Must I Convert Liquid Pump Head Before Selecting a Pump When It Is Not Water ?

It's not that simple. It depends on many factors.

I won't try to explain it myself.

Centrifugal Pumps and Viscosity - Engineering ToolBox

Centrifugal Pumps - Engineering ToolBox

The article perhaps does not express things very well.

If you have calculated correctly, and you know that the system requires a pump that will give 10 metres head for the required flow, then you don't need to convert. If the pump is selected for 10m head, then it will give 10m head whatever the liquid (only the pressure and absorbed power will change, according to the S.G.). If you do the conversion and select a pump that will give 7.5m of water, it will also only give 7.5m on the oil. If you consider the pump discharge connected to a tall vertical column, the selected pump would raise the fluid to a height of 10m irrespective of the fluid s.g, but you would get a pressure of about 0.75 bar on oil and about 1.86 bar, for example, on concentrated sulphuric acid.

In fact, you have already done the conversion when you did the system calculations. This is usually done in terms of pressure (rather than head), and for the actual fluid to be pumped, to ensure that the calculated pressure losses are correct. This gives you the pressure drop across the system for a given flow, and thus the pump discharge pressure you require. You then convert this pressure to head of the actual fluid to be pumped, and this gives the required pump head. You can then select from pump curves.

If it's only a question of fluid SG, you need to select the pump that gives 10m water. That will give 10m oil (or anything else). The power drawn will be 0.75 x the figure for water.

But as Lyn says, if there are other things to consider, particularly viscosity in the case of oil, you need to go into it a bit more.

Viscosity will affect the system calculations and the absorbed power, but not the pump head.

Not according to Lyn's first link. This gives viscosity correction factors for various pump parameters. Though as far as I can see it doesn't give a way of evaluating these factors, so it's not much help in practice.

You (and Lyn) are quite right. Unlike the viscosity, I sit corrected (it's a long time since I did viscosity corrections).

Dear Friend,

It is essential or a MUST to convert the liquid column equivalent to water column and hence Secific Gravity or Density of the liquid comes in to picture.

In other words (Liquid Column in Metres) x Sp. Gravity (or) Density of the Liquid to e pumped = Water Column in Metres.

DHAYANANDHAN.S

That tells you that 10m oil gives same pressure as 7.5m water, which is correct. But if you mean he should select the pump for 7.5m water, that's wrong! It will then give only 7.5m oil. See earlier posts.

Since the density of the oil is less than the density of water, the pressure available from the pump, rated at 10m water, should support a head of 10/0.75 = 13.3m of oil.

That will confuse him IMHO. If the impeller and pump speed are good for 10m fluid it will give 10m oil (or any other fluid, ignoring viscosity differences as discussed). On water it gives pressure rise ~ 1bar, but on oil 0.75bar, which is what he wants.

Sorry, but can't agree with that. Would it give 10m of mercury? (That's 13.3 bar).

A Group 1 ISO standard chemical process pump would probably handle mercury equally as well as oil, since you'd likely be running at 4-pole speed for 10m head it would have plenty in hand for the extra power requirement (and shaft radial loading). I'd be inclined to use a mag-drive though, I don't fancy a driptray full of mercury!

I still can't go with this constant head idea (with irrespective of SG).

So what do you think the answer is? To go back to your #10, a pump rated for 10m water will not support a 13.3m head of oil, because the centrifugal force from the impeller is smaller on oil, due to the lower SG. It will support a 10m head of oil (or other fluid).

I can't agree with that. If a pump designed for water draws say 10kW on water, it will draw 136kW on mercury. If you just fitted a motor with that power I doubt that the pump shaft would handle the extra torque. They're not built with that much strength in hand! And unless specified for mercury, the casing unlikely to take the pressure.

I don't have stuff to hand at the moment, and little time, but if you look at a set of curves for a typical ISO 5199 chemical process pump, you will find that a size 50-32-160 will probably do the duty. Possibly a 65-40-160 or 80-50-160, depending on flow, but it will certainly do the 10m head at 4-pole speed. It will draw about 0.6kw on oil, so about 7kw on mercury. That very same pump is designed to operate at much greater heads at 2-pole speeds, and is sized to be fitted with motors probably greater than 15kw. On a volute type centrif, radial shaft loading might be a bigger concern than motor power anyway.

For a pump rated for 10m head I'd think a typical maximum casing pressure would be 10 barg, but it would depend on the model. 13.6 barg likely to be pushing it IMHO. Not sure where you get the flow and hence power from, as I don't think the OP gave it.

You could be right, but I still think if you had a pump rated at 10m and you asked the supplier if he was happy for you to run it on mercury (with the appropriate power input) he wouldn't be too keen! Be good to hear a supplier's comments if there's any on the forum.

You start to pick on details, where the OP was a general question.

I can assure you that there are pumps that only deliver a 10m head but can withstand hunderds of bars. They are called circulators.

I'm sure there are, but in referring to a circulator you are picking on a specific type of pump, unlikely to be what the OP had in mind, specially a 100+ bar job!

If it's a circulator specified for water, with a built-in motor, as they often are, it won't have enough power to handle mercury, even if the casing can take it. If it's a separate motor it's unlikely the shaft can take the higher stress. That's all I'm saying.

ISO 5199 (DIN24256) chemical process pumps are rated for 16bar (hydrotested to 24bar), so it will easily handle the pressure.

I would have been wary about pumping mercury, purely because it's a bit different and I had not heard of it being done before. I doubt if you will find a supplier who has experience of pumping mercury with a centrif. I never did in 7 years, and we pumped all sorts of nasty stuff, and who is ever going to need to pump the volume of mercury that would require a centrif?

OK I didn't start this one about pumping mercury, it was JohnDG in #12!

I think we've kicked it to death, but thanks for the discussion . Between us we've answered the OP's question.

True, it's all just hypothetical anyway. Considering it probably costs £10,000 per litre, nobody's going to test it out just for fun.

As long as you guys don't get lost between statics and dynamics, we are OK. The bottom line, yes you must convert the head to water head. - Especially if you are starting from what you want A pump to do, and do not have THE pump as yet. If the pump will create a 10m dynamic head of water at a given flow rate, it will also create the same dynamic head with the oil, but at a different flow rate, based upon viscosity, different power input requirement, etc. There may have to be allowances made for heat dissipation or not, depending on the liquid properties, and any pump "slippage" which amounts to a certain quantity of mechanical energy input becoming heat.

Pretty much agree, but when you say "If the pump will create a 10m dynamic head of water at a given flow rate, it will also create the same dynamic head with the oil, but at a different flow rate, based upon viscosity......." the "but at a different flow rate" risks causing confusion IMHO. A different flow rate could be caused by change in viscosity (as discussed earlier), but if viscosity does not change, the pump Q/H curve (H in head of fluid) does not change with fluid SG. I'm sure you know this, but maybe not everybody .

Thanks for the clarification, and the vote of ~confidence.

Two of the main things to consider in any pumping set up:

1) viscosity (represents resistance to flow, and will create heat in the line (given the favorable dynamic conditions for that, extra work being done to pump the fluid, and thus a higher motor operating temperature (can be a huge problem and a headache) if the motor is not upsized (along with the control circuitry) appropriately for the application.

2) specific gravity (relative to water as the definition goes). Higher specific gravity fluids always require more work than s.g. 1 (water). Lower s.g. products require less work when all the dynamic conditions are similar (flow, pressure, viscosity).

Second option, the first one will limit you to that head and nothing more.

No! If he selects a pump for 7.5m (water or oil or anything) on oil it will give 7.5m oil. He wants 10m oil.

He asked

"do I choose directly my pump with 10 meters of water (column of water) on the performance curve or should I first convert my head of liquid into head of water (10 meters of oil = 7.5 m of water) and do my selection with 7.5 m?"

He needs the first option, as more than one poster has already said.

So how does the pump "know" when it's got to 10m (or 7.5m, or whatever)?

The particular impeller and speed give a specific head vs flow curve (head in m fluid). Where on the curve the head and flow finish up depends on the system resistance. But the curve does not depend on the fluid SG.

I would use the provided chart information and opt for the pump sized at 10 meters of water column head so that there is enough pump capacity (headspace) to allow better flow control and higher resolution tuning.

This will also allow some room to offset unknown and unplanned system dynamics that negatively affect flow.

Operating a centrifugal pump at the upper end of it's flow capacity often causes unreliable operation and a myriad of problems.

If you are unsure perhaps you should consider converting the head into pressure (BAR/PSI) then use the derived number to determine the horsepower and pump size required to lift the oil the distance required.

Dear friends,

Thank you for your comments. From what I can understand from the different responses, it sounds that I have to select a pump at 10 m. I have to admit that this is surprising for me but trusting the experience of the community, I would support now the idea of keeping 10 m.

It is not because 10 m of water is sufficient for 10 m of oil; because one day, I would have to pump a fluid with higher S.G than water. Then, it is important to have a clear understanding of what have been explained here.

Please, does somebody has an explanation of why the head of a centrifugal pump is constant? Does somebody where I could find viscosity head coefficient?

Regards

It is a very surprising fact and one that confused and surprised me at the start of my career. My simplistic understanding of the issue is

Pumps do not generate pressure but dynamic head. They do this by converting mechanical energy from the motor via the impellor to kinetic energy of the fluid.

The outlet of the pump then converts the kinetic energy to a head

1/2 m v^2 = m g h => h = v^2 / 2g

where h is the head of the pump.

This head then has an equivalent pressure based on the density of the fluid.

In principle yes the pump will generate 10m of head for any fluid of any SG. The limit will occur at the motor. As the SG increases the energy required to raise the dynamic head increases and for any fluid with an SG > 1.5 times the original fluid you are likely to run out of power. (Bigger motors are more expensive and so you dont want to put in a motor that is any bigger thanyou need). There will also be problems at higher SGs with the torque and the load on the driveshaft - so if you tried to pump mercury although the pump would in theory deliver the same head either you would red line the motor or twist the driveshaft out of shape.

If you know that you will want to use this pump for oil now at SG = 0.75 Head = 10m and in the future for SG = 1.2 Hd = 10m tell the vendor today to ensure that the motor and dive shaft are adequately sized

Note that these comments ONLY apply to centrifugal pumps

Please, does somebody has an explanation of why the head of a centrifugal pump is constant? Basically, it's because of Einstein's principle of equivalence. Inertial force, which the pump uses to generate head, and causes pipe losses, cannot be distinguished from gravitational force, which causes pressure from a column of fluid.

Does somebody know where I could find viscosity head coefficient? Not offhand, but if you include it in your enquiry to a supplier they should allow for it.

I used to use 'Cameron's Hydraulic Data' for viscosity corrections, which is based on data acquired by the Hydraulics Institute (summarised here: http://www.pumpfundamentals.com/help10.html ), but there are probably several online sources these days, like this: http://www.engineeringpage.com/calculators/pumps/viscosity_correction.html

All pump curves are plotted using a SG of 1, as the pump is tested to produce the pump curve using water.

So using various calculations you can work out the Head, power, flow rate and efficiency of the pump from the base pump curve

So the above curve is the standard produced for the pump, the next curve or curves are showing the changes depending upon the frequency supplied via a Variable Speed Drive, showing head, flow rate, range and efficiency.

I'm sure in the previous answers someone has added the calculation for the change in SG, however I will look back in my files for a graph that does it all for you.

And before I forget, in answer to the question "must I convert before selection the pump"?

Na.... I've yet to find a Application Engineer so inclined to do that calculation as the change in SG between oil (SG 0.87 - 0.9) and water is so small, its not worth bothering about.

If you would have been present in the fluidomechanics classes, which in my country as obligatory for every bachelor in engineering (all specialities), you would have known this.

Sorry to sound rude but this is even not a homework question anymore.

"You have calculated the head.." In my day the Process Engineer calculated the pressure drop or range of the equipment downstream of the pump. He then wrote up a Process sheet showing the max.-min specific gravity of the fluid, temperature of the fluid, its viscosity, inlet or suction pressure with NPSHA and the delta p needed or required. From this information the required head was calculated (2.31 x delta press/S.G.)(English units). The TDH is based on the impeller design, its diameter and its rpm. The TDH is independent of S.G. but the delta pressure developed is based on the S. G. and that is generally what the process requires. The horsepower curves shown are based of 1.0 S.G. of water and therefore also need to be de-rated to the actual S.G. of the process fluid. The viscosity correction generally was a minimal factor unless the fluid was very viscous at lower temperatures in which case a positive displacement pump would be more efficient and was often utilized.

Dear friends,

Thank you all for your useful comments. They prove me that the question was not so stupid as I initially tought. I retain that the height will be the same, as long as the pump receive the sufficient power and this is only valid for a centrifugal pumps.

Kind regards,