Question

The data below are taken from a test of a petrol engine for a motor car.

Power output	150 kW
Fuel consumption	20 litres per hour
The energy content of the fuel	40 M J per litre

What is the ratio \[\xrightarrow[\text{power input}]{\text{power output}}\]?
\[\begin{align}
  & A.\,\dfrac{150\times {{10}^{3}}}{40\times {{10}^{6}}\times 20\times 60\times 60} \\
 & B.\,\dfrac{150\times 60\times 60}{20\times 40\times {{10}^{3}}} \\
 & C.\,\dfrac{150\times {{10}^{3}}\times 40\times {{10}^{6}}\times 20}{60\times 60} \\
 & D.\,\dfrac{150\times {{10}^{3}}\times 20}{40\times {{10}^{3}}\times 60\times 60} \\
\end{align}\]

Answer 1

Hint: The ratio of power output by the power input is called efficiency. We have to multiply the factors that are used as the input source. This ratio will be the output released by the petrol engine to the input consumed by the petrol engine.
Formula used:
\[\eta =\dfrac{{{P}_{o}}}{{{P}_{i}}}\]

Complete answer:
From the given information, we have the data as follows.
Power output equals 150 kW
Fuel consumption is 20 litres per hour.
The energy content of a fuel is 40 M J per litre
Firstly, we will compute the overall input consumed by the petrol engine, so, we have,
Power input equals the product of the fuel consumed and the energy content of the fuel.
\[{{P}_{i}}=Fuel\times Energy\]
Substitute the values in the above formula.
\[{{P}_{i}}=20\times 40\times {{10}^{6}}\,{J}/{h}\;\]
Now we have to convert the unit from Joule per hour to Joule per second.
\[{{P}_{i}}=\dfrac{20\times 40\times {{10}^{6}}\,}{60\times 60}{J}/{s}\;\]……. (1)
Now, we will compute the overall output power of the petrol engine, so, we have,
\[\begin{align}
  & {{P}_{o}}=150\,kW \\
 & \therefore {{P}_{o}}=150\times {{10}^{3}}\,W \\
\end{align}\]……. (2)
Divide the equation (2) by (1) to get the ratio of the power output by the power input of the petrol engine. So, we have,
\[\begin{align}
  & \dfrac{{{P}_{o}}}{{{P}_{i}}}=\dfrac{150\times {{10}^{3}}}{{}^{20\times 40\times {{10}^{6}}}/{}_{60\times 60}} \\
 & \Rightarrow \dfrac{{{P}_{o}}}{{{P}_{i}}}=\dfrac{150\times {{10}^{3}}\times 60\times 60}{20\times 40\times {{10}^{6}}} \\
 & \therefore \dfrac{{{P}_{o}}}{{{P}_{i}}}=\dfrac{150\times 60\times 60}{20\times 40\times {{10}^{3}}} \\
\end{align}\]
Thus, the ratio of power output by the power input is \[\dfrac{150\times 60\times 60}{20\times 40\times {{10}^{3}}}\].
The ratio of power output by the power input is called efficiency.
\[\therefore \] The ratio of the power output by the power input is \[\dfrac{150\times 60\times 60}{20\times 40\times {{10}^{3}}}\].

Thus, option (B) is correct.

Note:
Instead of the ratio of power output by power input, they can ask for the efficiency of the petrol engine. The efficiency of the engine will be unitless, as it is a ratio of 2 quantities having the same unit.