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ELEC 3105 Basic EM and Power Engineering
Rotating DC Motor PART 2 Electrical
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Motor / Generator Action
Loop Equivalent circuit Expression of Vemf Slide extracted from linear motor and modified for loop motor.
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Motor / Generator Action
Slide extracted from linear motor and modified for loop motor. Linear relation between speed and torque Current flows in a direction to charge the battery. Stall torque Generator Motor Link
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ E = motor voltage Ra = armature resistance La = armature inductance V = Applied motor voltage Ia = armature current ο = magnetic flux Rf = field resistance Lf = field inductance Vf = Field voltage If = field current
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ In steady state operation: FIELD SIDE πΌ π = π π π
π Ξ¦= 2 ππΌ π π Magnetic circuit reluctance
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ In steady state operation: ARMATURE SIDE πΈ= π πΈ Ξ¦ π π Back emf Motor constant
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ In steady state operation: ARMATURE SIDE π πππ£ = π π Ξ¦ πΌ π Developed torque Motor constant π πΈ = π π Same motor constant in emf and developed torque
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ Power flow: ARMATURE SIDE; Conservation of energy KVL π= πΌ π π
π +πΈ Armature copper loss POWER π πΌ π = πΌ π 2 π
π + πΌ π πΈ Power developed Power in
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ Power flow: ARMATURE SIDE; Conservation of energy π πππ£ Power developed π ππ π πππ π π πππ£ = πΌ π πΈ= π π π πππ£ Electrical Mechanical copper
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ Power flow: ARMATURE SIDE π ππ’π‘ = π πππ£ β π πππ‘ π ππ’π‘ π πππ£ π πππ‘ π ππ π πππ π Rotational loss Copper Electrical Mechanical
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Electrical Equivalent
πΏ π π
π π
π ο πΏ π π π , π πππ£ Motor sequence π ππ’π‘ = π πππ£ β π πππ‘ π πΌ π π πππ£ π π πΈ π πππ£ Speed of rotation limiting loop
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Shunt Connected Field πΏ π π
π π= πΌ π π
π +πΈ π πππ£ = π π Ξ¦ πΌ π R π
π
π= πΌ π π
π +πΈ π πππ£ = π π Ξ¦ πΌ π R π
π πΈ= π πΈ Ξ¦ π π π π , π πππ£ πΏ π ο π πππ£ = πΞ¦π π
π β π 2 Ξ¦ 2 π
π π π Rotation rate Developed torque
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Shunt Connected Field πΏ π π
π R π
π π π , π πππ£ πΏ π ο
π πππ£ = πΞ¦π π
π β π 2 Ξ¦ 2 π
π π π Similar type of graph
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Shunt Connected Field πΏ π π
π π πππ£ = πΞ¦π π
π β π 2 Ξ¦ 2 π
π π π R π
π
π πππ£ = πΞ¦π π
π β π 2 Ξ¦ 2 π
π π π R π
π π π , π πππ£ πΏ π ο π πππ£ πΎΞ¦π π
π Force motor to spin backwards Force motor to spin to fast π πΞ¦ Generator Motor Generator π π
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Series Connected Field
π
π πΏ π πΏ π π
π ο π π , π πππ£ Universal motor design: works for D.C. and for A.C. π= πΌ π (π
π + π
π )+πΈ π πππ£ = π π Ξ¦ πΌ π π πππ£ = π β² πΌ π 2 Since Ξ¦β πΌ π πΈ= π πΈ Ξ¦ π π
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Series Connected Field
π
π πΏ π πΏ π π
π ο π π , π πππ£ Universal motor design: works for D.C. and for A.C. π πππ£ 1 π π 2 π πππ£ = π β² π π
π + π
π + π β² π π 2 π π
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Maximum Power Transfer
πΏ π π= πΌ π π
π +πΈ π
π π
π ο πΏ π π π , π πππ£ Power developed in the motor π πππ£ =πΈ πΌ π π πππ£ =πΈ πβπΈ /π
π Find maximum with respect to the motor voltage π πππ£ = πΈπβ πΈ 2 π
π
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Maximum Power Transfer
πΏ π π= πΌ π π
π +πΈ π
π π
π ο πΏ π π π , π πππ£ For extremes of a function, take derivatives and set to zero π πππ£ =πππ₯πππ’π π€βππ πΈ= π 2 π πππ£ {πππ₯ = π£ 2 4π
π
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Calculation example Solution provided in class π= πΌ π π
π +πΈ πΏ π π
π
π= πΌ π π
π +πΈ πΏ π π
π π
π π πππ£ = π π Ξ¦ πΌ π ο πΏ π π π , π πππ£ πΈ= π πΈ Ξ¦ π π A 120 volt dc motor has an armature resistance of 0.70 β¦. At no-load, it requires 1.1 A armature current and runs at 1000 rpm. Find the output power and torque at 952 rpm output speed. Assume constant flux. Solution provided in class
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Calculation example Solution provided in class π= πΌ π π
π +πΈ πΏ π π
π
π= πΌ π π
π +πΈ πΏ π π
π π
π π πππ£ = π π Ξ¦ πΌ π ο πΏ π π π , π πππ£ πΈ= π πΈ Ξ¦ π π A permanent magnet dc motor has the following information: 50 hp, 200 V, 200 A, 1200 rpm and armature resistance of 0.05 β¦. Determine the output power if the voltage is lowered to 150 V and the current is 200 A. Assume rotational losses are proportional to speed. Determine the rotational loss, armature resistance, no-load rpm, machine constant, efficiency? Solution provided in class
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Calculation example Solution provided in class π= πΌ π π
π +πΈ πΏ π π
π
π= πΌ π π
π +πΈ πΏ π π
π π
π π πππ£ = π π Ξ¦ πΌ π ο πΏ π π π , π πππ£ πΈ= π πΈ Ξ¦ π π An 80 V dc motor has constant field flux, separately excited, and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 A and the no-load current is 0.5 A. Assume constant rotational losses. Solution provided in class
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