JNTU
JNTU B.Tech II ELECTRICAL MACHINES I Previous Question Paper (October - 2020 - R13)
Answer any five questions. All questions carry equal marks.
Main
Show that the torque developed in a doubly excited magnetic system is equal to the rate of increase of field energy with respect to displacement at constant currents.
Answer
In a doubly excited magnetic system, the energy stored in the magnetic field is a function of the currents in the two windings (i1, i2) and the mechanical displacement (θ). The field energy Wf is given by the area under the magnetization curves or the integral of the flux linkages with respect to currents.
The total energy balance equation for a magnetic system undergoing a virtual displacement dθ is:
Where dWe is the electrical energy input, dWf is the change in stored field energy, and dWm is the mechanical work done by the system. The electrical energy input is given by:
The mechanical work done is defined as:
Substituting these into the energy balance equation:
Since Wf is a state function, its total differential is:
For a linear magnetic system, the flux linkages are λ1 = L11i1 + L12i2 and λ2 = L21i1 + L22i2. The field energy is Wf = 0.5L11i12 + L12i1i2 + 0.5L22i22. Substituting the differentials of λ1 and λ2 into the energy balance equation and holding currents constant (di1 = 0, di2 = 0), the electrical input becomes:
Because the change in field energy at constant current is dWf = 0.5i12dL11 + i1i2dL12 + 0.5i22dL22, comparing the terms shows that the mechanical work Tedθ must equal the remaining change in field energy. Thus, the electromagnetic torque is:
For a singly excited magnetic system, derive the relation for the magnetic stored energy in terms of reluctance.
Answer
In a singly excited magnetic system, the magnetic stored energy Wf is defined as the energy stored in the magnetic field when current flows through the coil. For a linear magnetic system, the energy stored is given by the area under the flux linkage (λ) versus current (i) curve.
The fundamental relation for magnetic stored energy is:
For a magnetic circuit with a core of reluctance R, the flux linkage λ is related to the current i by the number of turns N and the magnetic flux φ:
From the definition of reluctance, the flux φ is related to the magnetomotive force (MMF) by:
Substituting this expression for φ into the flux linkage equation:
Now, substitute this expression for λ back into the energy equation Wf = 1/2 λi:
Alternatively, since the MMF is defined as F = Ni, the stored energy can be expressed as:
This expression shows that the magnetic stored energy is directly proportional to the square of the MMF and inversely proportional to the reluctance of the magnetic path.
Explain about demagnetizing Ampere turn per pole and Cross magnetizing Ampere Turn per pole.
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An 8 pole DC generator has per pole flux of 30 mWb and winding is connected in lap with 960 conductors. Calculate the generated EMF on open circuit when it runs at 500 rpm. If the armature is wave wound at what speed must the machine be driven to generate the same voltage.
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Derive the expression for EMF generated in DC generator.
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A 10 kW, 240 V, 6-pole, 1200 r.p.m lap-connected d.c. generator has 500 armature conductors. At rated voltage and current, armature ohmic losses are 200 watts. Compute the useful flux per pole? Take 3 V as the brush drop at full load?
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What is critical field resistance? How do you calculate critical field resistance from magnetic characteristics in the laboratory?
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Explain the procedure of parallel operation of d.c. generators.
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Six DC generators are running in parallel, each having an armature resistance of 0.15 ohm, running at the same speed and excited to give equal induce e.m.f(s). All generators share load equally at a terminal voltage of 500V. The total load is 360KW. If the field current of one generator is raised by 5% and the speed remains constant, Calculate new Terminal Voltage.
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Describe and compare various methods of speed control of dc motors.
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A 400 V d.c. shunt motor takes 5 A at no-load. Ra=0.5 Ω and Rsh=200 Ω. Estimate the kW output and efficiency when the motor takes 50 A on full load.
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