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# Pump head in closed system

## Pump head in closed system

(OP)
I'm a little confused about the static height in a closed system.
For an assignment we have to calculate the pump head for a closed system, but everywhere I read I see that static head for closed system is 0. Normally you would take into account the difference between the gauges/manometers to correct for height difference is this not needed for a closed system?

From the attached figure I cannot see why I should not correct for it as they are on different levels.

Should I correct for the height difference of the gauges? And if not why not?

### RE: Pump head in closed system

I assume that the gray box with a cooling coil is a sort of vessel? And that there is a head space above the liquid level where you pump vacuum?

In this case its not a closed system since the vessel disengages the pump inlet from the pump outlet. So yes, use the elevation difference between pump outlet as part of the discharge head - and the elevation difference for lowest liquid level for the suction pressure.

--- Best regards, Morten Andersen

### RE: Pump head in closed system

"An assignment"??

Student posts have their own forum..

what people mean is that when calculating the head required by the pump to pump fluid around the circuit the static head is zero as the head to the top of the circuit is balanced by the head from the top back to the pump. Hence all the pump needs to do is circulate the water around the loop. This assumes that the gray box is full of liquid. Not quite sure why you have a vacuum pump on this circuit - normally just a header / expansion tank to keep it full of water.

TDH is usually measured relative to the pump suction header or the pump centre line so yes, you need to account for the head difference between the gauges. IMHO.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Pump head in closed system

(OP)
The vacuum is to deaerate the system and to control the 'atmospheric' pressure there. Discharge is well under water.

### RE: Pump head in closed system

Hi,
TDH = head losses due to flow in a closed system. = Head discharge -Head suction = head losses in the loop, expecting V d = V s, same pipe size. If not perform the adjustment due to velocity change.
If you want to use the Pressure gauges, you need to define a reference point (center line of the pump) and perform the adjustment due to position of the gauges.
H suction= P suction (gauge)/ (Ro*g) +Z1
H discharge = P discharge (gauge) /(Ro*g) +Z2

Pierre

### RE: Pump head in closed system

It is true that the total elevation head is zero for that system. However you are measuring the differential head between the suction and the discharge at the locations of the pressure gages which is equal to the differential head produced by the pump. In this case the differential head just looks at those two points in the system and takes the difference in the pressure head, velocity head and static elevation head at those two point to get the pump differential head. Note also the the pump differential head also includes the friction loss between the Ps and the inlet of th pump and the pump discharge and Pd that is not included in the equations you show.

### RE: Pump head in closed system

RegiSTR, great question and you are absolutely right. Gauge readings even in a closed system should be corrected based on difference in height of the gauges (P = rho * g * h). But there are exceptions depending on how the gauges are piped. There are often 2 or 3 pipes to a common gauge for a pump, or if gauges are on the same level, there would be no correction. I whipped up an attachment as an example.

### RE: Pump head in closed system

(OP)
Thank you for confirmation all.

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