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ELEC 3105 Basic EM and Power Engineering

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Presentation on theme: "ELEC 3105 Basic EM and Power Engineering"β€” Presentation transcript:

1 ELEC 3105 Basic EM and Power Engineering
Rotating DC Motor PART 2 Electrical

2 Motor / Generator Action
Loop Equivalent circuit Expression of Vemf Slide extracted from linear motor and modified for loop motor.

3 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

4 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 E = motor voltage Ra = armature resistance La = armature inductance V = Applied motor voltage Ia = armature current  = magnetic flux Rf = field resistance Lf = field inductance Vf = Field voltage If = field current

5 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 In steady state operation: FIELD SIDE 𝐼 𝑓 = 𝑉 𝑓 𝑅 𝑓 Ξ¦= 2 𝑛𝐼 𝑓 𝔑 Magnetic circuit reluctance

6 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 In steady state operation: ARMATURE SIDE 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š Back emf Motor constant

7 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 In steady state operation: ARMATURE SIDE 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž Developed torque Motor constant π‘˜ 𝐸 = π‘˜ 𝑇 Same motor constant in emf and developed torque

8 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 Power flow: ARMATURE SIDE; Conservation of energy KVL 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 Armature copper loss POWER 𝑉 𝐼 π‘Ž = 𝐼 π‘Ž 2 𝑅 π‘Ž + 𝐼 π‘Ž 𝐸 Power developed Power in

9 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 Power flow: ARMATURE SIDE; Conservation of energy 𝑃 𝑑𝑒𝑣 Power developed 𝑃 𝑖𝑛 𝑃 π‘™π‘œπ‘ π‘  𝑃 𝑑𝑒𝑣 = 𝐼 π‘Ž 𝐸= πœ” π‘š 𝑇 𝑑𝑒𝑣 Electrical Mechanical copper

10 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 Power flow: ARMATURE SIDE 𝑃 π‘œπ‘’π‘‘ = 𝑃 𝑑𝑒𝑣 βˆ’ 𝑃 π‘Ÿπ‘œπ‘‘ 𝑃 π‘œπ‘’π‘‘ 𝑃 𝑑𝑒𝑣 𝑃 π‘Ÿπ‘œπ‘‘ 𝑃 𝑖𝑛 𝑃 π‘™π‘œπ‘ π‘  Rotational loss Copper Electrical Mechanical

11 Electrical Equivalent
𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 Motor sequence 𝑃 π‘œπ‘’π‘‘ = 𝑃 𝑑𝑒𝑣 βˆ’ 𝑃 π‘Ÿπ‘œπ‘‘ 𝑉 𝐼 π‘Ž 𝑇 𝑑𝑒𝑣 πœ” π‘š 𝐸 𝑃 𝑑𝑒𝑣 Speed of rotation limiting loop

12 Shunt Connected Field 𝐿 π‘Ž 𝑅 π‘Ž 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž R 𝑅 𝑓
𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž R 𝑅 𝑓 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐿 𝑓  𝑇 𝑑𝑒𝑣 = π‘˜Ξ¦π‘‰ 𝑅 π‘Ž βˆ’ π‘˜ 2 Ξ¦ 2 𝑅 π‘Ž πœ” π‘š Rotation rate Developed torque

13 Shunt Connected Field 𝐿 π‘Ž 𝑅 π‘Ž R 𝑅 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐿 𝑓 
𝑇 𝑑𝑒𝑣 = π‘˜Ξ¦π‘‰ 𝑅 π‘Ž βˆ’ π‘˜ 2 Ξ¦ 2 𝑅 π‘Ž πœ” π‘š Similar type of graph

14 Shunt Connected Field 𝐿 π‘Ž 𝑅 π‘Ž 𝑇 𝑑𝑒𝑣 = π‘˜Ξ¦π‘‰ 𝑅 π‘Ž βˆ’ π‘˜ 2 Ξ¦ 2 𝑅 π‘Ž πœ” π‘š R 𝑅 𝑓
𝑇 𝑑𝑒𝑣 = π‘˜Ξ¦π‘‰ 𝑅 π‘Ž βˆ’ π‘˜ 2 Ξ¦ 2 𝑅 π‘Ž πœ” π‘š R 𝑅 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐿 𝑓  𝑇 𝑑𝑒𝑣 𝐾Φ𝑉 𝑅 π‘Ž Force motor to spin backwards Force motor to spin to fast 𝑉 π‘˜Ξ¦ Generator Motor Generator πœ” π‘š

15 Series Connected Field
𝑅 𝑓 𝐿 π‘Ž 𝐿 𝑓 𝑅 π‘Ž  πœ” π‘š , 𝑇 𝑑𝑒𝑣 Universal motor design: works for D.C. and for A.C. 𝑉= 𝐼 π‘Ž (𝑅 π‘Ž + 𝑅 𝑓 )+𝐸 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž 𝑇 𝑑𝑒𝑣 = π‘˜ β€² 𝐼 π‘Ž 2 Since Φ∝ 𝐼 π‘Ž 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š

16 Series Connected Field
𝑅 𝑓 𝐿 π‘Ž 𝐿 𝑓 𝑅 π‘Ž  πœ” π‘š , 𝑇 𝑑𝑒𝑣 Universal motor design: works for D.C. and for A.C. 𝑇 𝑑𝑒𝑣 1 πœ” π‘š 2 𝑇 𝑑𝑒𝑣 = π‘˜ β€² 𝑉 𝑅 π‘Ž + 𝑅 𝑓 + π‘˜ β€² πœ” π‘š 2 πœ” π‘š

17 Maximum Power Transfer
𝐿 π‘Ž 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 Power developed in the motor 𝑃 𝑑𝑒𝑣 =𝐸 𝐼 π‘Ž 𝑃 𝑑𝑒𝑣 =𝐸 π‘‰βˆ’πΈ /𝑅 π‘Ž Find maximum with respect to the motor voltage 𝑃 𝑑𝑒𝑣 = πΈπ‘‰βˆ’ 𝐸 2 𝑅 π‘Ž

18 Maximum Power Transfer
𝐿 π‘Ž 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝑅 π‘Ž 𝑅 𝑓  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 For extremes of a function, take derivatives and set to zero 𝑃 𝑑𝑒𝑣 =π‘šπ‘Žπ‘₯π‘–π‘šπ‘’π‘š π‘€β„Žπ‘’π‘› 𝐸= 𝑉 2 𝑃 𝑑𝑒𝑣 {π‘šπ‘Žπ‘₯ = 𝑣 2 4𝑅 π‘Ž

19 Calculation example Solution provided in class 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž
𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š A 120 volt dc motor has an armature resistance of 0.70 Ω. 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

20 Calculation example Solution provided in class 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž
𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š A permanent magnet dc motor has the following information: 50 hp, 200 V, 200 A, 1200 rpm and armature resistance of 0.05 Ω. 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

21 Calculation example Solution provided in class 𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž
𝑉= 𝐼 π‘Ž 𝑅 π‘Ž +𝐸 𝐿 π‘Ž 𝑅 π‘Ž 𝑅 𝑓 𝑇 𝑑𝑒𝑣 = π‘˜ 𝑇 Ξ¦ 𝐼 π‘Ž  𝐿 𝑓 πœ” π‘š , 𝑇 𝑑𝑒𝑣 𝐸= π‘˜ 𝐸 Ξ¦ πœ” π‘š 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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