Question

QUESTION 2
A single cylinder, single acting reciprocal air compressor takes in air 2 m of air per minute at 102 kPa. Air is delivered at 780 kPa and the compressor runs at 300 revolutions per minute. The free volume is 3% of the stroke volume. Index of compression n= 1,35
Determine:
5.1
Stroke volume in m'.
5.2
Power in kW.
5.3
Work done in kJ.

Answer 1

To solve the problem, we'll go through each sub-question step by step.

### Given Data
- Air intake flow rate (\( Q \)) = 2 m³/min
- Inlet Pressure (\( P_1 \)) = 102 kPa = 0.102 MPa
- Delivery Pressure (\( P_2 \)) = 780 kPa = 0.78 MPa
- Revolutions per minute (\( n_r \)) = 300 RPM
- Index of compression (\( n \)) = 1.35
- Free volume = 3% of the stroke volume

### 5.1 Stroke Volume (\( V_s \))

The stroke volume can be calculated using the formula for the flow rate of a single acting compressor:

\[
Q = \frac{V_s \times n_r}{60}
\]

Rearranging this formula to find \( V_s \):

\[
V_s = \frac{Q \times 60}{n_r}
\]

Substituting the values:

\[
V_s = \frac{2 \, \text{m}^3/\text{min} \times 60 \, \text{sec/min}}{300 \, \text{rev/min}} = \frac{120 \, \text{m}^3}{300} = 0.4 \, \text{m}^3
\]

Thus, the **stroke volume** is **0.4 m³**.

### 5.2 Power in kW

The power required can be calculated using the formula:

\[
\text{Power (P)} = \frac{(P_2 - P_1) \times Q}{\eta}
\]

However, we'll assume that the compressor is ideal (η = 1 for simplicity) and use the work done per cycle to estimate the power.

We can firstly calculate the work done for each cycle, which will help us:

1. Calculate the mass flow rate (\( \dot{m} \)).
2. Use it to find the power.

Using the ideal gas law, we can find the density of air at \( P_1 \):
\[
\rho = \frac{P_1}{RT}
\]
Assuming \( R = 287 \, \text{J/(kg·K)} \) and \( T \) = 300 K (approximate)

\[
\rho = \frac{0.102 \times 10^6 \, \text{Pa}}{287 \times 300} \approx 1.18 \, \text{kg/m}^3
\]

Now, we can calculate the mass flow rate:

\[
\dot{m} = \rho \times Q = 1.18 \, \text{kg/m}^3 \times 2 \, \text{m}^3/\text{min} = 2.36 \, \text{kg/min} \approx 0.0393 \, \text{kg/sec}
\]

Now, we can calculate the work done per cycle:

\[
W = \frac{(P_2 - P_1)V_s}{n}
\]
\[
W = \frac{(0.78 - 0.102) \times 0.4}{1.35} = \frac{0.678 \times 0.4}{1.35} \approx 0.202 \, \text{kJ}
\]

Now, the power can be expressed as:

\[
P = W \times n_r
\]

Convert RPM to cycles per second:

\[
n_r = \frac{300}{60} = 5 \, \text{rev/sec}
\]

Thus:
\[
P = 0.202 \, \text{kJ} \times 5 \, \text{rev/sec} = 1.01 \, \text{kW}
\]

### 5.3 Work Done in kJ

As calculated above, the work done per cycle for the compressor is:

\[
W \approx 0.202 \, \text{kJ}
\]

To find the total work done in a minute, we multiply by the number of cycles in a minute (which is \( n_r \)):
\[
W_{total} = W \times n_r = 0.202 \times 300/60 = 1.01 \, \text{kJ}
\]

So the answers are:
- **5.1 Stroke Volume in \( m^3 \)**: \( 0.4 \, \text{m}^3 \)
- **5.2 Power in kW**: \( 1.01 \, \text{kW} \)
- **5.3 Work Done in kJ**: \( 1.01 \, \text{kJ} \)