Finding efficiency of transformer

[visage]

Homework Statement

10kVA, 200:400 V transformer gave these test results:
open circuit test (LV winding supplied): 200 V, 3.2 A, 450 W
short circuit test (HV winding supplied): 38 V, 25 A, 600 W

Homework Equations

Calculate the efficiency when the transformer delivers its rated kVA at 0.85 power factor lagging.

3. The Attempt at a Solution

Calculated in part a of the question:

Open Circuit
R_c = V²/P = 200²/450 = 88.9ohms
S = VI = 200 x 3.2 = 640VA
Q = sqrt(640² - 450²) = 455VAR
X_m= V²/Q = 200²/455 = 87.81ohms

Short Circuit
R_t = P/I² = 600/25² = 0.96ohms
Z_t = V/I = 38/25 = 1.52ohms
X_t = sqrt(1.52² - 0.96²) = 1.178ohms
Referring to LV side
R_t' = 0.96(200/400)² = 0.24ohms
X_t' = 1.178(200/400)² = 0.295ohms

Part b
Efficiency = Power out / Power in
Power out = S x pf = 10000 x 0.85 = 8500W
Power in = 8500W + copper loss + iron loss
Copper/winding loss, P_c = I² x R_t' = 25² x 0.24 = 150W
Iron/core loss, P_i = V²/R_c = 38²/88.9 = 16.24W

Efficiency = Power out / Power in = 8500/(8500+150+16.24) = 98.08%

Need help to check the calcution in red. anyone?

 

[KataKoniK]

Your calculations seem to be correct. However, it is always a good idea to double check your calculations and make sure all units are consistent. Also, you may want to include a brief explanation of your calculations and how you arrived at your solution. This will make it easier for others to follow and understand your work. Additionally, you may want to label your equations and show your units to make it more organized and clear.

Overall, your solution looks good. Keep up the good work!