Question

A heat engine receives an amount of energy Qh= 790 kJ by heat transfer from a high temperature thermal reservoir at Th=950 K. Energy is rejected by heat transfer to a lower temperature thermal reservoir at T1=590 K. If waste heat in the amount of Q1=160 kJ is rejected to the low temperature thermal reservoir during each cycle.

a) Solve for the maximum theoretical efficiency that an engine in this situation could operate with. ANSWER: 0.379

b) Solve for actual efficiency that the engine is operating with.

c) Which of the following best describes the manner in which the cycle is operating...

-Reversibly or Impossibly?

Answer 1

Answer:

(a) $\eta_{max}=37.895\%$

(b) $\eta=79.75\%$

(c) Impossibly

Explanation:

Given:

• temperature of source reservoir, $T_H=950\ K$

• temperature of sink reservoir, $T_L=590\ K$

• heat absorbed by the engine, $Q_H=790\ kJ$

• heat rejected by the engine, $Q_L=160\ kJ$

(a)

Now the maximum theoretical efficiency of the engine:

$\eta_{max}=\frac{T_H-T_L}{T_H}\times 100\%$

$\eta_{max}=\frac{950-590}{950}\times 100\%$

$\eta_{max}=37.895\%$

(b)

Actual efficiency of the heat engine:

$\eta=\frac{Q_H-Q_L}{Q_H}\times 100\%$

$\eta=\frac{790-160}{790}\times 100\%$

$\eta=79.75\%$

(c)

This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.

Answer 2

The answers are:

(a) the maximum theoretical efficiency is 37.9%

(b) the actual efficiency is 79.7%

(c) The cycle is operating impossibly

The following data is provided to us:

The temperature of the source, $T_1=950K$

The temperature of the sink, $T_2=590K$

Heat absorbed by the engine, $Q_1=790kJ$

Heat rejected by the engine, $Q_2=160kJ$

(a) Now, the maximum efficiency of the engine is:

$\eta _{max}=1-\frac{T_2}{T1}\\ \\\eta_{max}=1-\frac{590}{950}\\ \\\eta_{max}=0.379$

maximum efficiency ( $\eta_{max}$ ) = 37.9%

(b) actual efficiency:

$\eta=1-\frac{Q_2}{Q_1}\\ \\\eta=1-\frac{160}{790}\\ \\\eta=0.797$

actual efficiency ($\eta$) = 79.7%

(c) it is not possible for a Carnot engine to have more efficiency than the maximum possible efficiency.

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