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ES2C6 Week 7 Examples - CWQEEW
Electromechanical System Design (ES2C6)
The University of Warwick
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ES2C6 Week 7 Examples: Electromagnetics
and Motors
Lecture 7: Magnetic Circuits
Question 1: A coil of wire with 𝑁 = 1000 turns is wrapped around a core of unknown material. The
cross-sectional area of the core is 𝐴 = 0𝑚 2 and its mean length is 𝑙 = 2𝑚. When a current of
𝑖 = 3 𝐴 flows through the coil, a flux of 𝜙 = 0 𝑊𝑏 flows through the core. Calculate the
reluctance of the core, and the relative permeability of the core material.
Solution 1: The first step is to relate the MMF and flux to the current using:
NI =
1000 3
1 10 At/Wb
0.
NI
= = =
Now we know the reluctance, we can find the unknown relative permeability using:
0 r
l
A
=
7 5
0
2
10610.
4 10 1 10 0.
r
l
A
= = − =
Question 2: The structure shown below has a uniform cross-sectional area of 𝐴 = 0 𝑚 2 , a mean
flux path of 𝑙 1 = 𝑙 5 = 0 𝑚, 𝑙 2 = 𝑙 4 = 2 𝑚, 𝑙 3 = 1 𝑚, and air-gap of 𝑙𝑔𝑎𝑝 = 40 𝑚𝑚. The relative
permeability of the core material is 𝜇𝑟 = 8,000, and the permeability of free space is assumed to be
𝜇 0 = 4𝜋 × 10−7 𝐻/𝑚. The coil of wire has 𝑁 = 2000 turns. What current is required to produce a
flux density of 𝐵 = 50 𝑚𝑇 in the air-gap.
l 1
l 2
l 3
l 4
l 5
lgap
Solution 2: The first step is to calculate the mean flux path in the core by summing the given lengths: 5 1
####### core n 0 0 2 2 1 6 m
n
####### l l
=
= = + + + + =
Now the reluctance of the air-gap and core can be calculated and combined to give the total reluctance of the circuit: 0 0 7 7
####### 0 6.
####### 4 10 0 4 10 8000 0.
####### 31830988 600809 32431797 At/Wb
gap core T g r
####### l l
A A
− −
####### = + = +
####### = +
#######
####### = + =
We can calculate the current required for a flux density of 𝐵 = 50 𝑚𝑇 using:
####### = NI = Hl = and =BA
####### NI BA
####### BA
####### I
####### N
####### =
####### =
####### 0 0 32431797.
####### 0 A
####### 2000
####### I
#######
####### = =
Question 3: The structure shown below has a uniform cross-sectional area of 𝐴 = 0 𝑚 2 , a mean flux path of 𝑙 1 = 𝑙 5 = 0 𝑚, 𝑙 2 = 𝑙 4 = 1 𝑚, 𝑙 3 = 1 𝑚, and air-gap of 𝑙𝑔𝑎𝑝 = 1 𝑚𝑚. The relative permeability of the core material is 𝜇𝑟 = 13,500, and the permeability of free space is assumed to be 𝜇 0 = 4𝜋 × 10−7 𝐻/𝑚. The coil of wire has 𝑁 = 1600 turns. What is the flux density in the air-gap is a current of 𝑖 = 500𝑚𝐴 flows though the coil. Solution 3: The first step is to calculate the mean flux path in the core by summing the given lengths: 5 1
####### core n 0 0 1 1 1 6 m
n
####### l l
=
= = + + + + =
Now the reluctance of the air-gap and core can be calculated and combined to give the total reluctance of the circuit: l 1 l 2 l 3 l 4 l 5 lgap
####### Hcore = 1200 Atm
The MMF across the airgap can be found with the application of the following equation:
= NI = Hl=
7 0
####### 1.
####### 0 1909 At
####### 4 10
gap
####### B
####### Hl l
−
####### = = =
#######
####### =
For the final part of the question, we need to find the current required to provide the required B field in the airgap. Once again, we can use the following equation:
####### = NI = Hl=
The total MMF in the system is :
####### Total = core + gap= 50 + 1909 =1959 At
Hence the required current is:
####### 1959.
####### 2 A
####### 800
####### I Total
####### N
####### = = =
Lecture 7: Simple Machines
Question 5: An electromagnet is used to lift and support a solid piece of steel. Calculate the starting current required to lift the load and the holding current required to keep the load in place once it has been lifted and is attached to the magnet. Use the following parameters: - Number of turns 𝑁 = 1000 - Permeability of free space 𝜇 0 = 4𝜋 × 10−7 𝐻/𝑚 - Relative permeability of core and bar 𝜇𝑟 = 12000 - Initial air gap 𝑥 = 0𝑚 - Mean length of core 𝑙 1 = 1𝑚 - Mean length of bar 𝑙 2 = 0𝑚 - Air gap cross sectional area = 2 × 10−3 𝑚 2 - Mass of load 𝑚 = 15 𝑘𝑔
Solution 5: The equation that related the force to the reluctance is:
2 ( )
####### 2
####### d x
####### F
####### dx
####### = −
We need to find an expression for the reluctance of the magnetic circuit, which consists of the reluctance of the core, bar and air gap:
( ) 1 2
0 0
####### 2
bar core gap r
####### l l x
####### x
A A
####### +
####### = + + = +
We then need to differentiate this expression with respect to 𝑥
( )
0
####### d x 2
dx A
####### =
The flux is related to the MMF and reluctance using Hopkinson's law:
( ) 1 2
0 0
####### 2
r
####### NI
####### x l l x
####### A A
####### = =
####### +
####### +
We can now put the terms for 𝜙 and 𝜕𝑅 𝜕𝑥 into the equation for Maxwell’s pulling force to yield: 2 2 2 1 2 0 0 0
####### 1
####### 2
r
####### N I
####### F
####### l l x A
####### A A
####### =
####### +
####### +
#######
The question requires us to find the current needed to generate the force required when the airgap is 𝑥=0𝑚. The required force must overcome the weight of the bar which is:
####### Fbar = mg= 15 9 =147 N
Hence the expression for the electromagnetic force is Core (Fixed) Bar (Moveable)
( )
0 0
####### 2 gap 2 gap
gap r r
####### l l
####### x
A ax
####### = = =
We then need to differentiate this expression with respect to 𝑥
( )
2 0
####### 2 gap
r
####### d x l
dx ax
####### = −
The flux is related to the MMF and reluctance using Hopkinson's law:
( )
0
####### 2
r gap
####### NI ax
####### x l
####### = =
We can now put the terms for 𝜙 and 𝜕𝑅 𝜕𝑥 into the equation for Maxwell’s pulling force to yield:
( ) ( ) ( )
2 2 2 0 0 2 2 0
####### 2
####### 2 8 4
r gap gap r gap
####### d x NI ax l a NI
####### F
####### dx l ax l
####### −
####### = = = −
We need to find the force required to move the plunger using
####### F = ka= 8 0 =0 N
Hence the force equation becomes
( )
2 0
####### 0.
####### 4 gap
####### a NI
####### F
####### l
####### = = −
Which can be rearranged to solve for MMF: 7 0
####### 1 1 0.
####### 319 At
####### 4 10 0.
####### l gap
####### NI
a −
#######
####### = = =
#######
Question 7: Determine the current required for the relay to make contact (i. pull in the plate) from a distance 𝑥. Assume that the reluctance of the core and the plate is negligible. Use the following parameters: - Air gap cross sectional area 𝐴𝑔𝑎𝑝 = 0𝑚 2 - Air gap length 𝑥 = 10 𝑚𝑚 - Force required to move plate 𝐹 = 1 𝑁 - Number of turns 𝑁=
Solution 7: The equation that related the force to the reluctance is:
2 ( )
####### 2
####### d x
####### F
####### dx
####### = −
We need to find an expression for the reluctance of the magnetic circuit, which only consists of the reluctance of air gap (we assume the core and plate reluctance are negligible):
( )
0 0
####### 2 gap 2
gap gap
####### l x
####### x
A A
####### = = =
We then need to differentiate this expression with respect to 𝑥 0
####### gap 2
gap
####### d
dx A
####### =
The flux is related to the MMF and reluctance using Hopkinson's law:
( )
0
####### 2
####### NI A gap
####### x x
####### = =
We can now put the terms for 𝜙 and 𝜕𝑅 𝜕𝑥 into the equation for Maxwell’s pulling force to yield:
( ) ( ) ( )
2 2 0 2 2 0
####### 2
####### 2 8 4
gap gap gap
####### d x NI A A NI
####### F
####### dx x A x
####### = = =
We are told that the force that the relay needs to overcome is 𝐹 = 1 𝑁. Hence:
( )
2 0
####### 2 1 N
####### 4
####### A gap NI
####### F
####### x
####### = =
Which solved for current gives: 2 2 7 0
####### 1 4 1 1 4 0.
####### 3 A
####### gap 200 4 10 0.
####### F x
####### I
N A −
#######
####### = = =
#######
End of Questions Mark Dooner, 2022
ES2C6 Week 7 Examples - CWQEEW
Module: Electromechanical System Design (ES2C6)
University: The University of Warwick
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