Question

An automobile engine consumes fuel at a rate of 22 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine. The efficiency of the engine is

Answer 1

Answer:

The efficiency of the engine is 27.9 %.

Explanation:

From Thermodynamics, we define the efficiency of the engine by the following ratio:

$\eta = \frac{\dot W_{out}}{\dot Q_{in}}$ (1)

Where:

$\dot W_{out}$ - Power delivered by the automobile engine, measured in watts.

$\dot Q_{in}$ - Heat energy rate from combustion, measured in watts.

By definition of combustion heat transfer, we expand the equation above:

$\eta = \frac{\dot W_{out}}{\rho\cdot \dot V\cdot L_{c}}$ (2)

Where:

$\rho$ - Density of the fuel, measured in kilograms per cubic meter.

$\dot V$ - Flow rate, measured in cubic meters per second.

$L_{c}$ - Latent heat of combustion of the heat, measured in joules per kilogram.

If we know that $\rho = 800\,\frac{kg}{m^{3}}$, $\dot V = 6.111\times 10^{-6}\, \frac{m^{3}}{s}$, $L_{c} = 44\times 10^{6}\,\frac{J}{kg}$ and $\dot W_{out} = 60000\,W$, then the efficiency of the engine is:

$\eta = \frac{60000\,W}{\left(800\,\frac{kg}{m^{3}} \right)\cdot \left(6.111\times 10^{-6}\,\frac{m^{3}}{s} \right)\cdot \left(44\times 10^{6}\,\frac{J}{kg} \right)}$

$\eta = 0.279$

The efficiency of the engine is 27.9 per cent.

Answer 2

The efficiency of the engine is 21.9%

The volume flow rate = 22L/h, heating value = 44,000 kJ/kg

The density = 0.8 g/cm³ = 0.8 kg/L

The mass flow rate of consumption = density * volume flow rate = 0.8 kg/L * 22 L/h = 22.4 kg/h

The rate of heat supply = mass flow rate * heating value = 22.4 kg/h * 44,000 kJ/kg = 985600 kJ/h = 273.78 kW

Thermal efficiency = power delivered / rate of heat supply = 60 kW/273.78 kW = 21.9%

The efficiency of the engine is 21.9%

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