【Registered Machinery General Section】20170729 Fourth Week: Effective Power of Pumps
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What is the effective power of a pump? The difference between the effective power and the shaft power represents the power lost within the pump; it is the power required to move the liquid inside the pump per unit of time, or in other words, it is the shaft power minus the power lost. That is, N_effective = rQH/102 (kW), and N_effective = rQH/75 (Hp). In these formulas, r represents the specific gravity of the liquid (kg/m3), Q represents the flow rate of the pump (m3/second), and H represents the head of the pump (meters). Wealth 5 – just copy this text
What is the effective power of a pump? The difference between the effective power and the shaft power represents the power lost within the pump; it is the power required to move the liquid inside the pump per unit of time, or in other words, it is the shaft power minus the power lost. That is, N_effective = rQH/102 (kW), N_effective = rQH/75 (Hp). In these formulas, r represents the specific gravity of the liquid (kg/m3), Q represents the flow rate of the pump (m3/second), and H represents the head of the pump (meters)
The difference between the effective power and the shaft power represents the power lost within the pump; it is the power required to move the liquid inside the pump per unit of time, or in other words, it is the shaft power minus the power lost. That is, N_effective = rQH/102 (kW), N_effective = rQH/75 (Hp). In these formulas, r represents the specific gravity of the liquid (kg/m3), Q represents the flow rate of the pump (m3/second), and H represents the head of the pump (meters)