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!

 

[Oldhydroman]

Presumably you know how to calculate the flow induced pressure drop in the pipework. So is this the situation (see attached)?

DOL

 

[nucleareng78]

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..

 

[Oldhydroman]

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

 

[BigInch]

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

 

[nucleareng78]

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.

 

[nucleareng78]

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?

 

[hydtools]

nucleareng78, the equation is from hydrostatics. One of your early engineering studies.

http://www.engineeringtoolbox.com/static-pressure-head-d_610.html

Ted

 

[hydtools]

Or this ref:

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Pressure/HydroStatic.html

Ted

 

[BigInch]

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

 

[nucleareng78]

^I'm not running a nuclear plant.

Simply a learning engineer

 

[BigInch]

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

 

[nucleareng78]

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??

 

[Artisi]

"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.)

 

[hydtools]

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

 

[Artisi]

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.)

 

[nucleareng78]

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?

 

[hydtools]

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

 

[nucleareng78]

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.

 

[hydtools]

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