EEE Job Math Solution-3

Induction Motor Math (Part-3): এখানে যে সব ম্যাথ দেয়া হয়েছে, সেগুলো খুব Important প্রশ্ন। আরো দেয়া হবে, আপনাদের সহযোগিতা চাই।

Question-6: A 400v, 4-Pole, 50z, 3-Phase, 10hp, Y-Connected induction motor has a no load slip of 1% and full load slip of 4%. Find the following:  (i) Syn. Speed   (ii) No load speed  (iii) Full load speed  (ix) Frequency of rotor current at full load  (v) Full load torque.
Solution-6:

Here,

Pout = 10hp = 10*746 = 7460Watt, Pole, P = 4, f = 50Hz, VL =400v

No load slip, s =0.01, Full load slip, sfl = 0.04

(i)  Syn. Speed   NS = 120f/P = 120*50/4 = 1500rpm

(ii)  No load speed  NO = NS(1-s) = 1500(1-0.01) = 1485rpm

(iii) Full load speed, NFL = NS(1- sfl) = 1500(1-0.04) = 1440rpm

(ix) Frequency of rotor current, fr = sfl .f = 0.04*50 = 2Hz

(v) Full load torque, Tsh = 9.55Pout/NFL = 9.55*7460/1440 = 48.78Nm

Question-7: The shaft output of a 3-phase 50z induction motor is 60kW. The friction and windage losses are 920W, the stator core loss is 4200W and the stator copper loss is 2600W. The rotor current and rotor resistance referred to stator are respectively 110A and 0.15 ohm. If the slip is 4%, what is the percent efficiency?
Solution-7:

Here, Pm = Output = 60kW, Friction and windage losses = 920W

Stator core loss = 4200W,  Stator copper loss = 2600W

Slip, s =4% =0.04

Gross Mechanical output = Pm + Friction and windage losses = 60kW+920W = 60.92kW

Rotor input/rotor gross output = 1/(1-s)

Rotor input = rotor gross output/(1-s) = 60.92kW/(1-0.04) =63.46kW

Stator input = rotor input + stator copper loss + stator core loss = 63.46kW + 2600W +4200W = 70.26kW

%η = (rotor output/stator input)*100 = (60/70.26)*100 = 85.4% (Ans.)

Question-8: A 6-Pole,  3-phase induction motor develop 30hp including mechanical loss totaling 2hp, at a speed of 950 rpm on 400v, 50z  lines. If the power factor is 0.85 and core loss negligible, calculate-
a) The slip b) The rotor copper losses c) The total input power if stator losses are 2kW d) The line current
Solution-8:

Here,

Pm = 32hp

Pole, P = 6, f = 50Hz, V = 400v, Power Factor, cos θ = 0.85, Nr = 950rpm

We know, Ns = 120f/P = 120*50/6 = 1000rpm
a) Slip, s = (Ns-Nr)/Ns = (1000-950)/1000 =0. 05
Rotor gross output = Pm + Mech. Losses = 30+2 =32hp

We know, Rotor cu loss  / Rotor gross output = s /(1-s)

b) Rotor cu loss = (Rotor gross output*s) /(1-s) = (32*0.05)/(1-0.05) = 68hp = 1.68*746 = 1256W

Rotor input = rotor gross output/(1-s) = 32hp/(1-0.05) = 33.68hp = 33.68*746 = 25128Watt
c) Total input = rotor input + stator cu loss + stator core loss = 25128+2000+0 = 27128Watt

d) Line current, IL = Total input / ( 3 VL cos θ ) = 27128/ (√3 *400*0.85) = 46A

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