Please answer the calculation method of circulating water pump efficiency and the test conditions of water pump head.

1. Calculation method of circulating water pump efficiency:

η={（ρg*Q*H)/(3600*P)}*100

η --Unit efficiency unit

ρg—9.81*1000

Q—Flow volume cubic/hour

H—Yang Cheng m

P—Input power W

2. The test conditions for the pump head should comply with the requirements of GB/T 3216-2016 "Level 1 and Level 2 Hydraulic Performance Acceptance Test for Rotary Power Pumps". The head of the centrifugal pump is also called It is the pressure head of the pump, 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 bending of the blades, etc.), and the rotation speed. At present, the pressure head of the pump cannot be calculated theoretically accurately, and is generally measured experimentally.

For example, the head of a centrifugal pump is measured. The working fluid is 20°C clean water. When the measured flow rate is 60m/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 tables is 0.45m. If the diameter of the suction pipe and the discharge pipe of the pump are the same, try to calculate the lift of the pump.

Solution:

Check 20. ℃,

h=0.45m 1Mpa is approximately equal to 100 meters of water column

pOutlet=0.47Mpa=0.47*100 meters of mercury=47 meters of water column

p Import=-0.02Mpa=0.02*100 meters mercury=2 meters water column

ρ is the density of the liquid

H=h (p outlet-p import)/ρ=0.45 ( 0.47*10e6-(-0.02*10e6))/(10e3*9.887)=50.5m