Question

A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.​

Answer 1

Answer:

I have attached my answer below

Explanation:

Use Power flow diagram of Induction Motor to solve this problem

Answer 2

Answer:

(i) stator core loss = $163.2W$

(ii) slip at full load =$10.8kW$

(iii) total rotor ohmic losses at full load =$227.86W$

(iv) full load speed =$1467.9 rpm$

Explanation:

$P_{out} =10kw\\V_{c}=V_{ph} =400v\\P=4\\f=50Hz\\Stator resistance, R_{1}=1.2 lph$Ω

Friction  and windge loss

No load =$P_{in} =660w$,$I_{L} =8A$

Full load=$P_{in} =11.2w$,$I_{L} =18A$

1) stator core loss

$P_{in} =P_{(st) code}+ P_{(st) cu}+friction and windge loss\\660=P_{(st) code}+76.8+420\\P_{(st) code}=163.2W$

2) slip load = $10.8kW$

3)total rotor ohmic losses at full load

$P_{mech(dev)} =P_{(out) }+ friction and windge loss\\10k=+0.42k\\=10.42kW\\w.k.t P_{mech(dev)} = (1-S) P_{(ag) }\\(1-S) _{(pl) =0.0214\\(1-S) _{(pl)} =0.0214*10.648* 10^{3} \\=227.86W$

4) Full load speed

$N_{FL)} =1-S_{(FL) }N_{(s)}\\=(1-0.0214)*1500\\=1467.9rpm$

To know more about ohmic,visit:

https://brainly.in/question/522731

To know more about speed,visit:

https://brainly.in/question/448356

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