Question III.4: (a) A rectangle shaped iron core, having cross-sectional are...
(a) A rectangle shaped iron core, having cross-sectional area of 9 cm², has an air gap of 0.01 cm and a length of 39.99 cm. The relative permeability of iron is 2000. A coil of 1000 turns is wound on one arm of the core. Calculate the amount of current that should flow in the winding to produce a flux of 1 mWb in the core. (permeabilty of air = 4π * 10^{−7} H/m)
(b) A 25 kVA, 2000/200 V transformer has iron loss of 350 W and full-load copper loss of 400 W. Calculate the efficiency of the transformer at full and half load at 0.8 power factor lagging.
(a) Total reluctance of the flux path = Reluctance of iron path + Reluctance of air gap,
\begin{aligned} i.e., \quad &S=S_{i}+S_{g} \\&S=\frac{1_{i}}{\mu_{0} \mu_{r} A}+\frac{1_{g}}{\mu_{0} A}\end{aligned}The iron path permeability is μ, which is equal to \mu_{0} \mu_{r} whereas for the air gap the permeability is \mu_{0} only.
Substituting the given values,
\begin{aligned}S &=\frac{39.09 \times 10^{-2}}{4 \pi \times 10^{-7} \times 2000 \times 9 \times 10^{-4}}+\frac{0.01 \times 10^{-2}}{4 \pi \times 10^{-7} \times 9 \times 10^{-4}} \\&=\frac{10^{6}}{4 \pi}\left[\frac{39.09}{18}+\frac{100}{10}\right]=\frac{295.45 \times 10^{5}}{36 \pi} \end{aligned}Flux, \phi=\frac{NI}{S}=\frac{1000 I}{S}
\therefore \quad I =\frac{\phi \times S}{1000}=\frac{1 \times 10^{-3} \times 295.45 \times 10^{5}}{36 \pi \times 1000}=\frac{29.545}{36 \pi}=0.26 A(b) Output in kVA = 25, output in kW = kVA cos Φ = 25 × 0.8 = 20 kW
Efficiency, \eta=\frac{\text { Output }}{\text { Input }} \times 100=\frac{\text { Output } \times 100}{\text { Output }+\text { Losses }}
Output in Watts = 20 × 1000 W
Core loss or iron loss = 350 W
Full-load copper loss = 400 W
\begin{aligned}\eta &=\frac{\text { Output } \times 100}{\text { Output }+\text { Core loss }+\text { Copper loss }} \\&=\frac{200 \times 1000 \times 100}{20 \times 1000+350+400}=96.4 \text { per cent }\end{aligned}At half load,
Output = 10 × 10³
Core loss = 350 W (remains constant at all loads)
Copper loss =\frac{400}{4} = 100 W (variable loss, varies as square of the load)
\eta=\frac{10 \times 1000 \times 100}{10 \times 1000+350+100}=95.7 per cent
