Question

Q2.) A 60KVA, Single phase 3300/400 V transformer gave the following test results. Table 2.1 Open circut test 3300V applied to primary, power taken 430W Short circuit test Primary Input power is 525 watts Calculate the efficiency at a) full load at 0.7 power factor b) Half load at 0.7 power factor ) An ideal transformer connected to a 240V mains, supplies a 12V, 150W lamp. Calculate a) Transformer turns ratio b) The current taken from the supply. [2] [2] [1] [2]

Answer 1

Answer for the Question-2(i)

Rating of transformer = 60KVA = 60000VA

From the open circuit test, we get constant losses = Wi = 430 Watts

From the short circuit test, we get full load copper loss = Wc = 525 Watts

If we assume x is the ratio of the amount of load connected to the rated load and pf is the power factor,

The efficiency of the transformer is given as,

$\eta =\frac{x*Rated...VA*pf}{x*Rated...VA*pf+W_i+x^2*W_c}*100$

(i)-(a) x = 1 and pf = 0.7

$\eta =\frac{1*60000*0.7}{1*60000*0.7+430+1^2*525}*100$

n = 97.77..........percent

(i)-(b) x = 0.5 and pf = 0.7

$\eta =\frac{0.5*60000*0.7}{0.5*60000*0.7+430+0.5^2*525}*100$

$\eta =97.39...percent$

Answer for the Question-2(ii)

Given,

Primary Voltage = V1 = 240 V

Secondary Voltage = V2 =12 V

Power at Secondary side = P2 = 150 W

(ii)-(a)

Turns Ratio = V1 / V2 = 240 / 12 = 20

(ii)-(b)

Secondary Current = I2 = P2 / V2 = 150 / 12 = 12.5 Amps

Primary Current = I2 / Turns Ratio = 12.5 / 20 = 0.625 Amps

Answer for the Question-4

Given data,

Load Power = PL = 22000 Watts

Since it is a Shunt generator, Terminal Voltage at field = Terminal Voltage at Load = VL = 220 V

Efficiency = n = 0.88

Shunt field resistance = Rsh = 110 Ohms

Armature resistance = Ra = 0.05 Ohms

Load current = IL = PL / VL = 22000 / 220 = 100 Amps

Shunt field Current = If = VL / Rsh = 220 / 110 = 2 Amps

Armature Current = Ia = IL - If = 100 - 2 = 98 Amps

(a)-answer

Shunt field copper loss = If2 * Rsh = 22 * 110 = 4 * 110 = 440 Watts

Armature copper loss = Ia2 * Ra = 982 * 0.05 = 480.2 Watts

Total Copper Loss = 440 + 480.2 = 920.2 Watts

(b)-answer

Efficiency = n = ( Load Power ) / (Load Power + Total Losses)

0.88 = ( 22000 ) / (22000+ Total Losses)

22000+ Total Losses = ( 22000 ) / 0.88

Total Losses = 25000 - 22000

Total Losses = 3000 Watts

Iron and Friction Loss = Total Losses - Total Copper Loss = 3000 - 920.2 = 2079.8 Watts