Efficiency of
Transformer
Output power and
efficiency of the input transformer and power transformer ratio. Skills η in
the directed
η = output / input
The Power input
transformer and the output power of the percentages expressed as the percent
efficiency.
% η = output / input ×
100
= Output / output +
Loss × 100
Or
% η = input - Loss /
input × 100
= (1 - Loss / input) ×
100
Transformer copper
loss is that it all depends on the coil current flow caused by resistance and
heating loss. The copper coil resistance and outlets are dependent attenuation.
Pse = I²pRep = I²Res resistance cable size and length depends on the coil and
the current load dependent. Load copper loss of the transformer depends on the
specific outlets.
If you say that, on
the transformer rating is KVA with kW not write this? Or on the transformer rating
is KVA you ?
Answer: - Los outlets on the transformer and copper and core loss
depends on voltage phase differences, but is not dependent on them. The phase
difference between voltage and current depends on the load type. It is the same
for Watt or kW load current may be different, but KVA Voltage and current are
the same on all the power factor. The rating transformer kW with KVA It is on.
transformer efficiency
is a maximum when the core loss and copper loss is equal.
That is, the core loss
= copper loss
Pc = Ph + Pe
transformer core loss and copper loss Pse = I ² pRep = I ² Res
Core loss is not
dependent on electrical outlets.
Considered the
direction of the primary input VpIp Cosθp
η = VpIp Cosθp - Loss
/ VpIp Cosθp
= 1 - I ² pRep / VpIp
Cosθp - Pc / VpIp Cosθp
= 1 - IpRep / VpIp
Cosθp - Pc / VpIp Cosθp
Differencent both parties relative to the Ip,
dη / dIp = 0 - Rep /
Vp Cosθp + Pc / VpI ² p Cosθp
Ip variable.
Therefore, the maximum efficiency dη / dIp = 0
For maximum efficiency
= -Rep/VpCosθp + Pc /
VpI ² p Cosθp = 0
Or Rep = Pc / I ² p
I ² pRep = Pc
Psc = Pc
Therefore, the maximum
efficiency of transformer core loss = copper loss.
For example of
mathematics
1. A 20KVA, 2300/230V transformer and full load copper loss 200W is calculated to -
a. 6KVA load of copper loss.
b. High Voltage in 4A
load of copper loss.
c. 10KW 0.8 power factor
load of copper in Loss.
Solution: -
a. Copper loss =
(6/20) ² × 200
=
18 w.
b. Full load current IFL = 20 × 1000/2300
=
8.69 A
Copper
loss = (4 / 8.69) ² × 200
=
42.38 W.
c. KW load = 10/0.8
=
12.5 KVA
Copper loss = (12.5 /
20) ² × 200
=
78.125 W. Ans.
2. A 5KVA, 2300/230V
transformer and 24 hours of low load calculation forms -
1 .1 ½ load power factor 0.8 to 1 hours.
3. 1 ¼ load power factor 0.8 to 2 hours
4. Rated load power
factor of 0.9 to 3 hours
5. ½ load power factor unity of 6 hours
6. ¼ load power factor unity of 8 hours
7. No load power
factor unity of 4 hours
Core Loss 40 Watt and Copper
loss 112 Watt now the skills of % η all day
Solution: -
24 Hour output
= (1 ½ × 5 × 0.8 × 1)
+ (1 ¼ × 5 × 0.8 × 2) + (5 × 0.9 × 3) + (½ × 5 × 1 × 6) + (¼ × 5 × 1 × 8)
= 54.5 KWh
24 hours in Los copper
= [(1 ½) ² × 112 × 6]
+ [(1 ¼) ² × 112 × 2] + [112 + 2] + [(½) ² × 112 × 6] + [(¼) ² × 112 × 8]
= 1162 Wh = 1.162 KWh
24 hour core loss
= 40 × 24 = 960Wh =
0.96 KWh
All day efficiency =
54.5 / 54.5 + 1.162 + 0.96 × 100%
= 96.25% Ans.
3. A 10KVA transformer
core loss and full load copper loss 55W 130W., Which follows 24 hours of
loading.
1 15KVA load power factor 0.85 to 2 hours
2. Rated load power
factor of 0.90 to 5 hours
3.7.5 load power
factor 0.95 to 6 hours
4.5KVA unity load power factor of 7 hours
5. No load of 4 hours
Transformer and
all-day skills that now.
Solution: -
Rating = 10 KVA
Pc = 55W
Psc = 130 W
% η all day =
?
24 hrs. output
15 × 0.85 × 2 = 25.5
KWh
10 × 0.9 × 5 = 45 KWh
7.5 × 0.96 × 6= 42.75
KWh
5 × 1 × 7 = 35 KWh
0 4 = 00
Total 24 hrs. = 148.25
KWh
Core loss = 55 × 24 =
1320 Wh = 1.32 KWh.
Copper loss = (15/10)²
× 130 × 2 = 585 Wh.
130 ×
5 = 650 Wh.
(7.5/10)² × 130 × 6 =
438.75Wh
(5/10)² × 130 × 7 = 227.5
Wh
Total = 1.9
KWh.
% η all day =
output / output + losses × 100%
= 148.25 / 148.25 + 1.32 +1.9
× 100
= 97.87% Ans.