Final answer:
The student's question is calculating the readings on two wattmeters connecting to a 420V, 3-phase induction motor with given output, power factor, and efficiency. The input power is first determined using the known output and efficiency, then the apparent power is calculated accounting for the power factor. Finally, the readings on the two wattmeters can be computed given the voltage, current, and phase angle.
Step-by-step explanation:
The student is asking to calculate the input power using two wattmeters for a 3-phase induction motor that has an output of 30 HP, operates with a power factor of 0.7 lagging, and has an efficiency of 87%. The induction motor is specified to be 420V. To find the input power, we first need to calculate the output power in watts.
Since 1 Horsepower (HP) is equal to 746 Watts, the output power (Pout) can be calculated as 30 HP × 746 W/HP = 22,380 W.
Next, the motor's input power (Pin) can be obtained by considering the efficiency (η) using the formula Pin = Pout / η. The efficiency is given as 87%, or 0.87 as a decimal. Thus the input power is Pin = 22,380 W / 0.87 ≈ 25,724.14 W.
For a 3-phase system, the total power is split between the phases. However, due to the power factor being less than 1, we have to account for this in our calculations. The apparent power (S) is S = Pin / power factor, so S = 25,724.14 W / 0.7 ≈ 36,748.77 VA.
Using the formula for total apparent power in a 3-phase system, S = √3 × Vline × I, we can solve for the line current (I) since the line voltage (Vline) is 420 V. This gives us I = S / (√3 × Vline) ≈ 50.58 A.
In a two wattmeter method, the readings of two wattmeters can be calculated by:
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- Wattmeter 1 (P1): P1 = √3 × Vline × I × cos(φ + 30°)
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- Wattmeter 2 (P2): P2 = √3 × Vline × I × cos(φ - 30°)
Where φ is the phase angle corresponding to the power factor, which is cos-1(power factor). Calculating these values will give the readings of the two wattmeters.