Tuesday, April 30, 2019
Monday, April 29, 2019
Wednesday, April 24, 2019
Wednesday, April 17, 2019
Monday, April 15, 2019
Sunday, April 14, 2019
Thursday, April 11, 2019
Monday, April 8, 2019
What is Useful flux, leakage factor, leakage flux ?
Q. With the help of neat sketch,
explain the following terms: Useful flux, leakage factor, leakage flux.
Ans.
Ans.
The flux which is available at the air gap and can be utilized for some useful
purpose is called as useful flux.
Useful flux (ϕu): The flux in an air gap is useful in many applications like motor, generator, moving coil, etc.
Leakage flux: The Total
flux produced by the coil does not pass through the air gap as some of the flux
leaks through the air surrounding the iron. This flux is called as leakage flux
and phenomenon is called magnetic leakage.
It is impossible to confine all the flux to the iron path only, as there is no perfect insulator for magnetic path.
It is impossible to confine all the flux to the iron path only, as there is no perfect insulator for magnetic path.
Leakage coefficient or leakage factor(λ) : The ratio of total flux (ϕt)to useful air gap flux(ϕu)is called leakage co-efficient.
Saturday, April 6, 2019
Friday, April 5, 2019
Numerical on Magnetic Circuits Hindi
Q. A Coil Of 500 turns resistance of 20 ohm is wound uniformly on an iron ring of mean circumference 50 cm. and cross- sectional area 4 cm2 . It is connected to 24 V d.c. supply, Relative permeability of the material = 800, Find (i) MMF (ii) Magnetising force (iii) Total flux (iv) Reluctance.
Ans: Given : No. of turns (N) = 500 Resistance (R) = 20 Ω Mean circumference = (pd) =50 cm = 0.5 m Cross Sectional area (a) = 4 cm2 =4x10-4 m2 Applied voltage (v) = 24 V Relative permeability (µr) = 800 To find : I) MMF IN ii) Magnetising force (H) iii) Total flux (ϕ) iv) Reluctance (S)
Soln. (i) mmf = IN I= V/R = 24/20 = 1.2 Amp
mmf = 1.2 x 500 mmf = 600 AT
(ii) Magnetising force H = mmf/l = 600/0.50 =1200 AT/ m
(iii) Total flux (ϕ) = mmf/s Mean circumference (pd) = l = 0.50 m
Reluctance (S) = l/(µoµr a) = 0.50/(4p x 〖10〗^(-7) x 800 x 4 x 〖10〗^(-4) ) = 1.24x106 AT/ Wb
(ϕ) = mmf/s = 1200 /1.24 x 106= 9.67 x 〖10〗^(-4) Webers = 0.967 x10-3 Wb = 0.967 mWb Total
flux (ϕ) = 0.967 mWb (iv) Reluctance (s) = 1.24x106 AT/ Wb
(i) MMF = 600 AT
(ii) Magnetising force =1200 AT / m
(iii) Total flux = 0.967 mWb
(iv) Reluctance = 1.24 x106 AT/ Wb
Ans: Given : No. of turns (N) = 500 Resistance (R) = 20 Ω Mean circumference = (pd) =50 cm = 0.5 m Cross Sectional area (a) = 4 cm2 =4x10-4 m2 Applied voltage (v) = 24 V Relative permeability (µr) = 800 To find : I) MMF IN ii) Magnetising force (H) iii) Total flux (ϕ) iv) Reluctance (S)
Soln. (i) mmf = IN I= V/R = 24/20 = 1.2 Amp
mmf = 1.2 x 500 mmf = 600 AT
(ii) Magnetising force H = mmf/l = 600/0.50 =1200 AT/ m
(iii) Total flux (ϕ) = mmf/s Mean circumference (pd) = l = 0.50 m
Reluctance (S) = l/(µoµr a) = 0.50/(4p x 〖10〗^(-7) x 800 x 4 x 〖10〗^(-4) ) = 1.24x106 AT/ Wb
(ϕ) = mmf/s = 1200 /1.24 x 106= 9.67 x 〖10〗^(-4) Webers = 0.967 x10-3 Wb = 0.967 mWb Total
flux (ϕ) = 0.967 mWb (iv) Reluctance (s) = 1.24x106 AT/ Wb
(i) MMF = 600 AT
(ii) Magnetising force =1200 AT / m
(iii) Total flux = 0.967 mWb
(iv) Reluctance = 1.24 x106 AT/ Wb
Tuesday, April 2, 2019
Monday, April 1, 2019
Subscribe to:
Posts (Atom)