DC MACHINES

Swinburne’s Test:

Figure 4.2

This test is a no load test and hence cannot be performed on series motor. The circuit connection is shown in Figure 4.2. The machine is run on no load at rated speed which is adjusted by the shunt field resistance.

ADVANTAGES

1. Economical, because no load input power is sufficient to perform the test

2. Efficiency can be pre-determined

3. As it is a no load test, it cannot be done on a dc series motor DISADVANTAGES

1. Change in iron loss from no load to full load is not taken into account. (Because of armature reaction, flux is distorted which increases iron losses).

2. Stray load loss cannot be determined by this test and hence efficiency is over estimated.

3. Temperature rise of the machine cannot be determined.

4. The test does not indicate whether commutation would be satisfactory when the machine is loaded.

IO = No load current; Ish = shunt field current

Iao = No load armature current = (Io - Ish)

V= Supply Voltage

DC MACHINES

No load input =VIo watts.

No load power input supplies

(i) Iron losses in the core

(ii) Friction and windings loss and (iii) Armature copper loss.

Let I = load current at which efficiency is required

Ia = I – Ish if machine is motoring; I + Ish if machine is generating Efficiency as a motor:

Input = VI; Ia2ra = (I- Ish)2ra

Constant losses Wc = VIo – (Io – Ish)2ra …………… 7

Total losses = (I- Ish)2ra + Wc

"'( )* – *"*,-./012 Therefore efficiency of motor = ; = ……. 8

%&

EFFICIENCY OF A GENERATOR:

Output = VI

Ia2ra = (I+Ish)2ra

Total losses = Wc+(I+Ish)2ra …………………………………….. 9

%&

Efficiency of generator = 0'( = %&0&0&34 #560 7 ……… 10

DC MACHINES Prof. Vasudevamurthy, Dr. AIT

1.A 220 V DC shunt motor at No-load takes a current of 2.5 A. the resistance of the armature and shunt field are 0.8Ω and 200Ω respectively. Estimate the efficiency of the motor when the input current is 32 A.

SOLUTION: No-load input = 220 x 2.5 = 500 W

8 = = 1.1 A;8:= 2.5 − 1.1 = 1.4 >;8:?: = 1.4) X 0.8 = 1.6 W

Constant losses = 550 – 1.6 = 548.4 W

When input current is 32 A, Ia = 32 – 1.1 = 30.9 A

8: ?: = 30.9) 0.8 = 764 ; Total losses = 764 + 548.4 = 1315 W

Input = 220 X 32

Therefore output = (220 X 32) – 1315 W

Therefore η = "'( = 81.3 %

2.A 440V D.C Shunt motor takes no load current of 25A. The resistance of shunt field and armature are550 Ω& 1.2Ω respectively. The full load line current is 32A. Determine the full load output &efficiency of the motor. SOLUTION:input on no load= VI=440×2.5=1100W8 = $

%34 = ZZ9

[[9 = 0.8A;8 = 8 + 8:, therefore, 8: =

8 − 8 = 2.5-0.8=1.7A ef gfh ?i?j ki gfll = 8:?:

= (1.7)2×1.2=3.468W

kfele gfll = meni fe ef gfh − efgfh ?i?j ki gfll= 1100-3.468=1096.532W8o=32A, 8: = 32-0.8=31.2A; Armature copper loss =

(31.2) 1.2 = 1168.128Total losses = 1168.128 + 1096.532 =2264.66Wη = tuvwx"yz,,,tuvwx = (ZZ9×|(ZZ9×|)"Z.) × 100 = 83.9 %

2. A 500 V DC shunt motor when running on No-load takes 5 A. Armature resistance is 0.5Ω and shunt field resistance is 250Ω. Find the output in kW and η of the motor when running on full load and taking a current of 50 A.