Calculating Power Output and Full Load Line Current

Motor input = electrical power

Output = mechanical power

Output = input x efficiency

Electrical power = VICosΦ Single phase = √3 VICosΦ (3 phase)

Mech power = Tω

Take care of units (rad/sec, N-m, ...)

P input (3P)=1.732* VL *IL* COS(ANGLE)

Pin/ (1.732 *415* 0.85) = IL;

Pout/Pin = EFFICIENCY

Pout = [(torque * rpm)/5252 ]*746 =13295

pin= 13295/0.85= 15641

IL= 25.6A

Recheck this with calculator.

Please dont give home work again :)) dont mind

mechanical power = torque x angular velocity

• P mech = T x ω
• = T x Nx2xpi/60
• = 65 x 1440 x 2 x 3.1415 /60
• = 9801 Watt
• this is the mechanical power at the output shaft of the motor

electrical active power x efficiency = mechanical power

• P elec activ x η = P mech
• P elec activ = P mech / η
• = 9801 / 0.85
• = 11531 Watt
• this is the electrical power we have to pay to the energy compagny

electrical active power = current x voltage x sqrt(3) x power factor

• P elec act = I x U x 1.73 x PF
• I = P elec act / (U x 1.73 x PF)
• = 11531 / (415 x 1.73 x 0.85)
• = 18.9 A
• this is the current which runs through the supply cable from the energy compagny to the motor

electrical total power x power factor = electrical active power

• P elec total x PF = P elec act
• P elec total = P elec act / PF
• = 11531 / 0.85
• = 13566 VA
• this is the size of the transformer the energy compagy has to have to feed the motor

electrical total power² = electrical active power² + electrical reactive power²

• P elec total² = P elec act² + P elec react²
• P elec react = sqrt (P elec total² - P elec act² )
• = sqrt (13566 ² - 11531 ²)
• = 7146 VAR
• this is the amount of power we could get out of a power factor correction capacitor installed near the motor. So this power will not flow through the supply cable from the energy compagny. The Power factor will be "one". In this case the energy compagny only has to deliver the activ part of the total power

current in supply cable when the energy compagny only delivers the activ part of the total power

• P elec act = I x U x 1.73 x PF
• I = P elec act / (U x 1.73 x PF)
• = 11531 / (415 x 1.73 x 1)
• = 16.06 A
• this means the surface of the supply cable can be smaller or a cheaper cable

pout(hp)=2piNT/33000, pin(watts)=pout(746)/efficiency, if your looking for the full load current, I=pin(watts)/1.737xVxpf

Thank you to everyone who has replied...

allyrat

Steps:

Efficiency: 85%, Power Factor 0.85lag, Supply voltage 415v, Torgue 65Nm, Speed 1440rpm, 50 hertz.

This I have already explained in #1 but again

You have an electro-mechanical system

Motor- Input power is electrical, output is mechanical (remember for generator it will be exactly opposite - for your next question )

The input power in electrical we express as

P = V.I - Note this is a phasor dot product (almost like vector)

= VI CosΦ ,

where Φ is the angle difference between the two phasors, but this angle is not in physical space but in time phase. So if you assume for simplicity the frequency is say 50 Hz

That means the voltage (and current) is oscillating 50 times every second. (Every single oscillation takes 1/50 = 0.02 seconds)

This 0.02 seconds in this phasor equals 360 degrees

If say the current zero is 0.001 seconds after the voltage zero, then we say

Current is lagging the voltage by 0.001/0.02 x 360 = 18 degrees.

As you see above, the ideal power is when both current and voltage are in phase (ie F = 0)

P = VI Cos 0 = VI

This factor (of power) , CosF is introduced due to the phase difference and is called the power factor.

In other words, the power factor is the factor by which the actual power is less than the apparent power. The actual power is the only part that can be converted to other form of power (eg mechanical, heat, light etc) and taken out of the system.

As said above ,

P = VICosF , is actually for single phase, ie when you have the line and the neutral connected to a load.

But if you are working with 3 phases (with or without neutral) , the power equation becomes

P = V₁I₁CosF₁ + = V₂I₂CosF₂ + = V₃I₃CosF₃

(1,2 3 are individual phase values of voltage, current and phase shifts)

If you have a balanced 3 phase, it simplifies to

P_elect = Ö3 VICosF (1)

(It is not 3 but Ö3 since in case of 3 phase supply, the line voltage itself may be Ö3 times the line to neutral voltage in case of star connection, and it is the current in case of delta connections)

Mechanical power, is simply calculated, as you have done,

P_mech = Torque x Angular speed. (2)

Then the third part

Output = efficiency x Input power (3)

Out of all these (1), (2) and (3)

You have all the required inputs (except single unknown- current) to calculate P_electr, P_mech and you know the efficiency.

Hope now you can solve it.

Note:

In case of motor, as is this one, Output is the mechanical, input is the electrical, so

P_mech = Efficiency x P_elect

In case of generator it will be reverse ie

P_elect = Efficiency x P_mech

Note-2

Remember also the caution of #1

The units, In SI,

Torque should be in N-m, Speed in radians per second (son you must convert RPM into radian per second)

Voltage in Volts (you may get questions with KV or MVs also) the current in amperes.

The power should be in watts, in case anything else (HP etc) convert into watts. Take care of any other such aspects.