Question

An ideal, Carnot engine operates between reservoirs at temperature 20 °C and 200 °C. If 10 kW of power is produced, the rate at which heat is rejected is approximately:

1. 15.3 kJ/ s
2. 12.3 kJ/s
3. 14.3 kJ/s
4. 16.3 kJ/s

Answer 1

Correct Answer - Option 4 : 16.3 kJ/s

CONCEPT:

• Carnot engine an ideal reversible engine that operates between two temperatures T1 (Source), and T2 (Sink).
• Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
• The steps of the Carnot cycle are
  1. Isothermal expansion
  2. Adiabatic expansion
  3. Isothermal compression
  4. Adiabatic compression
• The efficiency of the Carnot engine is defined as the ratio of net work done per cycle by the engine to heat absorbed per cycle by the working substance from the source. 
• The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

​\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

SOLUTION:

Given - T1 = 200° C = 473 K, T2 = 20° C = 293 K and W = 10 kW = 10 kJ/sec

• The efficiency is given by

​\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{{{293}}}{{{473}}}=0.374\)

• The amount of heat absorbed by the Carnot engine can be calculated is 

\(⇒ Q_1= \frac{W}{\eta}=\frac{10}{0.374}=26.73\; kW\)

• The amount of heat rejected to the sink will be:

⇒ Q2 = Q1 - W

⇒ Q2 = 26.73 - 10 = 16.73 kW = 16.73 kJ/sec