Question

A 500-kVA, 3-phase, 50-Hz transformer has a voltage ratio (line voltages) of 33/11−kV and is delta/star connected. The resistances per phase are: high voltage 35Ω, low voltage 0.876 2 and the iron loss is 3050 W. Calculate the value of efficiency at full-load and one-half of full load respectively (a) at unity p.f. and (b) 0.8 p.f.

Answer 1

The efficiency of a transformer can be calculated using the formula Efficiency = (Output Power / Input Power) * 100%. At full-load and unity power factor, the efficiency is 100%. At one-half of full load and unity power factor, the efficiency is 50%. At full-load and 0.8 power factor, the efficiency is 80%. At one-half of full load and 0.8 power factor, the efficiency is 40%.

Step-by-step explanation:

The efficiency of a transformer can be calculated using the formula:

Efficiency = (Output Power / Input Power) × 100%

At full-load and unity power factor (pf = 1), the output power is equal to the input power. Therefore, the efficiency at full-load and unity pf is 100%.

At one-half of full load and unity pf, the output power is equal to half of the input power. Therefore, the efficiency at one-half of full load and unity pf is 50%.

At full-load and 0.8 power factor (pf = 0.8), the output power is equal to the input power multiplied by the power factor. Therefore, the efficiency at full-load and 0.8 pf is (Output Power / Input Power)× 100% = (0.8 × 100) = 80%.

At one-half of full load and 0.8 pf, the output power is equal to half of the input power multiplied by the power factor. Therefore, the efficiency at one-half of full load and 0.8 pf is (Output Power / Input Power) × 100% = (0.4 × 100) = 40%.

Answer 2

Final answer:

The efficiency of the transformer at full load and one-half of full load can be calculated using the formulas. At unity power factor, the efficiency is 99.39%, while at a power factor of 0.8, the efficiency is 124.05%.

Step-by-step explanation:

The efficiency of a transformer can be calculated using the formula:

Efficiency = (Output power / Input power) x 100%

a) At unity power factor (p.f.), the power factor angle (θ) is 0°. Therefore, the apparent power (S) equals the active power (P).

For the given transformer, the output power (Pout) is 500 kVA and the iron loss (Piron) is 3050 W. The input power (Pin) can be calculated as:

Pin = Pout + Piron = 500,000 VA + 3050 W = 500,000 VA + 3.05 kW = 503,050 VA = 503.05 kVA

Efficiency = (Pout / Pin) x 100% = (500 kVA / 503.05 kVA) x 100% = 99.39%

b) At a power factor (p.f.) of 0.8, the power factor angle (θ) is 36.87°. Therefore, the active power (P) is calculated using the formula:

P = Pout x cos(θ) = 500 kVA x cos(36.87°) = 500 kVA x 0.8 = 400 kVA

Using the same formula as in part a, the input power (Pin) is:

Pin = P + Piron = 400 kVA + 3050 W = 400 kVA + 3.05 kW = 403,050 VA = 403.05 kVA

Efficiency = (Pout / Pin) x 100% = (500 kVA / 403.05 kVA) x 100% = 124.05%