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Brake HP and Motor HP calculation/understandingHelpful Member!(3) 

naturalgas01 (Chemical) (OP)
16 Feb 09 12:44
Hi all,

I have confusion in power calculation for the pump. I know that Motor HP is the indicator for power consumption/energy bill when running the pump.

In one of my project for selection of the pump, I know following things :
(1) my motor HP required is 50
(2) my brake HP is depending on flowrate and head varies between 30-40.

Now, when you do power calculation or how much will the energy bill per year you consider the motor HP, which is 50. Then why people talk about BHP. when you start the pump, depending on your flowrate and head your motor will consume electricity. then what is the significant of BHP for energy calculation/or how much money you can save for sizing pump of having 20% safety factor or 10% safety factor.

please explain me by doing calculation.
thanks in advance for your valuable answers.

naturalgas01
786392 (Petroleum)
16 Feb 09 13:20
Dear
although there are much more learned/ experienced colleagues on board I take the small lead.

Usually Motors are considered giving 80% efficency or slightly less available HP for the pump

i.e 80% of 50(0.8X50)= 40HP then what is the issue?

Moreover if we consider same value then what about coupling/ gearing/ power transmission system consumption(losses)

Also it is never 100% energy tranformation as such we may invite motor burn out.

hope this helps

Best Regards
Qalander(Chem)

Helpful Member!(3)  BigInch (Petroleum)
16 Feb 09 13:51
Power needed by the fluid.
Pump_hydraulicHP = Q_cfs * 62.4 * SG * PumpDiffHead_ft / 550

Power needed by the pump.
PumpBHP = pump_hydraulic_HP / Pump_Efficiency

Delivered Motor Power (often called motor shaft power) must be greater or equal to your maximum PumpBHP, plus a small safety margin of 10 to 15%, often achieved by rounding up to the next commonly available standard commercially available motor power rating.  In your case with a max of 40 PumpBHP, we find a motor available with a rating of 50 HP, so we select that.  That's 25% higher than needed, so it is a rather large "safety" margin, but in your case, that's what's available, so we select it.

Power needed by the motor.
If you want to find the electrical power consumed at the motor, you must adjust the power delivered by the motor by the motor's efficiency.  If you have a new energy efficient motor, its efficiency might be about 94%, when running at its rated speed.  To get power consumed at the electrical meter, divide the power delivered by the motor (motor shaft power) by the motor's efficiency 94%.
Motor_consumed_Power = Motor shaft power / Motor_Efficiency

To give you a calculation using your values, a minimum of 30 BHP and a maximum of 40 BHP is required.  I assume its BHP (you didn't say if it is fluid hydraulic power, or BHP).

Let's take the 40 BHP case.
PumpBHP = 40
so a 50 HP rated motor is OK, because that is a rated motor shaft power which will be delivered to the pump.

Motor Power Consumed = Motor shaft power / motor_efficiency
= 40/0.94 = 42.55 HP
Convert to kilowatts
42.55 * 0.746 = 31.74 kW

That's almost the power registered at the electrical meter box, but you should assume some additional losses from wiring, maybe 3 to 5% and any other losses, such as 5% from a VFD if you have one, or from a transformer, or any other equipment you might have between the motor and the electrical meter.  Ask an electrical engineer to review that part.

When you are at minimum flow, your motor's power consumption, assuming the same motor efficiency (it could vary a bit), is,
= 30/0.94 = 31.91 HP
Convert to kilowatts
42.55 * 0.746 = 23.80 kW
power draw at the meter box will be that plus any additional losses at the lower current draw.





 


 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

naturalgas01 (Chemical) (OP)
16 Feb 09 13:52
thanks 786392 (Petroleum),

For energy bill, we consider the MHP. And it depends on Voltage, Amperage, PF and efficiency of motor. So, when you start pump/motor the which parameters remain constant are: Voltage, efficiency and PF ?

So, depending on the head and flowrate generated by pump, your motor will have different Amperage, is it right ? and because of your motor will have different MHP and also energy bill? please explain me if i am right or wrong, this is my understanding.

So, question is how do you calculate motor amperage and energy bill for head and flowrate generated my pump without using amperage meter ?

any detail explanation will be great
thanks


 
BigInch (Petroleum)
16 Feb 09 14:00
Yes power consumed varies with flowrate.
To calculate amperage, take the power in watts being consumed by the motor at any given flowrate and divide by motor voltage and PF.

A = W / V / PF

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

naturalgas01 (Chemical) (OP)
16 Feb 09 14:03
Thanks BigInch (Petroleum)

please see my comments and things that confused me in reply to 786392.

In my plant, my pump BHP require is only 25. But my connected motor HP is 50. When I asked to my electrical department in the plant, "he told me that it doesn't matter what kind of motor size/HP is connected to the pump. My motor will consume HP depends on the pump requirement, which is only 25 BHP.your motor will draw amperage depending on your flowrate,head required by the system (which is BHP, maximum only 25)" Is it right ? i am confused and can't explain it engineering way.

it will nice if someone explain me correlation between H-Q generated by pump and motor HP(electric bill) for different BHP, and how it will impact on motor amperage.

thanks
BigInch (Petroleum)
16 Feb 09 14:05
Motor efficiency is not exactly a constant, but remains relatively constant when the motor RPM is 50% of its rated RPM.  Assuming we are talking typical induction motor.  Lower than 50% RPM, efficiency drops off rapidly.  Lightly loaded motors also might not run at their best efficiency either.  In some cases, motor's run slightly more efficiently at slightly higher loads than their rated load, but best not to count on that.

You can see when starting motor, and RPM is low, efficiency is low.  If voltage drops, its worse.  Amps go high in either case and very high in both cases.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

BigInch (Petroleum)
16 Feb 09 14:11
Your EE is basically right, although as I said, with light loads, motor efficiency can be reduced somewhat, but usually its not very much of an efficiency reduction.  A 50% loaded motor might be one of those cases where efficiency reduction could be checked.  Ask your EE to look into it.  

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

waross (Electrical)
16 Feb 09 14:44
It is possible to estimate the energy used by a motor with an ammeter but such a discussion may again lead to a lengthy and confusing thread.
A 50 HP pump can deliver 50 HP safely, regardless of efficiency.
A hypothetical motor with 100% efficiency will require 37300 Watts to develop 50 HP.
A motor with 95% efficiency will require 39263 Watts to develop 50 HP.
A motor with 90% efficiency will require 41444 Watts to develop 50 HP.
Issue #1.
Efficiency is affected by losses. Some losses are fixed such as windage losses, bearing losses and iron losses.
Often the largest loss at full load is the I2R loss. It is not uncommon for the best efficiency of a motor to be at less than full load due to the square factor of current losses.
Issue #2. Motor current is not proportional to motor load.
Motor current is made up of Reactive current and Active current. Reactive current is fairly constant. Active current varies as the load plus losses. The Actual current is measured by an ammeter. It is determined by the formula,
Actual current squared equals (Active current squared plus Reactive current squared).
Issue #3. The reactive current will change somewhat as the voltage changes.
Here is a suggested solution to your problem;
Pumping hydraulic load = 40HP
Pump load = Pumping hydraulic load/pump efficiency. = 40HP/0.8 = 50HP
Energy consumption = Pump load/motor efficiency = 50hp/.9 = (55.6HP x 746 Watts/HP) = 41.4 kw.

Use the average pumping HP demand.
Put in your own best guess for the pump efficiency.
Use the listed motor efficiency.

This will give an acceptable estimate of the power.
If you need a more accurate figure, use a Wattmeter. That is not the only way to determine Watts but it is one of the most accurate.
The Wattmeter will indicate the actual power demand and will inherently correct any inaccurate assumptions as to pump efficiency, inaccurate pump differential pressure calculations, increased piping fouling and anything else that makes the world of pumps and motors less than perfect.

Bill
--------------------
"Why not the best?"
Jimmy Carter

naturalgas01 (Chemical) (OP)
16 Feb 09 15:41
Thanks waross (Electrical)& BigInch (Petroleum),

can you explain this scenario :

how my Motor HP(and electric bill) will vary suppose(considering pump can generate maximum 400 gpm flow(1100 ft head) and maximum 1650 ft head(@zero flow) :
(1) if I close the pump discharge valve, ie at zero-flow
(2) if i open the valve at 50%
(3) if i open the valve at 100%

thanks
BigInch (Petroleum)
16 Feb 09 17:11
Power needed by the fluid.
Pump_hydraulicHP = Q_gpm / 60 * 7.4805 * 62.4 * SG * PumpDiffHead_ft / 550

Power needed by the pump.
PumpBHP = pump_hydraulic_HP / Pump_Efficiency

You just need to substitute the values for pump efficiency at 200 gpm and 400 gpm (assumed =50% and 100% valve opening, however that may not be the case).

Then proceed with those in the rest of the equations for motor shaft and electrical power.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

naturalgas01 (Chemical) (OP)
16 Feb 09 17:26
thanks BigInch (Petroleum),

but,when valve is 100% close, head is maximum and flow is zero. So from the hydraulic hp = 0 ?
then MHP = 0 ? how do you explain this ?


thanks
BigInch (Petroleum)
16 Feb 09 17:34
Basically all that power goes to into just recirculating the fluid trapped in the pump, and its getting very hot in there.  You're making thermal heat instead of doing mechanical work.  

 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

TenPenny (Mechanical)
17 Feb 09 13:45
At 0 flow, your pump will still consume some power.  If you're looking at a performance curve with a power line on it, you will see that at 0 flow, power is not 0.  

Most of this power simply becomes heat in the liquid trapped in the pump.  I know of two cases of hot water pumps operated with both suction and discharge valves closed, where the pump exploded due to the heat turning the already hot water into steam.

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