WHY CIRCLE DIAGRAM IS NOT DRAWN FOR 1-PHASE INDUCTION MOTORS

1. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 1
CONTENTS:
SL NO TITTLE PAGE NO.
01 INTRODUCTION 03
02 APPARATUS REQUIRED 04
03 PROCEDURE 05
04 CIRCUIT DIAGRAM 06
05 TABULAR COLUMN 07
06 CALCULATION 11
07 CONCLUSION 13

2. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 2
AIM: carry out an experiment to study the performance parameters of a single
phase induction motor by conducting no load and blocked rotor test ( circle
diagram ). Draw the conclusion based on observation made.
INTRODUCTION
NO LOAD TEST
In this test the motor is made to run without an load i.e no load condition. The
speed of the motor is very close to the synchronous speed but less than
synchronous speed. The rated voltage is applied to the stator . The input line
current and total input power is measured.The two wattmeter method is used to
measure the total input power in case of three phase induction motor but as in
our casesingle phase induction motor input power can be measured using single
wattmeter. As the motor is on no load the power factor is very low which is less
than
BLOCKED ROTOR TEST
In block rotor test the rotor is held fixed so that it will not rotate. A reduced
voltage is applied to limit the shortcircuit.This voltage is adjusted with help of
autotransformer so that the rated currentflows through the main winding.
The input voltage , currentand power are measured by connecting voltmeter ,
ammeter and wattmeter respectively. These readings are denoted as Vsc , Isc
and Wsc.

3. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 3
System under consideration: single phase capacitor start induction motor of
rating 750 watts/1H.P ,230v ,7.8 amps ,1425 rpm .
APPARATUS REQUIRED:
SL.NO APPARATUS/INSTRUMENTS/NOS RANGE/TYPE
1 1-phase dimestat 230v/10A
2 Voltmeter
0-150V (MI)
0-300V (MI)
3 Ammeter
0-10A (MI)
0-5A (MI)
4 Wattmeter (1 nos)
(1 nos)
300V/5A LPF
150V/10A UPF

4. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 4
CIRCUIT DIAGRAM FOR NO LOAD TEST:
PROCEDURE:- NO LOAD TEST
1. Make connection as per circuit diagram as shown in Fig 1.
2. Switch on the supply and vary the dimmestat till a rated voltage (230V )
is applied.
3. Take all the meter reading and switch off the supply.

5. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 5
CIRCUIT DIAGRAM FOR BLOCKED ROTOR TEST:
PROCEDURE:-BLOCKED ROTOR TEST
1. Make connection as per circuit diagram as shown in Fig 2.
2. Hold the rotorshaft firmly i.e block the rotor.
3. Switch on the supply and vary the dimmestat till a rated current (7.8A)
flows.
4. Take all the meter reading and switch off the supply.

6. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 6
CIRCUIT DIAGRAM FOR LOAD TEST:
PROCEDURE:- LOAD TEST
1. Make connection as per circuit diagram as shown in Fig 3.
2. Switch on supply and vary the dimmestat till rated voltage (230V) is
applied.
3. Apply the brake load in step and note down the reading of voltmeter ,
ammeter , wattmeter , speed and tangential forces (S1 and S2) etc.
4. Switch off the supply.

7. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 7
TABULAR COLUMN
NO LOAD TEST
Voltmeter (V0)
volts
Ammeter (I0) amps Wattmeter (W0)
watts
230 5.8 74
Wattmeter constant is 4.
BLOCKED ROTOR TEST
Voltmeter
(Vsc) volts
Ammeter (Isc) amps Wattmeter (Wsc) watts
59 7.8 150
Wattmeter constant is 2.

8. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 8
LOAD TEST
Radius of brake drum (r) =0.07m
SL.NO Voltmeter
(V) volts
Ammeter
(I) amps
Wattmeter
(W) watts
Speed
(N)rpm
Load
(kg)
(S1) (S2)
1 212 4.5 45 1493 0 0
2 212 5.3 150 1464 6 1.5
3 212 7 210 1430 8 2
4 212 7.9 310 1412 12 6
Wattmeter constant is 4.

9. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 9
CALCULATION:- (No Load and Block RotorTest)
The circle diagram is drawn by following the below given procedure.
Step1:calculate the No load power factor angle by equation.
𝜃𝑜 = cos−1
[
𝑊𝑜
𝑉𝑜 × 𝐼𝑜
]
𝜃𝑜 = cos−1
[
296
230× 5.8
] = 77.17°
And draw a no load current line OA (Choosecurrent scale) at angle𝜃𝑜
measured from vertical axis.
Step2: Calculate the short circuit power factor angle by equation.
𝜃𝑠𝑐 = cos−1
[
𝑊𝑠𝑐
𝑉𝑠𝑐 × 𝐼𝑠𝑐
]
𝜃𝑠𝑐 = cos−1
[
300
59 × 7.8
] = 49.31°
Calculate normal short circuit current by equation
𝐼𝑠𝑐𝑛 = (
𝑉𝑁
𝑉𝑠𝑐
) × 𝐼𝑠𝑐
𝐼𝑠𝑐 = (
230
59
) × 7.8 = 30.40𝐴
Where, VN = 230V
And draw a normal short circuit line OB at an angle 𝜃𝑠𝑐 measured from the
vertical axis.
Step3: Join A and B called as output line and divide the line and locate point C.
Step4: With C as center point and AC as radius draw a semicircle (point B also
lies on the semicircle).
Step5: Draw a vertical line from point B and locate the point D and E.

10. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 10
Step6: Divide the line BD (assuming equal rotor and stator copperloss)and
locate the point F. Then join AF (Torque line).
𝐴𝐹 = 2 × 9.8 = 19.6
Step7: Calculate the power scale as below,
Measure BE
BE=10.2 cm
Calculate Normal short circuit loss Pscn= (𝑉𝑁/𝑉𝑠𝑐)2
× (Wsc)
𝑃𝑠𝑐𝑛 = (
230
59
)2
× (300) = 4559.03𝑤𝑎𝑡𝑡
Then power scale 𝑥 =
𝑃𝑠𝑐𝑛
𝐵𝐸
(w/cm)
𝑥 =
4559.03
10.2
= 446.96(
𝑤
𝑐𝑚
)
Step8: Draw output power line GB of length calculated as,
𝐺𝐵 = 𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 (𝑤𝑎𝑡𝑡𝑠)/𝑥
𝐺𝐵 =
746
446.96
= 1.7 𝑐𝑚
Where full load output power (Pm) = 746 watts
Step9: Draw a line HG parallel to AB (output line), intersecting the semicircle
at point H.
Step10: Draw vertical line from point H, intersecting the output line,torque
line,horizontal line from A and x axis at point I,J,K,and L respectively. Also
draw a line OH.
Step11: Performance parameters at full load are calculated as below,
𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑃𝑖 = 𝐻𝐿 × 𝑥
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑃𝑖 = 3 × 446.96 = 1340.88 𝑤𝑎𝑡𝑡

11. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 11
𝑃𝑚 = 𝐻𝐼 × 𝑥 = 2.1 × 446.96 = 938.61 𝑤𝑎𝑡𝑡
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 ƞ =
𝑃𝑚
𝑃𝑖
× 100
Ƞ = (
938.61
1340.88
)× 100 = 70.04%
𝐴𝑖𝑟 𝑔𝑎𝑝 𝑝𝑜𝑤𝑒𝑟 𝑃𝑔 = 𝐻𝐽 × 𝑥
𝑃𝑔 = 2.3 × 446.96 = 1028.008 𝑤𝑎𝑡𝑡
𝑅𝑜𝑡𝑜𝑟 𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠 𝑃𝑐𝑟 = 𝐼𝐽 × 𝑥
𝑃𝑐𝑟 = 0.2 × 446.96 = 89.392 𝑤𝑎𝑡𝑡
𝑆𝑙𝑖𝑝 𝑠 =
𝑃𝑐𝑟
𝑃𝑔
𝑆 =
89.392
1028.008
× 100 = 8.69%
𝑆𝑡𝑎𝑡𝑜𝑟 𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠 𝑃𝑐𝑠 = 𝐽𝐾 × 𝑥
𝑃𝑐𝑠 = 0.2 × 446.96 = 89.392 𝑤𝑎𝑡𝑡
𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑙𝑜𝑠𝑠 = 𝐾𝐿 × 𝑥
𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑙𝑜𝑠𝑠 = 0.6× 446.96 = 268.17 𝑤𝑎𝑡𝑡
𝑆𝑡𝑎𝑡𝑜𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 = 𝑂𝐻 × 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑠𝑐𝑎𝑙𝑒
𝐼 = 4.5 × 2 = 9𝐴
Input power factor pf = cos (angle between vertical axis and line OH)
𝑝𝑓 = cos(47°) = 0.681

12. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 12
CIRCLE DIAGRAM:
CALCULATION:- ( Load Test)
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑃𝑖 =
𝑊
735.5
𝑃𝑖 =
1240
735.5
= 1.685𝑤𝑎𝑡𝑡
𝑇𝑜𝑟𝑞𝑢𝑒 𝑇 = 9.81× ( 𝑆1 − 𝑆2) × 𝑟

13. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 13
𝑇 = 9.81× (12− 6) × 0.07 = 4.120 𝑁𝑚
𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑃𝑜 =
2𝜋𝑁𝑇
60× 735.5
𝑃𝑜 =
36542.05
44130
= 0.828𝑤𝑎𝑡𝑡
%𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦( ƞ ) = (
𝑃𝑜
𝑃𝑖
) × 100
ƞ = (
0.828
1.685
) × 100 = 49.13%
𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑝𝑓 =
𝑊
𝑉𝐼
𝑝𝑓 =
1240
1674.8
= 0.74
𝑆 =
𝑁𝑠 − 𝑁
𝑁𝑠
× 100 =
1500 − 1429
1500
× 100 = 4.73%
CALCLULATION TABLE
Test Torque
(Nm)
Input
power(W)
Out put
power(W)
Efficiency Slip
Circle
diagram
19.6 1340.88 938.61 70.04% 8.69%
Load
test
4.12 1.685 0.828 49.13% 4.73%

14. CIRCLE DIAGRAM ON 1-PHASE INDUCTION MOTOR
Dept of EEE Page 14
CONCLUSION
o We obtained the efficiency (ƞ) of single phase induction motor
from circle diagram is 70.04% where as, from full load test it is
49% hence from circle diagram method, we cannot obtain accurate
efficiency for single induction motor.
o Because normally torque for single phase induction motor is less
for load test, where as we obtained from circle diagram is very high
hence the efficiency is high in case of circle diagram compare to
load test. As torque is directly proportional to output power.
o So generally circle diagram is not drawn for single phase induction
motor to find efficiency.