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Induction Motor Performance Calculation

The document provides the parameters and solution to a problem involving calculating various values for a 480V, 60Hz, 6-pole, three-phase, delta-connected induction motor. It calculates: 1) the synchronous speed is 1200 rpm, 2) the slip is 0.025, 3) the line current is 80.6 A, 4) the input power is 62.2 kW, 5) the airgap power is 59.2 kW, 6) the torque developed is 471 Nm, 7) the output power is 74.1 Hp, and 8) the efficiency is 88.9%.

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115 views4 pages

Induction Motor Performance Calculation

The document provides the parameters and solution to a problem involving calculating various values for a 480V, 60Hz, 6-pole, three-phase, delta-connected induction motor. It calculates: 1) the synchronous speed is 1200 rpm, 2) the slip is 0.025, 3) the line current is 80.6 A, 4) the input power is 62.2 kW, 5) the airgap power is 59.2 kW, 6) the torque developed is 471 Nm, 7) the output power is 74.1 Hp, and 8) the efficiency is 88.9%.

Uploaded by

ronyiut
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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EEE 259

Class Test 2 Date: 24.04.2016

Student ID: Time: 20 minutes

Problem: A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor


has the following parameters:

R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm=30.74 Ω

Rotational losses are 2450W. The motor drives a mechanical load at a speed of
1170 rpm. Calculate the following information:

1. Synchronous speed in rpm


2. Slip
3. Line Current
4. Input Power
5. Airgap Power
6. Torque Developed
7. Output Power in Hp
8. Efficiency

Solution:

This machine has no iron loss resistance, so the equivalent circuit is as follows:

1. Synchronous speed is given by:

Therefore

ns = 1200 rpm

2. Slip is given by
Using the rpm equation,

s = (1200-1170)/1200 = 0.025

3. Now, phase current is given by

where phase impedance is given by

Using the above equation, Zin = 9.57 + j3.84 Ω


And noting that the machine is delta connected, V1 = VLL = 480V

I1 = 43.1 - j17.4 A. |I1| =46.6 A, θ = -21.9°


Therefore IL = √3 × 46.6 = 80.6 A

4. Input power is given by:

Therefore:

Pin = 62.2 kW

5. To find airgap power, There are two possible approaches:


a. Airgap power is the input power minus stator losses. In this case the core losses
are grouped with rotational loss. Therefore

Pgap = 62.2 kW - 3× 46.62 × 0.461


Pgap = 59.2 kW

b. Airgap Power is given by


This approach requires rotor current to be found. With no core loss resistance:

Giving I2 = 43.7 A. Substituting into the power equation

Pgap = 59.2kW

6. Torque developed can be found from

where synchronous speed in radians per second is given by

giving

τ = 471 Nm

7. Output power in horsepower is the output power in Watts divided by 746. (there are 746
W in one Hp).

and

Therefore output power in Watts is:Pout = 55.3kW

Pout = 74.1 Hp
8. Efficiency is given by

Therefore

η = 55.3/62.2 = 88.9%

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