Shaft Power / Absorbed Power

Sure.

The difference between shaft power and absorbed power;

.

shaft power - absorbed power = hft - borbed

.

Of course, it probably isn't a great idea to let someone else do your homework....especially someone you don't know.

It depends on your local semantics. Usually absorbed hp, shaft hp, and brake hp are the same thing; namely, the power required by the driven device. The chosen motor will typically have a somewhat larger hp capability. For instance, if you have a compressor whose rating table says it consumes 92 bhp (brake horsepower) at design operating conditions, you might select a 100 hp motor. (Or maybe even 125 hp, depending on how much reserve factor you want.)

Does he mean the the absorbed power of the shaft creating the BHP? If he does mean that way, then it is in the same magnitude with the shaft power.

If shaft can not absorbed such power, it will mechanically fail by torsion.

That is just complete drivel.

Oops, my bad. Lately, I just realized, when shaft will absorbed the power,if and only if it has to undergo deformation. Sorry.

That was an excellent reply which i required. Thank you very much.

It also depends on whether the sales or engineering department produced the ratings.

The absorbed power must be calculated from the motor in actual operation, or just guesstimated from data....to calculate the absorbed power

P = η·√3·V·I·cosφ

where
P is absorbed power in watts
η is motor efficiency
V is applied voltage
I is absorbed current in amps
cosφ is the power factor

The applied voltage and current in amps must be measured, then the efficiency and the power factor must be determined. Electric motor catalogs usually state the efficiency and power factor at full load and at various part loads(for example at no load, 25%, 50%, and 75%) sometimes found on the motor...so if the motor is to be run at say 75% load or at varying loads at known intervals, then that has to be taken into account....

..and the "shaft horsepower" is the mechanical power output available from the motor...

I have provided a link here to various formulas for various uses...

http://www.iprocessmart.com/techsmart/formulas.htm

You're going beyond the OP's question there, IMHO, with a risk of causing confusion.

He already has the absorbed power of the driven equipment, as in #2. That's all he needs to select the motor. He would only need to measure voltage and current if he suspects there's a problem. For selection of electrical gear the rated current is on the nameplate, he doesn't need efficiency and power factor, unless he feels like checking, for the hell of it.

Define the absorbed power....

See #2. Also OP says "I was given an input kW from mechanical (pump) section and they say its absorbed power" which indicates to me it's the shaft power. And your formula gives motor output mechanical power (= pump input shaft power) and you defined this as absorbed power.

Dear Mr. Sarkana

The absorbed power is actually drawn from the mains and it can vary from No-Load Losses when the equipment is just revolving at rated speed - to rated.Full-Load, since load may vary.

The Shaft Power indicates the the Maximum Safe Power that can be transmitted while the equipment is on full load.

DHAYANANDHAN.S

You can't say that as a general statement, it depends how absorbed power is defined - shaft output (= pump input) as in this case I believe, or electrical input. Whoever's specifying it should make it clear.

Thanks a lot. I found your explanation very helpful.

I still think in the context of this thread, absorbed power and shaft power mean the same, the mechanical input power to the pump, and that's all he needs for motor selection. For electrical input, could say absorbed electrical kW, but just electrical kW is self-explanatory.

As I'm sure you know, "brake" refers to the brake dynamometer used to measure output of engines, motors etc. It was the convention at one time to refer to the shaft power required by a pump etc as brake hp (bhp) or bkW, though I don't see how you would use a brake dynamometer to measure that, perhaps it's just me, I'm happy to be corrected. But I haven't seen that used over here for some years (good thing IMO). Nowadays usually called shaft kW or input kW, or something else to make it clear.

In your example of selecting a motor for a pump, (assuming you're buying a bare-shaft pump) you also need, crucially, the pump efficiency. Easier to get shaft power from the supplier, which includes it, or better still pump performance curves, so the motor can be selected to cover maximum absorbed power, if appropriate. Afraid you've got your hp to kW conversion the wrong way round .

I tend to disagree, with Codemaster on this one. Absorbed Power is the power drawn from the MCC by IEEE definition. BHP is the power the machine can develop (at a specified condition of output of the machine, and shaft power is the mechanical power the motor can deliver at specified loading % (can be estimated as by others here), but for the most critical applications, should be determined by measurements on site installed, as a means of verifying the installation.

If the BHP can exceed shaft limits, the motor can stall, or the shaft could be broken (depending on the torsional limit). IF the BHP exceeds allowable absorbed power from the MCC, then the OCPD should trip the breaker. Never reset the breaker at the MCC in case of trips until the cause of trip (as in short circuit overload, or time delayed overcurrent, etc.) are determined in accordance with (at least in the U.S.) the NFPA 70E 2015 code, as you are asking for an arc flash, arc blast event to take place when diagnostics are not done, due diligence is not done, and all safety aspects of the work have not been done.

I hope you engineer fellows are impressed that a mere steam plant chemist would know all of this.

I wouldn't know, so I'll take your word for it that the IEEE definition of absorbed power is electrical power, but you would expect the IEEC to focus on the electrical aspects. I still think it's more likely, from the OP's question and later posts, that his mechanical (pump) section meant pump shaft power. Presumably they're buying a bare-shaft pump or it wouldn't be an issue. They won't even know the motor efficiency, hence electrical power, until after they've selected the motor.

When you say BHP is the power the machine can develop I assume you mean power input ("develop" sounds like a power output device). From the question asked and information given, it seems way OTT to talk about overload conditions, let alone shaft breakage. He's entitled to assume the pump and motor people will get that right, above his pay grade, as I believe you say over there!

BTW since you ask I'm impressed by your erudition .

Unfortunate choice of wording by me with respect to BHP. It is in fact not power developed, it is power input requirement for a level of physical output of the machine (whatever the machine might be, in this case of point, a pump).

In Washington, D.C. they are all about "above my pay grade", and apparently none of them are paid anything, because every question seems to have that answer, unless it is one of which codex, chapter, title, paragraph, section, and subsection some little known, and less understood or understandable regulation is to be found.

OK, I hadn't realised "above my pay grade" had become quite such a cliché in US!

OP is confused as we are of what he meant absorbed power, he must thought about turbine, not pump.

Your formula as written Absorbed Power : Volts X Amps X Power Factor X Cos q is wrong. It's usually cosφ, not cosq, but that is power factor, you don't need it twice.

380 X 27,1 X 0,85 X 1,732 is in the right form, but one of the factors (apart from 1.732 = √3) must be wrong. A 15kW motor has typical efficiency 90%, so absorbed power = 15/0.9 = 16.7kW.

PF 0.85 is typical, so assuming 380volt supply, amps = 16.7*1000/380/0.85/√3 = 29.8. If voltage is different, current changes.

You have to decide which parameters you know and which you want to calculate. You can't put in figures at random and be surprised to get unexpected results.

He is wrong.

Are you sure you and your "owner" are talking about the same thing? If he means power absorbed by the pump, this must be less than the motor rated (shaft output) power. You can calculate hydraulic power as per 67model in #33, and pump absorbed power = hydraulic power/pump efficiency.

On the other hand, he might not understand motors. Some people, even some engineers, think motor rated power is the power drawn from the mains. It's not, it's the shaft output power. This has been discussed on CR4 a few times. If that's the case, you can educate him by getting a motor catalogue and showing him the figures agree with amps = rated power (watt)/efficiency/PF/√3.

please,

assume if there is loss in shaft, if will elaborated in to heat only.

almost 99.99 the power will be transmitted.

the heat what we see is heat transmitted from sources in contact with shaft.

any have loss proportional top the angle shifted (very very negligible). if the power become more the shaft will shear due to increased angle of shift.

truly

kittamani

As you have seen, sarakana, the meaning of absorbed power is different to different experienced engineers. So you have good reason to go to the pump department and ask them what they mean for your better understanding of the answer.

My opinion is that a power for a pump would be the input mechanical shaft power - it is not their business to guess what losses are involved in delivering that shaft power.

I would ask to understand for what speed; fluid flow volume, density, temperature and pressure conditions the number applies. In my experience, fluid temperature (hot or cold) can have have a major effect and, when cold, a pressure relief valve may give a delivery pressure far higher than with hot fluid - consequently much greater shaft power input.

Will this cause a temporary overload of the motor - if it is just sized for hot conditions?

Is the overload acceptable for the warm-up period?

What will the load be if the motor speed (supply frequency) is at the supply maximum?

Some pumps, like centrifugal are easy to start, while piston pumps can require high drive torque at zero speed (electric motors cannot always give full speed torque at standstill and generally cannot if supply voltage is low, bearing in mind that supply capacity and cable voltage drop may give only 80% volts at motor during start).

A conversation with the pump people to understand issues like plant start could avoid problems or your worry.

<...the difference between absorbed power and shaft power...> is the efficiency of the motor.