Formula for calculating water pump power

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· What is the power of a water pump? Power: The amount of work done by a machine per unit of time is called power. It is usually denoted by the symbol N. Common units include: kilogram·meter/second, kilowatt, and horsepower. Typically, the power unit of motors is expressed in kilowatts ； The power unit of diesel or gasoline engines is expressed in horsepower. The power transmitted by the prime mover to the water pump shaft, that is, the power delivered by the prime mover to the pump shaft, is called shaft power, denoted by P. It can be regarded as the input power of the water pump; generally, when people talk about the power of a water pump, they are referring to shaft power. The effective power of a pump, also known as the output power, is denoted by Pe. It is the effective energy acquired by the liquid pumped out from the pump per unit of time. Since head refers to the useful energy acquired by a unit weight of liquid as it leaves the pump, the product of head, mass flow rate, and gravitational acceleration represents the useful energy obtained by the liquid discharged from the pump per unit time – in other words, the effective power of the pump. The equation is: Pe = ρgQH (W) = γQH (W), where ρ is the density of the liquid being pumped (kg/m³) ； γ — specific gravity of the liquid pumped by the pump (N/m3) ； Q —— Pump flow rate (m3/s) ； H —— Pump head (m) ； g — gravitational acceleration (m/s2). The difference between the shaft power P and the useful power Pe is the power lost within the pump, and its magnitude is measured by the pump’s efficiency. The efficiency of a pump is the ratio of useful power to shaft power, denoted by η. The more water that is pumped out, the more work is done; less water pumped out means less work is done, and thus the electric current should also be lower. The power of a water pump (in KW) is calculated as: head (in m) × flow rate (in m3/s) × specific weight of the fluid (1000 kg/m3 for water) ÷ 102 (power conversion factor) ÷ efficiency of the pump (around 70%). The power of the motor, on the other hand, is expressed as P = I × U; a higher power level corresponds to a higher current. Due to the frictional resistance of the bearings and fillers ； Friction between the impeller and water as it rotates ； Causes include vortices in the water flow within the pump, leakage backflow, inflow and outflow, and impingement at the inlet/outlet. Some power is inevitably lost, so the water pump cannot convert all of the power supplied by the prime mover into useful power; there is always some power loss. In other words, the useful power of the water pump plus the power lost within the pump equals the shaft power of the pump. · Formula for calculating water pump power? 1. The formula for calculating the power of a centrifugal pump is P = flow rate × head × 9.81 × specific gravity of the fluid ÷ 3600 ÷ pump efficiency. The unit of flow rate is cubic meters per hour, and the unit of head is meters. This can be simplified to P = 2.73HQ/η, where H represents the head in meters, Q represents the flow rate in m3/h, and η represents the pump efficiency. P represents the shaft power in KW. In other words, the shaft power of the pump is given by P = ρgQH/1000η (KW), where ρ = 1000 Kg/m3 and g = 9.8. The unit of specific gravity is Kg/m3, the unit of flow rate is m3/h, and the unit of head is meters. Since 1 Kg = 9.8 Newtons, then P = specific gravity × flow rate × head × 9.8 Newtons/Kg = Kg/m3 × m3/h × m × 9.8 Newtons/Kg = 9.8 Newtons × m/3600 seconds = Newtons × m/367 seconds = Watts/367. The above explanation covers the origin of the units; this formula is used to calculate the power required for water handling. The shaft power is then divided by the efficiency to obtain the actual power consumption. Let Ne represent the shaft power, P represent the motor power, and K represent a coefficient (the reciprocal of efficiency). The motor power P = Ne × K (K takes different values depending on Ne; see below): Ne ≤ 22: K = 1.25; 22 < Ne ≤ 55: K = 2.

2. The formula for calculating the shaft power of a slurry pump is as follows: Flow rate Q in m3/h, head H in meters of water, efficiency n in %, slurry density A in Kg/m3. The shaft power N is given by N = H × Q × A × g / (n × 3600). When calculating the motor power, it is also necessary to take into account the transmission efficiency and safety factors. Generally, for direct coupling, the value is 1; for belt drive, it is 0.96. The safety factor is 1.2. 3. Calculation of the shaft power of centrifugal pumps and motor power: After determining the flow rate and head of a centrifugal pump, engineering designers must determine another important parameter—the power of the pump’s motor. In many cases, by following the sample and based on the flow rate and head parameters, it is possible to determine the motor model and power rating of the water pump. We can derive the technical formula for the water pump motor based on the principle of energy conservation. The useful work done by the water pump is W = Mgh (where a certain mass of fluid is lifted to a certain height h; h represents the head). Here, M is the mass of water, given by m = ρV (ρ is the density of the fluid, and V is its volume). V can also be expressed as V = Qt, where Q is the flow rate of the pump and t is the operating time of the pump. Therefore, the useful work done by the pump is W = ρQtgH. The effective power of the pump is P = W/t = ρQgH. The shaft power of the pump (i.e., the actual output power) is P1 = ρQgH/η, where η represents the efficiency of the pump. The actual power required by the motor is P2 = γP1, with γ representing the safety factor for the motor (the value of γ ranges from 1.1 to 1.3, with 1.2 being the typical value). If the fluid being pumped is water, then the formula for calculating the motor power becomes P2 = (1.2QgH)/(3600*η). The unit of flow rate Q is m3/h, while the unit of head H is meters. It should be noted that the value of P2 calculated using these formulas may not exactly correspond to the actual motor power. For example, if the calculated value is 33.56 kW, then a 37 kW motor should be selected; if it’s 30.56 kW, then a 30 kW motor is appropriate. * Through the above formula derivations, we can conclude the following: 1. Regardless of the manufacturer, when the flow rate and head are fixed, the actual useful work remains constant. 2. The amount of electricity consumed by a water pump does not depend on the power rating of its motor; rather, it depends on the pump’s shaft power. With the same 30kw motor, one unit has an shaft power of 20.56kw while another has 25.18kw; clearly, the 20.56kw version is more energy-efficient. The key factor determining the magnitude of shaft power is the efficiency of the water pump.