Effect of head loss in pump power calculation 2

[Koros]

Hello,

The output pump power(Watt) based on chaning the head (h) is:

Power=ro*g*h*Q

but how to take into account the effect of headloss(m) in the equation.

 

[quark]

This is too basic to discuss here. Go to

http://www.mcnallyinstitute.com

and come back with specific questions, if any.

 

[ione]

I am inclined to think you’re missing something.
Pump curves without system curves are useless. Once you’ve drawn your system curve, its intersection with the pump curve will give you the operating point of the pump (Q,h). h is the head your pump has to produce to overcome the head loss of the system at the operating flow rate.

 

[BigInch]

There is no head loss in the pump power equation. Actually quite the opposite; H is a differential head and an increase in head across the pump, which equals discharge head - suction head.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[ione]

Head loss enters the formula via the total discharge head Hd

Ht = Hd – Hs (considering suction head)

With
Ht = Total differential head
Hd = Total discharge head
Hs = Total suction head

And

Hd = static discharge head + velocity discharge head + friction head in the discharge line

 

[Koros]

Thank you all for your comments

Let me raise again my question in this way:

For a pump-turbine in pump mode with maximum power input of 270MW, and 50 m3/s it is required to pump water to level i.e. 420m, with total head loss about, say 10m,
How to calculate efficiency of pump just base on above assumption?

I ‘ve calculated in this way: I subtract head loss of total required head!

P out= ro*g*h*Q=1000*9.81*(420-10)*50=201.105 MW

so

e=P out/Pin= (201.1/270)*100= 74.48%

Please correct me.

 

[tigerbEIT]

I am new here so please forgive me if I am wrong. I believe you should add the head loss of 10m to your total head (Ht). Therefore your output power will be higher and so will your efficiency.

The reason for this is because the head loss really creates more head for your pump to have to overcome (kind of ironic).

 

[BigInch]

Pump eff = output power / input power

input power is that provided by the motor at the pump shaft.
Output power required can be calculated by your first power formula, if H is the differential head.

Nothing to do with inlet and outlet pressure as you've shown it. That could only be related to the equivalent differential head. As you can see, you've missed accounting for flowrate and density and the pump characteristics important for determining its ability to convert input energy into useful output energy into the fluid.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[Koros]

Thank you all again for your kind comment,

Dear "Biginch" the equation which I've wrote is based on Power input/ouput not pressure (Pin and Pout)

As tigerbEIT mentioned I think I need to add the head loss to the total head required to remedy the head friction in calculating the delived output power? What is your Idea Biginch?

 

[BigInch]

Then its still not totally correct, as your method will yield overall efficiency because you have determined the electrical power input at the motor, not mechanical power input to the pump and then used the hydraulic power as the output power. You said you're looking for "efficiency of pump", not the overall efficiency, right?

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[tigerbEIT]

BigInch is right about this being the overall efficiency of the system. Seeing the information you have provided I believe this is the efficiency that you are looking for.

Please be more specific next time as there are other efficiencies that involove the pump (i.e. volumetric, mechanical, hydraulic, etc.).

 

[Koros]

Let rephrase the problem in this way: how to calculate the efficiency of pump-turbine in pump mode when we know, maximum static head (H) (which should water to be pumped up), max flow rate in pump mode (Q), maximum power input into pump (P), head loss (h)

Is it required to have more information for determining efficiency of pump-turbine in pump mode?
Thank you again

 

[BigInch]

You can't calculate PUMP efficiency unless you have the power input at the shaft of the pump and know the hydraulic power being output by the pump to the fluid. Usually its a value you find on the pump curve supplied by the vendor, but ...

Assuming you know the hydraulic power,
HP = ro * g * h * Q,

where h is the "differential head" of the pump,
NOT "head loss"

Differential head can be found from,
(pump outlet pressure - inlet pressure) / fluid_density

Then PUMP Eff = HP / Shaft_input_Power

Now have that read at Quarks' link.


"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[Artisi]

If you have an accurate measure of power input to the motor it is possible to calculate the overall pump/motor efficiency provided you also have accurate flow and head measurements. Also, if you have accurate performance data for the electric motor at full, 3/4 and 1/2 load or a speed/ power / efficiency curve it is possible to calculate the pump efficiency - but without this information you can only calculate overall efficiency.