An engine does 10J of work and exhausts 15J of waste heat. a. Draw an energy transfer diagram for this engine
An engine does 10J of work and exhausts 15J of waste heat. a. Draw an energy transfer diagram for this engine
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0a6cd48-f835-40c9-801b-cc7ae68f01ed%2F7594e5a3-f053-492c-8ffc-d5100a51e3e4%2F5sy7wr7_processed.jpeg&w=3840&q=75)
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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