Question

Can anyone help me please.
A 42 kW engine is operating between 227 degree C and 177 degree C. Calculate
i)efficiency of the engine
ii)the amount of heat absorbed and the amount of heat rejected per second.

Answer 1

My answer for 1st part is 0.1.
for the 2nd part
heat absorbed=420kW
heat rejected=378kW

Answer 2

I'll assume this is a Carnot Heat engine.

The efficiency of a carnot engine is simply\[\eta = 1 - {T_L \over T_H}\]

We can find the heat absorbed from the hot reservoir from the expression for thermal efficiency. \[\eta = 1 - {T_L \over T_H} = {W \over Q_H}\]We know \(T_L\), \(T_H\), and \(W\) therefore we can solve for \(Q_H\).

From first law analysis of the cold airstandard carnot cycle, we can show that\[{Q_H \over Q_C} = {T_H \over T_L}\] (Refer to this link for a derivation of this expression (scroll to the bottom): http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm)

Therefore, we can solve for \(Q_C\), which is the heat rejected.

Answer 3

Thanks a lot