Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
You must login to add an answer.
b) 1.5-1.7 Wb per m2
c) 1.15 to 1.2
a) area of cross section of pole body = 0.98 * axial length of the pole *
breadth of the pole
b) copper area = full load field mmf / current density in the field winding
c) total space = copper area / space factor
a) 0.8-0.9
a) height of winding = total winding area / depth of winding
d) radial length of the pole shoe = height of winding + height of pole
shoe + 0.02
a) height of pole body = height of the winding + 0.02
b) 0.3-1.5
a) true
a) true
a) true
b) area per pole of damper pass = 0.2 * specific electric loading * pole
pitch / current density in damper bars
a) 3-4 A per mm2
c) 20%
a) pole arc = number of bars per pole * stator slot pitch * 0.8
a) length of each damper bar = 1.1 * axial length
c) area of cross section of each damper bar = total area of bars per pole
/ number of damper bars per pole
b) area of each ring short-circuiting the bars = (0.8-1) * area of damper
bar
a) 59 mm2
b) height of pole shoe = 2 * diameter of damper bars
a) true
a) true
a) 7
d) hysteresis loss and eddy current loss
d) slot opening, air gap length, number of slots and speed of machines
c) 25-70 % of iron loss
c) copper loss per phase = current per phase2 * dc resistance
a) total copper loss = 3 * average value of the eddy current constant *
current per phase2 * dc resistance
a) stray field
a) 1 V
d) construction, speed, rating of the machine
c) 0.3-0.4 % of kVA rating
c) temperature rise of the surface = Surface area * cooling coefficient /
dissipating surface
c) speed of the machine and configuration of the surface
c) speed of the machine and configuration of the surface
a) true
a) true
d) hysteresis, eddy and heating losses
a) kVA output = output coefficient * diameter2 * length * synchronous
speed
c) output coefficient = 11 * specific magnetic loading * specific electrical
loading * winding space factor * 10-3
b) output = 1.11* specific magnetic loading * specific electrical loading *
winding space factor * 10-3 * peripheral speed2 *Length / synchronous
speed
c) 5
a) choice of magnetic loading is directly proportional to the iron loss
b) air gap density is indirectly proportional to the voltage
a) air gap density is directly proportional to the short circuit curren
a) air gap density is directly proportional to the short circuit curren
Answer: a
Explanation: The minimum value of the frequency constant used in the output equation
of P.H Tricky is 0.86. The maximum value of the frequency constant used in the output
equation of P.H Tricky is 1.0.
Answer: d
Explanation: The length and the diameter are the 2 main dimensions of the single phase
induction motor. The most economical relation between length and diameter is length =
0.63 * diameter.
Answer: a
Explanation: The reluctance motor is a small synchronous motor with salient pole rotor.
The single phase reluctance motors have the phenomenon of hunting
Answer: b
Explanation: The electromagnetic torque is maximum at the angle of 45°. The range of
operation of the reluctance motor lies in the range of 0-45°.
a) air gap density is directly proportional to the short circuit curren
b) false
b) false
a) 0.52-0.65 Wb per m2
c) 4
a) high specific electric loading gives high copper losses and high
temperature rise
a) true
c) specific electric loading is high, leakage reactance is high, giving high
synchronous reactance
a) 20,000-40,000 A per m
a) 1.7-1.8 Wb per mm2
b) minimum width of tooth = flux / pole proportion * (number of stator
slots / number of poles) * length * 1.8
a) parallel sided
a) teeth is tapered and minimum width is across the medium
d) maximum permissible width = slot pitch – minimum width of the teeth
b) three times
b) to reduce the short circuit current
a) length of mean turn = 2*length + 2.5*pole pitch + 0.06 kV + 0.2
a) length of mean turn = 2*length + 2.5*pole pitch + 0.06 kV + 0.2
a) true
c) depth of armature core = flux / 2 * length of the iron core * flux density
c) outer diameter = inner diameter + 2*(depth of the slots + depth of
armature core)
c) it represents the fraction of the armature circular area for conductors
c) it represents the fraction of the armature circular area for conductors
b) 0.1-0.2
a) armature resistance = (Diameter + length)*total number of armature
conductors/1.2 * 104 * number of parallel paths in the armature2
b) diameter = 2.32 * diameter * (number of armature teeth embraced by
one coil/total number of armature teeth)
b) geometry of the magnet, airgap, associated non-portions of the
magnetic circuit
c) 4-8
c) 6
b) false
d) d = angle/720
c) 1.15
a) magnetic to electrical boarding ratio = number of poles * permeance
coefficient * flux per pole/number of conductors * armature current
b) high magnetic to electrical boarding ratio gives low copper volume
and high iron volume
b) high magnetic to electrical boarding ratio gives low copper volume
and high iron volume
b) high magnetic to electrical boarding ratio gives low copper volume
and high iron volume
b) false
a) flux density = residual flux density / 1 + (1.11/permeance coefficient)
d) 11
d) volume of air gap and magnet decreases, armature diameter
increases
c) 4.3-4.6 * 106
d) volume of air gap and magnet decreases, armature diameter
increases
d) volume of air gap and magnet decreases, armature diameter
increases
c) 50
b) area of magnet = flux / 4.95 * residual flux density
a) true
a) brush shift is directly proportional to the demagnetization effect
c) 2
b) low motor torque
b) insulation failure
a) higher ventilation arrangement
a) higher ventilation arrangement
c) 3
a) spur gears
c) worm gears
b) helical gears
a) 2
a) 2
b) 3
a) 2
b) solenoidal, mechanical movement
d) ac or dc supply
d) ac or dc supply
d) track switches, bells, buzzers
a) holding magnet
b) only dc supply
b) 3
a) flat-faced armature type
b) cast steel
a) magnetic in series and mechanical in parallel
a) magnetic in series and mechanical in parallel
d) lifting magnets, magnetic clutches, magnetic chucks
c) force is directly proportional to the square of the air gap length
a) true
a) 1
c) force is directly proportional to the square of the air gap length
d) iron, nickel, cobalt
c) silicon, chromium, molybdenum
c) silicon, chromium, molybdenum
a) true
a) true
a) true
c) 40 mm
d) 3 mm
d) used to withstand the centrifugal forces
d) used to withstand the centrifugal forces
d) mean diameter at the position of centre of gravity = Inner diameter –
diameter of stator wires
a) 350 NM per m2
b) 600 NM per m2
a) head of air inside the machine = hydrodynamic resistance * volume
of air passing2
c) total head produced = ∑ coefficient of hydrodynamic resistance *
volume of air passing per second2
d) sharp or projecting inlet edges, inlet corners, variations in crosssections of air paths
b) 40-60 * 10-3
b) 30 * 10-3
a) 12-20 * 10-3
d) area of cross section and hydrodynamic coefficients
c) 5
a) 7
a) volume of air = 0.9 * losses in kW / difference of air temperature at
inlet and outlet
c) 12-160C
a) area of outlet opening = maximum air passing per second / 0.42 *
peripheral speed
d) width of fan = area of outlet opening / 2.88 * outside diameter *
coefficient of utilization
c) number of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of
fan
c) number of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of
fan
a) maximum air passing = 2 * volume of air passing per second
d) zero starting torque
c) 3
Answer: b
c) starting or auxiliary windings
b) running winding displaces the starting winding by 90°
b) running winding displaces the starting winding by 90°
d) by connecting a suitable impedance
b) when the motor speed reaches 75 % of the full load speed
b) part of the pole is shaded by short circuited copper ring
c) displacement between shaded and unshaded portion varies between
20°-30°
a) for outputs below 60 watt
b) when high starting torque is required
c) cage winding is replaced by commutator windings
b) the commutator windings are dominant
a) 0.37-50
b) 4–6
c) 0.2-0.3
c) 5
c) cast iron or die cast aluminium alloy frame
c) cast iron or die cast aluminium alloy frame
Answer: d Explanation: A thermal circuit in electrical machine design deals with the mode and media to dissipate heat produced in the machine. The heat produced in the machine leads to losses.
(D) C2C
Consumer to consumer
a) length of mean turn = 2*length + 2.5*pole pitch + 0.06 kV + 0.2
Answer: d