In both the operations, the engine rejects heat to a thermal sink at 20oC. Determine the operation in which the engine delivers more power.

Mumbai university > Mechanical Engineering > Sem 3 > Thermodynamics

Marks: 12M

Year: May 2016

A reversible engine operates with two sources separately:

Operation 1:

$T_H1=400+273=673K,T_L=20+273=293K,Q_H1=12000kW,$

Operation 2:

$T_H2=100+273=373K,Q_H2=25000kW$

To find: operation in which the engine will develop more power

Analysis:

Operation I:

The efficiency of a reversible heat engine

$η_1=\frac{1-T_L}{T_H1} =\frac{1-293}{673}=0.5646$

Power developed by the engine in the operation 1

$P_1=η_1×Q_H=0.5646×12000=6775.63kW$

Operation 2:

The efficiency of a reversible heat engine

$η_2=\frac{1-T_L}{T_H2} =\frac{1-293}{373}=0.2144$

Power developed by the engine in the operation 2,

$P_1=η_2×Q_H=0.2144×25000=5361.93kW$

The engine will develop more power in the operation 1 all engines develop more power when heat is supplied at higher temperature. Heat energy at higher temperature has more work capability.