Effect of Increasing Vacuum on Recirculating Pump

You will find some of the links below, helpful for understanding your problem.

Effect of Increasing Vacuum on Recirculating Pump

You may like to check the vacuum pump curves with respect to flow and vacuum and then compare with the performance. The flow will decrease with increase of vacuum as the load is more with more vacuum requirement and hence removal of moisture.

I suspect the problem is caused by higher viscosity of the slurry that effect pump performance curve and line pressure drop.

Thanks for your answer. It helps me a lot to realize the problem. Effectively was at the very edge and when the vacuum was increased its capacity dropped to a one where the NPSH requierment was satisfied.

Other question:

You said that i should sustract the vacuum applied to the static head, in ft. If you have a vacuum of 0.4 bar you should substract 0.6 bar (in ft) to the static head.

But if the pressure of the effect is 1.2 bar abs, you should consider also the 0.2 bar above the atmospheric pressure in the estimations?

i retract, at the bottom of the evaporator the pressure above the liquid is at its vapor pressure, so it only takes subtract the vacuum

The primary idea here is that most of the types of pumps employed cannot handle two-phase flow, and thus will suffer from cavitation when NPSH is below the design point. This means that vapor bubbles instantly form and violently collapse on the surfaces of the pump impeller that are at speed.

Consider inspecting the pump impeller if pump problem remains. Have you considered the use of a jet ejecter to help re-pressurize the fluid being pumped?

in the calculation to convert from pressure to ft or meter, used the density of slurry which is higher than that of water. The vapor pressure is the same as the evaporator bottom, that is correct. Normally you can hear cracking sound if cavitation occur.

you are correct, I totally ignored the specific gravity of the solution in any adjustments. I didn't bother looking it up.

I have the next doubt

We Know that

NPSH=P2-Pv

P2: Pressure at the pump entrance

PV=vapour pressure of juice

*** P1/pg + h1 + (v1)^2/2g = P2/pg + h2 + (v2)^2/2g + L ***

P1/pg + (h1-h2) + [(v1)^2-(v2)^2)]/2g - L = P2/pg

where:

1: surface of liquid

2: pump entrances

h1,h2: elevations of points 1 & 2.

v1,v2: velocity of the fluids at point 1,2

L: friction losses

the friction losses and velocity heads are neglible

and so P1/pg + (h1-h2) = P2/pg

or P1 + (h1-h2) = P2

expressing the pressure as height

and NPSH = (P1-PV)+(h2-h1)

I belive that P1=PV, because it's a liquid at equilibrium with its vapour (it should be some difference due to elevation of ebullition point - liquid with solids-), but i don't think that this doesn't affect much.

I read a comment above that the vacuum should be erased from this. Is that correct. The system that i am studying is the next in the link:

http://imgur.com/D5tn1

as I said before this is the 3rd effect of an countercurrent evaporator and the recirculating pump drops its capacity when it is set a vacuum pressure less than 0.5 bara

Uhhhh....algebra mistake, my friend.

P1/pg + (h1-h2) = P2/pg this reduces as follows:

P1-P2 + (h1-h2)pg = 0

you did not multiply all terms by pg.

Neglecting frictional effect

NPSH available = h1 - h2 in meters of liquid

Go back and think again. The system is not an open system at ambient pressure.