Centrifugal pumps selection

(ahssan) #1

The pump is an essential component of an irrigation system. Proper selection of pumping equipment that will provide satisfactory performance requires good understanding of existing conditions. Design restrictions, operating conditions of the irrigation system, and required flexibility in system operation must be understood before an efficient pump can be selected for a given system.
Capacity, head, power, efficiency, required net positive suction head, and specific speed are parameters that describe a pump’s performance.
The capacity of a pump is the amount of water pumped per unit time. Capacity is also frequently called discharge or flow rate (Q). In English units it is usually expressed in gallons per minute (gpm). In metric units it is expressed as liters per minute (l/min) or cubic meters per second (m3/sec).
Head is the net work done on a unit weight of water by the pump impeller. It is the amount of energy added to the water between the suction and discharge sides of the pump. Pump head is measured as pressure difference between the discharge and suction sides of the pump.
Pressure in a liquid can be thought of as being caused by a column of the liquid that, due to its weight, exerts a certain pressure on a surface. This column of water is called the head and is usually expressed in feet (ft) or meters (m) of the liquid. Pressure and head are two different ways of expressing the same value. Usually, when the term “pressure” is used it refers to units in psi (pounds per in2) in the English system or kilopascals (kPa) in metric units, whereas “head” refers to ft in English units or meter’s (m) in meteric units. A column of water that is 2.31 ft high will exert a pressure of 1 psi.
Power Requirements
The power imparted to the water by the pump is called water power. To calculate water power, the flow rate and the pump head must be known. In English units water power can be calculated using the following equation: WP = (Q x H)/3960(1) where: WP=water power in horsepower units Q=flow rate (pump capacity) in gpm H=pump head in ft In metric units water power is expressed in kilowatts, pump capacity in liters per minute, head in meters, and the constant is 6116 instead of 3960.
In any physical process there are always losses that must be accounted for. As a result, to provide a certain amount of power to the water a larger amount of power must be provided to the pump shaft. This power is called brake power (brake horsepower in English units). The efficiency of the pump (discussed below) determines how much more power is required at the shaft.
BP = WP/E(2) where E is the efficiency of the pump expressed as a fraction, BP and WP are brake power and water power, respectively.
Pump efficiency is the percent of power input to the pump shaft (the brake power) that is transferred to the water. Since there are losses in the pump, the efficiency of the pump is less than 100% and the amount of energy required to run the pump is greater than the actual energy transferred to the water. The efficiency of the pump can be calculated from the water power (WP) and brake power (BP): E% = (WP/BP) x 100(3) The efficiency of a pump is determined by conducting tests. It varies with pump size, type and design. Generally, larger pumps have higher efficiencies. Materials used for pump construction also influence its efficiency. For example, smoother impeller finishes will increase the efficiency of the pump.
Net Positive Suction Head
The required net positive suction head (NPSHr) is the amount of energy required to prevent the formation of vapor-filled cavities of fluid within the eye of impeller. The formation and subsequent collapse of these vapor-filled cavities is called cavitation and is destructive to the impeller.
The NPSHr to prevent cavitation is a function of pump design and is usually determined experimentally for each pump. The head within the eye of the impeller, also called net positive suction head available (NPSHa), should exceed the NPSHr to avoid cavitation.