Fluid Mechanics Questions

If the high speed pump is first in line, it will 'force feed' the slower pump.

If the slow pump feeds the high speed pump, the high speed pump will be starved and the input will operate at a reduced head.

Did they not supply any indication of how the pumps are arranged.?

hi, they did not mention how the pumps are arranged. but shouldn't the flow rate be the same anyway?

for the first question, if we equate the original and final head coefficient Ch1 = Ch2 --> gH1/(N^2 D^2) = gH2/((2N)^ 2 D^2) --> H2 = 4H1. but is it fine if i simply equate the two original and final head coefficient?

for the second question, can we assume the flow coefficient Cq remains unchanged. Does homologous pump mean that the Cq would be the same too? Cause if it is true, then I can equate the original and final Cq and get Q2 = 8Q1, which will give me option d.

A question from a student......

1.) B

And for Number two, I'll give you a hint.

Your answer lays in a pump curve,... go find one, any one for a centrifugal pump.

so from this graph, i know that when flow rate increases, the efficiency changes. but from the question, will i be able to tell how much Q increases by? cause i'm not sure if Cq is the same.

The pump curve is used for visual effects.

are you familiar with pump affinity laws?

The volume capacity of a centrifugal pump can be expressed below.

You know the original diameter and flow rate

q₁ / q₂ = (n₁ / n₂) (d₁ / d₂)

where

q = volume flow capacity (m³/s, gpm, cfm, ..)

n = wheel velocity - revolution per minute - (rpm)

d = wheel diameter (m, ft)

Now work the equation.

yea i would say so. so when the flow coefficient is the same, q1/q2 = (d1/d2)^3. but is it really just as simple as finding out that q2 = 8q1 and substituting it into the original n equation to get n= 2q - 2.5q^2?

I been popping in and out, but yes, affinity laws I use as a rule of thumb you could say.

In meetings when flow rates examples are given and were hashing things out (pumps size, pipe size. I will use affinity laws it as a rule of thumb when an 'educated guess' is required for a quick answer

wait but the n in the question is referring to efficiency, not wheel velocity though.

if the n in the question is efficiency, i don't think it's just about the pump affinity law?

Efficiency is an empirical measurement. It might be possible to approximate it by a quadratic for a specific pump, but I'm not sure that's a lot of use.

You haven't said what range of flow you have in mind, but from your

a. n = 128(Q - 10Q^2) if Q > 0.1, n is negative, which can't be right.

b. n = 16(Q - 10Q^2) Changing the impeller dia by 2x isn't going to change efficiency by a factor 8.

BTW, what is a homologous centrifugal pump?

homologous means 'geometrically similar' pumps ..if impeller Dia increase by 10% allother dimensions will change ...such that pump remains in precise geometric ratios. In practise this isn't valid because volute casing of a pump remains same while impeller cut may vary

I take it you want to learn... and questioning I think that's great...

Pardon me for copy clipping this......

But when the changing the impeller size, if wheel velocity is constant a change in impeller diameter simplifies the affinity laws to

Volume Capacity

q₁ / q₂ = d₁ / d₂

Head or Pressure

dp₁ / dp₂ = (d₁ / d₂)²

Power

P₁ / P₂ = (d₁ / d₂)³

hmm that's something i absolutely would not have related. so i presume the q2 = 2q1 and thus the new n will be n= 8q - 40q^2, which gives none of the options.

I'll let that sink in for now.

Normally I copy clip it, I'll cite it, but I refrain from doing this for now until the fundamentals are behind us.

Can I ask why if I equate the 1st Cq and the 2nd Cq, and with N (wheel velocity) being the same, I will get q1/q2 = (d1/d2)^3 instead. what's the difference between equating the two Cq and the affinity laws and when should I use which?

I was going to post but realised I'd made an error. I'd have liked to just cancel the post but can't see a way to do that

Quite the game

Efficiency is dimensionless.

Reynolds number

Until they can get empirical data

In the oil field and we have wells with a high GOR, I have in the past used a higher volume pump feeding a lower volume pump.

This helps to keep the gas in solution while in the centrifugal pump. The speed (RPM) of the two dissimilar rated pumps is the same as they are powered by one motor and are physically connected.

The flow rate, head and efficiency is calculated on the smaller volume pump.

As previously stated the configuration where the lower volume pump feeds the higher volume pump will result in cavitation.

Using the Laws of Affinity are a must when you are varying the speed of a pump or pumps, and the speed of the pump will also be the velocity of the fluid through the pump, or "specific suction speed". This will have an effect on your limits of RPM and flow rates.

It is worth noting that pumps are tested to produce their respective curves using water so from the produced curve, and change in viscosity will determine the type of impeller.

This will also have other effects, if the viscosity is higher, on the power and discharge pressure

Makes sense...Using it as a form of a stuffer keeping the gas 'dissolved'. It all depends on your process.

the submission for the assignment was yesterday lol. it was these 2 questions that i was unsure of. still very unsure and confused though haha.

At first it can be, as you working with it you gain practical experience. There are more competent experts on this site than me for sure. I'm a little surprised they didn't chime in.

we call the configuration a "tapered pump", where a larger pump feeds into the smaller pump set above it.

Other applications call for a "Gas Handler" which is a higher volume pump of about 5 to ten stages (one stage being a impeller and defuser), to compress the fluid keeping the gas in solution before the fluid enters the pump above it!

'Tapered Pump' Thanks for the terminology ,....

The second new term I learned in a (2) weeks.

The first was 'Ragging', I never heard it before which we are applying to one of our separation processes, and later in the same week (quite a coincidence) I received and signed up for a conference called, 'Pump Selection for Ragging and Solids.'

Looking forward to it.

Those of you who wish to get to know tapered pumps...

just dig around a little

is that your personal collection?

No, Imelda's...

Let's simply the problem. You have a "fixed" system with "turbulent flow" characteristics.

A centrifugal pump creates only "Total" pressure. The system in which the pump is operating determines "Velocity and Static Pressure". Agreed?

Two pumps in series will have identical "Velocity" and "Static Pressure" characteristics.

Two pumps in series wil deliver less flow than their individual ratings for that "fixed" system. In otherwords, you cannot have twice the individual pump flow rate for a "fixed" system with "turbulent flow" characteristics if the pumps are piped in series.

Make sense?

WOW! That was heavy!!

datf,

I'm not normally a pump guy, but have worked with a couple interesting cases. In question #1, the definition of series requires identical flow rates through both pumps at steady-state conditions. The second part of Question #1 depends greatly on the relative position of the two pumps. If The pump at speed N is the second pump, then all the (hidden) assumptions are reasonable, and your pressure drops of 4N and N are OK. If, however, the pump at speed N is the first pump, then the stated conditions in the question leave the answer in considerable doubt.

• if the 2N pump's discharge pressure is below its maximum head, the flow rate through it could cause its inlet pressure to be low enough to cause cavitation, so then the pressure drops cannot be correlated to the ratio of pump speeds. Indeed, the flow rate through the first pump could be above its performance curve also, resulting in cavitation on its inlet as well.
• If the 2N pump's discharge pressure is limited by the pump's performance curve (thus it is at the maximum head), the pump may or may not be in cavitation, and again the pressure drops cannot be correlated to the ratio of pump speeds.
• Consider what happens as the total flow is reduced towards zero by a discharge flow restriction. The first pump's discharge pressure is below but quite close to the maximum head pressure, and the second is limited by its maximum head pressure (a situation that meets the conditions in the question). Now the ratio of pressure drops could be as low as N:0.1N or lower.

Of course, plotting all this on actual pump curves will be much more instructive. Have fun with your (homework) questions.

--JMM

Hint: is there a branch pipe in between pump 1 and pump 2?

Regarding flow rates.... Because a pump can not move more fluid then is made available to it; flow rates are equal.

only if you match the pump output to the flow rate input, but that is a rare occassion.

Speaking from the point of view that in the oil patch we match the pump the PI of the well, but it is only matched at the right side of the curve, the BEP is usually less than the BHP and inflow rate!

Eh? How can the flowarates through each pump differ? Liquids are effectively incompressible. Even if there is air entrainment the mass flowrates are equal, except perhaps on a very short timescale (and the difference probably impossible to measure)

if you are talking about TWO pumps, one feeding another, then the smaller pump which is usually being fed from a larger pump will decide the final flow rate.

If you are talking about a tapered pump, the same applies.

In a oil well, using a submersible pump (centrifugal type) the RPM (and supply Hz)will decide the flow rate of the pump using the laws of affinity.

Now, designing the size (flow rate) of the pump, will depend on several things: bottom hole pressure or reservoir pressure, inflow from the reservoir and the static fluid level in the well.

What you do not want to do is to 'pump off' the well by using a pump that is to big, in fact you want a smaller pump to match, within its operating range, the PI of the well bearing in mind that you need to maintain the dynamic level in the well at least 600 to 1000ft above the pump intake at the BEP or slightly to the right of the curve if the choke is opened to produce more fluid or the VSD Hz is increased.

I wouldn't disagree with any of that, but it doesn't alter the fact that with 2 pumps (or anything else) in series, the flow through each is the same. Obviously the intermediate pressure, and where each pump falls on its performance curve, depend on the characteristics, and it's necessary to plan the complete system and select appropriate pumps.

I agree it's not unusual to have 2 pumps in series, to provide higher head or because the 2nd pump has high NPSH_R. I used to be involved with N. Sea oil platforms, with secondary recovery using seawater injection. The injection pumps were multi-stage centrifs, about 200bar generated head if I remember right, and NPSH_R about 6bara. This was provided by an upstream pump, sometimes fed from a vacuum deaerator, so had to a quite low NPSH_A.

totally agree with you, however, when designing a tapered pump and only used in wells of high GOR, or using a Gas Handler, the discharge rate is that of the smaller pump as the smaller pump rotating at the correct RPM, operating at its BEP will only produce as per its curve.

Please remember that is this is a special application only (to the best of my knowledge) used in Submersible pump applications in the oil field.

OK thanks, I think we've kicked this to death