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What will be the result if the pipe losses is greater than head produced by the pump? What will be the result if outlet is closed?

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    $\begingroup$ I wonder if Engineering SE would be a better place to ask this? I don't know, but if you haven't looked there yet you might check it out. $\endgroup$ – uhoh Jul 17 '18 at 8:43
  • $\begingroup$ Surely @uhoh, i'll check there as well. $\endgroup$ – Amar Jul 17 '18 at 10:20
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If the pipe losses are greater than the head produced by the pump, then it's an extremely crappy pump, and the result will be the pump will be replaced by a better one. It's really not worth discussing effect of situation as thoroughly faulty as this other than how to remedy it ASAP. Pipe losses are normally a couple orders of magnitude lower than the head produced by the pump, and facing a situation like this "how will the system/pump react" is the least of your worries. (well, probably the pump will break; no biggie, it was crap anyway.)

Now for the other half of your question, this happens. A valve meant to prevent a backflow is wrongly activated, or a nozzle is plugged by a contaminant, or such. This is a fault situation that needs to be foreseen.

For very high throughput, high pressure pumps, this has usually disastrous results; as the throughput is blocked, pressure rapidly (often impulsively) rises and may lead to the entire pump exploding - or make the pipe burst, or cause other damage. In more robust solutions, a safety valve will engage, the pump will be shut down due to overpressure, and the system will enter maintenance mode to fix the problem. In simple, 'homegrown' scenarios, the pump will struggle against the backflow and fail to move the medium. The motor might stop and either have the coils burn through, or a fuse will break the circuit (this causes major overcurrent) - durable, high-pressure solutions - or it will spin at equilibrium against backflow - typical to weak propeller pumps.

Since you asked this on space.SE, I guess you'd be asking about rocket turbopumps. Well, in that case, the pumps are very strong, and can't afford a lot of pressure redundancy due to weight constraints, never mind short paths, huge flow, no room or mass budget for advanced safety solutions, plus the flammable/explosive nature of media pumped... blocking the outlet rapidly while the pump is running will result in a RUD of the rocket.

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  • $\begingroup$ That's good enough explanation. But can you clarify, if in case of both rotodynamic and positive displacement pumps the above scenario (blocking of the discharge) applicable or only in case of positive displacement pumps. $\endgroup$ – Amar Jul 17 '18 at 12:09
  • $\begingroup$ @Amar: With sufficient flow and pressures, both rotodynamic and positive displacement will be as explosive. Rotodynamic of similar flow rate are less likely to build up dangerous pressures and just start spinning ineffectually, and the "stop mode" for both will differ (rotodynamic will speed up, positive displacement will stop or slow down), but the end effect is similar. BTW, spacecraft pumps are rotodynamic and most definitely very explosive. $\endgroup$ – SF. Jul 17 '18 at 12:41
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"What will be the result if the pipe losses is greater than head produced by the pump?"

The flow rate produced by the pump will decrease. Lower flow rates mean less pipe losses, so the pipe losses also decrease. The pump pipe system will come to an equilibrium point where energy supplied by the pump is equal to the losses in the pump pipe system.

Every pump has a relationship between the pressure and flow rate it can produce. A higher flow rate means less pressure. When we graph this relationship, we call it a pump curve. Systems of pipes, valves, etc have a different relationship called a system curve; A higher flow rate means the system removes MORE pressure. When we graph these, at intersection of these two curves the pressure applied by the pump is equal to the pressure removed by the system. This is called the operating point. (Beginner level reading about pump and system curves)

If there is suddenly a partial blockage, the system curve suddenly requires much more pressure to produce the same flow rate. The pump and system will find a new operating point with a lower flow rate and a higher pressure.

The system curve is due to the losses from pipes, valves, blockages, etc. The pipe losses specifically are one component of this system. The conclusion is still the same, when losses are greater than power the flow rate drops.

"What will be the result if outlet is closed?"

Flow rate will drop to zero and the pump's pressure will go to its maximum pressure. If this pressure is enough to break the pipes, the pipes will break. If this is not enough to break the pipes, the pump may overheat. Hopefully the pump is equipped with an auto shutoff and will switch off before its seals melt.

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