2.1 Definition of the Unit Pascal
The pascal (symbol: Pa) is the SI unit of pressure, defined as one newton per square meter. In other words, it measures how much force is applied to a given area. Therefore, \(1 , \text{Pa} = 1 , \text{N/m}^2\).
2.2 Pump Calculations
Given Data
- Diameter of each plunger (\(D\)) = 6 cm = 0.06 m
- Stroke length (\(L\)) = 20 cm = 0.2 m
- Pressure (\(P\)) = 1.2 MPa = 1,200,000 Pa
- Speed of motor (\(N\)) = 150 r/min
- Efficiency of motor (\(\eta\)) = 75% = 0.75
- Slip = 2%
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Calculate the Area of Each Plunger
The area \(A\) of the circular plunger can be calculated using the formula: \[ A = \pi \left( \frac{D}{2} \right)^2 \] Substituting the values: \[ A = \pi \left( \frac{0.06}{2} \right)^2 = \pi \left( 0.03 \right)^2 \approx 0.002827 , \text{m}^2 \]
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Calculate the Volume Delivered per Pumping Stroke (per plunger)
The volume \(V\) delivered per stroke is given by: \[ V = A \times L \] Substituting the values: \[ V = 0.002827 , \text{m}^2 \times 0.2 , \text{m} \approx 0.0005654 , \text{m}^3 \]
Since there are three plungers, the total volume delivered per stroke is: \[ V_{\text{total}} = 3 \times 0.0005654 , \text{m}^3 \approx 0.0016962 , \text{m}^3 \]
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Calculate the Volume Delivered per Minute (Without Slip)
The number of strokes per minute: \[ \text{Strokes per minute} = N = 150 , \text{strokes/min} \]
Therefore, the volume delivered per minute is: \[ V_{\text{delivered}} = \text{Strokes per minute} \times V_{\text{total}} = 150 \times 0.0016962 , \text{m}^3 \approx 0.25443 , \text{m}^3/\text{min} \]
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Adjust for Slip (2%)
The effective volume delivered after accounting for 2% slip: \[ V_{\text{effective}} = V_{\text{delivered}} \times (1 - \text{Slip}) \] \[ V_{\text{effective}} = 0.25443 , \text{m}^3/\text{min} \times (1 - 0.02) \approx 0.24935 , \text{m}^3/\text{min} \]
2.2.1 Calculate the Power Required to Drive the Pump
The hydraulic power \(P_h\) required to drive the pump can be calculated using the formula: \[ P_h = P \times Q \]
Where \(Q\) is the volumetric flow rate (in m³/s). To convert our volume flow from m³/min to m³/s: \[ Q = \frac{V_{\text{effective}}}{60} = \frac{0.24935}{60} \approx 0.0041558 , \text{m}^3/s \]
Now plug this value into the formula for hydraulic power: \[ P_h = 1,200,000 , \text{Pa} \times 0.0041558 , \text{m}^3/s \approx 4996.96 , \text{W} \approx 4.997 , \text{kW} \]
To find the power required from the motor accounting for efficiency: \[ P_{\text{motor}} = \frac{P_h}{\eta} = \frac{4996.96}{0.75} \approx 6662.61 , \text{W} \approx 6.663 , \text{kW} \]
Summary of the Results
- 2.2.1 The power required to drive the pump at 150 r/min with an efficiency of 75% is approximately 6.663 kW.
- 2.2.2 The volume of water delivered per minute, accounting for 2% slip, is approximately 0.24935 m³/min.