Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
Sheet (7)
DC Machines (Motors) Solution
1) A 20hp, 250V shunt motor with Ra=0.22 Ω, Rf=170 Ω. At no-load and rated voltage, the speed
is 1200 rpm and the armature current is 3 A. At full-load and rated voltage, the line current is
55A. What is the full-load speed?
Solution:
𝒂𝒕 𝒏𝒐 𝒍𝒐𝒂𝒅:
𝑰𝒇 =
𝑽𝒕 𝟐𝟓𝟎
=
= 𝟏. 𝟒𝟕 𝑨
𝑹𝒇 𝟏𝟕𝟎
𝑵𝒏𝒍 = 𝟏𝟐𝟎𝟎 𝑹𝑷𝑴
𝑬𝒏𝒍 = 𝑽𝒕 − 𝑰𝒂 𝑹𝒂 = 𝟐𝟓𝟎 − 𝟑𝒙𝟎. 𝟐𝟐 = 𝟐𝟒𝟗. 𝟑𝟒 𝒗
𝒂𝒕 𝒇𝒖𝒍𝒍 𝒍𝒐𝒂𝒅:
𝑰𝒇 =
𝑽𝒕 𝟐𝟓𝟎
=
= 𝟏. 𝟒𝟕 𝑨
𝑹𝒇 𝟏𝟕𝟎
𝑰𝒂 = 𝑰𝑳 − 𝑰𝒇 = 𝟓𝟓 − 𝟏. 𝟒𝟕 = 𝟓𝟑. 𝟓𝟑 𝑨
𝑬𝒇𝒍 = 𝑽𝒕 − 𝑰𝒂 𝑹𝒂 = 𝟐𝟓𝟎 − 𝟓𝟑. 𝟓𝟑 𝒙 𝟎. 𝟐𝟐 = 𝟐𝟑𝟖. 𝟐𝟐 𝒗
𝑰𝒇 𝑵𝒇𝒍
𝑬𝒇𝒍
𝟐𝟑𝟖. 𝟐𝟐 𝟏. 𝟒𝟕 𝑵𝒇𝒍
= ( 𝒏𝒍
)→
=
𝑬𝒏𝒍
𝑰𝒇 𝑵𝒏𝒍
𝟐𝟒𝟗. 𝟑𝟒 𝟏. 𝟒𝟕 𝟏𝟐𝟎𝟎
𝒇𝒍
𝑵𝒇𝒍 = 𝟏𝟏𝟒𝟔. 𝟓 𝑹𝑷𝑴
Good Luck 😊
Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
2) A 230V shunt motor delivers 30hp at the shaft at 1120rpm. If the motor has an efficiency of
87% at this load, determine:
a) The total input power.
b) The line current.
Solution:
a) 𝑷𝒐𝒖𝒕 = 𝟑𝟎 𝒙 𝟕𝟒𝟔 = 𝟐𝟐. 𝟑𝟖 𝒌𝑾
𝜼=
𝑷𝒐𝒖𝒕
𝟐𝟐. 𝟐𝟑
= 𝟎. 𝟖𝟕 =
𝑷𝒊𝒏𝒑𝒖𝒕
𝑷𝒊𝒏𝒑𝒖𝒕
𝑷𝒊𝒏𝒑𝒖𝒕 = 𝟐𝟓. 𝟕𝟐𝟒 𝒌𝑾
b) 𝑰𝑳 =
𝑷𝒊𝒏𝒑𝒖𝒕
𝑽𝒕
=
𝟐𝟓𝟕𝟐𝟒
𝟐𝟑𝟎
= 𝟏𝟏𝟏. 𝟖𝟒 𝑨
3) A separately excited motor runs at 1045rpm with a constant field current. The armature current
is 50 A at 120 V. The armature resistance is 0.1 Ω if the load on the motor changes such that
it now takes 95 A at 120 V, determine the motor speed at this load.
Solution:
𝑨𝒕 𝒍𝒐𝒂𝒅 𝟏:
𝑵𝟏 = 𝟏𝟎𝟒𝟓 𝒓𝒑𝒎
𝑬𝟏 = 𝑽𝒕 − 𝑰𝒂𝟐 𝑹𝒂 = 𝟏𝟐𝟎 − 𝟓𝟎 ∗ 𝟎. 𝟏 = 𝟏𝟏𝟓 𝒗
𝑨𝒕 𝒍𝒐𝒂𝒅 𝟐:
𝑬𝟐 = 𝑽𝒕 − 𝑰𝒂𝟐 𝑹 − 𝒂 = 𝟏𝟐𝟎 − 𝟗𝟓 ∗ 𝟎. 𝟏 = 𝟏𝟏𝟎. 𝟓 𝒗
𝑰𝒇𝟐 𝑵𝟐
𝑬𝟐
=(
)
𝑬𝟏
𝑰𝒇𝟏 𝑵𝟏
𝟏𝟏𝟎. 𝟓
𝑵𝟐
=
𝟏𝟏𝟓
𝟏𝟎𝟒𝟓
𝑵𝟐 = 𝟏𝟎𝟎𝟒. 𝟏 𝒓𝒑𝒎
Good Luck 😊
Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
4) The speed of 500 V shunt motor is to be raised from 700 rpm to 1000 rpm by field weakening
while the total torque remained unchanged. The armature and the shunt field resistance are 0.8
Ω and 750 Ω, respectively. If the supply current at 700 rpm is 12 A, calculate the additional
shunt field resistance required.
Solution:
𝑪𝒂𝒔𝒆 𝟏:
𝑵𝟏 = 𝟕𝟎𝟎 𝒓𝒑𝒎
𝑰𝒇𝒍 =
𝑽𝒕 𝟓𝟎𝟎
=
= 𝟎. 𝟔𝟕 𝑨
𝑹𝒇 𝟕𝟓𝟎
𝑰𝒂𝟏 = 𝑰𝑳𝟏 − 𝑰𝒇𝟏 = 𝟏𝟐 − 𝟎. 𝟔𝟕 = 𝟏𝟏. 𝟑𝟑 𝑨
𝑬𝟏 = 𝑽𝒕 − 𝑰𝒂𝟏 𝑹𝒂 = 𝟓𝟎𝟎 − 𝟏𝟏. 𝟑𝟑 𝒙 𝟎. 𝟖 = 𝟒𝟗𝟎. 𝟗𝟑 𝒗
𝑪𝒂𝒔𝒆 𝟐 ∶
𝑵𝟐 = 𝟏𝟎𝟎𝟎 𝒓𝒑𝒎
𝑻𝟐 𝑰𝒇𝟐 𝑰𝒂𝟐
=
𝑻𝟏 𝑰𝒇𝟏 𝑰𝒂𝟏
𝑰𝒇𝟐 𝑰𝒂𝟐
𝟕. 𝟓𝟗
= 𝟏 → ∴ 𝑰𝒂𝟐 =
→ (𝟏)
𝟎. 𝟔𝟕 𝟏𝟏. 𝟑𝟑
𝑰𝒇𝟐
𝑰𝒇𝟐 𝟏𝟎𝟎𝟎
𝑬𝟐 𝑰𝒇𝟐 𝑵𝟐
𝑬𝟐
=
→
=
𝒙
→ 𝑬𝟐 = 𝟏𝟎𝟒𝟔. 𝟕𝟔 𝑰𝒇𝟐 → (𝟐)
𝑬𝟏 𝑰𝒇𝟏 𝑵𝟏
𝟒𝟗𝟎. 𝟗𝟑 𝟎. 𝟔𝟕 𝟕𝟎𝟎
𝑬𝟐 = 𝑽𝒕 − 𝑰𝒂𝟐 𝑹𝒂 → 𝟏𝟎𝟒𝟔. 𝟕𝟔𝟏 𝑰𝒇𝟐 = 𝟓𝟎𝟎 −
𝟕. 𝟓𝟗
𝒙𝟎. 𝟖 → (𝟑)
𝑰𝒇𝟐
𝒇𝒓𝒐𝒎 (𝟑) ∶
𝑰𝟐𝒇𝟐 − 𝟎. 𝟒𝟕𝟕𝟔 𝑰𝒇𝟐 + 𝟓. 𝟖 𝒙 𝟏𝟎−𝟑 = 𝟎
∴ 𝑰𝒇𝟐 = 𝟎. 𝟒𝟔𝟓𝟏 𝑨 (𝒂𝒄𝒄𝒆𝒑𝒕𝒆𝒅 ) 𝒐𝒓 𝑰𝒇𝟐 = 𝟎. 𝟎𝟏𝟐 𝑨 (𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅)
𝑰𝒇𝟐 =
𝑽𝒕
𝟓𝟎𝟎
→ 𝟎. 𝟒𝟔𝟓𝟏 =
→ 𝑹𝑬𝒙𝒕 = 325 Ω
𝑹𝒇 + 𝑹𝒆𝒙𝒕
𝟕𝟓𝟎 + 𝑹𝒆𝒙𝒕
Good Luck 😊
Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
5) A 500 V, 10 HP, shunt motor has full load efficiency of 85%. For the same torque, an additional
armature resistance is inserted to reduce the motor’s speed by 30%. The resistance of the field
and armature are 400 Ω and 0.25 Ω. Assuming that the variation of all power losses except
copper losses is directly proportional to the speed, calculate the value of the inserted resistance
and the efficiency of the motor when running at the reduced speed.
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
𝑪𝒂𝒔𝒆 𝟏:
𝑰𝒇𝟏 =
𝑽𝒕 𝟓𝟎𝟎
=
= 𝟏. 𝟐𝟓 𝑨
𝑹𝒇 𝟒𝟎𝟎
𝑷𝒊𝒏 =
𝑷𝒐𝒖𝒕 𝟏𝟎𝒙𝟕𝟒𝟔
=
= 𝟖𝟕𝟕𝟔. 𝟒𝟕 𝑾
𝜼
𝟎. 𝟖𝟓
𝑰𝑳𝟏 =
𝑷𝒊𝒏 𝟖𝟕𝟕𝟔. 𝟒𝟕
=
= 𝟏𝟕. 𝟓𝟓 𝑨
𝑽𝒕
𝟓𝟎𝟎
𝑰𝒂𝟏 = 𝑰𝑳𝟏 − 𝑰𝒇𝟏 = 𝟏𝟕. 𝟓𝟓 − 𝟏. 𝟐𝟓 = 𝟏𝟔. 𝟑 𝑨
𝑬𝟏 = 𝑽𝒕 − 𝑰𝒂𝟏 𝑹𝒂 = 𝟓𝟎𝟎 − 𝟏𝟔. 𝟑𝒙𝟎. 𝟐𝟓 = 𝟒𝟗𝟓. 𝟗𝟑 𝒗
𝑷𝒄𝒖𝟏 = 𝑰𝟐𝒂𝟏 𝑹𝒂 + 𝑰𝟐𝒇𝟏 𝑹𝒇 = (𝟏𝟔. 𝟑)𝟐 𝒙 𝟎. 𝟐𝟓 + (𝟏. 𝟐𝟓)𝟐 𝒙 𝟒𝟎𝟎 = 𝟔𝟗𝟏. 𝟒𝟐 𝑾
𝑷𝒄𝒐𝒓𝒆𝟏+𝒎𝒆𝒄𝒉𝟏 = 𝑷𝒊𝒏 − 𝑷𝒄𝒖𝟏 − 𝑷𝒐𝒖𝒕 = 𝟖𝟕𝟕𝟔. 𝟒𝟕 − 𝟔𝟗𝟏. 𝟒𝟐 − 𝟕𝟒𝟔𝟎 = 𝟔𝟐𝟓 𝑾
𝑪𝒂𝒔𝒆 𝟐:
𝑰𝒇𝟐 =
𝑽𝒕 𝟓𝟎𝟎
=
= 𝟏. 𝟐𝟓 𝑨
𝑹𝒇 𝟒𝟎𝟎
𝑬𝟐 𝑰𝒇𝟐 𝑵𝟐
𝑬𝟐
=
→
= 𝟏 𝒙 𝟎. 𝟕 → 𝑬𝟐 = 𝟑𝟒𝟕. 𝟏𝟓𝟏 𝒗
𝑬𝟏 𝑰𝒇𝟏 𝑵𝟏
𝟒𝟗𝟓. 𝟗𝟑
𝑻𝟐 𝑰𝒇𝟐 𝑰𝒂𝟐
𝑰𝒂𝟐
=
→ 𝟏 = 𝟏𝒙
→ ∴ 𝑰𝒂𝟐 = 𝟏𝟔. 𝟑 𝑨
𝑻𝟏 𝑰𝒇𝟏 𝑰𝒂𝟏
𝟏𝟔. 𝟑
𝑬𝟐 = 𝑽𝒕 − 𝑰𝒂𝟐 (𝑹𝒂 + 𝑹𝒆𝒙𝒕 ) → 𝟑𝟒𝟕. 𝟏𝟓𝟏 = 𝟓𝟎𝟎 − 𝟏𝟔. 𝟑 𝒙(𝟎. 𝟐𝟓 + 𝑹𝒆𝒙𝒕 )
𝑹𝒆𝒙𝒕 = 𝟗. 𝟏𝟐𝟕 𝛀
Good Luck 😊
Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
𝑷𝒄𝒖𝟐 = 𝑰𝟐𝒂𝟐 (𝑹𝒂 + 𝑹𝒆𝒙𝒕 ) + 𝑰𝟐𝒇𝟐 𝑹𝒇 = (𝟏𝟔. 𝟑)𝟐 𝒙 (𝟎. 𝟐𝟓 + 𝟗. 𝟏𝟐𝟕) + (𝟏. 𝟐𝟓)𝟐 𝒙𝟒𝟎𝟎
= 𝟑𝟏𝟏𝟔. 𝟒 𝑾
𝑷𝒄𝒐𝒓𝒆+𝒎𝒆𝒄𝒉 𝟐 =
𝑵𝟐
𝑷
= 𝟎. 𝟕 𝒙 𝟔𝟐𝟓 = 𝟒𝟑𝟕. 𝟓 𝑾
𝑵𝟏 𝒄𝒐𝒓𝒆+𝒎𝒆𝒄𝒉 𝟏
𝑷𝒊𝒏 = 𝑽𝒕 𝑰𝒂𝟐 = 𝟓𝟎𝟎 𝒙 𝟏𝟔. 𝟑 = 𝟖𝟕𝟕𝟔. 𝟒𝟕 𝑾
𝑷𝒐𝒖𝒕 = 𝑷𝒊𝒏 − 𝑷𝒄𝒖𝟐 − 𝑷𝒄𝒐𝒓𝒆+𝒎𝒆𝒄𝒉 𝟐 = 𝟖𝟕𝟕𝟔. 𝟒𝟕 − 𝟑𝟏𝟏𝟔. 𝟒 − 𝟒𝟑𝟕. 𝟓 = 𝟓𝟐𝟐𝟐. 𝟓𝟕 𝑾
𝜼=
𝑷𝒐𝒖𝒕 𝟓𝟐𝟐𝟐. 𝟓𝟕
=
= 𝟓𝟗. 𝟓 %
𝑷𝒊𝒏
𝟖𝟕𝟕𝟔. 𝟒𝟕
Good Luck 😊
Faculty of Electrical Engineering
Electric Machines & Drives (ELCT403)
Dr. Eng. Noha Shouman
Eng. Omar Ghazal
6) A separately excited DC motor has the following specifications:
Terminal voltage
Field voltage
Armature resistance
Field resistance
250 V
250 V
0.03 Ω
250 Ω
Initially the motor was running at a speed of 1103 rpm while supplied by the rated terminal
voltage. Assume the torque is kept constant and the armature current is 120 A. Find the speed
of the motor if the terminal voltage is reduced to 200 V.
Good Luck 😊