Page 236 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

Subtract no-load armature Cu loss from the total input,then we get constant losses. Constant losses Wc= total
input - no-load armature Cu loss = VI0 − (I − I ) Ra
2
0 sh
From the constant losses of the machine, its efficiency at any other load can be determined
Eb
Let / = load current at which efficiency is required.
α N
Φ
Then, armature current is la = I - Ish...if machine is motoring
Eb V Ia Ra
= I + Ish ...if machine is generating
V Ia Ra
α N
Efficiency when running as a motor
Φ
Input = VI, Armature Cu loss = la²Ra = (1 - Ish) Ra
2
Constant losses = W c α N 1
2 Φ
Total losses = (1 - Ish) Ra + W c
output Input Losses
Efficiency
Input Input

VI [(I Ish) 2 Ra Wc]

VI

Efficiency when running as a generator
Let / = load current at which efficiency is required.
Then, armature current is la = I - Ish ...if machine is motoring
= I + Ish ...if machine is generating
Output = VI, Armature Cu loss = la²Ra = (1 + Ish) Ra
2
Constant losses = W
c
Total losses = [(1 + Ish) Ra + Wc]
2
Output
Efficiency
Output Losses

VI
VI [(1 I sh ) 2 R a W c

Advantages of swinburne’s test
• Convenient and economical

• Power required to test a large machine only the no-load input power.
• The efficiency can be predetermined at any load because constant-losses are known.
Disadvantages of swinburne’s test
• Armature reaction causes flux distortion

• In some cases, by as much as 50%.
• This is not taken care by Swinburne’s test
• Impossible to know whether full-load commutation would be satisfactory
• Temperature rise properties at full load not k

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