KW to HP Formula Conversion - Electric motors & generators engineering (2024)

Hi DickDV

try this convertor:-

http://www.mr2ownersclub.com/converter.htm

desertfox

Here is how I do it.When you have a formula with something like "P in horsepower", it represents a unitless quantity P/hp.Susbstitue P/hp and similar for other quantities and carry out standard unit analysis.

(P/hp) = (T/ft*lbf) * (N/rpm) / 5252

Multiply by conversions equal to 1 in square brackets:
(P/hp) * [hp/0.746kw] = (T/ft*lbf) *[0.74*ft-lbf/N-m] * (N/rpm) / 5252

(P/kw) = 0.746 * 0.74 * (T/N-m) * (N/rpm) / 5252

(P/kw) =* (T/N-m) * (N/rpm) / [5252/0.746 * 0.74]

Multiply out the quantity in brackets

(P/kw) = (T/N-m) * (N/rpm) / 9565
Pretty close to 9549... more decimal places on the conversion would probably recreate the number more exactly

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Correction: need extra brackets <> to clarify the meaning:
(P/kw) =* (T/N-m) * (N/rpm) / [5252/<0.746 * 0.74>]

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Did I get this right?
I ft. lb. = 1.3558179483 Newton meters
5252/1.356 = 3874

(N-m x rpm/9549)
Should this be (N-m x rpm/3874)

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

Quote:

Should this be (N-m x rpm/3874)

No.The two relationships stated by DickDv are correct. I provided above the conversion between the two relationships.Also we can start from first principlesand derive each of them:

P = T * 2 * Pi * N
Divide each side by hp
P / hp = T * 2 * Pi * N / hp
Multiply rhs by two items both equal to 1:
P / hp = T * (ft-lbf/ft-lbf)* 2 * Pi * N * (min/min) / hp
regroup terms:
P / hp = T/ft-lbf* 2 * Pi * N /min^-1* (ft-lbf / minute) *1/ hp
add conversion:
P / hp = T/ft-lbf* 2 * Pi * N /min^-1* (ft-lbf / minute) *(1/ hp)* <hp/<33000 ft-lbf/minute>
collect constants:
P / hp = T/ft-lbf* N /min^-1/ <33000/[2*Pi]>
compute constant:
P / hp = T/ft-lbf*N /min^-1/ 5250
Write in words:
Power in horsepower = Torque in ft-lbf* speed in rpm / 5250

P = T * 2 * Pi * N
divide each side by kw
P / kw = T * 2 * Pi * N / kw
Multiply rhs by two items both equal to 1:
P / kw = T * (N-m/N-m)* 2 * Pi * N * (min/min) / kw
regroup terms:
P / kw = T/N-m* 2 * Pi * N /min^-1* (N-m / minute) *(1/ kw)
collect constants:
P / kw = T/N-m* 2 * Pi * N /min^-1* (N-m / minute) *(1/ kw ) *<kw/[1000*N*m/sec]> (min/<60sec>)
P / kw = T/N-m* N /min^-1/ [1000*60/<2*pi>]
compute constant:
P / kw = T/N-m* N /min^-1/ 9549
Write in words:
Power in kw = Torque in N-m * Speed in rpm / 9549

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Nope.
5252 evolves from the amount of foot-lbs/min of 1 HP which is 33,000.
When used on the HP-Torque formula:

So, we divide 33,000 ft-lb/min by 2pi, the constants in the formula to simplify the equation, hence approx. equal to 5252.
WE have the formula:

P is in HP; Torque in foot-lbs

The same thing when doing it the metric way:
9550 is the result of converting the "minute" portion of RPM into "seconds" and the 2pi constant. 9550 comes from dividing 60 seconds/ 2pi. The result is 9.54929, say 9.55. If you will express the power in kW, then you need to multiply 9.55 by 1000, hence you will have 9550 as a constant of conversion.
P = Torque(N-m) X RPM/9550

A European perspective:

The European Horsepower is defined as 75 lifting 75 kg 1 m vertically in 1 second.

75 kg in Earth's gravity field is 9.81.., so 75 kg of mass equals 736 N. From which can be derived that 1 European Horsepower equals 736, sometimes 735 watts (gauge marks on slide rules have 735). It depends on how many decimals you use in g0.

If that is because European horses are weaker than the American horses or because a "convenient" number like 75 was chosen for the conversion, I don't know.

But I do know that this little discrepancy has caused some discussions over the years. We never use Horsepower for power - always watts. So, we don't bother to use the 736 factor either.

Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

First, thanks to those who replied.

I'm having a bit of trouble seeing why the hp and Ft-lbf terms are "unitless".That has the effect of inverting the .746 and .74 multipliers and, as a result, 5252 gets divided by .746 and .74 instead of multiplied.

Hmmmm!I'm going to let my subconscious work on that awhile.

Thanks again to each who responded.

Dick

A "forensic" technique I use when I get results that don't match is to compare my error to the numbers in the equations. I instantly applied it to your problem. You got a factor of 2899 when you should have gotten one of 9549. So:

2899/9549 = 0.304

Your two factors are 0.746 and 0.74. Their product is 0.552.

Now, note that 0.552^2 = 0.304. Coincidence? I think not!

This is a strong indication that you multiplied when you should have divided or vice versa (this accounts for the squaring).

The next question is why you made this mistake. I looked at your "formulas" for HP (ft-lb*rpm/5252) and kW (N-M*rpm/9549). I have grown to despise these types of "formulas" with constants in them because they are so prone to confusion. Let's look at them more carefully.

When you say that the formula for HP is (ft-lb*rpm/5252), what you really are saying is that

1 HP = 5252 ft-lb * rpm

(burnt2x provided the confirming derivation for this.) Note that 5252 is in the numerator, not the denominator.

Now we start our conversions very carefully.

1 HP * (0.746 kW / HP) = 5252 ft-lb * rpm

0.746 kW = 5252 ft-lb * rpm

1 kW = 7040 ft-lb * rpm

1 kW = 7040 ft-lb * rpm * (N-m / 0.74 ft-lb)

1 kW ~= 9549 N-m * rpm

Your problem came from the fact that this type of expression of units is fundamentally misleading. I have made the same type of mistake with formulas like this, which is why I could spot yours quickly.

Curt Wilson
Delta Tau Data Systems

I did not say P is unitless, I said P/hp is unitless.
P isa physical varaiable representing power.
I can choose to express a given power levelin many different units, but that does not change the physical variable.

For example I can write:
P = 10 KW = 7.5HP
The equals sign indicate that all three are the same even if I express them differently.

Now divide the equation by 1HP.

P/hp = 10 KW/hp = 7.5HP/hp = 7.5

I hope you agree 7.5 is unitless.The equals signs indicates everything is the same. So P/hp is a unitless quantity also.(Like I said. )

Consider now there are two different ways to write an equation.
1 - The standard way.P = 2*Pi*f * T.We can use any consistent system of units we want and the equation is still true (we dont' have to argue about the best units... just pick the ones you like).If we plug the units in for each quantity as we go, it provides a "double-check" that the units on each side match.(or else we know that we need to add more conversions).Most people that went to engineering school are very familiar and comfortable with this approach.

2 - The way that is common among textbooks geared for non-engineers:P = T * N / 5250 where P in hp, T in ft-lbf, N in rpm.The where and the fact that you are told what units to use are clues that this is a non-standard form.Someone tried to make the formula easier to doing the unit conversions etc for you.As you have discovered it can make lift a lot harder.

The bridge between the two forms is putting the 2nd form in a rational consistent unit basis.It states the numerical value of P in hp is equal to the numerical value of T in ft-lbf times the numerical value of speed in rpm divided by 5250.Each of these numerical quantities can also be viewed as a dimensionless quantity (a number).How do I calculate the number corresponding to "P in hp"?Take the physical variable P and divide by 1 hp (P/hp).That is the bridge between the two types of equations and I have provided several examples above how to apply it.

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Hey, Pete!

What kind of horsepower is that?

I know that an American HP is a little bit more than a European HP. But never heard of one that is 1.333 kW

Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

Yes, there was an obvious typo. Should be:

Quote:

P = 7.5 KW = 10HP
The equals sign indicate that all three are the same even if I express them differently.

Now divide the equation by 1HP.

P/hp = 7.5 KW/hp = 10HP/hp = 10.0

I hope you agree 10.0 is unitless.The equals signs indicates everything is the same. So P/hp is a unitless quantity also.(Like I said. )

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Eng-tips forums: The best place on the web for engineering discussions.

You were close to overunity there, Pete. Big time.

Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

DickV:

I believe ft-lb=0.74*Nm and NOT Nm/.74.

This relationship is derived from the very fact that .746kW=HP

or 0.746(ft-lb* rpm/5252)=Nm*rpm/9549

This leads to Nm=1.35 ft-lb or ft-lb=0.74 Nm.

There are other factors involved that makes the division of rpm by different numbers in the two systems to come up with the same amount of power. I think your direct substitution is not appropriate.

My point is you need two independent equations to solve two variables. You have only one.

I'm still having difficulty making the numbers work out correctly.For example, we know that a 1hp 1750rpm motor develops 3ft-lbf torque.Converting the 1hp to kw makes a .746kw motor develop 3ft-lbf torque.

Converting the torque to metric units gives us a .746kw motor developing 3/.74N-m of torque which is 4.05Nm.

Now, using the metric formula1kw=Nm x rpm/9549 substituting the rpm we have 1kw=Nm x 1750/9549.Solving for the torque is 1 x 9549/1750 =5.46Nm.Reducing the motor power to .746kw reduces the torque in the same proportion so a .746kw motor develops 5.46 x .746 = 4.07Nm

Well, look at that!I think I just persuaded myself!Thanks guys, I think I've got it.Kind of a twisted way of coming into it but it works for me.

Another reason to go metric.

1 pound = .4536 kg = 4.448 N
1 foot = .3048 m
1 foot*1 pound = .3048*4.448 = 1.356 Nm or, inverted, 1 Nm = 0.7375 foot-pound

Data from the CRC Handbook section F.

Vive la Revolution Francaise! Without it, we would still be using feet and pounds

Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

This discussion reminds me of an assignment I had in junior high school physics class when we were studying power. We had to weigh ourselves, get timed in the 50-yard dash, then calculate how much power we generated doing so.

My calculations went something like this:

(150ft * 100lbs) / 8 sec * (1HP / 550ft-lb/sec) = 3.4HP

This was the type of calculation the teacher wanted, but it left me scratching my head, because I knew that a "horsepower" was fundamentally based on the power a horse could generate, and here I was, the proverbial 98-lb weakling (I was a runt in junior high) generating more power than 3 horses! It wasn't until some years later that I figured out the problem with the calculations.

Actually, it was a useful "life lesson" as it sensitized me to issues of mass and force, and taught me that authority figures often get it all wrong.

Curt Wilson
Delta Tau Data Systems

David,

The horsepower unit was originally set by having a horse walking horizontally but with a rope pulling a large weight vertically up a mineshaft. The power the horse expended in moving its own mass horizontally was relatively trivial, and so not accounted for.

Similarly, in my stupid junior high experiment, I did not have to exert anything close to 100 pounds-force to move my 100 pounds-mass horizontally, and was not generating anywhere near 3.4HP. So the biggest issue was not gross versus net.

When I later figured out what was wrong with my teacher's analysis, I thought of two variants of the experiment that would have been more accurate. First, we could have been timed climbing a ladder up a wall of the school (it had one). Second, we could have walked on the roof of the school while tied to a weight we were pulling up the side of the school over a pulley, just like that original horse. Sadly, I don't think either of those experiments would have been feasible.

Curt

Actually, it's another reason to use something like Mathcad, which is unfortunately very expensive at its list price, but it does unit conversions completely transparently to the calculation.

type: hp = , result is 745.7W

change units to ft*lbf*2*pi/min, results in 5252.1
change units to N*m*2*pi/min results in 7120.9
N*m = 0.7376 ft*lbf

This is sort of related to a similar posting about engineering software programs.The fact that people state that they are continually making conversion errors is really a sign of a seriously missing tool in the panoply of accessible engineering tools.With a good tool, the conversions are irrelevant, and the unit system can be pretty much anything anyone wants, like:

hp = 71362 N*furlongs*2*pi/fortnight

The effort to this conversion is, in something like Mathcad, exactly the same as using any other unit system, other than the extra typing keystrokes for the longer unit names.