Answer to Question #305383 in Molecular Physics | Thermodynamics for Tobias Felix

Question #305383

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


1
Expert's answer
2022-03-07T11:45:53-0500

Solution;

Given;

T1=1000F=810.928KT_1=1000F=810.928K

T2=100F=310.928KT_2=100F=310.928K

P=10hp=7.457kWP=10hp=7.457kW

(a)

Thermal efficiency;

η=1TCTH\eta=1-\frac{T_C}{T_H}

η=1310.928810.928=0.61658\eta=1-\frac{310.928}{810.928}=0.61658

η=0.61658\eta=0.61658 or 61.65861.658 %

(b)

Heat supplied;

η=PQi\eta=\frac{P}{Q_i}

Q=Pη=7.4570.61658=12.09kJ/sQ=\frac{P}{\eta}=\frac{7.457}{0.61658}=12.09kJ/s

(c)

Change in entropy;

Heat rejected;

Qo=QiPQ_o=Q_i-P

Qo=12.097.457=4.633kJ/sQ_o=12.09-7.457=4.633kJ/s

Δs=QoT2=4.633×103310.298=14.98J/sK\Delta s=\frac{Q_o}{T_2}=\frac{4.633×10^3}{310.298}=14.98J/sK



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment