Homework Solution

TP 4-11

Given:

    Efficiency of plant, e_max = 0.38

    T_H = 550 ^oC

    T_C = 20 ^oC

Find:

    percentage of maximum possible efficiency

Science:

e_max = (1 - T_C /T_H) x 100%

where the temperatures must be expressed in kelvins

Solve:

 T_H = 273 K + 550 K = 823 K

T_C = 273 K + 20 K = 293 K

e_max = (1 - 293 K / 823 K) x 100%

e_max = 64.4%

% of max efficiency = (0.38)/(0644) x 100% = 59.0%_ANS