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Give step-by-step solution with explanation and final answer:1) A system consists of 2 kg of carbon dioxide gas initially at state 1, where pi = 1 bar, T1 = 3

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:1) A system consists of 2 kg of carbon dioxide gas initially at state 1, where pi = 1 bar, T1 = 300

K. The system undergoes a power cycle consisting of the following processes:

- Process 1-2: Constant volume to p> = 4 bar.

- Process 2-3: Expansion with pv!2$= constant.

- Process 3-1: Constant-pressure compression.
Assuming the ideal gas model and neglecting kinetic and potential energy effects,
sketch the cycle on a p—V diagram and calculate the work and heat transfer for each
process. Finally, calculate the thermal efficiency of the cycle.

VIPSolutions · Accepted Answer

Let's solve the problem step by step.

---

# **Step 1: Given Data**

- Mass, $ m = 2 $ kg
- Gas: CO$_2$ (ideal gas assumption)
- State 1: $ p_1 = 1 $ bar $ = 100 $ kPa, $ T_1 = 300 $ K
- Process 1–2: Isochoric (constant volume), $ p_2 = 4 $ bar $ = 400 $ kPa
- Process 2–3: $ p v^{1.28} = \text{constant} $ (polytropic expansion)
- Process 3–1: Isobaric (constant pressure) compression

The specific gas constant for CO$_2$:  
$ R = .1889 $ kJ/kg·K

---

# **Step 2: Find State Properties**

## **At State 1:**
Use the ideal gas law:
$$
p_1 V_1 = m R T_1
$$
$$
V_1 = \frac{m R T_1}{p_1}
$$
$$
V_1 = \frac{2 \times .1889 \times 300}{100} = \frac{113.34}{100} = 1.1334 \text{ m}^3
$$

## **At State 2:**
- $ V_2 = V_1 $ (isochoric)
- $ p_2 = 400 $ kPa

$$
T_2 = \frac{p_2 V_2}{m R}
$$
$$
T_2 = \frac{400 \times 1.1334}{2 \times .1889} = \frac{453.36}{.3778} \approx 120 \text{ K}
$$

---

# **Step 3: Find State 3 Properties**

## **Polytropic Expansion ($ p v^{1.28} = \text{constant} $)**

Let $ V_3 $ be the volume at 3, $ p_3 $ the pressure at 3.

$$
p_2 V_2^{1.28} = p_3 V_3^{1.28}
\implies \left(\frac{V_3}{V_2}\right)^{1.28} = \frac{p_2}{p_3}
$$

### **But Process 3–1 is Constant Pressure ($ p_3 = p_1 $)**

$$
p_3 = p_1 = 100 \text{ kPa}
$$
$$
\left(\frac{V_3}{V_2}\right)^{1.28} = \frac{400}{100} = 4
\implies \frac{V_3}{V_2} = 4^{1/1.28}
$$
Calculate $ 1/1.28 \approx .78125 $:

$$
\frac{V_3}{V_2} = 4^{.78125} \approx e^{.78125 \ln 4} \approx e^{.78125 \times 1.3863} \approx e^{1.083} \approx 2.953
$$

$$
V_3 = 2.953 \times V_2 \approx 2.953 \times 1.1334 = 3.349 \text{ m}^3
$$

Now, find $ T_3 $:
$$
T_3 = \frac{p_3 V_3}{m R}
= \frac{100 \times 3.349}{2 \times .1889}
= \frac{334.9}{.3778} \approx 886 \text{ K}
$$

---

# **Step 4: Sketch p-V Diagram**

```
Y-axis: Pressure (p)
X-axis: Volume (V)
1: (1 bar, 1.13 m³)
2: (4 bar, 1.13 m³) [vertical up: process 1-2]
3: (1 bar, 3.35 m³) [curve down/right: process 2-3]
Back to 1: (horizontal left: process 3-1)
```

---

# **Step 5: Work and Heat Calculations**

## **Process 1–2: Isochoric (ΔV = )**
$$
W_{1-2} = 
$$
$$
Q_{1-2} = m c_v (T_2 - T_1)
$$
For CO$_2$ (ideal): $ c_v \approx .655 $ kJ/kg·K

$$
Q_{1-2} = 2 \times .655 \times (120 - 300) = 2 \times .655 \times 900 = 2 \times 589.5 = 1179 \text{ kJ}
$$

## **Process 2–3: Polytropic Expansion ($ n = 1.28 $)**

$$
W_{2-3} = \frac{p_3 V_3 - p_2 V_2}{1 - n}
$$
$$
= \frac{100 \times 3.349 - 400 \times 1.1334}{1 - 1.28}
$$
$$
= \frac{334.9 - 453.36}{-.28}
= \frac{-118.46}{-.28} = 423.8 \text{ kJ}
$$

$$
Q_{2-3} = m c_v (T_3 - T_2) + W_{2-3}
$$
$$
= 2 \times .655 \times (886 - 120) + 423.8
= 2 \times .655 \times (-314) + 423.8
= 2 \times (-205.57) + 423.8
= -411.14 + 423.8 = 12.7 \text{ kJ}
$$

## **Process 3–1: Isobaric Compression**
$$
W_{3-1} = p_1 (V_1 - V_3)
= 100 \times (1.1334 - 3.349) = 100 \times (-2.2156) = -221.56 \text{ kJ}
$$
$$
Q_{3-1} = m c_p (T_1 - T_3)
$$
For CO$_2$ (ideal): $ c_p \approx .844 $ kJ/kg·K

$$
Q_{3-1} = 2 \times .844 \times (300 - 886)
= 2 \times .844 \times (-586)
= 2 \times (-494.78) = -989.56 \text{ kJ}
$$

---

# **Step 6: Cycle Summary**

| Process | $ W $ (kJ) | $ Q $ (kJ) |
|---------|--------------|--------------|
| 1–2     |             | 1179         |
| 2–3     | 423.8        | 12.7         |
| 3–1     | -221.56      | -989.56      |

### **Net Work Done**
$$
W_\text{cycle} =  + 423.8 - 221.56 = 202.24 \text{ kJ}
$$

### **Net Heat Added**
$$
Q_\text{in} = Q_{1-2} + Q_{2-3} = 1179 + 12.7 = 1191.7 \text{ kJ}
$$
(Only heat input during 1-2 and 2-3.)

---

# **Step 7: Thermal Efficiency**

$$
\eta = \frac{W_\text{cycle}}{Q_\text{in}} = \frac{202.24}{1191.7} = .1697 = 16.97\%
$$

---

# **# Final Answers**

### **a) Work and Heat for Each Process**
- **Process 1–2:** $ W =  $ kJ, $ Q = 1179 $ kJ
- **Process 2–3:** $ W = 423.8 $ kJ, $ Q = 12.7 $ kJ
- **Process 3–1:** $ W = -221.56 $ kJ, $ Q = -989.56 $ kJ

### **b) Net Work Output:**  
**202 kJ**

### **c) Thermal Efficiency:**  
**16.97%**

---

### **d) p-V Diagram Sketch**
```
  |
p |      2
  |      |
  |      |
  |      |
  |     / 
  |    /
  |___/__________ 
     1          3
        V
```
- 1→2: vertical (constant V, up)
- 2→3: curve down/right (polytropic)
- 3→1: horizontal (constant p, left)

---

**Let me know if you need clarifications or further steps!**

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