Question

There are developed 10hp by a Carnot engine between 1000F and 100F.

Determine;

a. The thermal efficiency

b. The heat supplied per seconds

c. The change in entropy each second during heat rejection

Answer 1

Solution;

Given;

$T_1=1000F=810.928K$

$T_2=100F=310.928K$

$P=10hp=7.457kW$

(a)

Thermal efficiency;

$\eta=1-\frac{T_C}{T_H}$

$\eta=1-\frac{310.928}{810.928}=0.61658$

$\eta=0.61658$ or $61.658$ %

(b)

Heat supplied;

$\eta=\frac{P}{Q_i}$

$Q=\frac{P}{\eta}=\frac{7.457}{0.61658}=12.09kJ/s$

(c)

Change in entropy;

Heat rejected;

$Q_o=Q_i-P$

$Q_o=12.09-7.457=4.633kJ/s$

$\Delta s=\frac{Q_o}{T_2}=\frac{4.633×10^3}{310.298}=14.98J/sK$