An engine does 10J of work and exhausts 15J of waste heat. a. Draw an energy transfer diagram for this engine

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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### Thermodynamics: Engine Efficiency and Energy Transfer

#### Problem 5

An engine performs 10J of work and exhausts 15J of waste heat.

**a. Draw an energy transfer diagram for this engine**

An energy transfer diagram typically shows the input energy, useful output energy (work done), and waste energy (heat exhaust). For this engine, the diagram would illustrate:

1. **Input Energy:**
   - Total input energy is the sum of work done and waste heat.
   - In this case, input energy = 10J (work) + 15J (waste heat) = 25J

2. **Output:**
   - Useful work done: 10J
   - Waste heat: 15J

*Energy Transfer Diagram:*
- Start with a box labeled "Engine".
- Draw an arrow labeled "25J Input Energy" pointing to the engine.
- Draw an arrow pointing out from the engine labeled "10J Useful Work".
- Draw another arrow pointing out from the engine labeled "15J Waste Heat".

**b. What is the engine’s efficiency?**

The efficiency (\(\eta\)) of an engine is calculated by the ratio of the work output to the total energy input.

\[ \eta = \frac{\text{Work Output}}{\text{Total Energy Input}} \]

\[ \eta = \frac{10J}{25J} \]

\[ \eta = 0.4 \]

Convert to percentage:

\[ \eta = 0.4 \times 100\% = 40\% \]

Therefore, the engine's efficiency is \(40\%\).

**c. If the cold reservoir temperature is \(20^\circ C\), what is the minimum possible temperature in °C of the hot reservoir?**

To determine the minimum possible temperature of the hot reservoir, we use the Carnot efficiency formula which relates the efficiencies to the temperatures of the hot and cold reservoirs. 

\[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \]

Where:
- \(\eta_{\text{Carnot}}\) is the maximum efficiency (in decimal form),
- \(T_{\text{cold}}\) is the absolute temperature of the cold reservoir, 
- \(T_{\text{hot}}\) is the absolute temperature of the hot reservoir.

Given:
- \(\eta = 0
Transcribed Image Text:### Thermodynamics: Engine Efficiency and Energy Transfer #### Problem 5 An engine performs 10J of work and exhausts 15J of waste heat. **a. Draw an energy transfer diagram for this engine** An energy transfer diagram typically shows the input energy, useful output energy (work done), and waste energy (heat exhaust). For this engine, the diagram would illustrate: 1. **Input Energy:** - Total input energy is the sum of work done and waste heat. - In this case, input energy = 10J (work) + 15J (waste heat) = 25J 2. **Output:** - Useful work done: 10J - Waste heat: 15J *Energy Transfer Diagram:* - Start with a box labeled "Engine". - Draw an arrow labeled "25J Input Energy" pointing to the engine. - Draw an arrow pointing out from the engine labeled "10J Useful Work". - Draw another arrow pointing out from the engine labeled "15J Waste Heat". **b. What is the engine’s efficiency?** The efficiency (\(\eta\)) of an engine is calculated by the ratio of the work output to the total energy input. \[ \eta = \frac{\text{Work Output}}{\text{Total Energy Input}} \] \[ \eta = \frac{10J}{25J} \] \[ \eta = 0.4 \] Convert to percentage: \[ \eta = 0.4 \times 100\% = 40\% \] Therefore, the engine's efficiency is \(40\%\). **c. If the cold reservoir temperature is \(20^\circ C\), what is the minimum possible temperature in °C of the hot reservoir?** To determine the minimum possible temperature of the hot reservoir, we use the Carnot efficiency formula which relates the efficiencies to the temperatures of the hot and cold reservoirs. \[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] Where: - \(\eta_{\text{Carnot}}\) is the maximum efficiency (in decimal form), - \(T_{\text{cold}}\) is the absolute temperature of the cold reservoir, - \(T_{\text{hot}}\) is the absolute temperature of the hot reservoir. Given: - \(\eta = 0
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