Introduction 2 to Machinery Principles (Contd.)
Mahesh S. Illindala
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Objectives
Synthesis (creating something, combining elements in novel ways):
Formulate an equivalent for parallel magnetic circuit with an air-gap
Comprehension (demonstrating understanding of terms and concepts):
Discuss Faraday's Law.
Describe production of induced (Lorentz) force and induced voltage on a linear dc
machine
Application (apply learned information to solve a problem):
Apply the Faraday's law, Lorentz force and induced voltage to solve problems
Knowledge (repeating from memory):
Define power relationships electrical and mechanical
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Class Exercise: Formulate the Magnetic Equivalent for this Core
Length `lo'
Length `lc'
Air Gap mr = 1
Area `A'
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Fringing Effect in Air Gap
Area `A'
Effective area of crosssection of the air gap will be larger than that of the core
Area `Aa'
Aa > A
Study Example 1-2 to better understand the fringing effect on the calculation of total reluctance.
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Induced Voltage from a Time-Changing Magnetic Field (Basis of Transformer Action)
(Faraday's Law) If a flux passes through a turn of a coil of wire, a voltage will be induced in the turn of wire that is directly proportional to the rate of change in the flux linkage with respect to time.
d eind = N dt
Negative sign is from Lenz's law
where eind is the voltage induced in N turns of the coil and is the flux passing through the turn.
Coil enclosing an increasing magnetic flux
Determining the resulting voltage polarity
(Lenz's Law) The direction of the voltage buildup in the coil is such that if the coil ends were short-circuited, it would produce current that would cause a flux opposing the original flux change.
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Faraday's Law Exercise Problem P1-16
Problem P1-16: The core shown in Figure P1-2 has the flux shown in Figure P1-12. Sketch the voltage present at the terminals of the coil.
Figure P1-2
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Figure P1-12
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Exercise Solution P1-16
Solution P1-16: By Lenz' Law, an increasing flux in the direction shown on the core will produce a voltage that tends to oppose the increase. This voltage will be the same polarity as the direction shown on the core, so it will be positive. The induced voltage in the core is given by the equation:
Positive sign with respect to the voltage polarity marked at the terminals
Figure P1-12: Flux profile
Figure P1-12: Plot of resulting induced voltage
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Production of Induced Force on a Wire (Basis of Motor Action)
(Lorentz Force) A current-carrying wire in the presence of a magnetic field has a force induced on it
Magnetic field is directed into the page
The force induced on the conductor is given by
F = i (l X B)
Vector cross product
The direction of the force is given by the right-hand rule: If the index finger of the right hand points in the direction of the vector l and the middle finger points in the direction of the flux density vector B, then the thumb points in the direction of the resultant force on the wire. The magnitude of the force is given by the equation
F = ilB sinq
where q is the angle between the wire and the flux density vector
341 - Electric Machinery Fundamentals
Right-hand rule for MI - 8 vector cross-product
Induced Force Exercise Problem and Solution P1-9
Problem P1-9: A wire is shown in Figure P1-6 which is carrying 2.0 A in the presence of a magnetic field. Calculate the magnitude and direction of the force induced on the wire.
Figure P1-6 Solution P1-9: The force on this wire can be calculated from the equation into the page
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Induced Voltage on a Conductor Moving in a Magnetic Field (Basis of Generator Action)
A moving wire in the presence of a magnetic field has a voltage induced in it
Magnetic field is directed into the page
The voltage induced in the wire is given by
eind = (v X B) l
Vector l points along the direction of the wire toward the end making the smallest angle with respect to the vector v X B. The voltage in the wire will be built up so that the positive end is in the direction of the vector v X B.
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Right-hand rule for MI - 10 vector cross-product
Induced Voltage Exercise Problem Solution and P1-11
Problem P1-11: A wire is shown in Figure P1-8 that is moving in the presence of a magnetic field. With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire.
Figure P1-8
Solution P1-11: The induced voltage on this wire can be calculated from the equation shown below. The total voltage is zero, because the vector quantity v X B points into the page, while the wire runs in the plane of the page.
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Linear DC Machine A Simple Example
Magnetic field is directed into the page
(battery)
Smooth frictionless rails Bar of conducting material
The force induced on the conducting bar
Find = i (l X B)
or the magnitude is
Find = ilB
to the right
The voltage induced on the wire moving in the magnetic field eind = (v X B) l
or the magnitude is
eind = vBl
positive upward
Right-hand rule for vector cross-product
Kirchhoff's voltage law for this machine
Newton's law for the bar across the tracks Fnet = ma
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Linear DC Machine Starting
Bar of conducting material Magnetic field is directed into the page
(battery)
Smooth frictionless rails
Closing the switch at t=0 produces a current flow
i = VB / R Find = ilB
The current flow produces a force on the bar given by
The bar accelerates to the right, producing an induced voltage eind as it speeds up. The induced voltage reduces the current flow
i = (VB - eind ) /R
The induced force is thus decreased (Find = i lB) until eventually Find = 0. At that point, eind = VB, i = 0, and the bar moves at a constant no load speed vss = VB /Bl.
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Linear DC Machine Motoring at Load
Bar of conducting material Magnetic field is directed into the page
Smooth frictionless rails
A force Fload is applied opposite to the direction of motion, which causes a net force Fnet opposite to the direction of motion The resulting acceleration a = Fnet /m is negative, so the bar slows down (v ) The induced voltage (eind = v Bl) falls, and so i = (VB - eind )/R increases
The induced force (Find = i lB) increases until |Find| = |Fload| at a lower speed v
An amount of electrical power (eind i ) is now being converted to mechanical power (Find v), and the machine is acting as a motor
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Linear DC Machine Generating at Applied Force
Bar of conducting material Magnetic field is directed into the page
Smooth frictionless rails
A force Fapp is applied in the direction of motion; Fnet is in the direction of motion The resulting acceleration a = Fnet /m is positive, so the bar speeds up (v ) The induced voltage (eind = v Bl) increases, and so i = (eind - VB)/R increases
The induced force (Find = i lB) increases until |Find| = |Fload| at a higher speed v
-ve direction
An amount of mechanical power (Find v) is now being converted to electrical power (eind i ), and the machine is acting as a generator
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Linear DC Machine Example
Bar of conducting material Magnetic field is directed into the page
Smooth frictionless rails
Study Example 1-10 in the textbook to better understand the linear dc machine
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Linear DC Machine HW P1-21(c)
Bar of conducting material Magnetic field is directed into the page
Smooth frictionless rails
P1-21(c) For calculating efficiency, consider the overall circuit including the battery and resistance in the determination of electrical input power
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Power Relationships Mechanical and Electrical
Mechanical For the linear dc machine, the mechanical power Pmech = Find v (watts)
In the case of a rotating dc machine, the mechanical power is given by Pmech = tindw (watts) where w is the angular velocity (in rad./s) and torque tind (in Nm) is defined as the product of the force applied and the smallest distance between the line of action of the force and the object's axis of rotation Torque on an object Electrical For a linear dc machine, power converted
(A)
Pelec = eindi
(watts)
(V) (W)
Or the power supplied to the dc load is simply the product of the voltage across the load and the current flowing through it. For the dc circuit shown on the right side, power generated from the voltage source V is given by Pelec = V I (watts)
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A dc voltage source supplying a load with resistance R
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