MODEL ANSWER FOR SECOND MID TERM
2. We need to determine the equivalent circuit impedances of a 20 kVA, 8000/240 V, 60 Hz transformer. The open-circuit and short-circuit tests led to the following data:
VOC = 8000 V / VSC = 489 VIOC = 0.214 A / ISC = 2.5 A
POC = 400 W / PSC = 240 W
Solution: The power factor during the open-circuit test is
The excitation admittance is
Therefore:
The power factor during the short-circuit test is
The series impedance is given by
Therefore: Req = 38.3 Ω; Xeq = 192Ω
The equivalent circuit
3. A 15-kVA, 23001230-V transformer is to be tested to determine itsexcitation branch components, its series impedances, and its voltage regulation. The followingtest data have been taken from the primary side of the transformer:
(a) Find the equivalent circuit of this transformer referred to the high-voltage side.
(b) Find the equivalent circuit of this transformer referred to the low-voltage side.
(c) Calculate the full-load voltage regulation at 0.8 lagging power factor, 1.0 power factor, and at 0.8 leading power factor.
(d) What is the efficiency of the transformer at full load with a power factor of 0.8
lagging?
Solution:
Solution: VT = 240 V, RA = 0.064Ω, RF = 93.6Ω
(a)The output power in watts Pout = 30hp*746 W/hp = 22,380 W
The line current IL =
the field current IF =
The armature current IA = IL – IF = 105.37 – 2.564 = 102.8 A
Therefore, at full load armature I2R losses are PA = I2ARA = (102.8)2*0.064 = 673.34 W
At full load field circuit I2R losses are Pf = I2FRF = (2.564)2*93.6 = 615.33 W
The output power Pout = Pin– PA – Pf – Prot - Pstray
Where Pstay is 1% of Pin
Prot = Pin – PA – Pf – Pstray - Pout = 25288 – 673.34 – 615.33 – 0.01*25288 – 22380
Prot= 1366.45 W
Percentage of Rotational losses =
(b)Let Rs be the series starter resistance, then
Ia =
Rs + 0.064 = 1.334
Rs = 1.27Ω
(c)
Q6) List and define three major losses in DC motors.
Q7) Compare between the three types of DC motors