Problem 4 In a long shunt compound generator (2024)

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Problem 4In a long shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A . Determine (i) induced emf, (ii)total power generated. Given that shunt field, series field, divertor and armature resistances are $92 \Omega, 0.015 \Omega, 0.03 \Omega$ and $0.032 \Omega$ respectively.

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#### Solution By Steps***Step 1: Calculate Voltage Drop in Series Field and Divertor***The total resistance in the series path (series field + divertor) is $0.015 \Omega + 0.03 \Omega = 0.045 \Omega$. The voltage drop across this path is $150 A imes 0.045 \Omega = 6.75 V$.***Step 2: Calculate Voltage Drop in Armature***The voltage drop across the armature is $150 A imes 0.032 \Omega = 4.8 V$.***Step 3: Calculate Total Voltage Drop***The total voltage drop in the series path and armature is $6.75 V + 4.8 V = 11.55 V$.***Step 4: Calculate Induced EMF***The induced EMF is the terminal voltage plus the total voltage drop, which is $230 V + 11.55 V = 241.55 V$.***Step 5: Calculate Total Power Generated***The total power generated is the product of the induced EMF and the current delivered, which is $241.55 V imes 150 A = 36232.5 W$.#### Final Answer(i) Induced EMF: $241.55 V$(ii) Total Power Generated: $36232.5 W$This problem involves calculating the induced electromotive force (EMF) and the total power generated in a long shunt compound generator, considering various resistances and current flows. Understanding these calculations is crucial for the design and operation of electrical generators.

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Example 4 A 100- kW 250- V 400-A long-shunt compound generator has an armature resistance (including brushes) of 0025 a series-field resistance of 0005 and the magnetization curve of Figure given below there are 1000 shunt-field turns per pole and three series-field turns per pole The series field is connected in such a fashion that positive armature current produces direct-axis mmf which adds to that of the shunt field Compute the terminal voltage at rated terminal current whenAnswered step-by-step 1 answer Example 2 A long-shunt compound generator delivers a load current of 50 A at 500 V and has armature series field and shunt field resistances of 005 003 and 250 respectivelyj Calculate the generated voltage and the armature current Allow 1 V per brush for contact drop EgAnswered step-by-step 1 answerQUESTION 2 A long-shunt compound generator delivers a load of 50 A at 500 V The generator has an armature series field and shunt field resistances of 005 003 and 250 respectively Allowing for a contact drop of 1 V per brush a sketch a labelled circuit diagram of the machine 07 marks b deternine the shunt and series field currents 07 marks c determine the armature current and the armature voltage drop 06 marks d determine the generated power 05 marksAnswered step-by-step 1 answer a) Explain the THREE major factors that determine the magnitude of em generated by a DC generator ( 3 marks) b) A long shunt compound generator delivers a load of 50 A at 500 V and has armature series field and shunt field resistance of 005 003 and 250 respectively Brush contact drop is negligible i Determine the value of armature current (3 marks) ii Calculate the generated voltage (3 marks) c) A 230 V D C shunt motor takes a current of 40 A and runs at 1100 rpm armature and shunt field resistances are 025 and 230 respectivel find i armature current (2 marks) ii back emf (3 marks) iii torque developed by the armature (3 marks) d) Explain the term Slip as applied to 3-phase induction motors (3 marks) e) A 3-phase induction motor is wound for 4 poles and is supplied from 50 Hz system Calculate i The synchronous speed (2 marks) ii The rotor speed when the slip is 4 (2 marks) f) A 2000/200V 20KVA transformer has 66 turns in the secondary Neglecting losses calculate i Primary turns (2 marks) ii Primary and secondary full load currents (4 marks)Answered step-by-step 1 answer 3) A 100- kW 250- V 400- A long-shunt compound generator has an armature resistance (including brushes) of 0025 a series-field resistance of 0005 and the magnetization curve of Fig 714 There are 1000 shunt-field turns per pole and three series-field turns per pole The series field is connected in such a fashion that positive armature current produces direct-axis mmf which adds to that of the shunt field (ie I s=I a ) Compute the terminal voltage at rated terminal current when the shunt-field current is 47 A and the speed is 1150 r / min Neglect the effects of armature reactionAnswered step-by-step 1 answer

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