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Pump head

浏览次数:146   发布日期:2020-05-15

 
Pump head
      The head of the   centrifugal pump is also called the head of the pump, which refers to the energy obtained by the pump per unit weight of fluid. The pump head depends on the pump structure, such as the size of the impeller diameter, blade bending, speed and so on. The pressure head of the pump can not be calculated accurately in theory, and it is generally determined by experimental methods.
  The head of the H (m) centrifugal pump is also called the head of the pump, which refers to the energy obtained by the pump per unit weight of fluid. The pump head depends on the structure of the pump, such as the diameter of the impeller, the bending of the blade, etc., and the rotation speed. The pressure head of the pump can not be calculated accurately in theory, generally measured by experimental methods.
The pump head can be determined by experiment, that is, a vacuum gauge is installed at the pump inlet and a pressure gauge is installed at the outlet. If the kinetic energy difference on the cross section of the two meters is not counted (ie Δu2 / 2g = 0), the energy between the cross sections of the two meters is not counted Loss (ie ∑f1-2 = 0), the pump head can be calculated using the following formula.
Note the following two points:
(1) where p2 is the reading of the pressure gauge at the pump outlet (Pa); p1 is the vacuum gauge at the pump inlet Reading (negative gauge pressure, Pa).
(2) Pay attention to distinguish the two different concepts of centrifugal pump head (head) and lifting height.
Head refers to the energy obtained by pumping fluid per unit weight. In a pipeline system, the Bernoulli equations are listed between two sections (including pumps) and sorted out. In the
formula, H is the lift, and the lift height refers to only Δz.
Example 2-1 Now determine the head of a centrifugal pump. When the working medium is clean water at 20 ° C and the measured flow rate is 60m ^ 3 / h, the pump inlet vacuum gauge reads 0.02Mpa and the outlet pressure gauge reads 0.47Mpa (gauge pressure). It is known that the vertical distance between the two gauges is 0.45m. The suction pipe of the pump has the same diameter as the pressure pipe. Try to calculate the head of the pump.
Solution:
Check the water density at 20 ° C 1.0 * 10 ^ 3kg / m 3 ,
h = 0.45m (1Mpa is approximately equal to 100m water column)
p outlet = 0.47Mpa (0.47 * 100m water column = 47m water column)
p inlet = -0.02Mpa (0.02 * 100m water column = 2m water column)
ρ is the density of the liquid
H = h + (p outlet-p inlet) /ρg=0.45+ (0.47 * 10 ^ 6-(-0.02 * 10 ^ 6)) / (10 ^ 3 * 9.8) = 50.45m (water column)
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