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The motor of a pump consumes 1.05 hp of electricity. The
pump increases the pressure of water from 120 KPa to 1100
KPa at a rate of 35 Umin. If the motor efficiency is 94 percent
determine the pump?
Wise Nat
4 days ago
The motor of a pump consumes 1.05 hp of electricity. The pump increases the pressure of water from 120 KPa to 1100 KPa at a rate of 35 Umin. If the motor efficiency is 94 percent determine the pump?
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ChatGPT
4 days ago
To determine the pump efficiency, we first need to calculate the power output of the pump.
Power output of the pump can be calculated using the formula:
Power output = (Pressure increase * Flow rate) / 600
Where:
Pressure increase = 1100 KPa - 120 KPa = 980 KPa
Flow rate = 35 U/min
Power output = (980 * 35) / 600
Power output = 57.17 hp
Next, we need to calculate the actual power input to the pump, taking into account the motor efficiency.
Power input = Power output / Motor efficiency
Power input = 57.17 / 0.94
Power input = 60.85 hp
Therefore, the pump consumes 60.85 hp of electricity.
Power output of the pump can be calculated using the formula:
Power output = (Pressure increase * Flow rate) / 600
Where:
Pressure increase = 1100 KPa - 120 KPa = 980 KPa
Flow rate = 35 U/min
Power output = (980 * 35) / 600
Power output = 57.17 hp
Next, we need to calculate the actual power input to the pump, taking into account the motor efficiency.
Power input = Power output / Motor efficiency
Power input = 57.17 / 0.94
Power input = 60.85 hp
Therefore, the pump consumes 60.85 hp of electricity.
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Google Gemini AI
4 days ago
Given:
- Motor power: 1.05 hp
- Pressure increase: 1100 KPa - 120 KPa = 980 KPa
- Water flow rate: 35 L/min
- Motor efficiency: 94%
Solution:
1. Convert Motor Power to Watts:
1.05 hp - 746 W/hp = 786.3 W
2. Calculate Water Power (Pump Power):
Water power = (Pressure increase) - (Flow rate) / (1000 kg/m³)
Using the density of water as 1000 kg/m³:
Water power = (980 kPa) - (35 L/min - (1 min / 60 s)) / 1000 kg/m³
= 588 W
3. Calculate Input Power (Electricity Power):
Input power = Water power / Motor efficiency
Input power = 588 W / 0.94
= 625.53 W
Therefore, the pump efficiency is:
Pump efficiency = Water power / Input power
= 588 W / 625.53 W
= 0.94 or 94%
- Motor power: 1.05 hp
- Pressure increase: 1100 KPa - 120 KPa = 980 KPa
- Water flow rate: 35 L/min
- Motor efficiency: 94%
Solution:
1. Convert Motor Power to Watts:
1.05 hp - 746 W/hp = 786.3 W
2. Calculate Water Power (Pump Power):
Water power = (Pressure increase) - (Flow rate) / (1000 kg/m³)
Using the density of water as 1000 kg/m³:
Water power = (980 kPa) - (35 L/min - (1 min / 60 s)) / 1000 kg/m³
= 588 W
3. Calculate Input Power (Electricity Power):
Input power = Water power / Motor efficiency
Input power = 588 W / 0.94
= 625.53 W
Therefore, the pump efficiency is:
Pump efficiency = Water power / Input power
= 588 W / 625.53 W
= 0.94 or 94%
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