Understanding Magnetic Circuits: Ampere's Law and Flux
School
The Hong Kong University of Science and Technology**We aren't endorsed by this school
Course
MECH 3630
Subject
Mechanical Engineering
Date
Dec 10, 2024
Pages
25
Uploaded by DukeLightningStork37
MECH3630 Electrical Technology Tutorial 4: Magnetic circuitTOM MOK1
Overview•Revisit and recap:➢Ampere’s circuit law➢B-H relationship➢Magnetic flux➢Magnetic equivalent circuit•Question solving2
Ampere’s circuit law3•From Ampere’s circuit law:➢𝑯? = 𝑵?➢𝐻: Field intensity➢?: Length of the magnetic core➢𝑁: Number of turns of the coil➢𝑖: CurrentToroid
Magnetic flux density –magnetic field intensityB-H relation4•The magnetic field intensity H produces a magnetic flux density B everywhere it exists•Magnetic flux density B equals to the product of magnetic field intensity H and permeability of a medium (?)➢? = 𝝁𝑯weber/??or tesla➢For instance, the permeability of free space is𝝁?= ?𝝅 × ??−?henry/meter•We can also express the permeability of ferromagnetic materials in terms of relative permeability (?𝑟):➢? = ?𝑟?0𝐻weber/?2or tesla•B-H relation can be found by graph!!
Magnetic flux5•The flux crossing the cross section of the toroid is:•𝚽 = ? 𝒅?•B: Magnetic flux density•A: Cross-sectional area of the toroid•For a toroid, B is actually a function of the radius r (We are going to look at an example later)•Therefore, integration is required for flux calculation!•Only if we know the average flux density (?𝑎𝑣𝑔), the following relationship is true:•𝚽 = ?𝒂𝒗??WbToroid
Equivalent magnetic circuit (Toroid)6•From the magnetic equivalent circuit:➢Magneto-motive force:o𝑁𝑖 = Φℜ➢Reluctance:oℜ=𝑙??ToroidMagneticequivalentcircuitElectric circuit analogy
Equivalent magnetic circuit(Magnetic circuit with air gap)7•From the magnetic equivalent circuit:•Magnetomotive force:➢𝑁𝑖 = 𝛷(ℜ?+ℜ𝑔)•Reluctance➢ℜ?=𝑙𝑐?𝑐?𝑐➢ℜ?=𝑙𝑔?𝑔?𝑔Equivalentmagneticcircuit
Inductance of an inductor8•A coil wound on a magnetic core, is frequently used in electric circuits•This coil may be represented by an ideal circuit element, called inductance•Which is defined as the flux linkage of the coil per ampere of its current
Question 1.159
Question 1.15The magnetic circuit for a saturable reactor is shown in Fig. P1.15. The B−Hcurve for the core material can be approximated as two straight lines as in Fig. P1.15.(a) If I1 =2.0 A, calculate the value of I2 required to produce a flux density of 0.6 T in the vertical limbs.(b) If I1 =0.5 A and I2 =1.96 A, calculate the total flux in the core.10
Question 1.15Draw the equivalent magnetic circuit:11Reluctance of the vertical limbsReluctance of the horizontal limbsMmf of coil 1Mmf of coil 2
(a) If I1 =2.0 A, calculate the value ofI2 required to produce a flux densityof 0.6 T in the vertical limbs.•From ampere’slaw:➢𝑁1𝑖1+ 𝑁2𝑖2= 𝐻𝑣?𝑣+ 𝐻ℎ?ℎ•To find the magnetic field intensity for the vertical limbs (𝐻𝑣) :➢We know the flux density of vertical limbs is ?𝑣= 0.6tesla➢Reading from B-H graph →𝐻𝑣=0.60.8× 400 = 300 At/m•To find the magnetic field intensity for the horizontal limbs (𝐻ℎ)➢The flux density of the horizontal limb (𝐻ℎ) is not known!➢From flux continuity:oΦ𝑣= Φℎo?𝑣?𝑣= ?ℎ?ℎo?ℎ=?𝑣?𝑣?ℎ=(0.6)(0.015×0.015)(0.01×0.015)= 0.9 tesla➢From the B-H graph →𝐻ℎ= 700At/m 12
(a) If I1 =2.0 A, calculate the value ofI2 required to produce a flux densityof 0.6 T in the vertical limbs.•We know:➢𝐻𝑣= 300 At/m➢𝐻ℎ= 700At/m •From ampere’slaw:➢𝑁1𝑖1+ 𝑁2𝑖2= 𝐻𝑣?𝑣+ 𝐻ℎ?ℎ➢2002 +100𝑖2=3000.14 +7000.15➢??= −?. ?? 𝐀13
(b) If I1 =0.5 A and I2 =1.96 A,calculate the total flux in the core.•From flux continuity:➢Φ𝑣= Φℎ➢?𝑣?𝑣= ?ℎ?ℎ➢?𝒗=?????𝒗=???.??×?.????.???×?.???=???.?------------(1)•From ampere’slaw:➢𝑁1𝑖1+ 𝑁2𝑖2= 𝐻𝑣?𝑣+ 𝐻ℎ?ℎ➢2000.5 +1001.96= 𝐻𝑣0.14 + 𝐻ℎ0.15➢??? = ?. ?? 𝑯𝒗+ ?. ?? 𝑯?--------------(2)•To solve this equation, we need to use trial and error method•To do this, we give a guess and hope our answer satisfy both equation (1) and (2), let’stry!14
(b) If I1 =0.5 A and I2 =1.96 A,calculate the total flux in the core.•The two equations to satisfy:➢?𝒗=???.?------------(1)➢??? = ?. ?? 𝑯𝒗+ ?. ?? 𝑯?--------------(2)15?𝒗??𝑯𝒗𝑯??. ?? 𝑯𝒗+ ?. ?? 𝑯?Trial 10.50.7525035387.95? =0.8400𝐻? = 3.333 × 10−4𝐻 + 0.6667
(b) If I1 =0.5 A and I2 =1.96 A,calculate the total flux in the core.•The two equations to satisfy:➢?𝒗=???.?------------(1)➢??? = ?. ?? 𝑯𝒗+ ?. ?? 𝑯?--------------(2)16?𝒗??𝑯𝒗𝑯??. ?? 𝑯𝒗+ ?. ?? 𝑯?Trial 10.50.7525035387.95Trial 20.71.053501150221.5? =0.8400𝐻? = 3.333 × 10−4𝐻 + 0.6667
(b) If I1 =0.5 A and I2 =1.96 A,calculate the total flux in the core.•The two equations to satisfy:➢?𝒗=???.?------------(1)➢??? = ?. ?? 𝑯𝒗+ ?. ?? 𝑯?--------------(2)17?𝒗??𝑯𝒗𝑯??. ?? 𝑯𝒗+ ?. ?? 𝑯?Trial 10.50.7525035387.95Trial 20.71.053501150221.5Trial 3(Sol.)0.81.24001600296? =0.8400𝐻? = 3.333 × 10−4𝐻 + 0.6667
(b) If I1 =0.5 A and I2 =1.96 A,calculate the total flux in the core.18?𝒗??𝑯𝒗𝑯??. ?? 𝑯𝒗+ ?. ?? 𝑯?0.81.24001600296? =0.8400𝐻? = 3.333 × 10−4𝐻 + 0.6667•We found that:•Total flux in the core:➢Φ = ?𝑣?𝑣=0.80.015 × 0.015= ?. ?? 𝐦𝐖𝐛= ?ℎ?ℎ
Question 1.1619
Question 1.16A toroidal core has a rectangular cross section as shown in Fig. P1.16a. It is wound with a coil having 100 turns. The B−H characteristic of the core may be represented by the linearized magnetization curve of Fig. P1.16b.(a) Determine the inductance of the coil if the flux density in any part of the core is below 1.0 Wb/m2.(b) Determine the maximum value of the current for the condition of part (a).(c) Determine the minimum value of the current for which the complete core has a flux density of 1.0 Wb/m2or greater.20
Question 1.16 (a)•From ampere’s law:➢𝑁𝑖 = 𝐻?➢𝑁𝑖 = 𝐻2𝜋𝑟➢𝑁𝑖 =??2𝜋𝑟➢? =𝑁𝑖?2𝜋𝑟•To find the inductance of the coil:➢𝐿 =?𝑖•From the definition:➢? = 𝑁Φ = 𝑁 ? ?? = 𝑁 𝑅1𝑅2?𝑡 ?𝑟➢? = 𝑁 𝑅1𝑅2𝑁𝑖?2𝜋𝑟𝑡 ?𝑟 =?𝑁2𝑖𝑡2𝜋𝑅1𝑅2?𝑟𝑟=?𝑁2𝑖𝑡2𝜋ln𝑅2𝑅1•Inductance of the coil:➢𝐿 =?𝑖=?𝑁2𝑡2𝜋ln𝑅2𝑅121
Question 1.16 (a)•We have derived that:➢Inductance of the coil:o𝐿 =?𝑁2𝑡2𝜋ln𝑅2𝑅1•For flux density of ? = 1 Wb/m2:➢From the graph, 𝐻 = 100 At/m➢? =?𝐻=1100H/m•Substitute into the inductance equation we derived:➢𝐿 =?𝑁2𝑡2𝜋ln𝑅2𝑅1=0.011002(0.03)2𝜋ln0.060.04➢𝐿 = 0.1936 H = ???. ? 𝐦𝐇22
Question 1.16 (b)•We have derived that:➢Magnetic field intensity:o? =𝑁𝑖?2𝜋𝑟•From the above equation, we know that the smaller the 𝑟, the higher is the magnetic flux intensity B•This means that the flux intensity B will saturate first at the core’s inner part➢? = 1 Wb/m2→𝐻 = 100 At/m(occurring at 𝑟 = 4??)•Determine the maximum value of the current:➢𝑖 𝑁 = 𝐻 ? = 𝐻2𝜋𝑟(Ampere’s law)➢𝑖 =𝐻2𝜋𝑟𝑁=(100)(2𝜋×4×10−2)100= ?. ???? 𝐀23
Question 1.16 (c)•We have derived that:➢Magnetic field intensity:o? =𝑁𝑖?2𝜋𝑟•From the above equation, we know that the larger the 𝑟, the smaller is the magnetic flux intensity B•To ensure that the complete core has a flux density of1.0 Wb/m2or higher, we should consider the outer part of the core (𝑟 = 6 ??)•Determine the maximum value of the current:➢𝑖 𝑁 = 𝐻 ? = 𝐻2𝜋𝑟(Ampere’s law)➢𝑖 =𝐻 (2𝜋𝑟)𝑁=(100)(2𝜋×6×10−2)100= ?. ???? 𝐀24