Convering inches of oil to NIPA

[nucleareng78]

I have a positive displacement pump at the top of a tank with a suction piping going into it. I know the oil operating level and have determined that the suction piping is 5 inches under the normal operating level.

My question is how do I convert the 5 inches of oil level to Net Inlet Pressure Avaliable? I've looked around online and cannot find a straight forward answer and I'm sure it's a simple equation. The tank does operate at a slightly negative pressure but normally it is atmospheric.

And NO this is not a homework question, it's a real issue!

 

[nucleareng78]

Okay this makes much much more sense now.

So L2 doesn't matter at all. L1 needs to be from the oil level to the suction flange of the PD pump or the middle of the screws? I need to check to see if there are any elbows towards the suction flange (shouldn't be, but need to text).

if the inlet pipe is 3" in diameter would it be worth calculating the head loss of that?

 

[BigInch]

There mayb e inlet opening losses to consider too.
Search for inlet nozzle loss coefficients.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[nucleareng78]

As in pressure losses across the suction strainer?

Since the strainer is below the oil level, this would impact how much net inlet pressure available there is?

 

[nucleareng78]

Tank air space pressure: Pt (the absolute value)
Minus the effective head of the fluid = L1 x r x g

I need to convert the effective head of the fluid to psia correct?

It is so weird I would use L1...I guess this is why PD pumps are so different than centrifugal pumps.

 

[BigInch]

Not across the strainer, although there will be a head loss there too, at the entrance to L2. Any fluid at rest in the tank, being accelerated into a pipe must use some potential energy to accomplish the acceleration. The energy used for acceleration between fluid at rest as it enters a pipe and picks up velocity is described by the entrance loss coefficient.

There is no difference between a PD and centrifugal in that regard. PDs, especially diaphram and reciprocating cylinder pumps have additional acceleration head losses, because each time the diaphram, or cylinder is full, the cylinder or diaphram reverses stopping all velocity for a second, so the velocity head must be energized again and again at the beginning of each new cylinder cycle. If there are 3 or more diaphrams, then one diaphram is always filling as each diaphram is offset in cyclic period by 2 pi/3, as opposed to 2 pi/2 for a 2 diaphram pump, or 2 pi/1 for a 1 diaphram pump. The dead zones are less pronounced as the number of cylinders, or diaphrams increase.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[nucleareng78]

I think you may be wrong on the strainer. You have to subtract the DP across the strainer because the pressure above the liquid level would be reduced due to the strainer.

Just seems like NIPA would decrease with a dirty strainer....

 

[nucleareng78]

just to clarify. Do we agree on the below equation?

Net Inlet Pressure Available = P(tank)-P(static)-DP(strainer)-DP(pipe)

P(tank) in this case is atmospheric, but sometimes a slightly negative pressure, due to fans inside.