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Bashyboy
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Calculating the Efficiency of A Rankine Cycle For Various Parameters
Calculate the efficiency of a Rankine cycle that is modified from the parameters used in the text in each of the following three ways (one at a time), and comment briefly on the on the results: (a) reduce the maximum temperature to 500 deg. C; (b) reduce the maximum pressure to 100 bars; (c) reduce the minimum temperature 10 deg. C
The parameters given the text are T_min = 20 C, T_max = 600, P_min = 0.023 bar, and P_max = 300 bars.
At vertex 3, the temperature of the gas is 500 deg. C, and the pressure is 300 bars. Therefore, the entropy of the gas is 5.791, according to table 4.2. Because the process described by the segment 3-4 is adiabatic, there will be no heat lost or gained, which means that entropy will not change due to this; furthermore, the volume changes is small enough to allow us to make the approximation that the entropy does not change very much. Therefore, the water at 3 should have approximately the same entropy as it has at 4. So, I can claim that
[itex]S_3 = S_4[/itex] and [itex]S_4 = S_g + S_l[/itex].
Apparently I am suppose to do some sort of interpolation using the tables, but I am unsure of how to do this.
Homework Statement
Calculate the efficiency of a Rankine cycle that is modified from the parameters used in the text in each of the following three ways (one at a time), and comment briefly on the on the results: (a) reduce the maximum temperature to 500 deg. C; (b) reduce the maximum pressure to 100 bars; (c) reduce the minimum temperature 10 deg. C
Homework Equations
The parameters given the text are T_min = 20 C, T_max = 600, P_min = 0.023 bar, and P_max = 300 bars.
The Attempt at a Solution
At vertex 3, the temperature of the gas is 500 deg. C, and the pressure is 300 bars. Therefore, the entropy of the gas is 5.791, according to table 4.2. Because the process described by the segment 3-4 is adiabatic, there will be no heat lost or gained, which means that entropy will not change due to this; furthermore, the volume changes is small enough to allow us to make the approximation that the entropy does not change very much. Therefore, the water at 3 should have approximately the same entropy as it has at 4. So, I can claim that
[itex]S_3 = S_4[/itex] and [itex]S_4 = S_g + S_l[/itex].
Apparently I am suppose to do some sort of interpolation using the tables, but I am unsure of how to do this.
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