Demonstrative Video
VIDEO
Problem Statement
The magnetic structure of a synchronous machine has the stator and
rotor iron of infinite permeability (\(\mu_r=\infty\)). Find
air-gap flux (\(\phi\))
flux density (\(B_g\)).
Consider \(I=10~A\), \(N=1000\) turns, \(g=1\) cm, and \(A_g=2000~\text{cm}^2\).
Solution
Solution-1
Notice that there are two air gaps in series, of total length
\(2 g,\) and that by symmetry the flux
density in each is equal.
Since the iron \(\mu_r=\infty\),
its reluctance is negligible and \(g\)
replaced by the total gap length \(2
g\) can be used to find the flux
\[\begin{aligned}
\phi&=\frac{N I \mu_{0} A_{g}}{2 g}=\frac{1000(10)\left(4 \pi \times
10^{-7}\right)(0.2)}{0.02}=0.13 \mathrm{~Wb} \\
B_{g}&=\frac{\phi}{A_{g}}=\frac{0.13}{0.2}=0.65 \mathrm{~T}
\end{aligned}\]
Problem Statement
The magnetic circuit consists of an N-turn winding on a magnetic core
of infinite permeability with two parallel air gaps of length \(g_1\) and \(g_2\), and areas \(A_1\) and \(A_2\), respectively. Find
the inductance of the winding and
the flux density \(B_1\) in
gap-1 when the winding is carrying a current \(i\)
Neglect fringing effects at the air-gap.
Solution
\[\begin{array}{c}
\phi_{1}=\dfrac{N i}{\mathcal{R}_{1}}=\dfrac{\mu_{0} A_{1} N
i}{g_{1}} \\
B_{1}=\dfrac{\phi_{1}}{A_{1}}=\dfrac{\mu_{0} N i}{g_{1}}
\end{array}\]
Solution-2
\[\phi=\frac{N
i}{\frac{R_{1} R_{2}}{R_{1}+R_{2}}} \quad \mathcal{R}_1 =
\dfrac{g_1}{\mu_0A_1} \quad
\mathcal{R}_2=\dfrac{g_2}{\mu_0A_2}\]
\[\begin{aligned}
L &=\frac{\lambda}{i}=\frac{N
\phi}{i}=\frac{N^{2}\left(\mathcal{R}_{1}+\mathcal{R}_{2}\right)}{\mathcal{R}_{1}
\mathcal{R}_{2}}
=\mu_{0}
N^{2}\left(\frac{A_{1}}{g_{1}}+\frac{A_{2}}{g_{2}}\right)
\end{aligned}\]
Problem Statement
Find
Solution
\[W=\frac{1}{2} L
i^{2}=\frac{1}{2}(0.56)(0.80)^{2}=0.18 \mathrm{~J}\]
\[\begin{aligned}
e &=\frac{d \lambda}{d t}=N \frac{d \varphi}{d t}=N A_{c}
\frac{d B_{\mathrm{c}}}{d t} \\
&=500 \times\left(9 \times 10^{-4}\right) \times(377 \times
1.0 \cos (377 t)) \\
&=170 \cos (377 t) \quad V
\end{aligned}\]
Solution-3
\[\begin{aligned}
L &=\frac{\lambda}{i}=\frac{N
\phi}{i}=\frac{N^{2}}{\mathcal{R}_{\mathrm{c}}+\mathcal{R}_{g}} \\
&=\frac{500^{2}}{4.46 \times 10^{5}}=0.56 \mathrm{H}
\end{aligned}\]
Problem Statement
\[\begin{aligned}
A_{c}&=A_{g}=9~\mathrm{cm}^{2}\\
g&=0.050 \mathrm{~cm}\\
l_{c}&=30 \mathrm{~cm} \\
N&=500~ \text{turns}
\end{aligned}\]
The core material has the dc magnetization curve as shown in the Fig.
Find the current \(I\) required to
produce \(B_c = 1 ~T\).
Solution
Solution-4
\[H_{\mathrm{c}}=11
\mathrm{~A} \cdot \text { turns } / \mathrm{m}\]
\[\mathcal{F}_{\mathrm{c}}=H_{\mathrm{c}}
l_{\mathrm{c}}=11(0.3)=3.3 \mathrm{~A} \cdot \text { turns
}\]
\[\mathcal{F}_{g}=H_{g} g=\frac{B_{g}
g}{\mu_{0}}=\frac{5 \times 10^{-4}}{4 \pi \times 10^{-7}}=396
\mathrm{~A} \cdot \text { turns }\]
\[i=\frac{\mathcal{F}_{\mathrm{c}}+\mathcal{F}_{\mathrm{g}}}{N}=\frac{399}{500}=0.80
\mathrm{~A}\]
Problem Statement
Given: Core \(A_{c}=1.8 \times 10^{-3}
\mathrm{~m}^{2},\) \(I_{c}=0.6
\mathrm{~m},\) \(g=2.3 \times 10^{-3}
\mathrm{~m}, ~N=83\) turns. Assume core is of \(\mu=\infty\) and neglect the effects of
fringing fields at the air gap and leakage flux.
Calculate \(\mathcal{R}_{\mathrm{c}}\) and \(\mathcal{R}_{\mathrm{g}}\)
total flux \(\phi\), flux
linkages \(\lambda\) of the coil, and
coil inductance L.
\(N\) required to achieve an
inductance of \(12
\mathrm{mH}\)
the inductor current which will result in \(B_c\) of \(1.0
\mathrm{~T}\).
Solution
\[\begin{aligned}
L&=\dfrac{N^{2}}{R_{g}}=12 \times 10^{-3} \mathrm{mH} \\
&\Rightarrow N=\sqrt{12 \times 10^{-3} \times 1.017 \times
10^{6}}=110.47 \Rightarrow N=110 \text { turns } \\
B_{c}&=B_{g}=1.0 \mathrm{~T} \Rightarrow \phi=B_{g} A_{c}=1.8
\times 10^{-3} \mathrm{~Wb} \\
i&=\dfrac{\lambda}{L}=\dfrac{N \phi}{L}=\frac{110 \times 1.8
\times 10^{-3}}{12 \times 10^{-3}}=16.5 \mathrm{~A}
\end{aligned}\]
Solution-5
\[\begin{array}{l}
R_{g}=\dfrac{g}{\mu_{0} A_{c}}=\dfrac{2.3 \times 10^{-3}}{4 \pi
\times 10^{-7} \times 1.8 \times 10^{-3}}=1.017 \times 10^{6}
\mathrm{~A} / \mathrm{Wb}
\end{array}\]
\(R_{c}=0 \text { since } \mu \rightarrow
\infty\)
Problem Statement
find the flux density and flux in each of the outer limbs and the
central limbs. Assume \(\mu_r\) for
iron of the core to be (a) \(\infty\)
(b) 4500.
Solution
\[\begin{aligned}
B_{1}&=\dfrac{0.785 \times 10^{-3}}{25 \times 10^{-4}}=0.314
\mathrm{~T} \\
\phi_{2}&=\dfrac{500}{0.3183 \times 10^{6}}=1571
\mathrm{mWb} \\
B_{2}&=\dfrac{157 \times 10^{-3}}{25 \times 10^{-4}}=0.628
\mathrm{~T} \\
\phi&=\phi_{1}+\phi_{2}=2.356 \mathrm{mWb} \\
B&=\dfrac{2.356 \times 10^{-3}}{50 \times 10^{-4}}=0.471
\mathrm{~T}
\end{aligned}\]
\(\mu_{\mathrm{r}}=4,500 .\) The
corresponding analogous electrical circuit is given in Fig. Effect of
air-gaps on iron path length is negligible.
\[\begin{aligned}
l_{\mathrm{cl}}&=l_{\mathrm{C} 2}=(40+5)+2 \times(30+5+2.5)=120
\mathrm{~cm} \\
R_{\mathrm{cl}}&=R_{\mathrm{C} 2}=\frac{120 \times 10^{-2}}{4
\pi \times 10^{-7} \times 4,500 \times 25 \times 10^{-4}}=0.085 \times
10^{6} \mathrm{AT} / \mathrm{Wb} \\
l_{\mathrm{c} 3}&=40+5=45 \mathrm{~cm} \\
R_{\mathrm{c} 3} &=\frac{45 \times 10-2}{4 \pi \times 10^{-7}
\times 4,500 \times 50 \times 10^{-4}}=0.016 \times 10^{6} \mathrm{AT} /
\mathrm{Wb} \\
R_{\mathrm{eq}} &=\left[\left(R_{\mathrm{cl}}+R_{\mathrm{g}
1}\right) \|\left(R_{\mathrm{c} 2}+R_{\mathrm{g}
2}\right)\right]+R_{\mathrm{c} 3} \\
R_{\mathrm{cl}}+R_{\mathrm{gl}} &=0.6366+0.084=0.7206 \times
10^{6} \\
R_{\mathrm{c} 2}+R_{\mathrm{g} 2} &=0.3183+0.085=0.4033 \times
10^{6}
\end{aligned}\]
\[\begin{aligned}
R_{\mathrm{eq}}&=\left[\dfrac{0.7206 \times
0.4033}{(0.7206+0.4033)}+0.016\right] \times 10^{6}=0.2746 \times 10^{6}
\mathrm{AT} / \mathrm{Wb} \\
\phi&=\frac{500}{0.2742 \times 10^{6}}=1823 \mathrm{mWb} \\
B&=\dfrac{1.823 \times 10^{-3}}{50 \times 10^{-4}}=0.365 \mathrm{~T}
\\
\phi_{1}&=1.823 \times \dfrac{0.4033}{1.1239}=0.654 \mathrm{mWb} \\
B_{1}&=\dfrac{0.653 \times 10^{-3}}{25 \times 10^{-4}}=0.261
\mathrm{~T} \\
\phi_{2}&=1.823 \times \frac{0.7206}{1.1239}=1.17 \mathrm{mWb} \\
B_{2}&=\dfrac{1.17 \times 10^{-3}}{25 \times 10^{-4}}=0.468
\mathrm{~T}
\end{aligned}\]
Solution-6
\[\begin{aligned}
\mu_{\mathrm{r}}&=\infty \\
N_{1}&=1000 \times 0.5=500 \mathrm{AT}\\
R_{\mathrm{g} 1}&=\frac{2 \times 10^{-3}}{4 \pi \times 10^{-7}
\times 25 \times 10^{-4}}\\
&=0.6366 \times 10^{6} \mathrm{AT} / \mathrm{Wb} \\
R_{\mathrm{g} 2}&=\frac{1 \times 10^{-3}}{4 \pi \times 10^{-7}
\times 25 \times 10^{-4}}\\
&=0.3183 \times 10^{6} \mathrm{AT} / \mathrm{Wb}
\end{aligned}\]
Problem Statement
In the magnetic circuit of Fig., the \(\mu_r\) of the ferromagnetic material is
1200. Neglect magnetic leakage and fringing. All dimensions are in
centimetres, and the magnetic material has a square cross-sectional
area. Determine the air gap flux, the air gap flux density, and the
magnetic held intensity in the air gap.
Solution
\[\begin{array}{c}
\Phi_{1}\left(\mathscr{R}_{\text {bafe
}}+\mathscr{R}_{\mathrm{be}}+\mathscr{R}_{\mathrm{g}}\right)+\Phi_{2}\left(\mathscr{R}_{\mathrm{be}}+\mathscr{R}_{\mathrm{g}}\right)=F_{1}
\\
\Phi_{1}\left(\mathscr{R}_{\mathrm{be}}+\mathscr{R}_{\mathrm{g}}\right)+\Phi_{2}\left({R}_{\text
{bcde }}+\mathscr{R}_{\mathrm{be}}+\mathscr{R}_{\mathrm{g}}\right)=F_{2}
\\
\Phi_{1}\left(13.34 \times 10^{6}\right)+\Phi_{2}\left(10.76
\times 10^{6}\right)=5000 \\
\Phi_{1}\left(10.76 \times 10^{6}\right)+\Phi_{2}\left(13.34
\times 10^{6}\right)=5000 \\
\Phi_{1}=\Phi_{2}=2.067 \times 10^{-4} \mathrm{~Wb}
\end{array}\]
The loop equations are
\[\Phi_{\mathrm{g}}=\Phi_{1}+\Phi_{2}=4.134 \times
10^{-4} \mathrm{~Wb}\]
\[H_{\mathrm{g}}=\frac{B_{\mathrm{g}}}{\mu_{0}}=\frac{1.034}{4
\pi 10^{-7}}=0.822 \times 10^{6} \mathrm{At} / \mathrm{m}\]
\[B_{\mathrm{g}}=\frac{\Phi_{\mathrm{g}}}{A_{\mathrm{g}}}=\frac{4.134
\times 10^{-4}}{4 \times 10^{-4}}=1.034 \mathrm{~T}\]
The air gap flux is
Solution-7
\[\mathscr{R}_{\text {bcde
}} =\mathscr{R}_{\text {bafe }}\]
From symmetry:
\[\begin{aligned}
\Re_{g} & =\dfrac{l_{\text{g}}}{\mu_{0}A_{g}}\\
= &
\dfrac{5\times10^{-3}}{4\pi\times10^{-7}\times2\times2\times10^{-4}}\\
= &9.94\times10^{6}~\text{At/Wb}
\end{aligned}\]
\[\begin{aligned}
\Re_{\text{be(core)}} &
=\dfrac{l_{\text{be(core)}}}{\mu_{c}A_{c}}\\
= & \dfrac{51.5\times10^{-2}}{1200\times4\pi\times
10^{-7}\times4\times10^{-4}}\\
= & 0.82\times10^{6}~\text{At/Wb}
\end{aligned}\]
