η= {(ρg * Q * H)/(3600 * P)} * 100%
η-unit efficiency%
ρg—9.8 1* 1000
Q— Flow cubic meter/hour
Zheng Haiyang
P- input power w
2. The test conditions of the pump head shall meet the requirements of GB/T 32 16-20 16 "Hydraulic Performance Acceptance Test of Rotary Power Pump 1 2". The head of centrifugal pump is also called pump head, which refers to the energy obtained by the unit weight of fluid through the pump. The lift of the pump depends on the structure of the pump (such as the diameter of the impeller, the curvature of the blades, etc.). , and speed. At present, the pump head can not be accurately calculated theoretically, but is generally determined by experimental methods.
For example, the head of a centrifugal pump can now be measured. When the working medium is clean water at 20℃ and the measured flow rate is 60m/h, the vacuum gauge reading at the pump inlet is 0.02Mpa, and the pressure gauge reading at the outlet is 0.47Mpa (gauge pressure). It is known that the vertical distance between two manometers is 0.45 meters. If the suction pipe and the extrusion pipe of the pump have the same diameter, please try to calculate the pump head.
Solution formula
Check 20℃,
H=0.45m 1Mpa is about equal to 100m water column.
P outlet = 0.47 MPa = 0.47 * 100 m Hg = 47m H2O.
P import =-0.02mpa = 0.02 *1200mhg = 2m H2O.
ρ is the density of the liquid.
H=h+(p exit -p entrance)/ρ = 0.45+(0.47 *10e6-(-0.02 *10e6))/(10e3 * 9.887) = 50.5m.