EEE Job Math Solution-2

Induction Motor Math (Part-2): এখানে যে সব ম্যাথ দেয়া হয়েছে, সেগুলো বিভিন্ন জব ও এডমিশন টেস্টে আসা প্রশ্ন। আরো দেয়া হবে, আপনাদের সহযোগিতা চাই।

Question 3: A motor has copper loss of 300 W and 4.5% slip. How much electrical power is converted to mechanical power?
Solution-3:

Here,
Copper loss, Pcr = 300 W.

Slip, s = 4.5% = 0.045.

Electrical Power converted to mechanical, Pm = ?

We know, Pcr / Pm= s /(1-s).

Pm = Pcr*(1-s) /s = 300 * (1- 0.045) / 0.045 = 6366.67 Watt. (Ans)

Question 4: What is the full load current of a 50 HP, 50Hz, 0.85 pf induction motor when the line voltage is 415 V
Solution-4:

Here,
P = 50HP = 50*746 = 37300 Watt

Power Factor, cos θ = 0.85

Line Voltage, VL = 415

P = √3 VL IL cos θ

IL = P / ( 3 VL cos θ ) = 37300/ (√3 *415*0.85) = 61.04A (Ans)

Question 5: Determine the rotor input and induction torque in a 400 V, 50 Hz, 4-pole, Y-connected induction motor operating at 2% slip with 240W power being lost in it’s rotor winding.
Solution-5:

Here,
VL = 400v, f= 50Hz, P = 4, s = 2% = 0.02,

Rotor Cu Loss, Pcr = 240W

NS = 120f/P = 120*50/4 = 1500rpm

Nr = NS (1-s) = 1500(1-0.02) = 1470rpm

We know, Pcr / Pm= s /(1-s).

Pm = Pcr (1-s)/s = 240(1-0.02)/0.02 = 11760Watt

Induction torque, Tg = 9.55*Pm/Nr = 9.55*11760/1470 = 76.4Nm (Ans)
 Rotor input, P2 = Pcr /s = 240/0.02 = 12000Watt. (Ans)

Induction Motor নিয়ে আরো ম্যাথ থাকছে…

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