3 ph Sq cage induction motor Calculation

 

Let's break down the complete calculation from the nameplate of your Bharat Bijlee 3-phase squirrel cage induction motor into power, current, losses, and RPM/slip, step-by-step.

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🔧 Nameplate Key Data:

Parameter	Value
Power	0.37 kW (or 0.5 HP)
Voltage	415 V
Frequency	50 Hz
Current	1 A
Efficiency	72.7%
Power Factor	0.68
RPM	1400
Synchronous Speed (Ns)	Unknown (we'll calculate)

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1️⃣ Input Power (Actual power drawn)

We are given output power (Pout) and efficiency (η):

$$P_{out} = 0.37\, \text{kW}$$ $$\eta = 72.7\% = 0.727$$

🔹 Formula:

$$P_{in} = \frac{P_{out}}{\eta}$$

🔹 Substituting:

$$P_{in} = \frac{0.37}{0.727} = 0.509 \, \text{kW} = 509 \, \text{Watts}$$

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2️⃣ Current Calculation (3-phase power formula)

Now we calculate the input current (I) using the formula:

$$P = \sqrt{3} \cdot V \cdot I \cdot \text{Pf}$$

Rearranged to calculate I:

$$I = \frac{P}{\sqrt{3} \cdot V \cdot \text{Pf}}$$

🔹 Substituting:

$$I = \frac{509}{\sqrt{3} \cdot 415 \cdot 0.68}$$ $$\sqrt{3} = 1.732$$ $$I = \frac{509}{1.732 \cdot 415 \cdot 0.68} = \frac{509}{489.5} \approx 1.04 \, \text{A}$$

✅ Nameplate shows 1 A — correct.

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3️⃣ Power Losses (Watts)

Losses = Input Power − Output Power

$$\text{Loss} = P_{in} - P_{out} = 509 - 370 = 139 \, \text{W}$$

These include:

• Copper losses (I²R)

• Iron/core losses

• Friction and windage losses

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4️⃣ Synchronous Speed (Ns)

The motor runs at 50 Hz. Assume a 4-pole motor (most common).

$$N_s = \frac{120 \cdot f}{P}$$ $$N_s = \frac{120 \cdot 50}{4} = 1500 \, \text{RPM}$$

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5️⃣ Slip Calculation (s)

Given actual speed N = 1400 RPM

$$s = \frac{N_s - N}{N_s} \times 100$$ $$s = \frac{1500 - 1400}{1500} \times 100 = \frac{100}{1500} \times 100 = 6.67\%$$

This slip is normal for small induction motors.

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6️⃣ Power Factor Check

$$\text{Power Factor (PF)} = 0.68$$

This is typical for small motors without correction capacitors. Lower PF means higher reactive power.

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7️⃣ Mechanical Output Torque (T)

We can calculate the output torque using:

$$T = \frac{9550 \cdot P_{out}}{N}$$ $$T = \frac{9550 \cdot 0.37}{1400} \approx 2.53 \, \text{Nm}$$

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✅ Summary of All Calculations:

Parameter	Value
Input Power	509 W
Output Power	370 W
Power Losses	139 W
Input Current	~1.04 A
Synchronous Speed	1500 RPM
Actual Speed	1400 RPM
Slip	6.67%
Power Factor	0.68
Torque	~2.53 Nm

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Let me know if you also want reactive power, kVAR, apparent power (kVA) or capacitor sizing!

Great! Let's now calculate the Apparent Power (kVA), Reactive Power (kVAR), and also estimate the Power Factor Correction Capacitor (in μF) needed to improve the power factor.

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⚡ Given:

• Input power (P) = 0.509 kW

• Voltage (V) = 415 V

• Current (I) = 1 A

• Power Factor (PF) = 0.68

• Frequency (f) = 50 Hz

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✅ 1. Apparent Power (S in kVA)

Apparent power $S$ is:

$$S = \sqrt{3} \cdot V \cdot I$$ $$S = 1.732 \cdot 415 \cdot 1 = 718.78 \, \text{VA} = 0.719 \, \text{kVA}$$

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✅ 2. Real Power (P in kW)

Already calculated:

$$P = 0.509 \, \text{kW}$$

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✅ 3. Reactive Power (Q in kVAR)

Use formula:

$$Q = \sqrt{S^2 - P^2}$$ $$Q = \sqrt{0.719^2 - 0.509^2} = \sqrt{0.517 - 0.259} = \sqrt{0.258} \approx 0.508 \, \text{kVAR}$$

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✅ 4. Power Factor Correction (Capacitor kVAR required)

To improve PF from 0.68 → 0.95 (typical target):

$$Q_{required} = P \cdot (\tan(\cos^{-1} \, PF_{initial}) - \tan(\cos^{-1} \, PF_{target}))$$ $$\cos^{-1}(0.68) = 47.2^\circ,\quad \tan(47.2^\circ) = 1.08$$ $$\cos^{-1}(0.95) = 18.2^\circ,\quad \tan(18.2^\circ) = 0.33$$ $$Q_c = 0.509 \cdot (1.08 - 0.33) = 0.509 \cdot 0.75 = 0.382 \, \text{kVAR}$$

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✅ 5. Capacitor Size (μF) for 415V, 50Hz

Use formula:

$$C(\mu F) = \frac{10^6 \cdot Q(kVAR)}{2 \pi f V^2}$$ $$C = \frac{10^6 \cdot 0.382}{2 \cdot \pi \cdot 50 \cdot 415^2} = \frac{382000}{2 \cdot 3.1416 \cdot 50 \cdot 172225}$$ $$= \frac{382000}{54114738} \approx 7.06 \, \mu F$$

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✅ Final Summary:

Parameter	Value
Apparent Power (S)	0.719 kVA
Real Power (P)	0.509 kW
Reactive Power (Q)	0.508 kVAR
Desired PF	0.95
Capacitor kVAR needed	0.382 kVAR
Capacitor value (at 415V, 50Hz)	7.06 μF

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