Centrifugal pump outlet pressure is reduced reason

Mondo Technology Updated on 2024-02-29

(1) The motor is reversed.

Due to the wiring, the steering of the motor will be the opposite of the actual desired steering of the centrifugal pump. In general, the steering of the pump should be observed at start-up. If the steering is reversed, any two wires on the motor binding posts should be replaced.

2) The working point is moved to a large flow rate and low head.

In general, centrifugal pumps have a continuous downward performance curve, and the flow rate gradually increases as the head decreases. During operation, due to some reason, the back pressure of the pump decreases, and the working point of the pump passively moves with the curve of the unit to the low head point flow, which will lead to a decrease in the head. In fact, this is caused by external factors such as changes in equipment and has no special relationship with the pump itself. At this time, as long as the back pressure of the pump increases, such as closing the outlet valve a little, etc. The problem can be solved.

3) Slow down. The important factors influencing the pumphead of a centrifugal pump are the impeller outside diameter and the pump speed. All other things being equal, the pump head is proportional to the power of the velocity. As can be seen, speed has a big impact on the pump head. Sometimes, if the speed of the pump is reduced due to some external reason, the pump head will be reduced accordingly. At this point, the speed of the pump should be checked. If the rotational speed is not enough, the cause should be checked and solved reasonably.

4) Cavitation at the entrance.

If the suction pressure of a centrifugal pump is too low and lower than the saturated vapor pressure of the pumped medium, cavitation can occur. At this point, check if the inlet piping system is clogged, or if the opening of the inlet valve is too small, or if the level of the suction tank is raised.

5) Internal leakage occurs.

Internal leakage occurs when the gap between the rotating and stationary parts of a centrifugal pump exceeds the design range, which is reflected in a drop in the discharge pressure of the pump, such as the gap between the impeller mouth rings and the interstage gap of a multistage pump. At this point, the corresponding disassembly and inspection should be carried out to repair or replace the parts that caused the excessive clearance.

6) The impeller channel is blocked.

If part of the flow channel of the impeller is blocked, the work of the impeller will be affected and the outlet pressure will drop. Therefore, it is necessary to disassemble the pump to inspect and remove foreign objects. To prevent this problem from happening again, if necessary, a filter can be added before the centrifugal pump inlet.

Read more:

How the shaft power of a centrifugal pump is calculated.

What is pump efficiency? How is the formula?

Refers to the ratio of the effective power of the pump to the shaft power. η=pe/p

The power of the pump usually refers to the input power, that is, the power transmitted by the prime mover to the pump shaft, so it is also called the shaft power, which is represented by P. The effective power is the product of the pump's head and mass flow and gravitational acceleration.

PE=PG QH W or PE=YQH 1000 (KW).

p: The density of the liquid conveyed by the pump (kg m3).

y: The gravity of the liquid delivered by the pump = pg (n m3).

g: acceleration due to gravity (m s).

Mass flow rate qm = pqt h or kg s

1. Calculation formula of centrifugal pump shaft power.

Flow head 981 medium specific gravity 3600 pump efficiency flow unit: cubic hour, head unit: m p = 273hq/n

where h is the head, the unit m, q is the flow, and the unit is m3 h, which is the efficiency of the pump. p is the shaft power, in kwThat is, the shaft power of the pump p=PGQH 1000?(kw), where p=1000kg m3, g=98. The unit of specific gravity is kg m3, the unit of flow is m3 h, the unit of head is m, 1kg=98 Newtons p = specific gravity * flow * head * 98 Newtons kg=kg m3*m3 h*m*98 Newton kg=98 Newtons * m 3600 seconds = Newtons * m 367 seconds = Watt 367

The above derivation is the origin of the unit, the above equation is the calculation of water power, and the shaft power is divided by the efficiency. Let the shaft power be Ne, the motor power be P, and K be the coefficient (efficiency reciprocal), and the motor power P=Ne*K (K has different values when Ne is different).

2. The formula for calculating the shaft power of the slurry pump is flow rate Q m3 h head h m H2O efficiency n %.

Slurry density a kg m3 shaft power n kwn=h*q*a*g (n*3600) motor power should also consider the transmission efficiency and safety factor. Generally, the direct connection takes 1, and the belt takes 096, safety factor 12

3. The efficiency of the pump and its calculation formula.

Refers to the ratio of the effective power of the pump to the shaft power. =PE P The power of the pump usually refers to the input power, that is, the power transmitted by the prime mover to the pump shaft, so it is also called the shaft power, which is represented by P. The effective power is the product of the pump's head and mass flow and gravitational acceleration.

PE=PG QH W or PE=YQH 1000 (KW).

p: The density of the liquid conveyed by the pump (kg m3).

y: The gravity of the liquid delivered by the pump = pg (n m3).

g: acceleration due to gravity (m s).

Mass flow rate qm = pqt h or kg s

Fourth, the empirical method to estimate the power of household pumps.

For home use, the smallest is 370w, and the largest is generally not more than 55kw, you can find a flashlight or something, or a chandelier to go down, directly see its size, and then estimate the power.

5. Estimate the power of the well pump.

Through the method of electrical measurement to rough calculation, you can turn on the power, find a clamp meter to measure the current, the current multiplied by the voltage multiplied by the power is the power (** pump housekeeper), you can consider the power factor is 08, so that the actual working power can be calculated, if there is water pumped up, you can consider 7 into the load design, and then divide by 08 is the rated power of the motor.

You can also turn off the water valve and so on, and directly let the motor with the pump idle, in this case, the estimated current will not exceed 40% of the rated current, measure the current, also calculate the power, and then divide by 04. To get the motor power, compare the above item to determine whether the motor power is wrong.

You can measure the distance between the pump and the ground, look at the height of the pumped water, you will know the head, and then find a bucket to see how much water it pumps up in 1 minute, measure the volume of water, and calculate the flow rate in turn, knowing the head and flow can also calculate the power in turn. Head* Flow 36702÷1.2. You can consider that the efficiency of the pump is 07, divided by 07, is the approximate motor power.

Water pumps and motors, long-term work in a humid environment, itself may have leakage problems, the current may be deviated, and the bearings and pipelines may have abnormal places, the power will also be very different, these methods can only be used for reference, the actual or down the well to deal with more appropriate, after all, the equipment should also be checked frequently.

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