 A heat engine receives heat from a source at 1200K at the rate of 500 Kw and rejects heat to a medium at 300K. The power output of the heat engine is 180 kW. Find irreversibility rate

To determine the **irreversibility rate** (also called the exergy destruction or the loss due to irreversibility), we can use the Second Law of Thermodynamics. The irreversibility is related to the entropy generation in the system.

### Given:
- Temperature of the heat source, \( T_H = 1200 \, \text{K} \)
- Temperature of the heat sink, \( T_L = 300 \, \text{K} \)
- Heat input from the source, \( Q_H = 500 \, \text{kW} \)
- Power output of the heat engine, \( W_{\text{out}} = 180 \, \text{kW} \)

### Step 1: Calculate the heat rejected
The first law of thermodynamics tells us that the difference between the heat input and the work output is the heat rejected to the sink:
\[
Q_L = Q_H - W_{\text{out}}
\]
Substitute the given values:
\[
Q_L = 500 \, \text{kW} - 180 \, \text{kW} = 320 \, \text{kW}
\]

### Step 2: Calculate the reversible (ideal) work
The reversible work can be calculated using the Carnot efficiency, which is the maximum efficiency of a heat engine operating between two thermal reservoirs:
\[
\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}
\]
Substitute the values:
\[
\eta_{\text{Carnot}} = 1 - \frac{300}{1200} = 1 - 0.25 = 0.75
\]
Now, the reversible work output is:
\[
W_{\text{rev}} = \eta_{\text{Carnot}} \times Q_H = 0.75 \times 500 \, \text{kW} = 375 \, \text{kW}
\]

### Step 3: Calculate the irreversibility (exergy destruction)
The irreversibility (I) is the difference between the reversible work and the actual work output:
\[
I = W_{\text{rev}} - W_{\text{out}} = 375 \, \text{kW} - 180 \, \text{kW} = 195 \, \text{kW}
\]

Thus, the irreversibility rate is **195 kW**.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the significance of the Carnot efficiency in heat engine analysis?
2. How does the temperature difference between source and sink affect irreversibility?
3. Can a real heat engine ever achieve Carnot efficiency? Why or why not?
4. How does entropy generation relate to irreversibility in thermodynamic systems?
5. What are the ways to reduce irreversibility in a practical heat engine?

**Tip:** Lowering the temperature difference between the heat source and sink can help reduce the irreversibility in real systems, improving overall efficiency.

A heat engine receives heat from a source at 1200K at the rate of 500 kW and rejects heat to a medium at 300K. The power output of the heat engine is 180 kW. Find irreversibility rate.

Learn how to calculate the irreversibility rate of a heat engine that receives heat at 1200K and rejects heat at 300K with a power output of 180 kW. This thermodynamic problem uses the Carnot efficiency formula and explores key concepts like exergy destruction. Suitable for undergraduate engineering students.

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