Page 205 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

Solution
Efficiency at no-load?

Stray losses = mechanical losses + iron losses = 700 W
Total constant losses at no-load=P stray + P Sh, Cu = 820 W
Efficiency at no-load will be zero.
Efficiency at 25 % Full - load?
Constant losses = 820 W
Arm. Cu. loss at 25 % of full-load = (25/100) x 600 = 37.5 W
2
Total losses = constant losses + armature copper loss = 857.5 W
Output at 25 % of full-load = 25 % of 20 KW = 5 KW
Total input = Output + total Losses = 5 KW + 857.5 W = 5857.5 W
Efficiency = output x 100 / input = 5000 x 100 / 5857.5 = 85.4 %

Problem # 2
A shunt generator delivers 195 A at terminal voltage of 200 V. The armature resistance and shunt field resistance
are 0.03 ohms and 40 ohms respectively. The iron and friction losses (stray losses) are equal to 900 watts.
Calculate the following:
• EMF generated?
• Copper losses?
• Output of the prime mover?

• Commercial efficiency?
• Mechanical efficiency?
• Electrical efficiency?
• EMF generated?
I SH = V / R f = 200 / 40 = 5 A

I a = Load current + field current
= 195 A + 5 A = 200 A
Armature voltage drop = I a Ra = 200 x 0.03 = 6 V
EMF generated = V + Ia Ra = 200 + 6 = 206 V

• Copper loss?
Armature copper losses = I aR a
2
= 200 x 200 x 0.03 = 1200 W
Shunt copper loss = P SH = V I SH = 200 x 5 = 1000 W
{P SH = I 2 R = 1000 W}
SH SH
Total copper loss = 1200 + 1000 = 2200 W

• Output of the prime mover?
Total losses = Total copper losses + stray losses
= 2200 W + 900 W
= 3100 W

Load Output = V I = 200 x 195
L
= 39000 W

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CITS : Power - Electrician & Wireman - Lesson 30-37