Question

A crane is being used to lift containers off a ship. One container has a mass of 14,000 kg and is being lifted vertically with a speed of $3.2\,m{{s}^{-1}}$. The electric motor used to supply the power to lift the container is using a current of 240 A at a potential difference of 2200 V. What is the efficiency of the system?
$A.\,8.1\%$
$B.\,8.5\%$
$C.\,48\%$
$D.\,83\%$

Answer 1

The efficiency of any system is basically calculated as the output by input. In this case, we will find efficiency in terms of power. The input power is the product of voltage and current. The output power is the energy generated by time. This value of the output power is, in turn, equal to the product of the mass, gravitational constant and the velocity.

Formula used: $P=VI$
$P=\dfrac{E}{t}$

Complete step by step answer:
The formula used to calculate the input power supplied to the electric motor of the crane is given as follows.
${{P}_{i}}=VI$
Where V is the voltage supplied and I is the current flowing.
From given, we have the values of the voltage supplied and the current flowing to be equal to 2200 V and 240 A respectively, so, we will substitute these values in the above equation.

& {{P}_{i}}=2200\times 240 \\\ & \Rightarrow {{P}_{i}}=528000\,W \\\ \end{aligned}$$ Therefore, the value of the input power is 528000 W. Now, we will continue with the calculation of the output power. The formula used to calculate the output power given by the crane to lift the container is given as follows. $$P=\dfrac{E}{t}$$ Where E is the energy required and t is the time taken. This energy can be represented as the product of force and displacement. Thus, $$P=\dfrac{F\times d}{t}$$ The ratio of the displacement by time gives the value of the velocity of the crane. Therefore, $${{P}_{o}}=F\times v$$ Where F is the force applied by the crane to lift the containers and v is the speed of the crane with which it lifts the container. Here, the force can be expressed as the product of mass and acceleration due to gravity. Thus, the final expression of the output power is, $${{P}_{o}}=mg\times v$$ From given, we have the values of the mass of the container and the speed to be equal to 14,000 kg and $$3.2\,m{{s}^{-1}}$$respectively, so, we will substitute these values in the above equation. $$\begin{aligned} & {{P}_{o}}=14000\times 9.8\times 3.2 \\\ & \Rightarrow {{P}_{o}}=439040\,W \\\ \end{aligned}$$ Therefore, the value of the output power is 439040 W. The efficiency of the system is given as the ratio of the output power by the input power. Thus, we have, $$\eta =\dfrac{{{P}_{o}}}{{{P}_{i}}}$$ Substitute the given values in the above equation. $$\begin{aligned} & \eta =\dfrac{439040}{528000} \\\ & \Rightarrow \eta =0.83 \\\ \end{aligned}$$ Now represent this value in terms of the percentage. $$\begin{aligned} & \eta \%=0.83\times 100 \\\ & \Rightarrow \eta \%=83\% \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** The units of the parameters should be taken care of. In this given case, in place of the acceleration of the crane, the acceleration due to gravity is considered. The reason being, the crane lifted the containers in the vertically upward direction.