Question

An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine the power output and the rate of heat rejection.

Answer 1

Answer on Question 58779, Physics, Mechanics, Relativity

Question:

An internal combustion engine uses fuel, of energy content $44.4 \, \text{MJ/kg}$, at a rate of $5 \, \text{kg/h}$. If the efficiency is $28\%$, determine the power output and the rate of heat rejection.

Solution:

a) Efficiency is defined as the ratio of useful work done to the heat energy absorbed by the engine:

$$\eta = \frac{\text{Work output}}{\text{Work input}} \cdot 100\%,$$

here, $\text{Work output}$ is the work done by the engine, $\text{Work input}$ is the heat energy absorbed by the engine.

Let's rewrite our efficiency formula in terms of power:

$$\eta = \frac{\text{Power output}}{\text{Power input}} \cdot 100\%,$$

here, $\text{Power input}$ is the amount of heat that is released during the combustion of fuel, $\text{Power output}$ is the useful power of the drive shaft of the engine without the power loss caused by gears, transmission or friction, for example.

Let's first calculate the $\text{Power input}$ by the formula:

$$\text{Power input} = \text{FC} \cdot \text{CV},$$

here, $\text{FC}$ is the fuel consumption, $\text{CV}$ is the calorific value of kilogram fuel.

Then, the $\text{Power input}$ will be:

$$\text{Power input} = \text{FC} \cdot \text{CV} = 5 \frac{\text{kg}}{\text{h}} \cdot \frac{1}{3600} \frac{\text{h}}{\text{s}} \cdot 44.4 \cdot 10^{6} \frac{\text{J}}{\text{kg}} = 61667 \, \text{W}.$$

As we know the $\text{Power input}$, we can find the $\text{Power output}$ from the efficiency formula:

$$\begin{array}{l} \text{Power output} = \frac{\eta}{100\%} \cdot \text{Power input} = \frac{28\%}{100\%} \cdot 61667 \, \text{W} = 0.28 \cdot 61667 \, \text{W} = \\ = 17267 \, \text{W}. \end{array}$$

b) The rate of heat rejection is equal to the difference of the power input and the power output:

$$Q = \text{Power input} - \text{Power output} = 61667\,W - 17267\,W = 44400\,W.$$

Answer:

a) Power output = 17267 W.

b) $Q = 44400\,W$.

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