A heat engine receives 1 kW heat transfer at 1000 K and gives out 400 W as work with the rest as heat transfer to the ambient. Find its first and second law efficiencies.
Question 8.99: A heat engine receives 1 kW heat transfer at 1000 K and give...
First law efficiency is based on the energies
\eta_{ I }=\dot{ W } / \dot{ Q }_{ H }=\frac{0.4}{1}= 0 .4The second law efficiency is based on work out versus exergy in
Exergy flux in: \dot{\Phi}_{ H }=\left(1-\frac{ T _{ o }}{ T _{ H }}\right) \dot{ Q }_{ H }=\left(1-\frac{298.15}{1000}\right) 1 kW =0.702 kW
\eta_{ II }=\frac{\dot{ W }}{\dot{\Phi}_{ H }}=\frac{0.4}{0.702}= 0 . 5 7Notice the exergy flux in is equal to the Carnot heat engine power output given 1 kW at 1000 K and rejecting energy to the ambient.
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