only solve question 1 completeleyER
) Tr — 3
KQ of steam at 5 bar, entropy 5.5 ky/(kg K), is heated revers, bly at constant pr
ntl the tem nperature is 300°C. Calculate the heat supplied and show o

Question

only solve question 1 completeleyER
) Tr — 3
KQ of steam at 5 bar, entropy 5.5 ky/(kg K), is heated revers, bly at constant pr
ntl the tem nperature is 300°C. Calculate the heat supplied and show on a T-S diagra i
rea whic! represents the heat flow. (15 marks) E —
b) Explain the application of the first and second laws of thermodynamics in a Rankine
~ cycle-based thermal power plant. (5 marks) Pr.

VIPSolutions · Accepted Answer

# Solution to Question 1

### Given Data

- **Mass of steam, $ m $ = 1 kg**
 **Initial pressure, $ P =5 $**
- ** entropy, $_1 =5.5 $ kJ/(kg·K)**
- **Final temperature, $ T_2 = 300^\circ C $**
- **Process: Reversible, constant pressure**

Let's solve step by step.

---

## Step 1: Find Initial and Final States

### **Initial State (State 1):**
- $ P_1 = 5 $ bar
- $ s_1 = 5.5 $ kJ/(kg·K)

Use steam tables to determine the properties at 5 bar:
- At 5 bar, saturated vapor:  
  - $ s_g = 6.821 $ kJ/(kg·K)  
  - $ T_{sat} = 151.86^\circ C $

Since $ s_1 < s_g $, it is a mixture or subcooled vapor. But $ s_1 $ is quite close to the typical entropy of vapor, so let's check quality:

$$
x_1 = \frac{s_1 - s_f}{s_g - s_f}
$$

Where:
- $ s_f $ at 5 bar = 1.939 kJ/(kg·K)

$$
x_1 = \frac{5.5 - 1.939}{6.821 - 1.939} = \frac{3.561}{4.882} \approx .73
$$

So, the initial state is a mixture with quality $ x_1 = .73 $.

### **Final State (State 2):**
- $ P_2 = 5 $ bar
- $ T_2 = 300^\circ C $

At 5 bar, the steam is **superheated** at 300°C.

---

## Step 2: Find Enthalpy at States

### **At State 1:**

$$
h_1 = h_f + x_1 (h_g - h_f)
$$

From steam tables at 5 bar:
- $ h_f = 640.09 $ kJ/kg
- $ h_g = 2748.7 $ kJ/kg

$$
h_1 = 640.09 + .73 \times (2748.7 - 640.09)
$$
$$
h_1 = 640.09 + .73 \times 2108.61
$$
$$
h_1 = 640.09 + 1549.29 = 2189.38 \text{ kJ/kg}
$$

### **At State 2:**

From superheated steam tables at $ P=5 $ bar, $ T=300^\circ C $:

- $ h_2 \approx 3045. $ kJ/kg

---

## Step 3: Calculate Heat Supplied

At constant pressure, the heat supplied ($ Q $) per kg is:

$$
Q = h_2 - h_1
$$
$$
Q = 3045. - 2189.38 = 855.62 \text{ kJ}
$$

---

## Step 4: T-s Diagram

- The area under the curve on a T-s diagram **(between $ s_1 $ and $ s_2 $ at constant pressure)** represents the heat supplied.

#### **Find $ s_2 $:**
From superheated steam tables at $ P=5 $ bar, $ T=300^\circ C $:

- $ s_2 \approx 7.013 $ kJ/(kg·K)

---

## Step 5: Final Answers

### **1. Heat Supplied:**
$$
\boxed{Q = 855.6 \text{ kJ}}
$$

### **2. T-s Diagram (Sketch):**

```
T
|
|         /------------------ (State 2: 300°C, s2)
|        /
|       /
|------/--------------------- (P constant)
|     /
|    /
|---/----------------------- (State 1: mixture, s1)
|
+----------------------------- s
   s1 (5.5)         s2 (7.013)
```

- **Area under curve between $ s_1 $ and $ s_2 $ at $ P=5 $ bar represents the heat supplied.**

---

## **Summary Table**

| State | Pressure (bar) | Temperature (°C) | Entropy (kJ/kg·K) | Enthalpy (kJ/kg) |
|-------|----------------|------------------|--------------------|------------------|
|  1    |      5         |     ≈152         |      5.5           |    2189.4        |
|  2    |      5         |     300          |      7.013         |    3045.        |

---

### **Final Answers**

- **Heat supplied:** $\boxed{855.6 \text{ kJ}}$
- **T-s diagram:** *See sketch above; area under the P=constant line from $ s_1 $ to $ s_2 $ represents heat supplied.*

---

If you need a hand-drawn T-s diagram or a more detailed one, let me know!

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