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Exercise 6.2

This document contains information about a 240-V series motor rated at 600 rpm and 80 A. It provides: 1) The efficiency is calculated to be 79.167% if 5% of power is lost to rotational losses. 2) The horsepower rating is calculated to be 20.37 hp. 3) When the current increases by 50% to 120 A, the speed decreases to 450 rpm and the torque increases to 458.367 Nm.

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Amir Dillawar
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241 views2 pages

Exercise 6.2

This document contains information about a 240-V series motor rated at 600 rpm and 80 A. It provides: 1) The efficiency is calculated to be 79.167% if 5% of power is lost to rotational losses. 2) The horsepower rating is calculated to be 20.37 hp. 3) When the current increases by 50% to 120 A, the speed decreases to 450 rpm and the torque increases to 458.367 Nm.

Uploaded by

Amir Dillawar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Exercise 6.

A 240-V series motor takes 80 A when driving its rated load at 600 rpm. The other parameters of the
motor are Ra = 0.2 Ω and Rs = 0.3 Ω.

(1) Calculate the efficiency if 5% of the power developed is lost as rotational loss. (79.167%)

(2) What is the horsepower rating of the motor? (20.37 hp)

(3) If a 50% increase in the armature current results in a 20% increase in the flux, determine

(a) the speed of the motor when the armature current is 120 A and (450 rpm)

(b) the torque developed by the motor. (458.367 Nm)

Solution

We can find the output power from the developed power since the rotational losses are 5% of the
developed power i.e., the output power is 95% of the developed power. Since 𝑃𝑑 = 𝐸𝑎 𝐼𝑎 then we need
to find 𝐸𝑎 .

From the circuit:

𝐸𝑎 = 𝑉𝑠 − 𝐼𝑎 𝑅 = 240 − 80(0.3 + 0.2) = 200 [𝑉 ]

Hence:

𝑃𝑑 = 𝐸𝑎 𝐼𝑎 = 200 × 80 = 16000 [𝑊 ]

Therefore:

𝑃𝑜 = 95% × 16000 = 15200 [𝑊 ]

The input power:


𝑃𝑖𝑛 = 𝑉𝑠 𝐼𝐿 = 240 × 80 = 19200 [𝑊 ]
This gives:
15200
𝜂= × 100 = 79.167 [%]
19200
The horsepower rating:
15200
𝑃𝐻𝑃 = = 20.37 [𝐻𝑃]
746
If there is a 20% increase in flux as a result of a 50% increase in current, this means that the flux will be
1.2 times higher at 120A than it was at 80A. Recall that 𝐸𝑎 = 𝑘𝑎 𝜙𝑝 𝜔𝑚 .

For a current of 120A:

𝐸𝑎 = 240 − 120(0.3 + 0.2) = 180 [𝑉]


Since the relationship between speed and angular velocity is completely linear we can re-state: 𝐸𝑎 =
𝑘𝑎 𝜙𝑝 𝑁𝑚 for purposes of comparison.

For a current of 80A:

200 [𝑉 ] = 𝑘𝑎 𝜙𝑝 × 600 [𝑟𝑝𝑚]

For a current of 120A:

180 [𝑉 ] = 𝑘𝑎 (1.2𝜙𝑝 ) × 𝑁𝑚(120)

Dividing these two equations to eliminate the unknowns:


200 600 600 × 180
= → 𝑁𝑚(120) = = 0.75 × 600 = 450 [𝑟𝑝𝑚]
180 1.2𝑁𝑚(120) 200 × 1.2

With a load current of 120A, the power developed would be:

𝑃𝑑(120) = 𝐸𝑎(120) × 𝐼𝑎(120) = 180 × 120 = 21600 [𝑊 ]

Which gives a developed torque of:


21600
𝑇𝑑(120) = = 458.366 [𝑁𝑚]
2𝜋
450 ×
60

END

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