10. Write the expression for the self inductance of a coil wound on an iron core
3.9 EFFICIENCY AND VOLTAGE REGULATION
( )
+ j = 0.082 + j 0.102
Z(LV)(pu) = 0 082 0 102 2
. + j . = 0.041 + j 0.051
Remark The earlier remark is, therefore, confirmed that pu values referred to either side of the transformer are the same so long as voltage bases on the two sides are in the ratio of transformation of the transformer.
3.9 EFFICIENCY AND VOLTAGE REGULATION
Power and distribution transformers are designed to operate under conditions of constant rms voltage and frequency and so the efficiency and voltage regulation are of prime importance.
The rated capacity of a transformer is defined as the product of rated voltage and full load current on the output side. The power output depends upon the power factor of the load.
The efficiency h of a transformer is defined as the ratio of the useful power output to the input power. Thus h = output
input (3.53)
The efficiency of a transformer is in the range of 96–99%. It is of no use trying to determine it by measuring the output and input under load conditions because the wattmeter readings are liable to have an error of 1–2%.
The best and accurate method of determining efficiency would be to find the losses from the OC and SC tests.
With this data efficiency can then be calculated as h = output
output losses+ = 1
-+ losses
output losses (3.54)
In Eq. (3.54) the effect of meter readings is confined to losses only so that the overall efficiency as obtained from it is far more accurate than that obtained by direct loading.
Various losses in a transformer have already been enumerated in Sec. 3.6 and the two important losses (iron-loss Pi and copper-loss Pc) are shown in Fig. 3.29 of a loaded transformer.
V1
Fig. 3.29(a) Transformer on load
The iron (core) losses depend upon the flux density and so on the induced emf. As E1ª V1 at all loads, these losses can be regarded as constant (independent of load) for constant primary voltage.
Copper losses in the two windings are
Pc = I21 R1 + I22 R2
= I22 Req(2)
where Req(2) = equivalent resistance referred to the secondary side. Thus it is found that copper losses vary as the square of the load current.
Transformer output, P0 = V2 I2 cos q2 where V2 is the rated secondary voltage. It varies slightly with the load but the variation is so small (about
3-5%) that it can be neglected for computing efficiency.
Equation (3.55) shows that for a given power factor, efficiency varies with load current. It can be written as
h = V
For maximum value of h for given cos q2 (pf), the denominator of Eq. (3.56) must have the least value. The condition for maximum h, obtained by differentiating the denominator and equating it to zero, is
I22 Req(2) = Pi (3.57)
or Copper-loss (variable) = core-loss (constant)
It means that the efficiency is maximum at a load when the copper-loss (variable loss) equals the core-loss (constant loss). Thus for maximum efficiency,
I22 = P
Thus the efficiency is maximum at a fractional load current given by Eq. (3.59a). Multiplying the numerator and denominator of Eq. (3.59a) by rated V2
k = V I V I fl
2 2
2 2 (3.59b)
Thus the maximum efficiency is given at a load k(V2I2fl) or kS2, where S2, is VA (or kVA) rating of the transformer. The expression for maximum efficiency is given by
hmax = kS
It can be easily observed from Eq. (3.60) that hmax increases with increasing pf (cos q2) and is the highest at unity pf. Also hmax = 0 when cos q = 0, i.e. at zero pf (purely reactive load). Therefore, knowledge of transformer losses is as important as its efficiency.
Power transformers used for bulk power transmission are operated near about full load at all times and are therefore designed to have maximum efficiency at full-load. On the other hand, the distribution transformers supply load which varies over the day through a wide range.
Such transformers are, therefore, designed to have maximum efficiency at about three-fourths the full load.
Normally transformer efficiency is maximum when the load power factor is unity. From Fig. 3.29(b), it is seen that the maximum efficiency
occurs at same load current independent of power I2(rated)
Load
factor, because the total core loss Pc, and equivalent resistance Req(2) are not affected by load power factor. Any way reduction of load power factor reduces the transformer output and the transformer efficiency also reduced.
The all-day efficiency of a transformer is the ratio of the total energy output (kWh) in a 24-h day to the total energy input in the same time. Since the core losses are constant independent of the load, the all-day efficiency of a transformer is dependent upon the load cycle; but no prediction can be made on the basis of the load factor (average load/peak load). It is an important figure of merit for distribution transformers which feed daily load cycle varying over a wide load range. Higher energy efficiencies are achieved by designing distribution transformers to yield maximum (power) efficiency at less than full load (usually about 70% of the full load). This is achieved by restricting the core flux density to lower values by using a relatively larger core cross-section. (It means a larger iron/copper weight ratio.)
EXAMPLE 3.11 For the transformer of Example 3.7 calculate the efficiency if the LV side is loaded fully at 0.8 power factor. What is the maximum efficiency of the transformer at this power factor and at what pu load would it be achieved?
SOLUTION
Power output = V2I2 cos q2
= 200 ¥ 100 ¥ 0.8 = 16000 W (independent of lag/lead)
Total loss PL = Pi + k2Pc
= 120 + 1 ¥ 300 = 420 W
\ h = 1 – P
P0+LPL
= 1 – 420
16000 420+ = 97 44%
For maximum efficiency
k = P
Pci = 120
300 = 0.632 i.e. at 0.632 pu load (this is independent of power factor). Now
hmax (cos q2 = 0.8) = 1 – 2
0 2 P P +iPi
= 1 – 2 120 16000 0 632 2 120
¥
¥ . + ¥
= 97.68%
EXAMPLE 3.12 A 500 kVA transformer has an efficiency of 95% at full load and also at 60% of full load; both at upf .
(a) Separate out the losses of the transformer.
(b) Determine the efficiency of the transformer at 3/4th full load.
SOLUTION
(a) h = 500 1
500 1
¥
¥ + +P Pi c = 0.95 (i)
Also 500 0 6
EXAMPLE 3.13 A transformer has its maximum efficiency of 0.98 at 15 kVA at upf. Compare its all-day efficiencies for the following load cycles:
(a) Full load of 20 kVA 12 hours/day and no-load rest of the day.
(b) Full load 4 hours/day and 0.4 full-load rest of the day.
Assume the load to operate on upf all day.
SOLUTION
Even though the load factor is the same in each case [(240/24)/20 = 0.5] the all day efficiencies still differ because of the difference in the nature of the two load cycles.
Voltage Regulation
Constant voltage is the requirement of most domestic, commercial and industrial loads. It is, therefore, necessary that the output voltage of a transformer must stay within narrow limits as the load and its power factor vary. This requirement is more stringent in distribution transformers as these directly feed the load centres. The voltage drop in a transformer on load is chiefly determined by its leakage reactance which must be kept as low as design and manufacturing techniques would permit.
The figure of merit which determines the voltage drop characteristic of a transformer is the voltage regulation. It is defined as the change in magnitude of the secondary (terminal) voltage, when full-load (rated load) of specified power factor supplied at rated voltage is thrown off, i.e. reduced to no-load with primary voltage (and frequency) held constant, as percentage of the rated load terminal voltage. In terms of symbols
% Voltage regulation = V V V
fl fl
20 2
2
- .
. ¥ 100 (3.61)
where V2, fl = rated secondary voltage while supplying full load at specified power factor V20 = secondary voltage when load is thrown off.
Figure 3.30(a) shows the transformer equivalent circuit* referred to the secondary side and Fig. 3.30(b) gives its phasor diagram. The voltage drops IR and IX are very small in a well-designed transformer (refer Example 3.6). As a result the angle d between V1 and V2 is of negligible order, so that**
V1ª OE
V1 – V2ª BE = I(R cos f + X sin f); f lagging (3.62a)
= I(R cos f – X sin f); f leading (3.62b)
V1
R X l
+
– –
+
Load O d
f
f
f B
IR IX
I
A
E
(a) Equivalent circuit referred to secondary (b) Phasor diagram (not proportional)
V2 V2
V1
Fig. 3.30 When the load is thrown off
V20 = V1
\ V20 – V2 = I(R cos f ± sin f) (3.63)
* In approximate equivalent circuit, the magnetizing shunt branch plays no role in determining voltages and hence is left out.
** V1 could be calculated from the phasor equation
V1 = V2 + I –f (R + jX)
but the approximate method is very much quicker and quite accurate (see Example 3.6).
where I is the full-load secondary current and V2, the full-load secondary voltage (equal to the value of V2 It is seen from Eq. (3.64) that the voltage regulation varies with power factor and has a maximum value when Equation (3.66) implies that voltage regulation is the maximum when the load power factor (lagging) angle has the same value as the angle of the equivalent impedance. From Eq. (3.64), the voltage regulation is zero when
R cos f – X sin f = 0
or tan f = R
X ; leading (3.67) For f (leading) larger than that given by Eq. (3.67), the voltage regulation is negative (i.e.
the secondary full-load voltage is more than the no-load voltage).
The complete variation of % regulation with power factor is shown in Fig. 3.31.
EXAMPLE 3.14 Consider the transformer with data given in Example 3.7.
(a) With full-load on the LV side at rated voltage, calculate the excitation voltage on the HV side. The load power factor is (i) 0.8 lagging, (ii) 0.8 leading. What is the voltage regulation of the transformer in each case?
(b) The transformer supplies full-load current at 0.8 lagging power factor with 2000 V on the HV side.
Find the voltage at the load terminals and the operating efficiency.
% Reg
Fig. 3.31 Percentage regulation versus power factor; R =
SOLUTION
(a) The HV side equivalent circuit of Fig. 3.26(a) will be used.
VL = 200 V, IL = 200 1000
200
¥ = 100 A
VL¢ = 2000 V, IL¢ = 10 A
Now VH = VL¢ + I¢L (RH cos f ± XH sin f); RH =: 30 W
XH = 5.2 W
(i) cos f = 0.8 lagging, sin f = 0.6
VH = 2000 + 10(3 ¥ 0.8 + 5.2 ¥ 0.6) = 2055.2 V Voltage regulation = 2055 2 2000
2000
. - ¥ 100 = 2.76%
(ii) cos f = 0.8 leading, sin f = 0.6
VH = 2000 + 10 (3 ¥ 0.8 – 5.2 ¥ 0.6) = 1992.8 V Voltage regulation = 1992 8 2000
2000
. - ¥ 100 = –0.36%
(b) IL(full-load) = 100 A, 0.8 lagging pf
V¢L = VH¢ – I¢L (RH cos f + XH sin f)
= 2000 – 10(3 ¥ 0.8 + 5.2 ¥ 0.6) = 1944.8 V
or VL = 194.48 V
Efficiency
Output, P0 = VLIL cos f
= 194.48 ¥ 100 ¥ 0.8 = 15558.4 W
PLOSS = Pi + Pc
Pi = 120 W (Ex. 3.7)
Pc = (10)2¥ 3 = 300 W
\ PLOSS = 420 W
h = 1 – 420
15558 4 420. ¥ = 97.38%
Example 3.14 is solved by writing the following MATLAB code.
clc clear S=20*1000;
V1=200;
V2=2000;
I1=S/V1;
I2=S/V2;
RH=3;
XH=5.2;
Cosine-phi=0.8;
Sin-phi=0.6;
VH=V2+I2*(RH*cosine-phi+XH*sin-phi) Vreg=(VH-V2)*100/V2
%% case2
VH=V2+I2*(RH*cosine-phi-XH*sin-phi) Vreg=(VH-V2)*100/V2
Il=100;
Vll=V2-I2*(RH*cosine-phi+XH*sin-phi);
Vl=Vll/10
Ploss=120+10*10*3;
Pop=Vl*Il*cosine-phi;
eff=(1-(Ploss/(Ploss+Pop)))*100 Answer:
VH = 2.0552e+003 Vreg = 2.7600 VH = 1.9928e+003 Vreg = –0.3600 Vl = 194.4800 eff = 97.3715
EXAMPLE 3.15 For the 150 kVA, 2400/240 V transformer whose circuit parameters are given in Example 3.8, draw the circuit model as seen from the HV side. Determine therefrom the voltage regulation and efficiency when the transformer is supplying full load at 0.8 lagging pf on the secondary side at rated voltage. Under these conditions calculate also the HV side current and its pf.
SOLUTION
R(HV) = 0.2 + 2 ¥ 10–3¥ (10)2 = 0.4 W X(HV) = 0.45 + 4.5 ¥ 10–3¥ (10)2 = 0.9 W The circuit model is drawn in Fig. 3.32.
I2( fl) = 150 1000 240
¥ = 625 A, 0.8 pf lagging
V2 = 240 V I2 = 625
10 = 62.5 A, 0.8 pf lagging V2¢ = 2400 V
Voltage drop = 62.5(0.4 ¥ 0.8 + 0.9 ¥ 0.6)
= 53.75 V
Voltage regulation = 53 75 2400
. ¥ 100 = 2.24%
V1 = 2400 + 53.75 = 2453.75 = 2454 V P(out) = 150 ¥ 0.8 = 120 kW
Pc(copper loss) = (62.5)2¥ 0.4 = 1.56 kW Pi(core loss) = (2454)
10 1000
2
¥ = 0.60 kW
PL = Pi + Pc = 0.60 + 1.56 = 2.16 kW
I1 +
I0
V¢2 V1
I¢2 0.4W 0.9W
+
–
10 kW
1.6 kW
– Fig. 3.32
h = 120