E.M.D-1 (170902) Numerical
[1] A 200 kVA, 6.6kV/440volts, 50Hz, three phase core type transformer has the following design data: Max. flux density : 1.25 wb/m
2
; Emf/turn : 10 volts; Stacking factor : 0.9 ; Window space factor : 0.32 ; Current density : 2.4 A/mm
2
.Overall width and overall height are same. If three stepped core is used determine overall dimensions. Also show them on a sketch.
[2] Calculate approximate overall dimensions for a 200 kVA, 6600/440V, 50 Hz, 3 phase core type transformer. The following data may be assumed: emf per turn=10V; maximum flux density=1.3
Wb/m
2
, current density=2.5 A/mm
2
,window space factor=0.3, overall height=overall width, stacking factor=0.9. Use a 3 stepped core. For a 3 stepped core, Width of largest stamping=0.9d and Net iron area= 0.6d
2
, where d=diameter of circumscribing circle.
[3] Determine the main core dimensions for a 250 KVA, 6600/500V, 50 Hz, 3-phase star/delta core type transformer from the following data: Window space factor = 0.27; Current density = 2.5
A/mm² ; Max. flux density = 1.25 Wb/m² ; Volts per turn = 8.5 V ; Use 4-stepped core limb section which has the area factor = 0.62. Height of window / width of window = 2
[4] Determine the main core & yoke dimensions for a 200 KVA, 50 Hz, 1-phase core type transformer. Window space factor = 0.32;Current density = 3 A/mm² ; Max.flux density = 1.1
Wb/m² ;Volts per turn = 14 V ;Stacking factor = 0.9; Net iron area = 0.56*d2; Cruciform core with distance between adjacent limbs = 1.6 times width of core.
[5] Estimate the main dimensions of complete core frame, winding conductor areas and no. of turns of a 3-phase core type delta-star transformer which is rated at 300 kVA, 6600/440 volts, 50 Hz. Use three stepped core with the diameter of circumscribing circle of 0.25 m.
Assume volts/turn = 8.5 ,current density of 2.5 A/mm
2
, window space factor of 0.28 , stacking factor of 0.9 and height to width ratio for window =3.
[6] Estimate the per unit regulation, at full load and 0.8 power factor lagging, for a 300 kVA, 50 Hz,
6600/400 V, 3 phase, delta/star, core type transformer. The data given is:
H.V. winding: outside diameter=0.36 m, inside diameter=0.29 m, area of conductor=5.4 mm
2
L.V. winding: outside diameter=0.26 m,inside diameter=0.22 m, area of conductor=170 mm
2
Length of coils=0.5 m, voltage per turn=8V, resistivity=0.21 Ω /m/ mm
2
[7] The tank of a 1.25 MVA natural oil cooled transformer has the dimensions of 155cm x 65 cm x 185 cm (length, width and height). The full load losses are 13100 watts. Estimate the number of cooling tubes required for this transformer. Assume: W/m
2
-
0
C due to radiation = 6 and due to
By PratikKabrawala 1

E.M.D-1 (170902) Numerical convection = 6.5; improvement in convection due to provision of tubes = 40%. ; Temperature rise=
40
0
C. ; length of each tube = 1 m ; diameter of tubes = 50 mm. Neglect top and bottom surfaces of the tank as regards cooling.
[8] Estimate the number of cooling tubes for a 250 KVA,6600/400 V, 50 Hz, 3-phase delta / star core type oil cooled transformer from the following data:Temp rise = 50° C; Total losses at 90° C are
5.0 KW; Tank size = 125 x 100 x 50 cms. (h x l x w) Oil level = 115 cms. Sp-heat dissipation of plain tank = 12.5 W/m²/°C Diameter of cooling tube = 5.0 cms. Show the tube arrangement by sketch
[9] A 40 Hz transformer is to be used on a 50 Hz system. Assuming the Steinmetz’s coefficient as 1.6 and losses at lower frequency 1.2%, 0.7% and 0.5% for I
2
R, hystersis and eddy current respectively. Find (a) losses on 50 Hz for the same supply voltage and current (b) Output at 50 Hz for the same total losses as on 40 Hz.
[10] An 11 kV, 25 Hz transformer has I
2
R ,hystersis and eddy current losses 1.6, 0.6 and 0.4 percent of the output. What will be the percentage losses if the transformer is connected to 22 kV, 50 Hz supply assuming the full load current to remain the same?
[11] A 400 kW, 500V, 500 rpm, 6 pole d.cgenerator is built with an armature diameter of 90 cm and core length of 36 cm. The lap wound armature has 700 conductors. Determine specific magnetic and electric loading of the machine.
[12] A 350 KW, 500 Volts, 450 rpm, 6 pole d.c. generator is built with an armature diameter of 87 cm and core length of 32 cm. The lap wound armature has 660 conductors. Determine specific magnetic and electric loading of the machine.
[13] Determine main dimensions(D and L) of a 12 kW, 230V, 2 pole, 1500 rpm D.C. shunt generator if the required data is : full load efficiency = 82% ; polearc/pole pitch = 0.63 ; avg. flux density = 0.4 Wb/sqr m. ; Ampere conductors / m =19000. Machine is designed to have square pole face.
[14] Calculate the main dimensions of the armature of a 400 KW, 500V, 180 rpm, 16 poles dc generator. Use square pole-face. Given data: Efficiency = 90 %; Pole-arc to pole pitch ratio = 0.7;
Average gap density = 0.6 Wb/m²; Ampere-conductors per metre = 35000.
[15] Determine the main dimensions, number of poles and length of air gap of a 600 kW, 500 V,
900r.p.m. generator. Assume average gap density as 0.6 Wb/m
2
and ampere conductors per metre as 35000. The ratio of pole arc to pole pitch is 0.75 and efficiency is 91 percent. The following are the design constraints: Peripheral speed: ≤ 40 m/s, frequency of flux reversals: ≤ 50 Hz, current per brush arm: ≤ 400 A and armature mmf per pole ≤ 7500 A. The mmf required for air gap is 50 percent of armature mmf and gap contraction factor is 1.15.
By PratikKabrawala 2

E.M.D-1 (170902) Numerical
[16] Calculate the diameter and length of armature for a 7.5 kW, 4 pole, 1000 r.p.m. 220 V shunt motor.
Given: full load efficiency=0.83; maximum gap flux density=0.9 Wb/m
2
; specific electric loading=30,000 ampere conductors per meter ; field formfactor=0.7. Assume that the maximum efficiency occurs at full load and field current is 2.5% of rated current. The pole face is square.
[17] Find out main dimensions and number of armature conductors for 350 KW, 440 Volts, 720 rpm, 6pole D.C. Generator. Assume square pole face, ratio of pole arc to pole pitch 0.66, efficiency
91%, 4% of rated voltage for brush drop. Use data obtained from the following machine. 250
KW, 500 Volts, 600 rpm DC generator with 720 lap connected armature conductors, 0.75m armature diameter and core length of 0.3 m.
[18] Calculate (i) the number of poles and (ii) air-gap length of a 600 KW, 500 V, 900 rpm dc generator from the following data: Average gap density = 0.6 Wb/m²; Ampere-conductors per metre =
35000; Pole-arc to pole-pitch ratio = 0.7; Gap contraction factor = 1.15; Diameter of armature core = 0.81 m; Length of armature core = 0.325 m; MMF for air-gap = 50% of armature MMF.
[19] Determine the main dimensions, no. of poles and thelength of air gap of a 500 kW, 500 V, 960 rpm d.c. generator. Assume: B av
= 0.58 wb/m
2
; ac = 36000; Pole arc/pole pitch = 0.75; Efficiency =
90% .The mmf required for air gap is 50% of armature mmf and gap contraction factor is
1.15. Design constraints (limiting values):Peripheral speed : 40 m/s ; Frequency of flux reversals:
50 Hz; Current per brush arm: 400 A; Armature mmf per pole: 7500 A.
[20] The output coefficient of a d.c. machine is 200 kVA/m
3
- r.p.s. Its armature power (Pa) is
1000 kW and speed is 300 rpm. Determine main dimensions (D and L) of the machine and also find main dimensions if: (1) Specific loadings are decreased by 10% each without changing speed. (2) Speed is decreased to 150 rpm without changing specific loadings. Assume L/D ratio
= 0.2 in all cases. Comment on your answers.
[21] A 4 pole generator supplies a current of 140 A. It has 480 armature conductors (a) wave connected,
(b) lap connected. The brushes are given an actual lead of 10˚. Calculate the cross and demagnetizing mmf per pole in each case. The field winding is shunt connected and takes a current of 10 A, find the number of extra shunt field turns to neutralize the demagnetization.
[22] Find out armature demagnetizing & cross magnetizing mmf per pole if brushes are given a lead of
9
0
electrical for a 500 KW, 375 RPM, 8 Pole, D.C. Generator with flux per pole of 88.5 mwb.
Assume that the power developed by armature is equal to rating of machine.
[23] The given details refer to the shunt field coil of a 440 volt , 6 pole d.c. machine. Mmf per pole =
7200 A ; Depth of winding = 50 mm; Length of inner turn = 110 cm ;Length of outer turn = 140 cm ; space factor = 0.6 ; resistivity = 0.02 Ω /m and mm
2
; loss radiated from outer surface excluding ends = 1400 W/m
2
.20% of the terminal voltage drops across the field regulator.
Determine: The diameter of wire, length of coil, number of turns and exciting current.
By PratikKabrawala 3

E.M.D-1 (170902) Numerical
[24] The following particulars refer to the shunt field coil for a 440 V, 6 pole, DC generator: Mmf per pole=7000 A; depth of winding=50 mm; length of inner turn=1.1 m; length of outer turn=1.4 m; loss radiated from outer surface excluding ends = 1400 W/m
2
; space factor=0.62; resistivity= 0.02
Ω /m and mm
2
. Calculate (a) the diameter of wire (b) length of coil (c) number of turns and (d) exciting current. Assume a voltage drop of 20 % of terminal voltage across the field regulator.
[25] Design a permanent magnet to have minimum volume. The magnet is to produce a flux density of
0.3 Wb/m
2
, in a parallel sided air gap 50 mm square and 6mm long. The demagnetization curve of the magnet material is given below:
[26] The armature of 12 pole ,500KW, 550 V, generator has a simplex lap winding consisting of 2484 conductors. There are 621 commutator segments & ratio of pole arc to pole pitch is 0.7.
(a) Calculate the demagnetizing & cross magnetizing mmf / pole at rated full load current if brushes are shifted through 3 segments from G.N.A.
(b) Calculate no. of conductors that must be provided in each pole face if a compensating winding is used
By PratikKabrawala 4

E.M.D-1 (170902) Numerical
By PratikKabrawala 5