EE362-Electromechanical Energy Conversion-II

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# EE-362 ELECTROMECHANICAL ENERGY CONVERSION-II

# Power and Torque in Induction Machines

## Ozan Keysan

[keysan.me](http://keysan.me)

Office: C-113 <span class="meta">•</span> Tel: 210 7586

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# Equivalent Circuit of Induction Motors

## The rotor can be referred to the stator side
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# Equivalent Circuit with Referred Rotor

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# Power Flow in Induction Motors
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- ## Total Power: \\(\quad P\_{in}=3V\_{1} I\_{1} cos(\theta)\\)
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- ## Stator Copper Loss: \\(\quad P\_{cu1}=3 I\_{1}^2 r\_1\\)
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- ## Core Loss: \\(\quad P\_{c}=3 \dfrac{E\_1^2}{R\_c} \approxeq 3 \dfrac{V\_1^2}{R\_c}\\)

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# Power Flow in Induction Motors

## Electrical Power Transferred to Rotor?
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- ## Air-gap Power: \\( P\_{ag} = P\_{in}-P\_{cu1}-P\_{c} \\)

## or 
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## \\(\quad P\_{ag} = 3 I\_{2}'^2 \dfrac{r'\_2}{s} \\)

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# Power Flow in Induction Motors

### Not all the air-gap power dissipates as heat, most of it is converted to mechanical power
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### The equivalent circuit (\\(\dfrac{r'\_2}{s} \\)) can be separated into two components:

### \\(\dfrac{r'\_2}{s} = r'\_2  + \dfrac{1-s}{s}r'\_2 \\)

###\\( P\_{ag} = P\_{cu2}+P\_{mech} \\)
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## Rotor Side Equivalent Circuit

### (\\(\dfrac{r'\_2}{s}\\)) is separated into two components

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# Equivalent Circuit of Induction Motors

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# Mechanical Power

## Gap Power: \\(\quad P\_{ag} = 3 I\_{2}'^2 \dfrac{r'\_2}{s} \\)
## Rotor Copper Loss: \\(\quad P\_{cu2}=3 I\_1^2 r'\_2\\)

## \\( P\_{cu2}\\) can also be expressed in terms of \\( P\_{ag}\\)

## \\( P\_{cu2} = s  P\_{ag} \quad  \\)

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# Mechanical Power

## Gross Mech. Power = Gap Power - Rotor Copper Loss
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## Gap Power: \\(\quad P\_{ag} = 3 I\_{2}'^2 \dfrac{r'\_2}{s} \\)

## \\(P\_{mech} = 3 I\_{2}'^2 \dfrac{(1-s)}{s}r'\_2 \\)

## \\( P\_{mech}  = (1-s)  P\_{ag}  \\)

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# Mechanical Power

## Gross mechanical power is the power converted into mechanical form

## Net power is the useful mechanical energy to drive the mechanical load
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## Net Power = Gross Power - Rotational Losses

## \\(\quad P\_{net} = P\_{mech} - P\_{rot}\\)

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# Power Flow in Induction Motors

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# Efficiency

# Output Power / Input Power
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# \\(\eta=\dfrac{P\_{out}}{P\_{in}}\\)

## Don't forget the rotational losses!

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# Example

## What is the efficiency of the motor in the last exercise if the rotor is running with a slip of s=0.05?

(Assume the parallel branch is moved to the primary side.)

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## You can download this presentation from: [keysan.me/ee362](http://keysan.me/ee362)