Electric Machines and Drives A 1Phase 2winding transformer i

Electric Machines and Drives

A 1-Phase, 2-winding transformer is rated at 10 kVA, 200 V (primary) - 400 V (secondary). The transformer efficiency is 98.6% when the transformer is delivering 20% above the rated load at 0.9 (lag) power factor. Calculate the power input and the losses in the transformer. Calculate the corresponding load current. Ans: P_i = 10953.3 W, P_loss = 153.3 W, I = 30A A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances: R_P = 32 ohm R_S = 0.05 ohm X_P = 45 ohm R_S = 0.06 ohm R_C = 250 k ohm X_M = 30 k ohm The excitation branch impedances are given referred to the high-volt age side of the transformer. Find the equivalent circuit of this transformer referred to the high-voltage side. Assume that this transformer is supplying rated load at 480 V and 0.8 PF lagging. What is this transformer\'s input voltage? What is its voltage regulation? What is the transformer efficiency under the conditions of part (b)?

Solution

Problem No-2)

Rated Load KVA=10 KVA

20% above rated means S=10(1+20%)=12 KVA

Power output Po=S*Cos(phi)=S*power factor=12*0.9=10.8 kW

Po=10.8 kW

Input Power Pi=Po/efficiency=10.8/0.986=10.95335 kW

Pi=10.95335 kW=10953.35 W

Pi=10953.35 W

------------------------------

Losses=Pi-Po=10953.35-10800=153.35 W

Losses=153.35 W

-------------------------

Secondary voltage V=400 V

Load Current I=Po/(V*pf)=10800/(400*0.9)=30A

I=30A

------------------

Only 1 question per post will be answered please

-------------------