Magnetic Circuits Calculations

Part 4: Examples and Calculations

Example 6.13.

A laminated soft iron ring of relative permeability 1000 has a mean circumference of 800 mm and a cross-sectional area 500 mm${}^2$. A radial air-gap of 1 mm width is cut in the ring which is wound with 1000 turns. Calculate the current required to produce an air-gap flux of 0.5 mWb if leakage factor is 1.2 and stacking factor 0.9. Neglect fringing.

Solution. $\text{Total AT reqd.}=\frac{{\Phi }_g{l}_g}{{\mu }_0{A}_g}+\frac{{\Phi }_i{l}_i}{{\mu }_0{\mu }_r{A}_i}$

Now, air-gap flux ${\Phi }_g=0.5\text{ mWb}=0.5\times 10^{-3}\text{ Wb}$; $l_g=1\text{ mm}=1\times 10^{-3}\text{ m}$; $A_g=500{\text{ mm}}^2=500\times 10^{-6}{\text{ m}}^2$ Flux in the iron ring, ${\Phi }_i=1.2\times 0.5\times 10^{-3}\text{ Wb}$ Net cross-sectional area $=A_i\times \text{stacking factor}=500\times 10^{-6}\times 0.9{\text{ m}}^2$ $\therefore \text{total AT reqd.}=\frac{0.5\times 10^{-3}\times 1\times 10^{-3}}{4\pi \times 10^{-7}\times 500\times 10^{-6}}+\frac{1.2\times 0.5\times 10^{-3}\times 800\times 10^{-3}}{4\pi \times 10^{-7}\times 1000\times (0.9\times 500\times 10^{-6})}=1644$ $\therefore I=1644/1000=1.64\text{ A}$

Example 6.14.

A ring has a diameter of 21 cm and a cross-sectional area of 10 cm${}^2$. The ring is made up of semicircular sections of cast iron and cast steel, with each joint having a reluctance equal to an air-gap of 0.2 mm. Find the ampere-turns required to produce a flux of $8\times 10^{-4}$ Wb. The relative permeabilities of cast steel and cast iron are 800 and 166 respectively. Neglect fringing and leakage effects. (Elect. Circuits, South Gujarat Univ.)

Solution. $\Phi =8\times 10^{-4}\text{ Wb}$; $A=10{\text{ cm}}^2=10^{-3}{\text{ m}}^2$; $B=8\times 10^{-4}/10^{-3}=0.8{\text{ Wb/m}}^2$

Air gap $H=B/{\mu }_0=0.8/4\pi \times 10^{-7}=6.366\times 10^5\text{ AT/m}$ Total air-gap length $=2\times 0.2\text{ mm}=0.4\text{ mm}$ $=4\times 10^{-4}\text{ m}$ $\therefore \text{AT required}=H\times l=6.366\times 10^5\times 4\times 10^{-4}=255$

Cast Steel Path (Fig. 6.34) $H=B/{\mu }_0{\mu }_r=0.8/4\pi \times 10^{-7}\times 800=796\text{ AT/m}$ path $=\pi D/2=21\pi /2\text{ cm}=0.33\text{ m}$ AT required $=H\times l=796\times 0.33=263$

Cast Iron Path $H=0.8/4\pi \times 10^{-7}\times 166=3,835\text{ AT/m}$; path $=0.33\text{ m}$ AT required $=3,835\times 0.33=1265$

Total AT required $=255+263+1265=1783.$

Example 6.17.

A rectangular iron core is shown in Fig. 6.35. It has a mean length of magnetic path of 100 cm, cross-section of $(2\text{ cm}\times 2\text{ cm})$, relative permeability of 1400 and an air-gap of 5 mm cut in the core. The three coils carried by the core have number of turns $N_a=335,N_b=600$ and $N_c=600$; and the respective currents are $1.6\text{ A},4\text{ A}$ and $3\text{ A}$. The directions of the currents are as shown. Find the flux in the air-gap. (F.Y. Engg. Pune Univ.)

Solution. By applying the Right-Hand Thumb rule, it is found that fluxes produced by the current $I_a$ and $I_b$ are directed in the clockwise direction through the iron core whereas that produced by current $I_c$ is directed in the anticlockwise direction through the core. $\therefore \text{total m.m.f.}=N_a{I}_a+N_b{I}_b-N_c{I}_c=335\times 1.6+600\times 4-600\times 3=1136\text{ AT}$ $\text{Reluctance of the air-gap }=\frac{l}{{\mu }_{0}A}=\frac{5\times 10^{-3}}{4\pi \times 10^{-7}\times 4\times 10^{-4}}=9.946\times 10^6\text{ AT/Wb}$ $\text{Reluctance of the iron path }=\frac{l}{{\mu }_0{\mu }_{r}A}=\frac{100-(0.5)\times 10^{-2}}{4\pi \times 10^{-7}\times 1400\times 4\times 10^{-4}}=1.414\times 10^6\text{ AT/Wb}$ $\text{Total reluctance }=(9.946+1.414)\times 10^6=11.36\times 10^6\text{ AT/Wb}$

The flux in the air-gap is the same as in the iron core. $\text{Air-gap flux }=\frac{\text{m.m.f.}}{\text{reluctance}}=\frac{1136}{11.36\times 10^6}=100\times 10^{-6}\text{ Wb}=100\mu \text{Wb}$

Example 6.19.

A magnetic circuit made of mild steel is arranged as shown in Fig. 6.36. The central limb is wound with 500 turns and has a cross-sectional area of 800 mm${}^2$. Each of the outer limbs has a cross-sectional area of 500 mm${}^2$. The air-gap has a length of 1 mm. Calculate the current required to set up a flux of 1.3 mWb in the central limb assuming no magnetic leakage and fringing. Mild steel required 3800 AT/m to produce flux density of 1.625 T and 850 AT/m to produce flux density of 1.3 T. (F.Y. Engg. Pune Univ.)

Solution. Flux in the central limb $=1.3\text{ mWb}=1.3\times 10^{-3}\text{ Wb}$ Cross section $A=800{\text{ mm}}^2=800\times 10^{-6}{\text{ m}}^2$ $\therefore B=\frac{\Phi }{A}=\frac{1.3\times 10^{-3}}{800\times 10^{-6}}=1.625\text{ T}$

Corresponding value of $H$ for this flux density is given as $3800\text{ AT/m}$. Since the length of the central limb is $120\text{ mm}$. m.m.f. required is $=H\times l=3800\times (120\times 10^{-3})=456\text{ AT/m}.$

Air-gap Flux density in the air-gap is the same as that in the central limb. $H=B/{\mu }_0=1.625/4\pi \times 10^{-7}=0.1293\times 10^7\text{ AT/m}$ Length of the air-gap $=1\text{ mm}=10^{-3}\text{ m}$ m.m.f. reqd. for the air-gap $=H\times l=0.1293\times 10^7\times 10^{-3}=1293\text{ AT.}$

The flux of the central limb divides equally at point $A$ in figure along the two parallel path $ABCD$ and $AFED$. We may consider either path, say $ABCD$ and calculate the m.m.f. required for it. The same m.m.f. will also send the flux through the other parallel path $AFED$.

Flux through $ABCD=1.3\times 10^{-3}/2=0.65\times 10^{-3}\text{ Wb}$ Flux density $B=0.65\times 10^{-3}/500\times 10^{-6}=1.3\text{ T}$ The corresponding value of $H$ for this value of $B$ is given at $850\text{ AT/m}$. $\therefore \text{m.m.f. reqd. for path }ABCD=H\times l=850\times (300\times 10^{-3})=255\text{ AT}$ As said above, this, m.m.f. will also send the flux in the parallel path $AFED$. Total m.m.f. reqd. $=456+1293+255=2004\text{ AT}$ Since the number of turns is 500, $I=2004/500=4\text{ A.}$

Example 6.21.

A cast steel magnetic structure made for a bar $8\text{ cm}\times 2\text{ cm}$ is shown in Fig. 6.35. Determine the current that the 500 turn-magnetising coil on the left limb should carry so that a flux of $2\text{ mWb}$ is produced in the right limb. Take ${\mu }_r=600$ and neglect leakage. (Elect. Technology Allahabad Univ. 1993)

Solution. Since path $C$ and $D$ are in parallel with each other w.r.t. path $E$ (Fig. 6.38), the m.m.f. across the two is the same. ${\Phi }_1{S}_1={\Phi }_2{S}_2$ ${\Phi }_1\times \frac{15}{\mu A}={\Phi }_2\times \frac{25}{\mu A}$ $\therefore {\Phi }_1=10/3\text{ mWb}$ $\Phi ={\Phi }_1+{\Phi }_2=16/3\text{ mWb}$

Total AT required for the whole circuit is equal to the sum of (i) that required for path $E$ and (ii) that required for either of the two paths $C$ or $D$.

Flux density in path $E=\frac{16\times 10^{-3}}{3\times 4\times 10^{-4}}=\frac{40}{3}{\text{ Wb/m}}^2$ $\text{AT reqd.}=\frac{40/3}{4\pi \times 10^{-7}\times 600}\times 0.25=4,420$

Flux density in path $D=\frac{2\times 10^{-3}}{4\times 10^{-4}}=5{\text{ Wb/m}}^2$ $\text{AT reqd.}=\frac{5}{4\pi \times 10^{-7}\times 600}\times 0.25=1658$ Total AT $=4,420+1,658=6,078;$ Current needed $=6078/500=12.16\text{ A}$

Example 6.22.

A ring of cast steel has an external diameter of 24 cm and a square cross-section of 3 cm side. Inside and cross the ring, an ordinary steel bar $18\text{ cm}\times 3\text{ cm}\times 0.4\text{ cm}$ is fitted with negligible gap. Calculating the number of ampere-turns required to be applied to one half of the ring to produce a flux density of $1.0{\text{ weber per metre}}^2$ in the other half. Neglect leakage. The B-H characteristics are as below:

$B$ in Wb/m${}^2$	1.0	1.1	1.2
Amp-turn/m	900	1020	1220
For Cast Steel

$B$ in Wb/m${}^2$	1.2	1.4	1.45
Amp-turn/m	590	1200	1650
For Ordinary Plate

(Elect. Technology, Indore Univ.)

Solution. The magnetic circuit is shown in Fig. 6.39. The m.m.f. (or AT) produced on the half $A$ acts across the parallel magnetic circuit $C$ and $D$. First, total AT across $C$ is calculated and since these amp-turns are also applied across $D$, the flux density $B$ in $D$ can be estimated. Next, flux density in $A$ is calculated and therefore, the AT required for this flux density. In fact, the total AT (or m.m.f.) required is the sum of that required for $A$ and that of either for the two parallel paths $C$ or $D$.

Value of flux density in $C=1.0{\text{ Wb/m}}^2$ Mean diameter of the ring $=(24+18)/2=21\text{ cm}$

Mean circumference $=\pi \times 21=66\text{ cm}$ Length of path $A$ or $C=66/2=33\text{ cm}=0.33\text{ m}$ Value of AT/m for a flux density of $1.0{\text{ Wb/m}}^2$ as seen from the given $B$-$H$ characteristics $=900\text{ AT/m}$ $\therefore \text{Total AT for path }C=900\times 0.33=297$. The same ATs are applied across path $D$. Length of path $D=18\text{ cm}=0.18\text{ m}$. $\therefore \text{AT/m for path }D=297/0.18=1650$ Value of $B$ corresponding to this $\text{AT/m}$ from given table $=1.45{\text{ Wb/m}}^2$

Flux through $C=B\times A=1.0\times 9\times 10^{-4}=9\times 10^{-4}\text{ Wb}$ Flux through $D=1.45\times (3\times 0.4\times 10^{-4})=1.74\times 10^{-4}\text{ Wb}$ $\therefore \text{Total flux through }A=9\times 10^{-4}+1.74\times 10^{-4}=10.74\times 10^{-4}\text{ Wb}.$ Flux density through $A=10.74\times 10^{-4}/9\times 10^{-4}=1.193{\text{ Wb/m}}^2$ No. of AT/m reqd. to produce this flux density as read from the given table $=1200\text{ (approx.)}$ $\text{Amp-turns required for limb }A=1200\times 0.33=396$ Total AT required $=396+297=693$