why?

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I was told 1 horse power = 746 watt. but a 1 horse power ,200 volt single phase motor,accroding to T 430.248 have a current of 9.2 amp.
what is the different in the two answer.
is there a formula? :?
 

iwire

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You need to include power factor and efficiency in the calculation.

Exactly, the heat you feel from a running motor is evidence that power is being used to do more than spin the motor, that is why the efficiency of the motor has to be included.
 
what is the value

what is the value

is there a base value that we use for the power factor and efficeincy?
200x9.2=1840va. this is the apparent power what is next?
 

cripple

Senior Member
Horse power is a unit of measurement of power it takes to do a unit of work and is expressed in the following formula

power = work/ time = force x distance/ time = (180)(2.4 x 2 Pi x 12 ft)/ 1 min = 32,572 ft lbf/ min

Power is expressed in watts, which has nothing to do with watts as used in electrical circuits.
 

rbalex

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The REAL Truth

The REAL Truth

The "full load" currents in Tables 430.248 through 430.250 have nothing to do with Watts (or horsepower actually), power factor or efficiency. They are based on the worst motor any NEMA manufacturer makes at the specified horsepower rating.

Edit add: I was called away when I posted. The Table values allow motors of the same frame and hp to be swapped without other major changes to the circuit. Theoretically, the only adjustment necessary, if any, would be to overload protection. See 430.6(A)(1) and (2).
 
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John120/240

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Olathe, Kansas
Horse power is a unit of measurement of power it takes to do a unit of work and is expressed in the following formula

power = work/ time = force x distance/ time = (180)(2.4 x 2 Pi x 12 ft)/ 1 min = 32,572 ft lbf/ min

Power is expressed in watts, which has nothing to do with watts as used in electrical circuits.

On a similar note Horsepower is the amount of energy expended by a horse

to raise 33,000 pounds one foot in one minute. Do not know if they used a

draft horse (Clydeasdale or Percheon or quater horse). My two cents.
 

al hildenbrand

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Minnesota
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Electrical Contractor, Electrical Consultant, Electrical Engineer
I was told 1 horse power = 746 watt. but a 1 horse power ,200 volt single phase motor,accroding to T 430.248 have a current of 9.2 amp.
what is the different in the two answer.
In a nutshell, 1 HP = 746 W is true for a theoretical "ideal" motor. The ideal motor has no losses. A "real" motor
  • is made of copper that is not a resistance-less super conductor, but, rather, has resistance.
  • has bearings that have drag that takes energy to overcome
  • has "windage", that is, the motor doesn't spin in a frictionless vacuum, but, rather, air is moved which requires energy proportional to the amount of air movement.
  • if the motor has brushes, has resistance in the brush and a voltage drop at the point of contact with the slip rings or commutator.
  • is made of magnetic metals that are not ideal, but, rather have hysteresis, a magnetic memory that has to be overcome to change the magnetic flux.
  • has air gaps across which the magnetic flux must move where some loss into the rest of the universe occurs.
  • has eddy currents flowing in the magnetic metal, currents that flow through the resistance of the metal creating heat, the heat being another amount of energy that must be supplied before what's left gets converted into spinning torque on the turning motor shaft.
is there a formula? :?
There is a "generalization" that I was given in school. 1 HP = 1 kW.

But, as you can tell, after giving the bulleted list above some thought, the physical construction of the real world motor will affect the loses of the motor.
 

Rick Christopherson

Senior Member
Al, your bullet list is spot-on correct. However, these items are accounted for in the full equation of motor power conversion. There are 2 terms in the equation that frequently get overlooked. Powerfactor is the obvious one, and it will vary depending on how much load the motor is sustaining. The efficiency factor is what is covered in your bullet list, and it will vary with the design of the motor.

The full equation is:
hp = I x V x pf x eff / 746

For most motors at nameplate loading, the powerfactor will be 0.80 to 0.85. For good quality motors, the efficiency will be 0.85 to 0.90. I have found that setting both of these to 0.85 gives fairly realistic values for most smaller motors.
 

al hildenbrand

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Minnesota
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Electrical Contractor, Electrical Consultant, Electrical Engineer
The efficiency factor is what is covered in your bullet list, and it will vary with the design of the motor.
Actually, the last three bullet points cover the reactance of the motor. I was trying to keep the jargon down to a minimum. Power Factor is hard to grasp at the beginning, when the difference between theoretical and physical assemblies is still being grasped.

There's a lot of science that I don't talk about directly. If kendrichchan wants to follow that rabbit down that rabbit hole, there will be no shortage of explanation.
 
One big reason that the numbers in the table are higher is that remember unlike an internal combustion engine, An electric motor will make more that its nameplate horsepower if loaded down. I think they are accounting for wost case scenarios, poorly designed drives and motors pushed beyond their nameplate in addition to the other things such as efficiency and power factor.
 

rbalex

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One big reason that the numbers in the table are higher is that remember unlike an internal combustion engine, An electric motor will make more that its nameplate horsepower if loaded down. I think they are accounting for wost case scenarios, poorly designed drives and motors pushed beyond their nameplate in addition to the other things such as efficiency and power factor.
I said in my post above what the real reason is. I was a menber of CMP 11 (1996) when NEMA proposed revising the Tables.
 

Electric-Light

Senior Member
For most motors at nameplate loading, the powerfactor will be 0.80 to 0.85. For good quality motors, the efficiency will be 0.85 to 0.90. I have found that setting both of these to 0.85 gives fairly realistic values for most smaller motors.

Maybe for large(r) three phase motors. Small single phase motors are not that efficient.

Baldor 1725RPM VL1309 1hp right off the spec sheet:
8.4@208v
1.25 SF
65% full load eff.
0.65PF

so this comes out to be about
output ~0.75kW
input pwr ~1.15kW
1.15kW/0.65 ~1.8kVA
1800/208 ~ 8.7A
 
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