Brake Horsepower 1

[robot6]

My boss says that you can quote the net electrical requirement of a system in brake horsepower.

Perhaps I am confused, but it seems like brake horsepower is a measure of the amount of horsepower presented to the shaft, in other words, what horsepower would be required to stop the shaft of the motor.

By reporting to the customer the net electrical requirement in brake horsepower, am I doing the customer a favor? Does it with evaluation of the kVA requirement? Or is it not well related, in the sense that there is still a loss between the shaft torque of the motor and the kVA required?

If I have a pump that is reported in bhp, and a compressor reported in kW of electrica energy use, how to I total these? Is it correct to just use the 0.745 factor, or is there something more involved?

Thank you.

 

[jraef]

What is the context here? It makes a difference.

Lets say you are selecting a pump, the pump has a performance curve related to flow and head which will tell you what the BHP required by the pump is at the working shaft of the pump any given point on that curve. That is the MINIMUM HP that the pump demands to do the job you are selecting it to do. You cannot however buy an electric motor that is going to deliver that EXACT amount of HP, they only come in standard sizes. Plus, there are always losses associated with the mechanical coupling of the motor, the SOURCE of torque and speed (aka HP) and the pump, which REQUIRES it. What stating the requirement in BHP does for you then, it to put the onus of HOW that BHP arrives at the pump shaft onto the person who selects the motor. The ACTUAL HP (and/or kW) that the pump consumes however, will be the BHP for the design point, plus whatever losses there are in the system getting it to the pump shaft.

So lets say for example that your pump requires 6.3BHP at the design point. You cannot use a 5HP motor, so you buy a 7.5HP motor. But the pump must run at 600RPM and rather than buy a 10 pole motor, you buy a 4 pole motor and use a belt drive to slow it down to 600RPM. The belt drive then has about 10% losses, so your 7.5HP motor still works, because even through the losses in the belt drive, you have 6.75 net HP at the pump shaft, and you needed 6.3BHP.

If you wanted to determine the kW in electrical usage for that pump, it would not be 6.3HP converted to kW however, because you lost some energy in getting it there. Your energy consumption would be the 6.75HP, plus any internal efficiency losses inside of that motor. If the motor eff rating was 80% at full load, then your consumption would be 6.75/.8 or 8.4375HP, / .743 = 11.35kW

When a compressor mfr tells you that the compressor power consumption is a certain amount of kW, they have done all of that for you (albeit usually by testing, not calculating). Reverse engineering the BHP at the compressor shaft from that is no easy calculation, because there will be a lot of unknowns.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[robot6]

Thanks for the detailed answer, but I guess I need to make the question more direct, if I may.

He wants me to report energy requirement for the proposal (which contains many motors) in units of BHP, because these are English units, and he wants the units to be consistent.

However, it seems folks measure electricity only in kW or kVA.

Is he right or not?

 

[LionelHutz]

I'm not sure if that is a standard definition, but brake horsepower typically refers to the output power available at the shaft of a motor.

Personally, I would say it's the wrong units to use when supplying the electrical power requirement to a customer. Use units of kW for the real power required from the system and kVA for the apparent power required from the system.

If you gave me brake horsepower then I would expect that number to represent the shaft power, not the input power or input VA and I would add an appropriate factor to convert to kVA. Typically, given no other information I would use 1kVA of source capacity per horsepower of motor.

 

[waross]

A more important question: Is he your boss or not?
I was once tasked by a mill manager to report on the connected horse-power of all the motors in a large mill, broken down into departments.
I started to explain that some motors were heavily loaded and some were lightly loaded, some motors ran intermittently and some ran steadily.
My manager explained that he wanted to break the power bill up and proportion it to the various departments.
He was aware of he shortcomings that I had pointed out.
Never-the-less, he wanted some way to apportion costs.
I did my job and reported as accurately as possible, the connected horsepower in each department.
Without knowing the use to which your boss will put the information, I really can't comment on the correctness of the method.
1 KVA per HP is a good rule of thumb for a quick estimate feeder and transformer size for a large number of motors per the Canadian Electrical Code.
However, one utility at one point in time, for a particular application, (A shingle mill) used a figure of 1/2 KVA per HP when sizing a supply transformer. Their experience with similar mills showed them that due to under-loaded motors and a diversity factor, 1/2 KVA per HP was a safe factor.
So: If the customer had a high voltage service and owned the transformer then under the Canadian Electrical Code, the transformer must be twice as large as if the customer had a low voltage service and the utility owned the transformer.
Your question is hard to answer definitively.

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

 

[robot6]

LionelHutz, you understand my confusion, thanks for the help!

And he is my boss, acts like a * and that's why I find it hard to approach him with questions. Thanks for participating in eng-tips!

Actually all your responses are great, but here's the scoop, I finally found what I am supposed to do. (Let's introduce a new unit, the Brake kiloWatt, BkW!)

When our company reports something in BkW or BHP to the customer, we are saying that we have summed up all the shaft power required to run the system and we report that number. We expect the customer to then divide by their favorite motor efficiency (say 0.96) to get what the motors are actually drawing, and use whatever multiplier they have (you suggest 1kVA/hp) to get to their line requirements. So we leave all that up to them.

This is different than adding up all the motor ratings, which we might do and report as "connected horsepower" or "connected kW". These units we expect the customer to use to size transformers and MCCs and the like.

So at the proposal stage, I'm supposed to report BHP or BkW to facilitate the customer's decision-making: how much is this going to cost the mill to operate, and how does the cost of operating this system compare to another system? At this pre-sales stage, this is the question, not how big a transformer he'll need, because operating costs nearly always overwhelm capital costs.

Seems odd to me, but it makes some sense so I will go with it. Thanks to all for the help.

 

[LionelHutz]

That makes a little sense, but if I was the customer I would want to know the electrical kW, not the mechanical kW so I could predict the operating cost. But then, I would also want to know the kVA so I could predict the demand charges every month since that can be a large portion of the hydro bill.