Page 139 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Example 4.1

Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of

waste heat rejection to a nearby river is 45 MW, determine the net work done and

the thermal efficiency for this heat engine.

Solution to Example 4.1

FURNACE

Q1 = 80 MW

HEAT W =?

ENGINE

Q2 = 45 MW

RIVER

A schematic of the heat engine is given in the diagram above. The furnace serves as the
high-temperature reservoir for this heat engine and the river as the low-temperature

reservoir.

Assumption: Heat lost through the pipes and other components are negligible.
Analysis: The given quantities can be expressed in rate form as;

Q1 = 80 MW

Q2 = 45 MW

From equation 9.1, the net work done for this heat engine is;

W = Q1 – Q2
= (80 – 45) MW

= 35 MW

Then from equation 4.2, the thermal efficiency is easily determined to be

W 35 MW
 = = = 0.4375 (or 43.75%)
Q 1 80 MW

That is, the heat engine converts 43.75 percent of the heat it receives to work.

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