considered in the solution to this problem, let us consolidate our solution procedure somewhat in this example. Using the notation of Fig. 4.11, we have:
All processes: Steady-state.
Model: Steam tables.
From the steam tables:
h1 = 3023.5 kJ/kg
h2 = 3002.5 kJ/kg
h3 = 225.9 + 0.9(2373.1) = 2361.7 kJ/kg
h4 = 188.4 kJ/kg
All analyses: No changes in kinetic or potential energy will be considered in the solution. In each case, the energy equation is given by Eq.
Equation 4.13
Now, we proceed to answer the specific questions raised in the problem statement.
a. For the control volume for the pipeline between the boiler and the turbine, the energy equation and solution are
1q2 + h1 = h2
1q2 = h2 − h1 = 3002.5 − 3023.5 = −21.0 kJ/kg
b. A turbine is essentially an adiabatic machine. Therefore, it is reasonable to neglect heat transfer in the energy equation, so that
h2 = h3 + 2w3
2w3 = 3002.5 − 2361.7 = 640.8 kJ/kg
c. There is no work for the control volume enclosing the condenser. Therefore, the energy equation and solution are
3q4 + h3 = h4
3q4 = 188.4 − 2361.7 = −2173.3 kJ/kg
d. If we consider a control volume enclosing the boiler, the work is equal to zero, so the energy equation becomes
5q1 + h5 = h1
A solution requires a value for h5, which can be found by taking a control volume around the pump:
h4 = h5 + 4w5
h5 = 188.4 − (−4) = 192.4 kJ/kg
Therefore, for the boiler,
5q1 + h5 = h1
5q1 = 3023.5 − 192.4 = 2831.1 kJ/kg
FIGURE 4.11 Simple steam power plant.
Questions & Answers
QUESTION:
Problem 44HP
Calculate the thermal efficiency of a Carnot cycle heat engine operating between reservoirs at 300°C and 45°C. Compare the result to that of Example 4.7.
Example 4.7
Consider the simple steam power plant, as shown in Fig. 4.11. The following data are for such a power plant where the states are numbered and there is specific pump work as 4 kJ/kg.
State |
Pressure |
Temperature or Quality |
1 |
2.0 MPa |
300°C |
2 |
1.9 MPa |
290°C |
3 |
15 kPa |
90% |
4 |
14 kPa |
45°C |
Determine the following quantities per kilogram flowing through the unit:
a. Heat transfer in the line between the boiler and turbine.
b. Turbine work.
c. Heat transfer in the condenser.
d. Heat transfer in the boiler.
Since there are several control volumes to be considered in the solution to this problem, let us consolidate our solution procedure somewhat in this example. Using the notation of Fig. 4.11, we have:
All processes: Steady-state.
Model: Steam tables.
From the steam tables:
h1 = 3023.5 kJ/kg
h2 = 3002.5 kJ/kg
h3 = 225.9 + 0.9(2373.1) = 2361.7 kJ/kg
h4 = 188.4 kJ/kg
All analyses: No changes in kinetic or potential energy will be considered in the solution. In each case, the energy equation is given by Eq.
Equation 4.13
Now, we proceed to answer the specific questions raised in the problem statement.
a. For the control volume for the pipeline between the boiler and the turbine, the energy equation and solution are
1q2 + h1 = h2
1q2 = h2 − h1 = 3002.5 − 3023.5 = −21.0 kJ/kg
b. A turbine is essentially an adiabatic machine. Therefore, it is reasonable to neglect heat transfer in the energy equation, so that
h2 = h3 + 2w3
2w3 = 3002.5 − 2361.7 = 640.8 kJ/kg
c. There is no work for the control volume enclosing the condenser. Therefore, the energy equation and solution are
3q4 + h3 = h4
3q4 = 188.4 − 2361.7 = −2173.3 kJ/kg
d. If we consider a control volume enclosing the boiler, the work is equal to zero, so the energy equation becomes
5q1 + h5 = h1
A solution requires a value for h5, which can be found by taking a control volume around the pump:
h4 = h5 + 4w5
h5 = 188.4 − (−4) = 192.4 kJ/kg
Therefore, for the boiler,
5q1 + h5 = h1
5q1 = 3023.5 − 192.4 = 2831.1 kJ/kg
FIGURE 4.11 Simple steam power plant.
ANSWER:
Solution
Step 1 of 2
We need to find out the thermal efficiency of the Carnot cycle.