Question

The core of a 100-kVA, 11000/550-V, 50-Hz, 1-phase, core type transformer has a cross-sectional area of 20 cm x 20 cm. The maximum core flux density is not to exceed 1.3 Tesla. Assume a stacking factor of 0.9. (a) Determine the turns ratio, a. (b) Determine the cross-section area, Ac. (e) Determine the peak flux in the core, Pmax (d) Determine the primary number of turns, N. (e) Determine the secondary number of turns, N2. (1) Determine the emf/turn. (g) What will happen if the primary voltage is increased by 10% on no-load?

Answer 1

Given:

Power of transformer$=100 kVA$

cross-sectional area$=20cm\times20cm$

flux density $B_{m}=1.3 T$

stacking factor $k_{s}=0.9$

Primary emf $E_{1}=110000$

Secondary emf $E_{2}=550 V$

To Find: (a) turns ratio, (b) cross section area, (c) peak flux (d) primary turns

(e) secondary turns (f) Emf/ turn (g) primary voltage is increased

Solution:

(a) Turns ratio

$a=\dfrac{E_{1} }{E_{2} }\\=\dfrac{11000}{550}\\=\dfrac{20}{1}\\=20:1$

(b) Area of core

$A_{g}=20\times 20\\=400\ cm^{2}$

Area of cross-section area

$k_{s}=\dfrac{A_{c} }{A_{g} }\\A_{c}=0.9\times 400\\=360\ cm^{2}$

(c) peak flux in the core

$\phi=B_{max} \times A_{n}\\\=1.3\times360\times10^{-4}\\=0.0468\ Wb$

(d) Primary number of turns

$E=4.44\times B_{m}\times A_{c}\times f\times N_{1} \\N_{1} =\dfrac{E_{1} }{4.44\times B_{m} \times A_{c} \times f}$

$N_{1}=\dfrac{11000}{4.44\times 1.3\times 360\times10^{-4}\times 50 } \\\\=1059$

(e) Secondary number of turns

$N_{2}=\frac{E_{2} }{4.44\timesB \times B\times A_{c}\times f }\\=\frac{550 }{4.44\times1.3\times (360\times10^{-4} )\times50}\\\\=53$

(f) The Emf/turn

$=\dfrac{11000}{1059}\\=10.387$

(g) The primary voltage is increased by 10% then iron losses get increased by more than 10%. Since the iron losses depend on the supply voltage and if the core reaches saturation then magnetizing current increases.

Answer 2

Answer:

⁹

Explanation: