Condition for Maximum Efficiency

1. Condition for Maximum Efficiency

       When a transformer works on a constant input voltage and frequency then efficiency varies with the load. As load increases, the efficiency increases. At a certain load current, it achieves a maximum value. If the transformer is loaded further the efficiency starts decreasing. The graph of efficiency against load current  I₂ is shown in the Fig.1

Fig. 1

       The load current at which the efficiency attains maximum value is denoted as I_2m and maximum efficiency is denoted as η_max.

       Let us determine,

1. Condition for maximum efficiency.

2. Load current at which η_max occurs.

3. KVA supplied at maximum efficiency.

       The efficiency is a function of load i.e. load current I₂ assuming cos Φ constant. The secondary terminal voltage V₂ is also assumed constant. So for maximum efficiency,

              dη /d I₂ = 0

Now      η = (V₂ I₂ cos Φ₂ )/(V₂ I₂ cos Φ₂ + P_i  + I₂² R_2e)

.^..       (V₂ I₂ cos Φ₂ + P_i  + I₂² R_2e)(V₂ cos Φ₂) - (V₂ I₂ cos Φ₂)(V₂ cos Φ₂ + 2I₂  R_2e) = 0

       Cancelling (V₂ cos Φ2) from both the terms we get,

          V₂ I₂ cos Φ₂ + P_i  +I₂² R_2e - V₂ I₂  Φ₂ - 2I₂² R_2e = 0

.^..       P_i  - I₂² R_2e= 0

.^..       P_i  = I₂² R_2e = P_cu

       So condition to achieve maximum efficiency is that,

               Copper losses = Iron losses

1.1 Load Current I_2m at Maximum Efficiency

For η_max,           I₂² R_2e = P_i  but   I₂ = I_2m

                          I_2m² R_2e = P_i

                         I_2m = √(P_i / R_2e)

      This is the load current at η_max,

Let        (I₂)F.L. = Full load current

.^..          I_2m /(I₂) F.L.= (1/(I₂) F.L.)√(P_i / R_2e)

.^..           I_2m /(I₂) F.L.=   √(P_i )/({(I₂) F.L.}² R_2e)

                                 =  √(P_i )/((P_cu) F.L.)

.^..            I_2m = (I₂ )F.L.√(P_i )/((P_cu) F.L.)

       This is the load current at η_max interms of full load current.

1.2 KVA supplied at maximum Efficiency

       For constant V₂ the KVA supplied is the function of.

       KVA at η_max = I_2m V₂= V₂ (I₂) F.L. x √(P_i) /((P_cu)F.L.)

      KVA at η_max = (KVA rating) x √(P_i) /((P_cu)F.L.)

      Substituting condition for in the expression of efficiency, we can write expression for η_max as,

Example : A 250 KVA single phase transformer has iron loss of 1.8 KW. The full load copper loss is 2000 watts. Calculate

i) Efficiency at full load, 0.8 lagging p.f.

ii) KVA supplied at maximum eficiency

iii) Maximum efficiency at 0.8 lagging p.f.

Solution : The given values are,

                P_i = 1800 W , (P_cu)F.L. = 2000 W

i)

                                                          = 98.135%

ii)                 KVA at = KVA rating x √(P_i) /((P_cu)F.L.)

                                = 250 x √(1800/2000)

                                 = 237.1708 KVA

iii)

where        P_cu = P_i =1800W

                                                                        = 98.137%

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