Home | | **Basic Electrical and Instrumentation Engineering** | | **Electrical Engineering and Instrumentation** | Solved Problems: Electrical Engineering and Instrumentation - Transformer

Electrical Engineering and Instrumentation - Transformer - Solved Problems: Electrical Engineering and Instrumentation - Transformer

**PROBLEMS**

** A 400 kVA transformer has a primary winding
resistance of 0.5 ohm and a secondary winding resistance of 0.001 ohm. The iron
loss is 2.5 Kw and the primary and secondary voltages are 5 kV and 320 V
respectively. If the power factor of the load is 0.85, determine the efficiency
of the transformer (i) on full load and (ii) on half load. **

**Solution**

Rated
output = 400 kVA = 400x10^{3} kVA

Full load secondary current, I_{2} = Rated
output/V_{2} = 1250 A

Total
resistance referred to secondary, r_{e2} = r_{2}+r_{1}(V_{2}/V_{1})^{2}
= 0.033 ohm Full load copper loss, P_{c} = I_{2}^{2} r_{e2}
= 51.5625 Kw

Iron
loss, Pi = 2.5 x10^{3} watts

**(i) Transformer efficiency at
full load and 0.85 pf**

= 86.2%

**(ii) Transformer efficiency at half load and 0.85
pf**

= 91.69%

**2. ****Find all day efficiency of a transformer having
maximum efficiency of 98% at 15 kVA at unity power factor. Compare its all day
efficiencies for the following load cycles:**

**a. Full load of 20 kVA, 12 hours
per day and no load rest of the day. b. Full load, 4 hours per day and 0.4 full
load rest of the day.**

**Assume the load to operate on upf
all day**

Solution

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

**Related Topics **

Copyright © 2018-2021 BrainKart.com; All Rights Reserved. (BS) Developed by Therithal info, Chennai.