Pump question

Well, unless 12 bars is the maximum operating pressure, then that isn't your limit but yea, the "pump head" is 2 bar, about 30psi and around 67 ftH2O.

But the liquid you are pumping will exit the pump with a total of 12 bar (175 psi, 405 ftH2O)

 

? I don't think I understand the question?

I assume this is a centrifugal pump? In which case yes, pumps can operate in any number of circumstances, one such "normal" operation is taking water under pressure (say, 10 bar) and spinning it up to 12 bar.

Hms is the suction head of the pump. I don't mean to sound pedantic, but do you know what head is? If you do disregard this next couple of sentences. Your 10 bar (~335 ft of head) is the height in feet that the water (or whatever medium, after accounting for SG) would rise to if you piped it straight up. The pump takes this water and spins it with the impeller, giving it a higher velocity. This higher velocity creates a vacuum between the faster liquid and the slower stuff at the suction port. This, combined with the pressure already imparted on the water at the suction (the 10 bar), sucks the water into the impeller to be propelled, continuing the cycle.
The energy imparted on the water by the pump results in a higher maximum vertical height, or head. Thus, higher pressure.

 

I don't see a problem. I see a pump operating exactly as it should. Taking liquid of one pressure, adding energy, and discharging the liquid at a higher pressure.

As I said, if your pump is taking a liquid at 10 bar and spinning it up to 12 bar then you have the following situation:

1) A pump that has a discharge head of 2 bar.
2) A pump that is designed to withstand and internal pressure that is greater than 12 bar.

 

It is for a closed or an open system ? Why don't you look at the pump curve?