boss_notes/T-18_Circle_Diagram.md

7.7 KB · commit d2f3b6e

← T-17b: Blocked Rotor Test | 🏠 Index | T-19: Starting Methods (3-Phase) →

T-18: Circle Diagram

Section: B | Priority: 🟠 HIGH | Exam Frequency: 4/7 years Sources: Theraja Ch-35 (Art. 35.9-35.14), Slides L-07

Why This Topic Matters

Circle diagram problems appeared in 4 out of 7 papers (2017, 2018, 2019, 2021). Each problem is worth 6-9 marks. The trend is declining (absent in 2023, 2024), but it remains in syllabus and can return. The construction is mechanical: once you know the steps, it is guaranteed marks.

📝 Key Definitions

Circle Diagram: "The locus of the stator current phasor of an induction motor, as the load changes from no-load to blocked rotor, is approximately a semicircle. This semicircle, called the circle diagram, can be used to predict the performance of the motor at any load." — Theraja, Art. 35.9

Data Required

You need two test results:

Test	Measurements	What It Gives
No-Load	$V_{0}$ , $I_{0}$ , $P_{0}$ (or $cos ϕ_{0}$ )	No-load point on circle
Blocked Rotor	$V_{sc}$ , $I_{sc}$ , $P_{sc}$ (or $cos ϕ_{sc}$ )	Short-circuit point on circle

Both must be referred to the same (rated) voltage.

Construction Steps

Step 1: Scale BR data to full voltage

If BR test was at reduced voltage $V_{sc}$ :

$I_{sc, full} = I_{sc} \times \frac{V _{rated}}{V _{sc}}$

Power factor stays the same: $cos ϕ_{sc} = P_{sc} / (3 V_{sc} I_{sc})$

Step 2: Calculate angles

$ϕ_{0} = cos^{- 1} (\frac{P _{0}}{3 V _{0} I _{0}})$

$ϕ_{sc} = cos^{- 1} (cos ϕ_{sc})$

Step 3: Plot points

On a graph with horizontal axis = active current (watts component) and vertical axis = reactive current:

No-load point $O^{'}$ :

$O_{x}^{'} = I_{0} cos ϕ_{0}$ , $O_{y}^{'} = I_{0} sin ϕ_{0}$

Short-circuit point $S$ :

$S_{x} = I_{sc, full} cos ϕ_{sc}$ , $S_{y} = I_{sc, full} sin ϕ_{sc}$

Step 4: Draw the circle

Draw a semicircle through $O^{'}$ and $S$ . The center lies on a horizontal line through $O^{'}$ .

Step 5: Draw the power and torque lines

Output line: Horizontal line from $O^{'}$ (no-load point)
Torque line (air-gap line): Connects $O^{'}$ to a point on the $S$ -line divided by the rotor/total Cu loss ratio.
The rotor Cu loss line divides the SC intercept in the ratio of rotor Cu loss to total Cu loss at standstill.

If rotor Cu loss = stator Cu loss at standstill (equal division), the line bisects the vertical intercept.

Step 6: Read performance

For any operating point on the circle:

Vertical distance below power line = output power
Vertical distance below torque line = developed torque (in sync watts)
Power factor = $cos$ of angle between current vector and voltage reference

Circle diagram construction

What the Circle Diagram Can Predict

Quantity	How to Read
Line current	Length of current vector from origin to operating point
Power factor	Cosine of angle between current vector and horizontal
Output power	Vertical height above output line, times voltage scale
Max output	Longest vertical intercept below output line
Max torque	Longest vertical intercept below torque line
Slip	Ratio of rotor Cu loss segment to total air-gap segment
Efficiency	Output segment / input segment

Circle diagram showing maximum quantities

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Circle diagram: 415V, delta, 50 Hz motor. NL: 415V, 21A, 1250W. BR: 100V, 45A, 2730W. Find line current and pf at rated output.

Appeared: 2017 Q3(b), 2019 Q6(a) — (9 marks)

Full Answer:

No-load data: $V = 415$ V (delta), $I_{0} = 21$ A (line), $P_{0} = 1250$ W

$cos ϕ_{0} = \frac{1250}{3 \times 415 \times 21} = \frac{1250}{15094} = 0.0828, ϕ_{0} = 85.25°$

Blocked rotor (scaled to 415V):

$I_{sc, full} = 45 \times 415/100 = 186.75 A$

$cos ϕ_{sc} = \frac{2730}{3 \times 100 \times 45} = 0.35, ϕ_{sc} = 69.5°$

Plot points $O^{'}$ and $S$ on the circle diagram. Draw the semicircle. Locate the operating point for rated output (29.84 kW).

Power scale: $3 \times 415 = 718.7$ W per ampere of active component.

From diagram: Line current at rated output $\approx 58$ A. Power factor $\approx 0.714$ lagging.

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, $cos ϕ_{0} = 0.087$ . BR: 100V, 12A, 720W. $R_{1} = 0.67 Ω$ , $R_{2} = 0.185 Ω$ . Find full-load current, slip, pf, max power.

Appeared: 2018 Q7(b) — (8 marks)

Full Answer:

Scale BR to full voltage:

$I_{sc} = 12 \times 400/100 = 48$ A

$cos ϕ_{sc} = 720/ (3 \times 100 \times 12) = 0.347$ , $ϕ_{sc} = 69.7°$

No-load point: $I_{0 x} = 6 \times 0.087 = 0.52$ A, $I_{0 y} = 5.98$ A

SC point: $I_{sc x} = 48 \times 0.347 = 16.66$ A, $I_{scy} = 45.02$ A

Rotor Cu loss line: Total SC Cu loss at full voltage $= 720 \times (400/100)^{2} = 11520$ W.

Stator Cu loss $= 3 \times 4 8^{2} \times 0.67 = 4631$ W. Rotor Cu loss $= 11520 - 4631 = 6889$ W.

Ratio: $6889/11520 = 0.598$ . The rotor Cu loss line divides the SC intercept at 59.8% from the power line.

From diagram: Full-load current $\approx 10.5$ A, slip $\approx 6.2%$ , pf $\approx 0.78$ lagging, max power $\approx 8.2$ kW.

🎯 Q3: Circle diagram: 14.92 kW, 400V, 6-pole. NL: 11A, pf=0.2. SC: 100V, 25A, pf=0.4. Rotor Cu = half total Cu at standstill.

Appeared: 2021 Q8(c) — (6 marks)

Full Answer:

No-load: $I_{0 x} = 11 \times 0.2 = 2.2$ A, $I_{0 y} = 10.78$ A

SC (scaled): $I_{sc} = 25 \times 400/100 = 100$ A, $cos ϕ_{sc} = 0.4$

$I_{sc x} = 40$ A, $I_{scy} = 91.65$ A

Plot $O^{'}$ and $S$ , draw semicircle. Since rotor Cu = stator Cu at standstill, the rotor Cu line bisects the SC intercept at 50%.

$N_{s} = 120 \times 50/6 = 1000$ rpm. Power scale: $3 \times 400 = 692.8$ W/A.

For 14.92 kW rated output, locate operating point:

From diagram: Line current $\approx 30$ A, slip $\approx 5%$ , efficiency $\approx 84%$ , pf $\approx 0.76$ lagging.

Maximum torque = longest vertical intercept below torque line (read from diagram in sync watts).

Exam Variants

Year	Question	Data	Marks
2017 Q3(b)	415V delta, 29.84 kW	NL+BR data	9
2018 Q7(b)	400V, 5.6 kW slip-ring	NL+BR+R1+R2	8
2019 Q6(a)	Same as 2017	Same problem	9
2021 Q8(c)	400V, 14.92 kW, 6-pole	NL+BR, equal Cu	6

⚡ Exam Tips & Common Mistakes

Always scale BR data to rated voltage. Current scales linearly with voltage: $I_{sc, full} = I_{sc} \times V_{rated} / V_{sc}$ . Power factor stays the same.
Choose a clear current scale. For example: 1 cm = 5 A. Label your scale.
Delta connection: $I_{phase} = I_{line} / 3$ . Star: $I_{phase} = I_{line}$ .
The power axis is horizontal. Active current on x-axis, reactive on y-axis.
The circle diagram is declining in frequency. Absent in 2023 and 2024. But still in syllabus.
Carry a protractor and compass to the exam for this question.

T-17a: No-Load Test — Provides no-load point
T-17b: Blocked Rotor Test — Provides SC point
T-14: IM Equivalent Circuit — Analytical alternative

← T-17b: Blocked Rotor Test | 🏠 Index | T-19: Starting Methods (3-Phase) →

boss_notes/T-18_Circle_Diagram.md

7.7 KB · commit d2f3b6e

← T-17b: Blocked Rotor Test | 🏠 Index | T-19: Starting Methods (3-Phase) →

T-18: Circle Diagram

Section: B | Priority: 🟠 HIGH | Exam Frequency: 4/7 years Sources: Theraja Ch-35 (Art. 35.9-35.14), Slides L-07

Why This Topic Matters

📝 Key Definitions

Circle Diagram: "The locus of the stator current phasor of an induction motor, as the load changes from no-load to blocked rotor, is approximately a semicircle. This semicircle, called the circle diagram, can be used to predict the performance of the motor at any load." — Theraja, Art. 35.9

Data Required

You need two test results:

Test	Measurements	What It Gives
No-Load	$V_{0}$ , $I_{0}$ , $P_{0}$ (or $cos ϕ_{0}$ )	No-load point on circle
Blocked Rotor	$V_{sc}$ , $I_{sc}$ , $P_{sc}$ (or $cos ϕ_{sc}$ )	Short-circuit point on circle

Both must be referred to the same (rated) voltage.

Construction Steps

Step 1: Scale BR data to full voltage

If BR test was at reduced voltage $V_{sc}$ :

$I_{sc, full} = I_{sc} \times \frac{V _{rated}}{V _{sc}}$

Power factor stays the same: $cos ϕ_{sc} = P_{sc} / (3 V_{sc} I_{sc})$

Step 2: Calculate angles

$ϕ_{0} = cos^{- 1} (\frac{P _{0}}{3 V _{0} I _{0}})$

$ϕ_{sc} = cos^{- 1} (cos ϕ_{sc})$

Step 3: Plot points

On a graph with horizontal axis = active current (watts component) and vertical axis = reactive current:

No-load point $O^{'}$ :

$O_{x}^{'} = I_{0} cos ϕ_{0}$ , $O_{y}^{'} = I_{0} sin ϕ_{0}$

Short-circuit point $S$ :

$S_{x} = I_{sc, full} cos ϕ_{sc}$ , $S_{y} = I_{sc, full} sin ϕ_{sc}$

Step 4: Draw the circle

Draw a semicircle through $O^{'}$ and $S$ . The center lies on a horizontal line through $O^{'}$ .

Step 5: Draw the power and torque lines

Output line: Horizontal line from $O^{'}$ (no-load point)
Torque line (air-gap line): Connects $O^{'}$ to a point on the $S$ -line divided by the rotor/total Cu loss ratio.
The rotor Cu loss line divides the SC intercept in the ratio of rotor Cu loss to total Cu loss at standstill.

If rotor Cu loss = stator Cu loss at standstill (equal division), the line bisects the vertical intercept.

Step 6: Read performance

For any operating point on the circle:

Vertical distance below power line = output power
Vertical distance below torque line = developed torque (in sync watts)
Power factor = $cos$ of angle between current vector and voltage reference

Circle diagram construction

What the Circle Diagram Can Predict

Quantity	How to Read
Line current	Length of current vector from origin to operating point
Power factor	Cosine of angle between current vector and horizontal
Output power	Vertical height above output line, times voltage scale
Max output	Longest vertical intercept below output line
Max torque	Longest vertical intercept below torque line
Slip	Ratio of rotor Cu loss segment to total air-gap segment
Efficiency	Output segment / input segment

Circle diagram showing maximum quantities

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Circle diagram: 415V, delta, 50 Hz motor. NL: 415V, 21A, 1250W. BR: 100V, 45A, 2730W. Find line current and pf at rated output.

Appeared: 2017 Q3(b), 2019 Q6(a) — (9 marks)

Full Answer:

No-load data: $V = 415$ V (delta), $I_{0} = 21$ A (line), $P_{0} = 1250$ W

$cos ϕ_{0} = \frac{1250}{3 \times 415 \times 21} = \frac{1250}{15094} = 0.0828, ϕ_{0} = 85.25°$

Blocked rotor (scaled to 415V):

$I_{sc, full} = 45 \times 415/100 = 186.75 A$

$cos ϕ_{sc} = \frac{2730}{3 \times 100 \times 45} = 0.35, ϕ_{sc} = 69.5°$

Plot points $O^{'}$ and $S$ on the circle diagram. Draw the semicircle. Locate the operating point for rated output (29.84 kW).

Power scale: $3 \times 415 = 718.7$ W per ampere of active component.

From diagram: Line current at rated output $\approx 58$ A. Power factor $\approx 0.714$ lagging.

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, $cos ϕ_{0} = 0.087$ . BR: 100V, 12A, 720W. $R_{1} = 0.67 Ω$ , $R_{2} = 0.185 Ω$ . Find full-load current, slip, pf, max power.

Appeared: 2018 Q7(b) — (8 marks)

Full Answer:

Scale BR to full voltage:

$I_{sc} = 12 \times 400/100 = 48$ A

$cos ϕ_{sc} = 720/ (3 \times 100 \times 12) = 0.347$ , $ϕ_{sc} = 69.7°$

No-load point: $I_{0 x} = 6 \times 0.087 = 0.52$ A, $I_{0 y} = 5.98$ A

SC point: $I_{sc x} = 48 \times 0.347 = 16.66$ A, $I_{scy} = 45.02$ A

Rotor Cu loss line: Total SC Cu loss at full voltage $= 720 \times (400/100)^{2} = 11520$ W.

Stator Cu loss $= 3 \times 4 8^{2} \times 0.67 = 4631$ W. Rotor Cu loss $= 11520 - 4631 = 6889$ W.

Ratio: $6889/11520 = 0.598$ . The rotor Cu loss line divides the SC intercept at 59.8% from the power line.

From diagram: Full-load current $\approx 10.5$ A, slip $\approx 6.2%$ , pf $\approx 0.78$ lagging, max power $\approx 8.2$ kW.

🎯 Q3: Circle diagram: 14.92 kW, 400V, 6-pole. NL: 11A, pf=0.2. SC: 100V, 25A, pf=0.4. Rotor Cu = half total Cu at standstill.

Appeared: 2021 Q8(c) — (6 marks)

Full Answer:

No-load: $I_{0 x} = 11 \times 0.2 = 2.2$ A, $I_{0 y} = 10.78$ A

SC (scaled): $I_{sc} = 25 \times 400/100 = 100$ A, $cos ϕ_{sc} = 0.4$

$I_{sc x} = 40$ A, $I_{scy} = 91.65$ A

Plot $O^{'}$ and $S$ , draw semicircle. Since rotor Cu = stator Cu at standstill, the rotor Cu line bisects the SC intercept at 50%.

$N_{s} = 120 \times 50/6 = 1000$ rpm. Power scale: $3 \times 400 = 692.8$ W/A.

For 14.92 kW rated output, locate operating point:

From diagram: Line current $\approx 30$ A, slip $\approx 5%$ , efficiency $\approx 84%$ , pf $\approx 0.76$ lagging.

Maximum torque = longest vertical intercept below torque line (read from diagram in sync watts).

Exam Variants

Year	Question	Data	Marks
2017 Q3(b)	415V delta, 29.84 kW	NL+BR data	9
2018 Q7(b)	400V, 5.6 kW slip-ring	NL+BR+R1+R2	8
2019 Q6(a)	Same as 2017	Same problem	9
2021 Q8(c)	400V, 14.92 kW, 6-pole	NL+BR, equal Cu	6

⚡ Exam Tips & Common Mistakes

Always scale BR data to rated voltage. Current scales linearly with voltage: $I_{sc, full} = I_{sc} \times V_{rated} / V_{sc}$ . Power factor stays the same.
Choose a clear current scale. For example: 1 cm = 5 A. Label your scale.
Delta connection: $I_{phase} = I_{line} / 3$ . Star: $I_{phase} = I_{line}$ .
The power axis is horizontal. Active current on x-axis, reactive on y-axis.
The circle diagram is declining in frequency. Absent in 2023 and 2024. But still in syllabus.
Carry a protractor and compass to the exam for this question.

T-17a: No-Load Test — Provides no-load point
T-17b: Blocked Rotor Test — Provides SC point
T-14: IM Equivalent Circuit — Analytical alternative

← T-17b: Blocked Rotor Test | 🏠 Index | T-19: Starting Methods (3-Phase) →

T-18: Circle Diagram

Why This Topic Matters

📝 Key Definitions

Data Required

Construction Steps

Step 1: Scale BR data to full voltage

Step 2: Calculate angles

Step 3: Plot points

Step 4: Draw the circle

Step 5: Draw the power and torque lines

Step 6: Read performance

What the Circle Diagram Can Predict

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Circle diagram: 415V, delta, 50 Hz motor. NL: 415V, 21A, 1250W. BR: 100V, 45A, 2730W. Find line current and pf at rated output.

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, cosϕ0​=0.087. BR: 100V, 12A, 720W. R1​=0.67Ω, R2​=0.185Ω. Find full-load current, slip, pf, max power.

🎯 Q3: Circle diagram: 14.92 kW, 400V, 6-pole. NL: 11A, pf=0.2. SC: 100V, 25A, pf=0.4. Rotor Cu = half total Cu at standstill.

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

T-18: Circle Diagram

Why This Topic Matters

📝 Key Definitions

Data Required

Construction Steps

Step 1: Scale BR data to full voltage

Step 2: Calculate angles

Step 3: Plot points

Step 4: Draw the circle

Step 5: Draw the power and torque lines

Step 6: Read performance

What the Circle Diagram Can Predict

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Circle diagram: 415V, delta, 50 Hz motor. NL: 415V, 21A, 1250W. BR: 100V, 45A, 2730W. Find line current and pf at rated output.

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, cosϕ0​=0.087. BR: 100V, 12A, 720W. R1​=0.67Ω, R2​=0.185Ω. Find full-load current, slip, pf, max power.

🎯 Q3: Circle diagram: 14.92 kW, 400V, 6-pole. NL: 11A, pf=0.2. SC: 100V, 25A, pf=0.4. Rotor Cu = half total Cu at standstill.

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, $cos ϕ_{0} = 0.087$ . BR: 100V, 12A, 720W. $R_{1} = 0.67 Ω$ , $R_{2} = 0.185 Ω$ . Find full-load current, slip, pf, max power.

🎯 Q2: Circle diagram: 5.6 kW, 400V, 4-pole, 50 Hz slip-ring IM. NL: 6A, $cos ϕ_{0} = 0.087$ . BR: 100V, 12A, 720W. $R_{1} = 0.67 Ω$ , $R_{2} = 0.185 Ω$ . Find full-load current, slip, pf, max power.