quiz - pump efficiency calc without flow 1

[electricpete]

Here is a quiz.

For a water pump. If given only the following:
DeltaH - Head increase suction to discharge.
DeltaT - Temperature increase suction to discharge

Can I calculate pump efficiency?

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[25362]

Only when assuming all the generated heat stays with the water, the efficiency of a centrifugal pump operating under steady-state conditions, as a decimal, using SI units, would be:

[η]=9.807 H/([δ]t . C_p)​

where

H = total head, m
C_p = water specific heat, J/(kg.^oC) from tables depending on temperature
[δ]t = ^oC

 

[d23]

Pete,

Only using the dt assumes that all the inefficiency is converted to heat. Some of the inefficiency is dissipated in other forms such as vibration.

As 25362 states it also assumes you are not dissipating any heat through the pump housing, shaft etc.

It may be as accurate to use the dp and a pump curve. To use dp you will need to calculate a Y equivalent for your motor and power system so you don’t include the motor efficiency in the calculation.

D23

 

[electricpete]

OK, you correctly guessed my answer was yes. In fact there is a vendor marketing equipment to evaluate pump condition (has it degraded from factory performance and in need of maintenance) using this technique when flow is unavailable or inaccurate. Also they use extremely accurate fluid temperature probes (claimed 0.002F… not my number so don’t argue with it) and claim better accuracy than almost any calculation using flow measurement.

I saw a slide with derivatio of their method which I didn’t absorb, but I think I have recreated it.

Yes it is assumed all losses are converted to heat. Does anyone think this is a significant error? My thought is no but interested to hear otherwise.

Let m’ = mass flow rate and deltaH and deltaT as mentioned above.
Pout = m’ * deltaH
Losses = m’ * c * deltaT
Pin = Pout + Losses = m’ * deltaH + m’ * c * deltaT

Eff = Pout / Pin = m’ * deltaH / [m’ * deltaH + m’ * c * deltaT]
Divide through numerator and denominator by m’ * deltaH gives
Eff = 1 / [1 + (c * deltaT) / (deltaH) ]

That is similar but to 25362, but not quite. I assume maybe you were just computing the Losses/Pin?

Anyway, it seemed to me initially a very surprising nonintuitive result that efficiency can be calculated without knowing flow or input power (even thought he derivation is relatively simple). That’s why I decided to share it.

D23 – you mention use of pump curve. If pump characteristics were known exactly there would be no need for calculation. The objective of this particular excercize is to evaluate actual performance of pump in field so use of factory data defeats the purpose in this particular case.

So we have efficiency and we’d like to know how the pump is doing. Efficiency doesn’t do us any good unless we can locate on the pump curve (since efficiency varies with operating point even for like-new pump from factory).

Another piece of data is required to locate the operating point. This would come from estimating motor power based on measurement of speed or measurement of electrical input power with assumption on motor efficiency. This creates another possible source of error.

Has anyone used method based on pump delta-T to evaluate pump performance?

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[quark]

The simplified equation for water pump is approximated as

t = (1-e)H/(427e)

(PS: derivation can be provided upon request)

where t is in ⁰C and H is head in meters.

Your formula is the basis for deriving much simplified versions and I feel 25362's equation can be yielded upon simplification.

Check this link

http://www.mcnallyinstitute.com/01-html/1-04.html

This method can be difficult to check with smaller pumps. For example, if you take a pump with 50m head and 75% efficiency, the temperature rise may be 0.039⁰C.

Regards,

 

[25362]

I've quoted a formula from the Pump Handbook by Karassik et al. The book brings also [δ]T=H/(778C_p[η]) in british units.
It also mentions a correction by Bush:

[δ]T=H/(778(1.0)[η]-C)​

where C_p=1.0 and C is a correction factor as function of T and
H/(778C_p) that can be obtained from a graph submitted therein.

I've also read that this (thermodynamic) efficiency formulae can be used for checking changes in the efficiency of pumps with large heads having appreciable temperature rises.

As I am aware of, the equations derive from the ratio of the isentropic enthalpy of compression to the effective enthalpy of compression. The latter includes the former plus the heat losses by friction.

Sulzer's Centrifugal Pump Handbook brings a formula similar to that of electricpete developed -it says- by Dr A. Whillier (The South African Mechanical Engineer, Oct. 1967):

[η]=1/[B+C([δ]T/[δ]p)]​

where B and C are constants representing water characteristics presented in graph form as functions of the water's inlet temperature and the discharge pressure.

 

[electricpete]

"I've also read that this (thermodynamic) efficiency formulae can be used for checking changes in the efficiency of pumps with large heads having appreciable temperature rises"

I can see the logic of that.... stemming from the same thought that quark expressed: lower-head higher flow pumps would have temperature rise too small to work with.

However, as mentioned this vendor has apparently developed very accurate temperature sensors.

In fact one application where the instrument is considered is our power plant circulating water system. Pipes are eight feet in diameter. Motor is 3500hp. Very high flow, low head. If memory serves me right 250,000 gpm, 35 feet head.

We don't have any flow indication. For some reason it is difficult to provide a flow indicator in this application (anyone know why?)

The vendor says it will work fine in this application in spite of the low delta-T.

Here is a link

http://www.yatesmeter.com/

Link to articles and case studies under "News"

http://www.yatesmeter.com/

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[rmw]

Pete,

Your surface condenser is a huge flow orifice, and if it is not scaled or otherwise plugged, mud, debris, mussels, it is a good cross check on pump curves, flow meters, etc.

Now, I want to throw a question into the mix. What if the pump casing does not lose heat to its surroundings, but gains heat from it.

What about pumping a 60F fluid, for example, a good cold lake, on a 90F day. The incremental heat increase due to the addition of pumping horse power might be influenced by the Q=h A dt equation of the casing surface????

Same thing might apply to losing casing or bowl heat to the sump and discharge piping in a submerged pump.

A point to ponder when working with temperature rises and measurements of this small of an amount.

Interesting concept. I am always looking for accuate ways to measure CW flow.

rmw

 

[PUMPDESIGNER]

I have been crushed at work and have not a moment hardly to devote to this, I am miserable about that.

Here are the problems that I see with the method, problems I would have to solve before I could swallow their results.

Centrifugal pump impartation of energy to the pumpage is almost or essentially adiabatic. This fact is confirmed by authorities such as Cooper writing in The Pump Handbook, and also by experience. Therefore, attempts to use temperature in analyzing pump performance would be limited in accuracy.

The accuracy of using ?T would be higher under some conditions than other conditions. The further below BEP, the more accurate ?T analyses would be. This is because the ratio of ?T to flow rate would be greater at low flow rates than at high flow rates, thus increasing the importance of ?T.

The accuracy of using ?T would be higher for certain types of pumps than for other pumps. The higher the Pump Specific Speed, and the higher the Suction Specific Speed, the more accurate the results would be.

One important part of the inaccuracy results from the difficulty in measuring such small changes in temperature under dynamic conditions where the pumpage at the discharge nozzle would probably not be homogenous yet in regards to temperature change in the pumpage. The formula requires a total change in enthalpy, not measured temperatures at various points in the pumpage. It appears to me that attempting to measure any such thing in the field would be not possible due to this one fact alone.

Also, a full accounting of temperature must include not only ?T of the pumpage, but also the pump casing temperature rise, just one more thing that would be difficult to do with any reliable or useful accuracy. Another factor would be heat lost to the shaft, by my experience considerable indeed.

Thus we have a very small component of the total energy transfer, which energy transfer is basically adiabatic, which cannot be accurately determined by any convenient method, being used to analyze the entire process where the total energy transfer is magnitudes larger than the tiny little itty bitty temperature component of the energy transfer.

If they can account for all that and more yet I could be convinced.

PUMPDESIGNER

 

[PUMPDESIGNER]

The ENG-TIPS software substituted ? for the delta in my previous post.

PUMPDESIGNER

 

[electricpete]

Your point about adiabatic process I would like to understand better. I am not a thermo guy but just looking at conservation of energy: if a pump is operating at 66% efficiency, where does the 34% of input power go if not heat? I have a hard time believing that energy to surroundings through vibration is significant.

I agree your point that certainly heat transfer from casing is important. In the case of circ water the cooling water temperature is not too far from ambient and my guess is this heat transfer will be small in comparison with the 34% input power causing deltaT.

Your point about mixing flow is valid and was discussed by the vendor as one of the biggest sources of the 1 or 2% effieicny measurement error number they quote. Probe placement measures temperature 1/7 of diameter into pipe. Turbulent flow is preferred due to better mixing which minimizes this error.



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[rmw]

Pete, go get a heat balance for one of the larger turbine generators in your plant, and look at the BFP. If they are done right, there will be a slight enthalpy rise of the BFW across the pump. When you divide the total flow by those few btu, you will see that it is essentially adiabatic, but, it is accounted for on the heat balance.

Without having any in front of me just now, I can remember BFP's of ~30,000 HP input, pumping 5 million gpm, and the enthalpy rise across the pump was single digit. This was preheated BFW, where the radiation and convection losses from the pump probably exceeded the heat rise of the power input to the pump. (a pure supposition on my part, said for demonstration only)

Part of the inefficiency goes to slippage around casing wear rings, from the discharge side to the suction side of the pump impeller/stage, but you are right, some of the mechanical HP put into the pump does heat the pumpage. Just shut the discharge valve, and it won't be long until you have boiled the water in the pump, and vapor locked it.

However, the same power that will boil the water at zero flow, when spread out across full or rated flow, becomes miniscule. Real, but miniscule, hence, my point, and pumpdesigner's point about the difficulty of practically measuring it in the field.

The theory is correct. The practicality is what bothers me.

rmw

 

[25362]

I've heard that tracers or flow dilution markers can be used to measure flow rates to +/-2% accuracy. A Google search may help.

 

[rmw]

Hello, duh...

That was 5 million lb/hr, not gpm, but I would note that it was at roughly 10,000 ft. head TDH in the example above that I cited.

Sorry if I caused any confusion.

And, yes, 25362, dye tests are commonly used in CW pumping circuits for flow testing.

rmw

 

[electricpete]

Good point - the dye test provides an alternate test for consideration for CW.

I still have a question for rmw and pumpdesigner concerning this term adiabatic. As I understand it means no increase in temperature of the working fluid. My question remains: where does the lost energy (1-Efficiency)*BHP go... if not temperature rise of the working fluid?

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[rmw]

Without looking up the dictionary meaning of adiabatic, it means that no heat is gained or lost by the fluid, as it, in this case, is acted upon by the pump.

Pumpdesigners words were "almost or essentially" and I would have used the word 'virtually.'

Pumps are not adiabatic, however. Just so nearly adiabatic, that the heat gain to the fluid, especially in the type of flow condition you describe, the large flow, low head situation typical of CW systems temperature rise is not usually considered.

The McNally link given above by Quark describes in detail all the places that the motor work goes in addition to into the fluid to raise it's temperature. Some of the mechanical friction described, seal/packing friction, bearing friction, etc., is actually transmitted down the shaft into the fluid, and thereby heats it.

My guess is that yours are vertical circulators, and all the line shaft bearings are cooled by the pumpage, so all that bearing heat goes into the fluid. Not much of the thrust bearing/seal work would, though.

Some heat is transferred from the bearing housings, and/or seal casings and/or shaft directly to the atmosphere, and never sees the pumpage. This would also need to be measured in order to have a conplete picture of pump efficiency losses. Pumpdesigner mentioned that as well. Ask your mechanical engineers if they have any seal coolers on any of their pumps.

There are hydraulic losses that have to be overcome by the powering device just to get the fluid into and out of the pump. That is an efficiency penalty to the pump itself.

But, yes, most of the heat that goes into the pumpage is from the friction of the fluid, as it is acted upon by the impeller, and as it enters, circulates around, through, and out of the pump.

That is why a dead headed pump will boil water in short order. More heat is being added by the friction of circulating the fluid in the pump than can be disapated by the casing surfaces.

Some inefficiencies that the driver has to be sized for are not attributable to the pump, coupling or belt/sheave losses for example.

So, that being said, if you were to assume that some significant fraction of the inefficiency went to heat the fluid, and you take your pump overall efficiency loss, ignore the coupling, and take that percentage of your 3500 Hp driver, assuming it delivers 3500 HP to the shaft, and spread it among 250,000 gpm, I bet you will calculate a number that is quite low. You have got me curious now, I will have to do that for myself next week. I just don't want to do it right now.

I hope this is a good explaination of your question.

rmw

 

[PUMPDESIGNER]

electricpete,
I am not ignoring this post, I just do not have the time this week, sorry.

In short, I am not negative on the idea you say they are using. I am not too old to learn new tricks. But as with anyone that stands accountable for my work, I must understand the accuracies and the process before I can use it.

Measuring enthalpy in field situations is a bummer, cannot be done accurately, or they would have to show me how they know their enthalpy estimate is accurate. When measuring such a small thing as the heat component in a centrifugal pump energy transfer from the driver, the enthalpy measurement must be dead on or all hope is lost.

If I get time I will work more on the formulas.

PUMPDESIGNER

 

[electricpete]

For a pump with 75% efficiency and 35 ft head, the delta T is calcaulated ~ 0.1F as follows:

Losses = m’ c dt = (1 – Eff) * m’ * H
dt = (1 – Eff) * H m’ / (m’c)
dt = (1 – Eff) * H * (1/c)
= (1-0.75) * (35 ft*lbf / lbm ) (lbm F / BTU)
= 0.25 * 35 F ft-lbf/BTU * [0.0013 BTU / (ft-lbf)]
= 0.25 * 35 * 0.0013 F
= 0.011 F

Which is exactly why I brought up the low head example (smallest deltaT). And no surprise that temperature measuremnt will be a key factor in limiting accuracy for this method. Surely will not be accurate with normal temperature instrumentation but as I said this vendor claims to have instrumentation to measure temperature to 0.002F. Even with that extremely accurate measurement 0.002F error would amount to 0.002F/0.011F ~ 20% error in estimating losses here? Hmmm. I have to bring that question back to them.

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[electricpete]

Karassik shows a formula dt = H / [778 C Eff] for calculating efficiency.

I come up with a much different expression. Derived as follows:

m' = mass flow rate
c = specific heat capacity
dt = temperature rise across pump
Eff = Efficiency
H = Head
BHP = brake horsepower = input power to pump
FHP = fluid horsepower = output of the pump = Eff* BHP
Losses = BHP – FHP = (1-Eff)*BHP

BHP (solved from FHP equation) = BHP (solved from Losses equation)
(1/Eff) * FHP = (1/[1-Eff]) * Losses

Substitute FHP = H * m’ and Losses= m’*c*dt
(1/Eff) * H*m’ = (1/[1-Eff]) * m’ * c * dt

Solve dt
Dt = [ (1-Eff)/Eff] * H*m’/c “Equation 1”

This is pretty close to what I used above except I left off a factor of Eff in denominator above.

But look at Karassik equation. He left off a factor of (1-Eff) in the numerator. (a bigger error for efficiencies above 50%!).

Karrasik’s cannot be right. If efficiency goes to 1, then losses go to zero and dt goes to zero right? That is predicted by my equation 1 but not by Karassik’s. So is electricpete right and Karassik wrong? Or more likely what is electricpete missing?

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[electricpete]

Correction:
Let me summarize the question of my previous message.

Karrassik shows: dt = H / [778 C Eff]

I derived: dt = [ (1-Eff)/Eff] * H/c “Equation 1”

1 - Why are these expressions different?
2 - Why doesn't Karrassik's dt go to zero as Eff goes to 1?

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