A heat engine develops 10 KW power when receiving heat at the rate of 2250 KJ/min. Determine thermal efficiency and the rate of heat rejection.
The correct answer and explanation is :
However, I can still manually calculate and explain the solution for you step-by-step and generate an image to represent the question. Let’s go through it:
Given:
- Power output (W) = 10 kW
- Heat input ( Q_{in} ) = 2250 kJ/min
Step 1: Convert Heat Input to kW
Since:
- 1 kW = 1 kJ/s
- 2250 kJ/min = ( \frac{2250}{60} ) kJ/s = 37.5 kW
Step 2: Calculate Thermal Efficiency
[
\text{Thermal Efficiency} = \frac{\text{Power Output}}{\text{Heat Input}} = \frac{10}{37.5} = 0.2667
]
[
\text{Thermal Efficiency (\%)} = 0.2667 \times 100 = \boxed{26.67\%}
]
Step 3: Determine Heat Rejection Rate
[
Q_{out} = Q_{in} – W = 37.5 – 10 = \boxed{27.5\text{ kW}}
]
Final Answers:
- Thermal Efficiency = 26.67%
- Heat Rejection Rate = 27.5 kW
Explanation (300+ words):
A heat engine is a system that converts heat (thermal energy) into mechanical work. The performance of a heat engine is primarily evaluated using thermal efficiency, which is the ratio of useful work output to the heat energy input.
In this case, the heat engine receives 2250 kJ of heat per minute from an external source. Since power and energy must be expressed in consistent time units for accurate efficiency calculations, we first convert this to kW (kilojoules per second), which is the standard unit for power. Doing so, we find that the engine receives 37.5 kW of thermal energy.
The engine produces 10 kW of useful mechanical power. The thermal efficiency is then calculated by dividing the power output by the heat input: ( \frac{10}{37.5} = 0.2667 ), or 26.67%. This indicates that only about a quarter of the supplied thermal energy is converted into useful work — a common occurrence in practical thermodynamic systems due to limitations imposed by the Second Law of Thermodynamics.
The remaining energy (i.e., the part of the input heat that is not converted into work) must be rejected, usually into the environment, in the form of waste heat. This is found by subtracting the useful work output from the total heat input: ( 37.5 – 10 = 27.5 ) kW. This rejected heat needs to be dissipated effectively to maintain continuous engine operation.
These calculations are essential in understanding the effectiveness of real-world engines and are foundational to fields like mechanical engineering, energy management, and environmental engineering.