Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal Sheet (6) Solution DC Machines (Generators) 1) A 400V, 8-pole, 600 rpm DC machine has 100 slots. Each slot contains 40 conductors. The flux per pole is 0.01 Weber. What type of winding is used? Solution: π¬= ππ π ∗ ∗π∗π΅ ππ ππ πππ = π πππ ∗ ππ ∗ ∗ π. ππ ∗ πππ ππ ππ ∴ ππ = π ∴ ππ = ππ = π ∴ π»ππ ππππππππ ππππ πππ ππ π³ππ ππππ πππ. 2) An 8-pole DC generator has 500 conductors on its armature and is designed to have 0.02 Wb of magnetic flux per pole. Solution: a) Find the voltage that will be generated at a speed of 1800 rpm if the armature is: i. Wave-wound π¬π = π∗π∗π΅ π ∗ ππ π¨ πππππ βΆ π = π. ππ , π = πππ , π¨ = π , π΅ = ππππ , π = π ∴ π¬π = π. ππ ∗ π ∗ ππππ πππ ) = ππππ π ∗( ππ π ii. Lap-wound π = π. ππ , π = πππ , π¨ = π , π΅ = ππππ , π = π ∴ π¬π = π. ππ ∗ π ∗ ππππ πππ ) = πππ π ∗( ππ π Good Luck Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal b) If allowable current per path is 5 A, calculate the power generated by the machine for each case in (a). i. Wave-wound πππ ππππ ππππ πππ πππππ πππ π ππππππππ πππππ ∴ π°π = π ∗ (πππππππ πππ ππππ) = π ∗ π°π = ππ π¨ ∴ π· = π¬π ∗ π°π = ππππ ∗ ππ = ππ ππΎ ii. Lap-wound πππ πππ ππππ πππ πππππ πππ 'p' ππππππππ πππππ π°π = π ∗ π°π = π ∗ π = ππ π¨ π· = πππ ∗ ππ = ππ ππΎ 3) A separately excited DC generator has a field resistance of 50 Ω, armature resistance of 0.125 Ω and brush voltage drop of 2 V. At no load, the generated voltage is 275 V while the full load current is 95 A. Given that the excitation voltage is 120 V, the friction and core losses are 1500 W. Calculate: Solution: a) The rated terminal voltage and output power. π½π = π¬π − π°π πΉπ − π½π© = πππ − (ππ ∗ π. πππ) − π = πππ. ππ π π·πππ = π½π ∗ π°π = ππ. ππ ∗ ππ = ππ. ππ ππΎ b) The efficiency at full load. π·ππ = π·πππ + π·ππππππ ππππ π) π (ππ. (ππππ) (ππ = ππ ∗ ππ + + ∗ π. πππ) + ( ) = ππ. πππ ππΎ ππ ππ. ππ ∴πΌ = = ππ. ππ % ππ. πππ Good Luck Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal 4) A separately excited DC generator has rated terminal voltage and full load armature current of 200 V and 50 A respectively and rated speed of 1500 rpm. The generator armature resistance Ra=0.1 Ω and field resistance Rf =100 Ω and field voltage of 150 V. Calculate: Solution: a) The induced emf and torque at full load. π°π = π°π³ = ππ π¨ (πππππππ ππππππππππ πππππππ πππππππππ) π¬ = π½π + π°π ∗ πΉπ = πππ + ππ ∗ π. π = πππ π π¬ ∗ π°π π¬ ∗ π°π πππ ∗ ππ = = = πππ. ππ π΅. π ππ π΅ ππ π ∗ ππππ ππ ππ b) The emf, armature current and terminal voltage at rated speed when the field voltage is changed to 100 V. π»π = π³ππ ππππ π ππππππ ππππ π½ππ = πππ π → π°ππ = ππππ π ππππππ ππππ π½ππ = πππ π → π°ππ = π½ππ πππ = = π. π π¨ πΉπ πππ π½ππ πππ = =ππ¨ πΉπ πππ π¬π ππ π΅π (π°ππ π΅π ) π°ππ = = = , π΅π = π΅π = π΅πππππ π¬π ππ π΅π π°ππ π΅π π°ππ π¬π π°ππ π¬π π = → = → π¬π = πππ. ππ ππππ π¬π π°ππ πππ π. π π¬π = π½ππ + π°ππ ∗ πΉπ πππ. ππ = π½ππ + π. π ∗ π°ππ → (π) β΅ ππ ππππππ ππ ππππ ππππππππππ ∴ πΉππππ = π½ππ =ππ π°ππ ∴ πΉππππ = π½ππ π½ππ → π= π°ππ π°ππ → (π) πΊππππππ (π) πππ (π): π½ππ = πππ. ππ π , π°ππ = ππ. ππ π¨ Good Luck Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal c) The developed torque at rated speed when the field voltage is changed to 120 V. π½ππ = πππ πππππ β΅ π»π π = β΅ π¬π ∗ π°ππ ππ π΅ ( ππ π ) π¬π π°ππ π΅π π°ππ = = , πππππ πππππ π΅π = π΅π = π΅πππππ π¬π π°ππ π΅π π°ππ π¬π π°ππ = π¬π π°ππ π°ππ = ∴ π½ππ πππ = = π. π π¨ πΉπ πππ π¬π π. π = → π¬π = πππ ππππ πππ π. π π°ππ = π¬π πππ = = ππ π¨ πΉπ + πΉππππ π. π + π π΅π = ππππ πππ ∴ π»π π = π¬π π°ππ πππ ∗ ππ = = ππ. ππ π΅π ππ π΅π ππ ∗ ππππ ππ ππ d) The emf, armature current and terminal voltage at half rated speed when the field voltage is changed to 200 V. π½ππ = πππ π , π΅π = β΅ ππππ = πππ πππ π π¬π ππ π΅π π°ππ π΅π = = π¬π ππ π΅π π°ππ π΅π π°ππ = π½ππ πππ = =ππ¨ πΉπ πππ Good Luck Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal π¬π π°ππ π΅π π¬π π ∗ πππ = → = → π¬π = πππ. ππ π π¬π π°ππ π΅π πππ π. π ∗ ππππ π°ππ = π¬π πππ. ππ = = ππ. ππ π¨ πΉπ + πΉππππ π. π + π π½ππ = π°ππ ∗ πΉππππ = ππ. ππ ∗ π = πππ. ππ π ` 5) A shunt generator has a field resistance of 60 Ω. When the generator delivers 6 kW, the terminal voltage is 120 V, while the generated EMF is 133 V. Determine: Solution: a) The armature resistance. π¬ = π½π + π° π ∗ πΉ π πΉπ = π¬ − π½π π°π π°π = π°π + π°π³ = πΉπ = π½π π·πππ πππ π ∗ ππππ + = + = ππ π¨ πΉπ π½π ππ πππ π¬ − π½π πππ − πππ = = π. ππ π π°π ππ b) The generated emf when the output is 2 kW at terminal voltage of 135 V. π¬ = π½π + π° π ∗ πΉ π π°π = π°π + π°π³ = π½π π·πππ πππ π ∗ ππππ + = + = π. ππ + ππ. π = ππ. ππ π¨ πΉπ π½π ππ πππ π¬ = π½π + π°π ∗ πΉπ = πππ + (ππ. ππ ∗ π. ππ) = πππ. ππ π½ Good Luck Faculty of Electrical Engineering Electric Machines & Drives (ELCT403) Dr. Eng. Noha Shouman Eng. Omar Ghazal 6) A long shunt compound DC generator delivers a load current of 150A at 230 V and has armature, series field and shunt field resistances of 0.032 Ω , 0.015 Ω and 92 Ω respectively. Calculate: Solution: a) The induced emf. π°π = πππ = π. π π¨ ππ π°π¨ = π°π + π°π³ = π. π + πππ = πππ. π π¨ πππππππ π πππ ππππππ ππππππ ππππ πππ: π°π¨ πΉπ = πππ. π ∗ π. πππ = π. ππππ π½ π¨πππππππ πππππππ π πππ = π°π¨ πΉπ¨ = πππ. π ∗ π. πππ = π. ππ π½ π¬π¨ = π½π» + π°π¨ πΉπ¨ + π°π¨ πΉπ = πππ + π. ππππ + π. ππ = πππ. ππππ π½ b) The power generated by the armature. π·π¨ = π¬π¨ π°π¨ = πππ. ππππ ∗ πππ. π = ππ. ππ ππΎ c) The copper losses. π·ππ = π°ππ¨ ∗ πΉπ¨ + π½π π°π + π°ππ¨ ∗ πΉπ = (πππ. π)π ∗ (π. πππ) + (πππ) ∗ (π. π) + (πππ. π)π ∗ (π. πππ) = ππππ πππππ Good Luck
