Question

A reversible heat engine, operation on Carnot cycle, between the temperature limits of 300 K and 1000 K produces 14 kW of power. If the calorific value of the fuel is 40,000 kJ/kg. The fuel consumption will be:
1. 1.4 kg/hr
2. 1.8 kg/hr
3. 2.0 kg/hr
4. 2.2 kg/hr

Answer 1

Correct Answer - Option 2 : 1.8 kg/hr

Concept:

For a reversible heat engine

\(\eta = 1 - \frac{{{T_{min}}}}{{{T_{max}}}}\)

where, T_min is the minimum temperature and T_max is the maximum temperature of the cycle.

Also, \(\eta = \;\frac{W}{Q}\)

where Q is the heat supplied to the engine and W is the work output of the engine

Q = ṁ_f × CV

Where, ṁ_f is the mass flow rate of fuel and CV is the calorific value of the fuel

Calculation:

Given, T_min = 300 K, T_max = 1000 K, W = 14 kW and CV of fuel = 40,000 kJ/kg

\(\eta = 1 - \frac{{300}}{{1000}} = 0.7\)

\(\eta = \;\frac{W}{Q}\)

\(Q = \frac{W}{\eta } =\frac{{14}}{{0.7}} = 20\;kW\)

Q = ṁ_f × CV

\({\dot m_f} = \frac{Q}{{CV}} = \frac{{20\;KW}}{{40000\;KJ/s}} = \frac{{20 \times 3600}}{{40000}}Kg/hr\)

ṁ_f = 18 kg/hr