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Convering inches of oil to NIPA

Convering inches of oil to NIPA

Convering inches of oil to NIPA

(OP)
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!

RE: Convering inches of oil to NIPA

(OP)
That picture is exactly what I'm referring to. But do you really need L1 in the equation if I know the height of the bottom of the suction pipe?

Since this pump is only 32 gpm I think the only other consideration we need to take into account is the pressure drop across the strainer, what is an approximate loss??

Also, can we not use this equation? Pressure=0.434*Head*SG where head is the 5 inches?

Maybe I'm only confusing myself..

RE: Convering inches of oil to NIPA

The loss of head caused by the weight of the fluid will be according to the dimension L1 (actually only the vertical distance between the fluid level and the inlet to the pump but I assumed the suction pipe was already vertical). The loss of head caused by fluid friction in the pipe will be caused by L1+L2, i.e., the total length of the pipe regardless of its orientation.

Can't really answer your question about the loss across the strainer: how big is the strainer? how fine is the mesh? what is the viscosity of your fluid? How clogged is the strainer? Does the strainer have a bypass valve? If so, how strong is the spring?

The flow rate is an important consideration: but calling it "only 32 gpm" isn't really fair, especially if your suction pipe is only 1/2" diameter (how big is the pipe?). If I were you, I wouldn't neglect the fluid friction in the suction line.

Post some numbers and I'll do a quick calculation for you. Please be specific about your units - which size "gallon" are you using?

DOL

RE: Convering inches of oil to NIPA

Net suction Pressure available?
You mean net suction head, that's 5" no matter if it's oil or water.
For a small pump strainer, 5 psi would be too much.
That's 5 psi * 62.4/144 / SG_oil in feet of head terms.

0.434 ft/psi is for water only. Divide that by SG of the oil to get feet head of oil.

Hope you're not running a nuc plant.

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

RE: Convering inches of oil to NIPA

(OP)
I didn't bring some of the dimensions and values from.work but will post on here Monday with the info about the suction pipe.

We are actually installing a vacuum dehydrator oil purification kidney loop to the lube oil tanks. Would this in theory improve the suction if the lube oil pump? Less sludge on suction strainer and less crude going into the screw pump. Thanks for the help so far.

RE: Convering inches of oil to NIPA

(OP)
Big Inch, can you verify that equation you listed?

I found out the the NIPR of the pumps (depends on the temperature of the oil), but since this is based in PSI, I was trying to convert it to inches of head.

If we are increasing the oil level above the suction of the pipe (inches of oil level), how would I can I convert this to psi?

RE: Convering inches of oil to NIPA

Head in feet.
62.4 pcf density of water
144 in2/ft2

Hope you're not running a nuc plant.

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

RE: Convering inches of oil to NIPA

(OP)
^I'm not running a nuclear plant.

Simply a learning engineer

RE: Convering inches of oil to NIPA

Great. Keep up the good work.

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

RE: Convering inches of oil to NIPA

(OP)
Pressure (psia)= (h*2.31)/SG where h= head (ft) SG=spec. gravity 2.31=conversion factor

For example if I have 17 inches (1.416 feet) of liquid between the bottom of the suction piping and top of the liquid level, that would give me 17 inches of head...

pressure = (1.416*0.9)/(2.31) = 0.551 psia

This doesn't seem right at all.....has to be way too low....do I need to add 14.7 psia (atmospheric)??

Big Inch, should I use 0.434 in place of the 2.31 like you said above??

RE: Convering inches of oil to NIPA

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

Maybe it should be (a homework question).

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Convering inches of oil to NIPA

If the pressure at the depth of 2.3 feet of water is 1 psi, why would you expect the pressure to be higher at 60% of the depth of 90% of the weight fluid?

Ted

RE: Convering inches of oil to NIPA

Sounds like more homework needed.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Convering inches of oil to NIPA

(OP)
0.5 psia for NIPA seems low doesn't it? If the pump there now has a NIPR of 3.5 psia then how does it even start up??

NIPR is in units of psia. That is why I'm thinking my 2.31 conversion number may in error....should it be the 0.434 number instead?

RE: Convering inches of oil to NIPA

At the fluid level including at the same level inside the suction pipe, pressure is 14.7psia
At the pump inlet 5" above the fluid level, pressure is 14.7 - (5/12)*(1/2.31)*0.9 = 14.7 - 0.16 = 14.54psia, static pressure. Does not account for flow losses when the pump is running.

1/2.31 = 0.43 like biginch said.

Ted

RE: Convering inches of oil to NIPA

(OP)
I'm sorry but you said 5" above fluid level?

I may just be confusing myself so here is a quick sketch of my situation to clear the air. Pressure at the bottom of the suction pipe would be Static Pressure PLUS Atmospheric....correct?

I read that NIPR is measured in PSIA, but I need to confirm.

RE: Convering inches of oil to NIPA

I should have looked at your original post and seen that you said the end of the suction pipe was 5" below the liquid level. My error for causing more confusion. The static pressure at the pump inlet port will be air pressure in the tank air space minus the static pressure of the length of fluid from the fluid level in the tank to the pump inlet. Go back to Oldhydromam's pdf and calculate his statement of tank air pressure in psia - L1/2.31 to calculate pump inlet static pressure. L1 in feet.

It is of no interest what the pressure is at the suction pipe inlet. The pressure due to fluid in the pipe back up to the tank fluid level is balanced(canceled) by the depth to which the pipe inlet is below that fluid level. Static pressure concern is only with the distance the pump inlet is above the tank fluid level. Go back and re-read other references given to you.

Ted

RE: Convering inches of oil to NIPA

(OP)
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?

RE: Convering inches of oil to NIPA

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

RE: Convering inches of oil to NIPA

(OP)
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?

RE: Convering inches of oil to NIPA

(OP)
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.

RE: Convering inches of oil to NIPA

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

RE: Convering inches of oil to NIPA

(OP)
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....

RE: Convering inches of oil to NIPA

(OP)
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.

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