Efficiency of a Transformer This Photo by Unknown Author is licensed under CC BY Efficiency of a Transformer ο΄ As in the case with other types of electrical machines, the efficiency of a transformer at a particular load and power factor is defined as the ratio of its power output, Po against its input power, Pi. ο΄ Po and Pi being measured in the same units (either watts or kilowatts). π0 πΈπππππππππ¦ = π = ∗ 100% ππ Efficiency of a Transformer ο΄ Due to high efficiency, it is found to be impractical to measure a transformer’s efficiency by accounting for its input and output. ο΄ A better method is to determine the losses. π0 ππ − ππππ π ππ π= ∗ 100% = ∗ 100% ππ + ππππ π ππ ππ ππππ π ππ = πππππ + πππππππ Efficiency of a Transformer ο΄ It must be noted that the efficiency of a transformer is based on Watts and not on VA. ο΄ Hence, at any volt-ampere load, the efficiency depends on power factor, being found to maximize at unity. ο΄ The core loss of a transformer may be found from a no-load or open-circuit test and the Cu loss from a short-circuit test. Problem: A 250/500-V transformer gave the following results: OC test: 250 V, 1 A, 80 W, measured on the low-voltage side SC test: 20 V, 12 A, 100W, with low-voltage winding short-circuited Determine the equivalent circuit parameters and calculate the efficiency when the current drawn by the load is 10 A at 500 V, 0.8 lagging power factor. Solution: From the Open-Circuit Test πππ πΌππ πππ π = πππ 250 ∗ 1 ∗ πππ π = 80 80 πππ π = = 0.32 250 πΌππ = πΌπ€ + πΌπ πΌπ€ = πΌππ πππ π = 1 ∗ 0.32 = 0.32 π΄ πΌπ = 2 2 = πΌππ − πΌπ€ 12 − 0.322 = 0.95 π΄ Solution: 2π 2 πππ = πΌπ€ ππ = 80 = 0.32 ∗ π ππ 80 π ππ = = 781.25 Ω = π π 2 0.32 πππ 250 πππ = = = 263.16 Ω = ππ πΌπ 0.95 From the Short-Circuit Test: ππ π 100 π π π = 2 = 2 = 0.69 Ω = π ππ2 πΌπ π 12 ππ π 20 ππ π = = = 1.67 Ω πΌπ π 12 ππ π = 2 2 ππ π − π π π = 1.672 − 0.692 = 1.52 Ω = πππ2 Solution: While Ro and Xo are referred to the primary, for simplicity, we need to transform Req2 and Xeq2 into their equivalent into the same side. πΈ1 250 π= = = 0.5 πΈ2 500 πππ1 2 π = πππ2 πππ1 = π2 πππ2 = 0.52 ∗ 0.69 + j1.52 = 0.17 + j0.38 Ω ∴ π ππ1 = 0.17 Ω πππ1 = 0.38 Ω Solution: ∴ π ππ1 = 0.17 Ω πππ1 = 0.38 Ω ∴ π π = 781.25 Ω ππ = 781.25 Ω Solution: For the efficiency, πππ’ = 102 ∗ 0.69 = 69 π πππ = 80 π ππ = 500 ∗ 10 ∗ 0.8 = 4000 π ππ 4000 π= ∗ 100% = ∗ 100% = 96.41% ππ + πππ’ + πππ 4000 + 69 + 80 More examples next meeting. THANK YOU. This Photo by Unknown Author is licensed under CC BY
