Air is compressed from
1
5
0
Kpa and
5
0
degrees Celsius to
1
5
0
0
Kpa
a
.

To optimise the final temperature of the gas, you are asked to use various polytrophic compression processes

Question

Air is compressed from 
1
5
0
Kpa and 
5
0
 ﻿degrees Celsius to 
1
5
0
0
Kpa
a
.

﻿To optimise the final temperature of the gas, you are asked to use various polytrophic compression processes from first principles and calculate the final temperature when the process is:

i
.

﻿Isothermal,
ii
.

﻿Polytropically 
(
n
=
1
.
2
)
,
iii.

﻿Adiabatically
iv
.

﻿Polytropically 
(
n
=
1
.
6
)
.
v
.

﻿Comment on your findings.
b
.

﻿From the polytrophic processes above 
(
i 
-
 ﻿iii
)
,
 ﻿apply the first law principles to derive the work and heat transfer for the thermodynamic processes of a perfect gas., Give complete answer

VIPSolutions · Accepted Answer

# Air Compression: Final Temperature & Thermodynamic Analysis

We are given:

- **Initial pressure**: $P_1 = 150\,\text{kPa}$
- **Final pressure**: $P_2 = 150\,\text{kPa}$
- **Initial temperature**: $T_1 50^\circ\text{C} = 323.15\,\text{K}$
- **Working fluid**: Air (assumed ideal gas)
- **Specific gas constant for air**: $R = .287\,\text{kJ/kg}\cdot\text{K}$
- **Specific heat at constant pressure** ($c_p$): $1.005\,\text{kJ/kg}\cdot\text{K}$
- **Specific heat at constant volume** ($c_v$): $.718\,\text{kJ/kg}\cdot\text{K}$
- **Ratio of specific heats** ($\gamma$): $1.4$

We are to find the **final temperature ($T_2$)** for:

1. Isothermal compression
2. Polytropic compression ($n=1.2$)
3. Adiabatic compression ($n=\gamma=1.4$)
4. Polytropic compression ($n=1.6$)

## 1. Final Temperature for Each Process

### 1.1. Isothermal Compression ($n = 1$)

For an **isothermal process** ($T_2 = T_1$):
$$
T_2 = T_1 = 323.15\,\text{K}
$$

---

### 1.2. Polytropic Compression ($n = 1.2$)

For a **polytropic process**:
$$
\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}
$$
$$
\frac{T_2}{323.15} = \left(\frac{150}{150}\right)^{\frac{1.2-1}{1.2}}
$$
$$
\frac{T_2}{323.15} = (10)^{\frac{.2}{1.2}} = 10^{.1667} \approx 1.4678
$$
$$
T_2 = 1.4678 \times 323.15 \approx 474.8\,\text{K}
$$

---

### 1.3. Adiabatic Compression ($n = \gamma = 1.4$)

For an **adiabatic process** ($n = \gamma$):
$$
\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}
$$
$$
\frac{T_2}{323.15} = (10)^{\frac{.4}{1.4}} = 10^{.2857} \approx 1.9307
$$
$$
T_2 = 1.9307 \times 323.15 \approx 623.8\,\text{K}
$$

---

### 1.4. Polytropic Compression ($n = 1.6$)

$$
\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}
$$
$$
\frac{T_2}{323.15} = (10)^{\frac{.6}{1.6}} = 10^{.375} \approx 2.3714
$$
$$
T_2 = 2.3714 \times 323.15 \approx 765.8\,\text{K}
$$

---

### **Summary Table**

| Process           | $n$  | $T_2$ (K) | $T_2$ (°C) |
|-------------------|--------|-------------|--------------|
| Isothermal        | 1.    | 323.2       | 50           |
| Polytropic (1.2)  | 1.2    | 474.8       | 201.7        |
| Adiabatic         | 1.4    | 623.8       | 350.7        |
| Polytropic (1.6)  | 1.6    | 765.8       | 492.6        |

---

## 2. Work and Heat Transfer (First Law) for Each Process

### Assumptions

- **One kg of air** (for per-mass basis)
- **Ideal gas behavior**

### 2.1. **Isothermal Compression**

#### **Work Done**

$$
W = nRT_1 \ln\left(\frac{V_1}{V_2}\right) = nRT_1 \ln\left(\frac{P_2}{P_1}\right)
$$

For 1 kg:
$$
W = RT_1 \ln\left(\frac{P_2}{P_1}\right) = .287 \times 323.15 \times \ln(10)
$$
$$
W = .287 \times 323.15 \times 2.3026 \approx 213.6\,\text{kJ}
$$

#### **Heat Transfer**

For isothermal, $\Delta U = $ (internal energy depends only on temperature):

$$
Q = W = 213.6\,\text{kJ}
$$

---

### 2.2. **Polytropic Compression** ($n = 1.2$)

#### **Work Done**

$$
W = \frac{nR(T_2 - T_1)}{1 - n}
$$
Or,
$$
W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
$$

Using temperatures:
$$
W = \frac{.287 \times (474.8 - 323.15)}{1 - 1.2}
$$
$$
W = \frac{.287 \times 151.65}{-.2}
$$
$$
W = \frac{43.51}{-.2} = -217.55\,\text{kJ}
$$

The negative value indicates work **done on the gas** (compression). Conventionally, compression work is positive, so:

$$
W = 217.6\,\text{kJ}
$$

#### **Change in Internal Energy**

$$
\Delta U = m c_v (T_2 - T_1) = .718 \times (474.8 - 323.15) = .718 \times 151.65 \approx 108.97\,\text{kJ}
$$

#### **Heat Transfer**

First law: $Q = \Delta U + W$

$$
Q = 108.97 + 217.6 = 326.6\,\text{kJ}
$$

---

### 2.3. **Adiabatic Compression** ($n = \gamma = 1.4$)

#### **Work Done**

$$
W = \frac{R(T_2 - T_1)}{1 - \gamma}
$$
$$
W = \frac{.287 \times (623.8 - 323.15)}{1 - 1.4}
$$
$$
W = \frac{.287 \times 300.65}{-.4}
$$
$$
W = \frac{86.29}{-.4} = -215.7\,\text{kJ}
$$
So,
$$
W = 215.7\,\text{kJ}
$$

#### **Change in Internal Energy**

$$
\Delta U = .718 \times (623.8 - 323.15) = .718 \times 300.65 \approx 216.1\,\text{kJ}
$$

#### **Heat Transfer**

Adiabatic: $Q = $

---

## 3. **Comments and Observations**

### Temperature Trends

- **Isothermal Compression** yields the lowest final temperature (no increase).
- **Adiabatic Compression** yields a significantly higher final temperature due to no heat loss.
- **Polytropic Compression** falls between isothermal and adiabatic, with final temperature depending on the polytropic exponent $n$. As $n$ increases, the process becomes more adiabatic, and $T_2$ increases.

### Work and Heat Transfer

- **Isothermal**: All the work done appears as heat transfer out of the system; no internal energy change.
- **Adiabatic**: All the work increases internal energy (temperature); no heat is transferred.
- **Polytropic**: Both internal energy and heat transfer change; split depends on $n$.

### Process Efficiency

- **Isothermal compression** is ideal for minimizing temperature rise (and thus, work input for a given p

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