|
r = Smax. X —— (I) |
r+R= (Smax)’. X —— (ii) |
From equation (i) and (ii), R = r(S’ max/Smax – 1) —— (iii) By definition of Smax, we get Smax = 1 – (Nmax/Ns) —— (iv) Putting S’max =1 in equation (iii), we get | ||
R = r. (1/Smax-1) —— (v) Let’s say, Ns = synchronous speed of 1000rpm and pull-out torque happens at 900 rpm, equation (iv) reduces to Smax = 0.1 (i.e., 10% slip) Substitute in equation (v), R = r. (1/0.1 – 1) | ||
R = 9. r | ||
We can mathematically say that, , As per DC (M) equation But in case of induction motor Suffix 1 -> used for stator/stator parts (qty) Suffix 2-> used for rotor/rotor parts(qty)
Therefore, ----(1)
= flux that induces the EMF in rotor = rotor current under running condition = P.F of rotor But,stator vtgalso i.e | ||||
----(2) Transformation ratio i.e. given by Therefore, and hence
also at slip ’S’ is given by And also at slip ‘S’ is =S. Hence in equation can be replaced by i.e. U =----(3)----() 4 =------(4) Substract (3) and (4) into (1) equation we get, | ||||
(N-m) , Torque equation of I.M.
At starting condition slip S=1 So, | ||||
9. A three phase 15hp,460V , 4-pole 60 Hz 1728 rpm induction motor delivers full output power to the load connected to its shaft. The wind age and friction loss of the motor is 750 W. Determine the
Full load shaft power = Pout = 15.746 Mechanical power = Pm = Pout + ∆ Pm = 11190 + 750 = 11940 W Air gap power Pavg = Pm / 1-s Ns = 120.6/4 = 1800 rpm S = ns – n /ns = 1800 – 1728 /1800 =0.04 Pag = 11940/1-0.04 = 12.437.5 W Rotor copper loss ∆Pα = 0.04 x 12.437.5 = 497.5 W |
Mativa) |Ia f.l| = S rated/ V rated = 10 x 10 3/ 230/ = 25.4 A | ||
25.1 < -36.87 E = Vt + Ia(Ra + jXs) = 230<0/ + (25.1)<-36.8 x (0.5 + 1.2j) = 161.7 < 5.8 V ꙴ = |E| - |Vt|/ |Vt| = 161.7 – 230/3 / 230/ = 21.89 %
b) Ia = 25.1 < 36.87 A
E = 230/ + 25.1 <36.8 < 0.5 + 1.2 j= 128.7 < 14.2 V | ||