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]

Electricpete, if I may venture somme comments.

As I see it: the definitions of efficiency are the reasons for the discrepancy. Your derivation includes mechanical losses that don't "affect" the pumped fluid. See, please:

BHP covers for:

a) Mechanical losses (bearing, seals, leakage) which do not impart heat to the pumped liquid;

b) Useful hydraulic power output;

c) Hydraulic power losses and frictional losses, all of which result in heat-up of the pumped fluid. These losses are part of the so-called internal efficiency [η]_i = (b)/[(b)+(c)].

Your definition of effy. is the total pump efficiency. Thus in eq.1, when effy=1, dt=0.

If one defines the internal efficiency as the ratio of the isentropic enthalpy change to the effective enthalpy change (=isentropic + heat losses), efficiency=1 would mean a totally isentropic compression.
Even an isentropic compression results in a fluid heat up and a temperature rise. In this sense Karassik's equation would be right.

In view of these considerations the accepted thermodynamic definition of efficiency as given by Sulzer:

[η] = 1/[B+C([δ]T/[δ]H)]​

Where B and C represent physical characteristcs of the fluid.

 

[electricpete]

Good point. I forgot that an ideal pump still creates a temperature rise. I remember seeing the constant pressure lines on the T-s diagram are distinct (different temperatures) even in the subcooled region. One of those thermodynamic oddities that seems nonintuitive to me. I will have to think about the rest of it for awhile. Thanks.

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[electricpete]

OK, I can very well accept splitting losses into categories a) and c) and if we wish to define a new quantity called internal efficiency to apply to only subset c) of total losses

Now I still can't understand Karassik's equation:

dt = H / [778 C Eff]

Throw away the 778 (why do we need to clutter the equation with unit coversions)

Multiply both sides by C*Eff to get

dt * C * Eff = H

Multiply both sides by m'

m' * dt * C * Eff = m' * H

Divide both sides by m' * dt * C

Eff = [m' * H] / [m' * dt * C]

It seems this equation should be consistent with what the definition du jour of efficiency has been chosen.

The numberator m' * H seems to me to represent fluid power. Makes sense since this is the intended useful output of the pump.

The denominator is m' * dt * C. This represents energy associated with temperature change of the fluid. On what universe is that the input?

In addition, wouldn't we expect m'H can very easily be greater than m' * dt * C? Wouldn't this mean efficiency greater than 1?

Hopelessly confused. Any comments welcome.





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[25362]

Electricpete: Let's start by saying I'm not a pump expert. Nevertheless, I'll try to find an explanation to Karassik's formula by a backwards analysis.

In your formula the numerator m'*H indeed represents fluid power, but the denominator should include among others that part of the flow rate lost as internal and external leakage and some recirculation, m'_L.

Internal leakage between wearing rings with a gap width of 0.01 in. are considered negligible. Larger gaps affect leakage more on those centrifugal pumps having low specific speeds (low flowrates and high heads).

Thus, the equation would (a grosso modo) become:

Effy.= (m'*H) / [(m'+ m'_L)*C_p*dt]​

When m'_L is negligibly small when compared with m', we come back, so to say, to Karassik's formula.

Sulzer's formula Effy. = 1/(B+C*dt/H) seems to better cater for inefficiencies which do not affect the fluid's temperature.

Kindly correct me if I'm wrong.

 

[electricpete]

I'm not sure I understand. The denominator of Efficiency should represent some form of input power, shouldn't it? What does a power term associated with fluid temperature rise have to do with input power?


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[25362]

Electricpete, I think you are absolutely right. The terms should be deducted from:

Internal effy. = [η] = dH/(dH + thermal losses)​

or better from

[η] = dH/(B*dH + thermal losses)​

where B is a correcting factor depending mainly on the suction temperature and the fluid under consideration.

Dividing all terms by dH:

[η] = 1/(B + thermal losses/dH)
when thermal losses are = C*dt, we obtain

[η] = 1/(B + Cdt/dH)​

which is Sulzer's equation

For higher temperatures and the type of fluid pumped it appears that B ==> 0 (don't ask me why) and we then get Karassik's formula:

[η] = dH/Cdt​

QED.

Not that I feel satisfied, but it seems to me now the only way to provide an answer to the puzzle.

 

[25362]

To electricpete, while looking for Karassik's articles in the past I came across his formula for the pumped liquid heat up in the Chemical Processing issue of April 1987. In this article the formula is like yours, namely

dT= (H/778 C) [(1/eff.)-1]​

which tells us that the formula appearing in the Pump Handbook is incorrect.

In the water pump example (designed for a flow of 550 gpm, at 250 deg F, against a head of 1800 ft) given therein by Karassik, the plotted heatup of the water above an efficiency of, say, 62% (~350 gpm) up to about 700 gpm, doesn't change much and stays at about 1 deg F.

 

[25362]

Electricpete, in my post of June 16, I should have added that the value of B==>0 happens at, or near, shutoff. In these particular cases all of the pumping energy is (conservatively) assumed to be expended in heating the fluid.

Thus, Karassik's "disputed" formula is apparently mostly used to determine the minimum continuous stable flow rate having decided what is the maximum temperature rise allowed from NPSH considerations:

Q_min= HP/(dT*[ρ]*C)​

using suitable units and conversion factors. Do you agree ?

 

[levadoux]

Hi everyone,

Thanks for this topic, I love it.
I am designing an experimence to test performance of ESP (electrical submersible pump) under viscous application. Esp are centrifugal pumps... I am now worried about the temperature rise. I can predict what is going to be the performance curve of the pump including efficiency. Then I would like to estimate what would be my temperature increase. I am using a formula based on Sulzer. I just made a simple balance considering all the losses going to the fluid (heat).

I have now 2 questions:

- I am working with viscous fluid. The efficiency is already corrected for the viscous fluid. Do I need any other correction? Sulzer uses some coefficient based on water...

- Then I know this approximation is not accurate but I would like to know where it would be more accurate. My feeling is that for low flowrate, the assumption that no heat is dissipated through the body of the pump might not be valid. I think the fluid is moving slowly, there is so some fluid ''staying'' in the pump then that would increase the heat transfer through the pump by conduction... More, with viscous fluid, the boundary layer is bigger with low velocity fluid so that the conduction transfer is bigger??????
For high flowrate I could imagine the assumption is good????

Thanks for your reply

Julien

 

[sailoday28]

25362 (Chemical) Jun 24, 2004 WRITES
To electricpete, while looking for Karassik's articles in the past I came across his formula for the pumped liquid heat up in the Chemical Processing issue of April 1987. In this article the formula is like yours, namely

dT= (H/778 C) [(1/eff.)-1]

which tells us that the formula appearing in the Pump Handbook is incorrect
I agree with above but note why 100% eff yields dt=0
MY DERIVATION
W= WORK IN TO SHAFT OF PUMP
H= SPEICIC ENTHALPY
E= SPEFIC INTERNAL ENERGY
V= SPECIFIC VOLUME
U= VELOCITY
Q= heat flow in
p= pressure
Z= elevation
dt=temp increase
Energy balance -steady state based on per unit mass flowing


(H+ V^2/2)in +W +Q= (H+ V^2/2)out
Assuming pump is insulated and H=E +pV
AND approximating fluid as incompressible
(E +pV+ V^2/2+Z)in +W +Q= (E +pV + V^2/2+Z)out

WITH Pump Head = (pV+ V^2/2+Z)out -(pV+ V^2/2+Z)in
Eout-Ein + Pump Head =W = cdt + Pump Head
EFFICIENCY =(PUMP HEAD)/W
AND Eout-Ein approx Cdt
effic*W=pump head
efficency*(cdt + Pump Head) = Pump Head
Which gives above formula from magazine.
Since liq is assumed incompressible and C= constant, a 100% eff pump gives dt=0.