4. Magnetic circuit The path of magnetic flux is called magnetic circuit Magnetic circuit of dc machine comprises of yoke, poles, airgap, armature teeth and armature core Flux produced by field coils emerges from N pole and cross the air gap to enter the armature tooth. Then it flows through armature core and again cross the air gap to enter the S pole Yoke Flux Path Pole Body Armature Core N S S N Magnetic Circuit of 4-Pole DC Machine h pl lyly lclc

5. Let B g – Max. flux density in the core K g - Gap contraction factor l c – Length of magnetic path in the core l y – Length of magnetic path in the yoke d s - Depth of the slot d c - Depth of core h pl - Height of field pole D m – Mean diameter of armature When the leakage flux is neglected magnetic circuit of a DC machine consists of following: i.Yoke ii.Pole and pole shoe iii. Air gap iv.Armature teeth v.Armature core Magnetic circuit

6. Total MMF to be developed by each pole is given by the sum of MMF required for the above five sections. MMF for air gap AT g =800000 B g K g l g MMF for teeth AT t =at t X d s MMF for core AT c =at c X l c /2 MMF for pole AT p = at p X h pl MMF for yoke AT y = at y X l y /2 at t, at c, at p, at y - are determined B-H curves l c = πD m /P = π(D – 2d s – d c )/P l y = πD my /P = π(D+ 2l g + 2h pl +d y )/P AT total =AT g + AT t + AT c + AT p +AT y

7. Design of field system  Consists of poles, pole shoe and field winding.  Types:  Shunt field  Series field  Shunt field winding – have large no of turns made of thin conductors,because current carried by them is very low  Series field winding is designed to carry heavy current and so it is made of thick conductors/strips  Field coils are formed, insulated and fixed over the field poles

8. Factors to be considered in design:  MMF/pole &flux density  Losses dissipated from the surface of field coil  Resistance of the field coil  Current density in the field conductors Design of field system

10. Let, AT fl -MMF developed by field winding at full load Q f - Copper loss in each field coil(W) q f - Permissible loss per unit winding surface for normal temperature rise(W/m 2 ) S f - Copper space factor ρ - Resistivity (  –m) h f - Height of winding(m) d f - Depth of winding(m) S - Cooling surface of field coil(m 2 ) L mt - Length of mean turn of field winding(m) R f - Resistance of each field coil (ohms) T f - Number of turns in each field coil A f - Area of each conductor of field winding(m 2 ) I f - Current in the field winding (A) δ f - Current density in the field winding(A/mm 2 ) Design of field system Design of field system Tentative design of field winding

11. Cooling surface of the field winding, S=2L mt h f -- (1) Permissible copper loss in each field coil, S qf =2L mt h f q f -- (2) Area of X-section of field coil=h f d f -- (3) Area of copper in each section=S f h f d f -- (4) i.e, T f a f =S f h f d f -- (5) Copper loss in each field coil, Q f =I f 2 R f =I f 2 (T f  L mt )/a f i.e., Copper loss   f 2 (Square of the current density) Design of field system

12. To have temperature rise within the limit, the copper loss should be equal to the permissible loss. Using Eqns. (2) & (6), 2L mt h f q f =  f 2  L mt (S f h f d f ) => MMF per metre height of field winding Design of field system

13.  Normal values:  Permissible loss, q f -700W/m 2  Copper Space factor, S f :  Small wires: 0.4  Large round wires: 0.65  Large rectangular conductors: 0.75  Depth of the field winding, d f : Design of field system Armature Dia (m)Winding Depth (mm) 0.230 0.3535 0.540 0.6545 1.0050 1.00 and above55

14. Height of field, Total height of the pole, h pl =h f +h s + height for insulation and curvature of yoke where, h s - Height of the pole shoe (≈0.1 to 0.2 of the pole height) Design of field system

15. Design of shunt field winding  Involves the determination of the following information regarding the pole and shunt field winding  Dimensions of the main field pole,  Dimensions of the field coil,  Current in shunt field winding,  Resistance of coil,  Dimensions of field conductor,  Number of turns in the field coil,  Losses in field coil.  Dimensions of the main field pole  For rectangular field poles o Cross sectional area, length, width, height of the body  For cylindrical pole o Cross sectional area, diameter, height of the body

16.  Area of the pole body can be estimated from the knowledge of flux per pole, leakage coefficient and flux density in the pole  Leakage coefficient (C l ) depends on power output of the DC machine  Bp in the pole 1.2 to 1.7 wb/m 2  Ф p = C l. Ф  A p = Ф p /B p  When circular poles are employed, C.S.A will be a circle  Ap = πd p 2 /4 Design of shunt field winding

17.  When rectangular poles employed, length of pole is chosen as 10 to15 mm less than the length of armature  L p =L –(0.001 to 0.015)  Net iron length L pi = 0.9 L p  Width of pole, b p = Ap/L pi  Height of pole body h p = h f + thickness of insulation and clearance  Total height of the pole h pl = h p + h s Design of shunt field winding

18.  Field coils are former wound and placed on the poles  They may be of rectangular or circular cross section depends on the type of poles  Dimensions – L mt, depth, height, diameter  Depth(d f ) – depends on armature  Height (h f ) - depends on surface required for cooling the coil and no. of turns(T f )  h f, T f – cannot be independently designed Design of shunt field winding

19.  L mt - Calculated using the dimensions of pole and depth of the coil  For rectangular coils  L mt =2(L p + b p + 2d f ) or (L o +L i )/2  Where L o – length of outer most turn & L i – length of inner most turn  For cylindrical coils  L mt = π(d p +d f )  No of turns in field coil: When the ampere turns to be developed by the field coil is known, the turns can be estimated  Field ampere turns on load, AT fl = I f. T f  Turns in field coil, T f = AT fl /I f Design of shunt field winding

20. Power Loss in the field coil: Power loss in the field coil is copper loss, depends on Resistance and current Heat is developed in the field coil due to this loss and it is dissipated through the surface of the coil In field coil design, loss dissipated per unit surface area is specified and from which the required surface area can be estimated. Surface area of field coil – depends on L mt, depth and height of the coil Design of shunt field winding

21. L mt – estimated from dimensions of pole Depth – assumed (depends on diameter of armature) Height – estimated in order to provide required surface area Heat can be dissipated from all the four sides of a coil. i.e, inner, outer, top and bottom surface of the coil Inner surface area= L mt (h f – d f ) Outer surface area = L mt (h f + d f ) Top and bottom surface area = L mt d f Total surface area of field coil, S= L mt (h f – d f )+ = L mt (h f + d f )+ L mt d f + L mt d f S= 2L mt h f +L mt d f = 2L mt (h f +d f ) Permissible copper loss, Q f =S.q f [q f -Loss dissipated/ unit area] Design of shunt field winding

22. Substitute S in Q f, Q f = 2L mt (h f +d f ).q f Actual Cu loss in field coil=I f 2 R f =E f 2 /R f Substituting R f =(  L mt T f )/ a f, Actual Cu loss in field coil=E f 2.a f /(  L mt T f )  Design of shunt field winding

23. Procedure for shunt field design Step1 : determine the dimensions of the pole. Assume a suitable value of leakage coefficient and B = 1.2 to 1.7 T Ф p = C l. Ф A p = Ф p /B p When circular poles are employed, C.S.A will be a circle A p = πd p 2 /4 : d p =Ѵ(4A p /π) When rectangular poles employed, length of pole is chosen as 10 to15 mm less than the length of armature L p =L –(0.001 to 0.015) Net iron length L pi = 0.9 L p Width of pole = A p /L pi

24. Step 2 : Determine L mt of field coil Assume suitable depth of field winding For rectangular coils L mt =2(L p + b p + 2d f ) or (L o +L i )/2 For cylindrical coils L mt = π(d p +d f ) Step 3: Calculate the voltage across each shunt field coil E f = (0.8 to 0.85) V/P Step 4 : Calculate C.S.A of filed conductor A f = ρL mt AT fl /E f Step 5:Calcualate diameter of field conductor d fc =Ѵ(4a f /π) Diameter including thickness d fci = d fc + insulation thickness Copper space factor S f = 0.75(d fc /d fci ) 2 Procedure for shunt field design

25. Step 6 : Determine no. of turns (Tf) and height of coil (hf) They can be determined by solving the following two equations 2L mt (h f + d f ) = E f 2 a f /ρL mt T f T f.a f = S f.h f.d f Step 7 : Calculate R f and I f : R f = T f. ρL mt /a f I f = E f /R f Step 8 : Check for δ f δ f = I f / a f δ f – not to exceed 3.5A/mm 2. If it exceeds then increase a f by 5% and then proceed again Procedure for shunt field design

26. Step 9 : Check for desired value of AT AT actual = I f. T f AT desired - 1.1 to 1.25 times armature MMF at full load When AT actual exceeds the desired value then increase the depth of field winding by 5% and proceed again. Procedure for shunt field design

27. Check for temp rise: Actual copper loss = I f 2 R f Surface area = S = 2L mt (h f + d f ) Cooling coefficient C = (0.14 to 0.16)/(1 + 0.1 V a )  m = Actual copper loss X (C/S) If temperature rise exceeds the limit, then increase the depth of field winding by 5% and proceed again.

28. Design of Series Field Winding Step 1: Estimate the AT to be developed by series field coil, AT /pole = (I z. (Z/2))/P For compound m/c, AT se = (0.15 to.25) (I z. Z)/2P For series m/c, AT se = (1.15 to 1.25) (I z. Z)/2P Step 2: Calculate the no. of turns in the series field coil, T se = AT se /I se (Corrected to an integer) Step 3: Determine cross sectional area of series field conductor, a se = I se /δ se Normally, δ se - 2 to 2.3 A /mm 2

29. Step 4 : Estimate the dimension of the field coil Conductor area of field coil = T se.a se Also Conductor area of field coil = S fse.h se.d se When circular conductors are used S fse = 0.6 to 0.7 For rectangular conductors, S fse – depends on thickness and type of insulation On equating above two expressions, T se.a se = S fse.h se.d se h se = (T se.a se )/(S fse.d se ) Design of Series Field Winding

32. Design of commutator and brushes  Commutator and brush arrangement are used to convert the bidirectional current to unidirectional current  Brushes are located at the magnetic neutral axis ( mid way between two adjacent poles)  The phenomenon of commutation is affected by resistance of the brush, reactance emf induced by leakage flux, emf induced by armature flux.

33. Classification of commutation process 1.Resistance commutation 2.Retarded commutation 3.Accelerated commutation 4.Sinusoidal commutation  Commutator is of cylindrical in shape and placed at one end of the armature  Consists of number of copper bars or segments separated from one another by a suitable insulating material of thickness of 0.5 to 1mm  Number of commutator segments = no. of coils in the armature  Materials used :  Commutator segments: Hard Drawn Copper or Aluminum Copper  Insulation :Mica, Resin Bonded Asbestos  Brushes :Natural Graphite, Hard Carbon, Electro Graphite, Metal Graphite Design of Commutator and brushes

34. Design formulae 1.No. of commutator segments, C = ½ u.S a where, u – coils sides/slot S a – no. of armature slots 2.Minimum no. of segments = E p /15 3.Commutator segment pitch = β c = πD c /C where, Commutator Diameter D c – 60% to 80% of diameter of armature βc ≥ 4mm 4.Current carried by each brush I b = 2I a /P  for lap winding I b = I a  for wave winding 5.Total brush contact area/spindle A b = I b /δ b 6.Number of brush locations are decided by the type of winding Lap winding: No of brush location = no. of poles Wave winding : No of brush location =2 Design of Commutator and brushes

35. 7.Area of each individual brush should be chosen such that, it does not carry more than 70A Let, a b – Contact area of each brush n b – Number of brushes / spindle  Contact area of brushes in a spindle, A b = n b. a b also a b = w b.t b  Ab = n b. w b.t b Usually, t b = (1 to 3) β c w b = A b / n b. T b = a b /t b 8.L c – depends on space required for mounting the brushes and to dissipate the heat generated by commutator losses L c = n b (w b + C b ) + C 1 + C 2 where, C b - Clearnace between brushes (5mm) C 1 - Clearance allowed for staggering of brushes (10mm, 30mm) C 2 – Clearance for allowing end play (10 to 25 mm) Design of Commutator and brushes

36. 9.Losses :  Brush contact losses: depends on material, condition, quality of commutation  Brush friction losses Brush friction loss P bf = μ p b A B.V c μ – Coefficient of friction p b -Brush contact pressure on commutator (N/m 2 ) A B - Total contact area of all brushes (m 2 ) A B =P A b (for lap winding) = 2 A b (for wave winding) V c – Peripheral speed of commutator (m/s) Design of Commutator and brushes

37. Design of Interpoles  Interpoles: Small poles placed between main poles  Materials Used: Cast steel (or) Punched from sheet steel without pole shoes  Purposes:  To neutralize cross magnetizing armature MMF  To produce flux density required to generate rotational voltage in the coil undergoing commutation to cancel the reactance voltage.  Since both effects related to armature current, interpole winding should be connected in series with armature winding  Average reactance voltage of coil by Pitchelmayer’s Equation is, E rav = 2T c ac V a.L.λ Inductance of a coil in armature =2T c 2.L.λ

38. Normally, Length of interpole = length of main pole Flux density under interpole, B gi = ac. λ.(L/L ip ) where, L ip - length of interpole In general, B gi = 2 I z. Z s. (L/L ip ). (1/V a.T c ).λ MMF required to establish B gi = 800000B gi.K gi.l gi Design of Interpoles

39. Losses and efficiency : 1.Iron Loss - i)Eddy current loss ii) Hysteresis loss 2.Rotational losses - Windage and friction losses 3.Variable or copper loss Condition for maximum efficiency : Constant Loss= Variable Loss Design of Interpoles