Question

A small diesel engine uses a volume of \[1.5 \times {10^4}c{m^3}\] fuel per hour to produce a useful power output of \[40kW\] . It may be assumed that \[34kJ\] of energy is transferred to the engine where it uses \[1.0c{m^3}\] of fuel.
What is the rate of transfer from the engine of energy that is wasted?
A. \[850kW\]
B. \[920kW\]
C. \[840kW\]
D. \[810kW\]

Answer 1

Hint: As we know that the rate of transfer of energy is known as power. And we are asked to calculate the wasted or dissipated power. First calculate the power generated by fuel in one hour. And then calculate the energy produced per second. As we are given the useful energy and we also know that total energy is the sum of useful energy and dissipated energy.

Complete step-by-step solution:he engine uses a volume of \[1.5 \times {10^4}c{m^3}\] of fuel per hour. This quantity is the volume per unit time (hour).
\[\dfrac{{volume}}{{time}} = 1.5 \times {10^4}c{m^3}{h^{ - 1}}\]
From the question, when \[1.0c{m^3}\] of fuel is used, \[34kJ\] of energy is transferred.
Total energy transferred \[1.5 \times {10^4}{\text{ }} \times {\text{ }}34{\text{ }} = {\text{ }}510{\text{ }}000{\text{ }}kJ\]
\[ = \;5.1 \times {10^5}\;kJ\]
Now energy produced in one second is \[\dfrac{{5.1 \times {{10}^5}\;}}{{60}}kJ/s = 850\;kW\]
As we know Total Energy = Wasted(dissipated) Energy + Useful Energy
So Wasted Energy = Total Energy – Useful Energy
\[ = (850 - 40)\;kW\]
\[\; = \;810\;kW\]
Hence, option D is correct.

Note:- We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that appliance. The power consumption rate is \[P = \dfrac{W}{t} = \dfrac{E}{t}\]