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:
- Calculate the mass flow rate (\( \dot{m} \)).
- 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: \( 1.01 , \text{kJ} \)
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