Chapter - 3 Design of Transformers 3.1 Introduction. A transformer is static electrical machine in which energy is transferred from one electrical circuit to another through magnetic field. The transformer is not an energy convesion device but it transforms electrical energy between primary and secondary circuits with changed values of voltage and current. It is used to match source and load impedance for maximum power transfer in electric and control circuits. The transformer is also used to provide electric isolation between the circuits. Transformer Construction: Main parts of the transformer are primary and secondary windings made up of /copper or aluminum conductors and iron core. The expression for emf induced E = 4.44 f Фm T Where f = frequency, Фm =mutual flux and T=number of turns in the primary/ secondary winding Usually the winding connected to supply is called primary winding and the other winding connected to load is called secondary winding. Depending upon the type of construction used, the transforms are classified into two categories as : (i) core type, and (ii) shell type Core type transformers: The magnetic core is built of laminations to form rectangular frame and the windings are arranged concentrically with each other around t legs or limbs of the core as shown in Fig 2.1 Shell type transformers: In shell type transformers the windings are put the central limb and the flux path is completed through two side limbs as shown in Fig2.2 The central limb carries total mutual flux while the side limbs forming a part parallel magnetic circuit carry half the total flux. Consequently, the cross-sectional area (and width) of the central limb is twice that of each of the side limbs. Comparison of core and shell types of transformers: S.No Core type Shell type 1 Easy in design and construction. Comparatively complex. High mechanical strength. 2 Has low mechanical strength due to nonbracing of windings. Reduction of leakage reactance 3 Reduction of leakage reactance. is not easily is highly possible. possible. It cannot be easily dismantled for repair work. 4 The assembly can be easily dismantled for Heat is not easily dissipated repair work. 5 Better heat dissipation from windings. from windings since it is 6 Has longer mean length of core and shorter surrounded by core. mean length of coil turn. Hence best suited for It is not suitable for EHV (Extra EHV (Extra High Voltage) requirements. High Voltage) requirements. 3.2 Output equation of a Single Phase transformer: The transformer output is rated in kilo volt ampere (KVA). It depends on flux density (i.e. core area) and ampere turns (AT) (i.e. winding area). The voltage induced in a transformer winding with T turns E = 4.44 f Фm T volts --E Emf per turn Et = = 4.44 f Фm volts --T Emf 1 2 Figure 3.1- core and shell types of transformers The window space factor Kw is defined as ·the ratio of copper area in window to total area of the window. . conductor area in. window Ac Kw = = -------- 3 Total area in. window Aw Conductor area in window = Ac = Kw * Aw -------- 4 Taking the current density δ is same in both the windings Is Ip Therefore, Current density, δ = = as ap Area of a cross – section of primary conductor = ap = Ip / δ -------- 5 ----- 6 Area of a cross – section of secondary conductor= as = Is / δ ----- 7 If we neglect magnetizing mmf then primary ampere turns is equal to secondary ampere turns. Therefore, Ampere turns = AT = Ip * Tp + Is * Ts ------ 8 The window in a single phase transformer contains one primary and one secondary winding. Total copper area in window Ac = copper area of primary + copper area of secondary =primary turns * area of primary conductor + secondary turns * area of secondary conductor Ac =Tp * ap+ Ts * as ------ 9 Substituting for ap and as from 6 and .7 in 9 we get Ac = Tp * (Ip / δ ) + Ts * (Is / δ ) --------- 10 Ac = 1/ δ (Tp * Ip + Ts * Is ) --------- 11 From 8 we have Ac = 1/ δ ( AT + AT ) --------- 12 Ac = 2 AT / δ --------- 13 Equating 4 and 13 we get, Kw * Aw = 2 AT / δ --------- 14 1 Now AT = ( Kw * Aw *δ) --------- 15 2 KVA output of the transformer = Q = Vp * Ip * 10 -3 --------- 16 Since Ep ≈ Vp we have = Ep * Ip * 10 -3 --------- 17 Dividing 17 by Tp and multiplying by Tp we get, = (Ep / Tp )* Tp Ip * 10 -3 --------- 18 = Et * AT* 10 -3 --------- 19 Substituting for Et from 2 and AT from15 we get, 1 Q = 4.44 f Фm * ( Kw * Aw *δ) * 10 -3 --------- 20 2 = 2.22 f Фm * ( Kw * Aw *δ) * 10 -3 --------- 21 But Bm = Фm / Ai --------- 22 Фm = Bm * Ai --------- 23 Substituting 23 in 21 we get, Q = 2.22 f Bm * Ai * ( Kw * Aw *δ) * 10 -3 --------- 24 Equation 24 is output equation of single phase transformer. EXAMPLE3.1: Calculate the core and window areas required for a 1000 kV A, 6600/400 V, 50Hz, single phase core type transformer. Assume a maximum flux density of 1.25Wb/m2 and a current density of 2.5 A/mm2. Voltage per turn = 30 V. Window space factor = 0.32. Given data: kVA = 1000 . f = 5Q Hz B = 1.25 Wb/m2 Vp = 6600V Vs =400V δ = = 2.5 A/mm2 Et = 30 V Kw =0.32 Core type Solution Emf per turn Et = 4.44 f Фm volts Фm = Et / 4.44 f = 30/4.44*50 = 0.1351 Wb Flux density Bm = Фm / Ai The net area of core = Ai = Фm / Bm = 0.1351 / 1.25 = 0.108 m2 The kV A rating of transformer Q = 2.22 f Bm * Ai * ( Kw * Aw *δ) * 10 -3 :. Window area Aw = Q / 2.22 f Bm * Ai * ( Kw *δ) * 10 -3 = 1000 / 2.22*50*1.25*0.108*0.32*2.5*106* 103 = 0.0834 m2 Result Net core area, Ai = 0.108 m2 = 0.108 x 106 mm2 Window area, Aw = 0.0834 m2 = 0.0834 x 106 mm2 3.3Output equation of Three phase transformer: The voltage induced in a transformer winding with T turns The induced Emf per phase E = 4.44 f Фm T volts 1 E Emf per turn Et = = 4.44 f Фm volts --2 T Three Phase Transformers, In the case of three window contains two primary and two secondary windings, The window space factor Kw is defined as ·the ratio of copper area in window to total area of the window. . conductor area in. window Ac = -------- 3 Kw = Total area in. window Aw Conductor area in window = Ac = Kw * Aw -------- 4 Taking the current density δ is same in both the windings Is Ip Therefore, Current density, δ = = -------- 5 as ap Area of a cross – section of primary conductor = ap = Ip / δ ----- 6 Area of a cross – section of secondary conductor= as = Is / δ ----- 7 If we neglect magnetizing mmf then primary ampere turns is equal to secondary ampere turns. Therefore, Ampere turns = AT = Ip * Tp + Is * Ts -----8 Three Phase Transformers, In the case of three window contains two primary and two secondary windings, Total copper area in window Ac = copper area of primary + copper area of secondary =2*primary turns * area of primary conductor + 2*secondary turns * area of secondary conductor Ac = 2*Tp * ap+ 2*Ts * as ------ 9 Substituting for ap and as from 6 and 7 in 9 we get, Ac =2*Tp * (Ip / δ ) + 2*Ts * (Is / δ ) --------- 10 Ac = 2/ δ (Tp * Ip + Ts * Is ) --------- 11 From 8 we have Ac = 2/ δ ( AT + AT ) --------- 12 Ac = 4 AT / δ --------- 13 Equating 4 and 13 we get, Kw * Aw = 4 AT / δ --------- 14 1 --------- 15 Now Ampere-turn AT = ( Kw * Aw *δ) 4 KVA output of the three phase transformer = Q =3* Vp * Ip * 10 -3 ----- 16 Since Ep ≈ Vp we have =3* Ep * Ip * 10 -3 Dividing 17 by Tp and multiplying by Tp we get, = 3*(Ep / Tp )* Tp Ip * 10 -3 --------- 17 --------- 18 = 3*Et * AT* 10 -3 --------- 19 Substituting for Et from 2 and AT from 15 we get, 1 Q =3* 4.44 f Фm * ( Kw * Aw *δ) * 10 -3 --------- 20 4 = 3.33 f Фm * ( Kw * Aw *δ) * 10 -3 --------- 21 But Bm = Фm / Ai --------- 23 Фm = Bm * Ai --------- 24 Substituting 24in 21 we get, Q = 3.33 f Bm * Ai * ( Kw * Aw *δ) * 10 -3 --------- 25 Equation 2.48 is output equation of three phase transformer. 3.4 Equation for Emf per turn: Selection of an appropriate value for emf per turn is most important step in transformer design . Hence an equation for emf per turn can be developed by relating output k V A, magnetic and electric loading. In transformers the ratio of specific magnetic and electric loading is specified rather than actual value of specific loadings. Let the ratio of specific magnetic and electric loading = r = Фm / AT ----- 1 AT = Фm / r ----- 2 The volt-ampere per phase of a transformer is given by the product of voltage and current per phase. Considering the primary voltage and current per phase we can write, kV A per phase, Q = Vp Ip x 10-3 Since Ep ≈ Vp we have = Ep * Ip * 10 -3 But Ep = 4.44 f Фm Tp Substituting 4 in 3 we get, Q = 4.44 f Фm Tp Ip* 10 -3 Q = 4.44 f Фm AT* 10 -3 From 2 we can write 7 as Q = 4.44 f Фm * (Фm / r) *10 -3 Q = 4.44 f (Фm2 / r) * 10 -3 Q*r Фm2 = 4.44 f * 10 − 3 ------3 --------- 4 ------ 5 ------ 6 ------ 7 ------ 8 ------- 9 = √ Q*r * 103/ 4.44 f ------ 10 E We knew that Emf per turn Et = = 4.44 f Фm volts --- 11 T Substituting for Фm in eqn11 from eqn10 we get, Et = 4.44 f * √ Q*r * 103/ 4.44 f ---- 12 Фm Et =√4.44 f *r* 103 * √ Q ---- 13 Et = K * √ Q ---- 14 Where K =√4.44 f *r* 103 ---- 15 K =√4.44 f *( Фm / AT )* 103 ---- 16 14 is Equation for Emf per turn In equation 14 , Q is kVA rating for single phase transformer and Q is kVA per phase for three phase transformer. Values of K for different types of transformer: The value of K depends on the type, service condition and method 'of construction of transformer. value of K for Single phase shell type 1.0 to 1.2 Single phase core type 0.75 to 0.85 , Three phase shell type 1.3 . Three phase core type, distribution transformer 0.45 Three phase core type, power transformer 0.6 to 0.7 3.5 Design of cores: Different types of cores: 1.Rectangular core 2. Square core 3. Stepped core. • For core type transformer the cross-section may be rectangular, square or stepped. • For shell type transformer the cross-section may be rectangular. When rectangular cores are used the coils ate also rectangular in shape. • The excessive leakage fluxes are produced during short circuit and over loads. This develops severe mechanical stresses on the coils. • On circular coils these forces are radial and there is no tendency for the coil to change its shape. • But on rectangular and square coils the forces are perpendicular to the conductors and tend to deform the shape of coil. • Hence circular coils are employed in high voltage and high capacity transformers. • When circular coils are employed the square and stepped cores are used. Figure 3.2- Square core and stepped core. In square cores the diameter of the circumscribing circle is larger than the. diameter of stepped cores of same area of cross-section. • Thus when stepped cores are used the length of mean turn of winding is reduced with consequent reduction in both cost of copper and copper loss. However with larger number of steps a large number of different sizes of laminations have to be used: This result in higher labour charges for shearing and assembling different types of laminations. Square core: Let d = diameter of circumscribing circle Also, d = diagonal of the square core a= side of square core Diameter of circumscribing circle, d = √ a 2 + a 2 = √2 a Side of square, a = d / √2 Gross core area, Agi = area of square = a2 = (d /√2)2 = 0.5 d2 Let stacking factor, Sf =0.9 The gross core area is the area including insulation area and net core area is e area of iron alone excluding insulation area. Hence Net core area, Ai = Stacking factor x Gross core area = 0.9 x 0.5 d2 = 0.45 d2 Area of circumscribing circle = ( Π/4) * d2 Netcorearea = 0.45 d2 / ( Π/4) * d2 = 0.58 The ratio, Areaofcircumscribingcircle Grosscorearea The ratio, = 0.5 d2 / ( Π/4) * d2 = 0.64 Areaofcircumscribingcircle Another useful ratio for the design of transformer core is core area factor. It is the ratio of net core area and square of the circumscribing circle Netcorearea Core area factor, Kc = Squareofcircumscribingcircle = Ai / d2 = 0.45 d2 / d2 = 0.45 Two stepped core or cruciform core: • In stepped cores the dimensions of the steps should be chosen, such as to a occupy maximum area within a circle. The dimensions of the two step to give maximum area for the core in the given area of circle are determined as follows [Also for given diameter of circle]. Let, a = Length of the rectangle b = Breadth of the rectangle d = Diameter of the circumscribing circle Also d = Diagonal of the rectangle θ = Angle between the diagonal and length of the rectangle. Figure 3.3- Two stepped core. The cross· section of two stepped core is shown in above figure. The maximum core area for a given d is obtained when θ is maximum value. Hence differentiate Agi with respect to θ and equate to zero to solve for maximum value of θ. a From above figure we get, Cos θ = ; a = d*Cos θ ---- 1 d b Sin θ = ; b = d*Sin θ ---- 2 d The two stepped core can be divided into three rectangles. The area of three rectangles gives the gross core area. With reference to above figure we can write, a−b a−b a−b Gross core area, Agi = ab + ( )*b + ( )*b = ab + 2* ( )*b 2 2 2 Agi = ab+ab-b2 = 2ab-b2 ----- 3 On substituting for a and b from equations 1 & 2 in equation 3 we get, Agi = 2(d cos θ)(d sin θ)-(d sin θ)2 = 2d2 cos θ sin θ - d2 sin2 θ = d 2 (2 sin θ cos θ - sin 2 θ ) = d 2 (sin2 θ - sin 2 θ ) = d2 sin2 θ -d2sin2 θ -4 To get maximum value of θ, differentiate Agi with respect to θ, and equate to zero, d/d θ Agi = 0 On differentiating equation 4 with respect to θ we get, d/d θ Agi = d2 cos 2 θ x 2- d2 2 sin θ cos θ . put d/d θ Agi = 0 d2 cos2 θ x 2 - d2 2 sin θ cos θ = 0 2 d 2 sin θ cos θ = d2 cos2 θ x 2 d2 2 sin 2θ = d2 cos2 θ x 2 sin 2 θ / cos2 θ = 2 2 θ = tan-12 θ = ½ tan-12 = 31.72° When θ = 31.72°, the dimensions of the core (a & b) will give the maximum area for core for a specified d. When θ = 31.72°, a = d cos θ = d cos 31.720 = 0.85d b = d sin θ = d sin 31.720 = 0.53d On substituting the above values of a & b in equation 3 we get, Gross core- area, Agi = 2ab- b2 = 2(O.85d)(O.53d)- (O.53d)2 = 0.618 d2 . Let stacking factor, Sf = 0.9 Net core-area, Ai = Stacking factor x Gross core area = 0.9 x 0.618 d2 = 0.56 d2 ---- 4 Netcorearea The ratio, = 0.56 d2 / ( Π/4) * d2 = 0.71 Areaofcircumscribingcircle Grosscorearea The ratio, = 0.618 d2 / ( Π/4) * d2 = 0.79 Areaofcircumscribingcircle Netcorearea Core area factor, Kc = Squareofcircumscribingcircle = Ai / d2 = 0.56 d2 / d2 = 0.56 --5 3.6 Choice of flux density in the core: • The flux density decides the area of cross-section of core and core loss. • Advantages of higher values of flux density: It results in smaller core area, lesser cost, reduction in length of mean turn of winding • Advantages of higher values of flux density : It results in higher iron loss and large magnetizing current. • The choice of flux density depends on the service condition (i.edistribution or transmission) and the material used for laminations of the core. • The laminations made with cold rolled silicon steel can work with higher flux densities than the laminations made with hot rolled silicon steel. • Usually the distribution transformers will have low flux density to achieve lesser iron loss. • Bm = 1.1 to 1.4 Wb/m2 For distribution transformers • Bm = 1.2 to 1.5 Wb/m2 For power transformers 3.7 Overall dimensions of the transformer: • The main dimensions of the transformer are Height of window (Hw) and Width of window Ww • The other important dimensions of the transformer are • width of largest stamping (a), • diameter of circumscribing circle (d), • distance between core centers (D), • height of yoke (Hy), depth of yoke (Dy) • Overall height of transformer frame (H) and overall width of transformer frame (W). • These dimensions. for various types of transformers are shown in figure (4.7), (4.8) and (4.9). ' • The figure 4.7 shows a vertical and horizontal cross-section of the core and winding assembly of a core type single phase transformer. • The fig 4.8 shows a vertical and horizontal cross-section of the core and winding assembly of a core type three phase transformer. • The fig 4.9 shows a vertical and horizontal cross-section of a shell type single phase transformer. Figure 3.4- Overall dimensions of the 1Ф transformer D = d + Ww H = Hw + 2Hy W=D+a --- 1 ---- 2 ---- 3 Figure 3.5- Overall dimensions of the 3Ф transformer Figure 3.6- Overall dimensions of the 1Ф shell type transformer 3.8 Design of winding: • The transformer has one high voltage winding and one low voltage winding. • The design of winding involves the determination of number of turns and area of cross-section of the conductor used for winding. • The number of turns is estimated using voltage rating and emf per turn (or by using ampere-turns and rated current). • The area of cross-section is estimated using rated current and current density. • Usually the number of turns of low voltage winding is estimated first using the given data and it is corrected to nearest integer. Then the numbers of turns of high voltage winding are chosen to satisfy the voltage rating of the transformer. • Number of turns in low voltage winding TLV = VLV / Et or AT / ILV ---- 1 where, V LV = Rated voltage of low voltage winding ILV = Rated current of low voltage winding • Number of turns in high voltage winding THV = TLV * (VHV / TLV ) where, V HV = Rated voltage of high voltage winding • Area of cross - section of primary winding conductor ap = Ip / δ • In transformers same current density is assumed for primary and secondary. • Area of cross - section of secondary winding conductor as = Is / δ • δ = 1.1 to 2.2 A/ mm2 for distribution transformers • δ = 2.2 to 3.2 A/ mm2 for large power transformers with self oil cooling or airblast. • δ = 5.4 to 6.2 A/ mm2 for large power transformers with forced circulation of oil or with water cooling coils. EXAMPLE 3.2: Estimate the main dimensions including winding conductor area of a 3- phase, ∆- Y core type transformer rated at 300 kV A, 6600/440 V, 50Hz. A suitable core with 3-steps having a circumscribing circle of 0.25 m diameter and a leg spacing of 0.4 m is available. Emf per turn = 8.5V, δ = 2.5A/mm2, Kw= 0.28, Sf= 0.9 (stacking factor). Solution Let 440V side be secondary and 6600V be primary. Here the secondary is star connected and primary is delta connected. Secondary voltage per phase, Vs = 440/√3 = 254V Also, Es ≈ Vs we have Emf per turn, Et = Es /Ts It is given that Et = 8.5 V. 254 Number of secondary turns per phase, Ts = = 29.88 ≈ 30 turns 8 .5 6600 Number of primary turns per phase, Tp = Ts *( Vp / Vs ) = * 30 = 779.5 turns 254 = ≈ 780 turns kV A rating of transformer =Q = √3* VLP ILP x 10-3 = √3* VLS ILS x 10-3 where, V LP Line voltage on primary side ILP Line current on primary side V LS Line voltage on secondary side ILS Line current on secondary side Line current on primary side, ILP = Q / √3* VLP x 10-3 = 300/ √3* 6600 x 10-3 = 26.24 A Since primary is delta connected, The phase current on primary I P= ILP = 26.24/ √3 = 15 15 A The area of cross-section of primary conductor ap = I P /δ = 15.15 / 2.5 =6.06mm2 Line current on secondary side, ILs = Q / √3* VLs x 10-3 = 300/ √3* 440 x 10-3 = 393.65 A Since secondary is star connected, The phase current on secondary, I s= ILs = = 393.65 A The area of cross-section of secondary conductor as = I s /δ = 15.15 / 2.5 =157.5mm2 The copper area in window, Ac = 2*Tp * ap+ 2*Ts * as =2 (6.06 x 780 +157.5 x30) = 18903.6 mm2 Window Area, Aw = Ac / Kw = 18903.6 / 0.28 = 67512.86 mm2 = 67512.86 x 1O-6m2 = 0.0675 m2 Area of circumscribing circle = ( Π/4) * d2 Grosscorearea = 0.84 Areaofcircumscribingcircle Gross core area, Agi = 0.84 x area of circumscribing circle = 0.84 x 0.049 = 0.041 m2 Net core area, A I = Sf x A gi = 0.9 x 0.041 = 0.0369 m2= 0.37 x 106 mm2 Given that, leg spacing = 0.45 m Width of window, W w = leg spacing = 0.45 m Height of window, H w = A w / Ww = 0.0675 / 0.45 = 0.15 m Result: Number of primary turns per phase T p = 780 Area of cross-section of primary conductor a p = 6.06 mm2 Number of secondary turns per phase T s = 30 Area of cross-section of secondary conductor a s = 157.5 mm2 Net core area, A i = 0.0369 m2 Window area A w =0.0675 m2 Height of window H w = 0.15 m Width of window, W w =0.45 m Example3.3: Determine the dimensions of core and yoke for a 200 KVA, 50 Hz single phase core type transformer. A cruciform core is to be used with distance between adjacent limbs equal to 1.6 times the width of core laminations. Assume voltage per turn is 14 volts, Bmax = 1.1 Tesla,Window space factor is 0.32. Current density is 3 A/mm2 and stacking factor is 0.9. Net iron area is 0.56 m2 where d is the diameter of the circumscribing circle of cruciform core. Width of largest stamping is 0.85d. (AnnaUni – Nov/Dec 2003) Solution: Emf per turn Et = 4.44 f Bm * Ai volts Given that Et = 14 V f = 50 Hz Bm = 1.1 Wb/m2 Substituting we get Net iron area = Ai = 14 / 4.44*50*1.1 = 0.573 m2 Netcorearea We know that, Core area factor, Kc = Squareofcircumscribingcircle = Ai / d2 = 0.56 d2 / d2 = 0.56 So, diameter of the circumscribing circle = d = √ Ai / 0.56 = 0.32m Width of the largest stamping = a = 0.85d = 0.85 * 0.32 = 0.272m It is given that, distance between adjacent limbs equal to 1.6 times the width of core laminations. Distance between core centers = 1.6 a = 1.6 * 0.272 = 0.435m Width of window, W w = D-d = 0.45 -0.32 = 0.115m For a single phase transformer, the output KVA is Q = 2.22 f Bm * Ai * ( Kw * Aw *δ) * 10 -3 Substituting,200 = 2.22*50*1.1*0.32*3*106 * Aw * 10 -3 Window area A w = 0.0293m2 Height of window H w = A w/ W w = 0.0298/0.115 = 0.26m Using the same stepped section for the yoke as for core, Depth of yoke = Dy = a = 0.272m Height of yoke = Hy = 0.272m Referring to figure 4.7, we have H = Hw + 2Hy =0.26 +2*272 = 0.804m For 3 stepped core, The ratio, W = D + a = 0.435 + 0.272 = 0.737m Example3.4: Compute the main dimensions of the core of a 5 KVA, 11000/400 volts 50 Hz single phase core type transformer. Window space factor = 0.2. The height of the window is 3 times its width. Current density = 1.4 A/mm2. Bmax = 1.0 Tesla. stacking factor = 0.9. Net conductor area in the window = 0.6 times the net cross sectional area of iron in the core. Also find no. of primary and secondary turns. Given that Net conductor area = 0.6* Net iron area Or Kw * Aw = 0.6 Ai Window area = Aw = (0.6 / Kw )* Ai = (0.6/0.2 )*Ai = 3 Ai For a single phase transformer, the output KVA is Q = 2.22 f Bm * Ai * ( Kw * Aw *δ) * 10 -3 Substituting, 5 = 2.22*50*1.0*0.2*1.4*106 * 3Ai * Ai 10 -3 We get net iron area Ai = 0.00732m2 Gross iron area Agi = Ai/Sf = 0.00732m2 / 0.9 = 0.00814m2 = a2 Width of the core = √ 0.0814 = 0.09 m, Gross iron area provided Agi = 0.0081m2 Net iron area provided Ai = 0.00729m2 Window area A w = 3 * 0.00729= 0.02187m2 Height of window H w = 3 W w But Aw = H w * W w So 3 W w2 = 0.02187 Hence Width of window, W w = 0.085m Height of window H w = 3* 0.085 = 0.255m The yoke has same gross area as the core, Gross area of yoke = Ay = 0.0081m2 Depth of yoke = Dy = a = 0.09m Height of yoke = Hy = 0.0081/ 0.09 = 0.09m Overall dimensions of core: Referring to figure 4.7, we have Distance between core centers D = a +Ww = 0.09 + 0.085 = 0.175m Length of the frame W = D + a = 0.175 + 0.09 = 0.265m Height of the frame H = Hw + 2Hy =0.255 +2*0.09 = 0.435m Flux Фm = Bm * Ai = 1.0 * 72.9*10-3 = 7.29* 10-3 wb Emf per turn, Et = 4.44 f Фm = 4.44*50*7.29* 10-3 =1.625V 400 = 246 turns Number of secondary turns per phase, Ts = 1.625 11000 * 246 = 6765 turns Number of primary turns per phase, Tp = Ts *( Vp / Vs ) = 400 Primary current = Q/KV = 5000/11000 = 0.455A The area of cross-section of primary conductor ap = I P /δ = 0.455 / 1.4 =0.384mm2 Using circular conductors, diameter of primary conductor = √0.324*4/Π = 0.642m secondary current = Q/KV = 5000/400 = 12.5A The area of cross-section of secondary conductor as = I s /δ = 12.5 / 1.4 =8.93mm2 Using square conductors, 3*3 mm2 The area of of secondary conductor as = 9 mm2 3.10 Cooling of transformers: • The losses developed in the transformer cores and windings = Pi + Pc • Pi = Iron loss, Pc = Copper loss The losses developed in the transformer cores and windings are converted into thermal energy and cause heating of corresponding transformer parts. • The heat dissipation in transformer occurs by Conduction, Convection and Radiation. • The paths of heat flow in transformer are the following • From the internal most heated spots of a given part (of core or winding) to their outer surface in contact with the oil. • From the outer surface of a transformer part to the oil that cools it. • From the oil to the walls of a cooler, eg. Wall of tank. • From the walls of the cooler to the cooling medium air or water. The various methods of cooling transformers are Air natural, Air blast, Oil natural, Oil natural-air forced , Oil natural-water forced, Forced circulation of oil , Oil forced-air natural , Oil forced-air forced , Oil forced-water forced • The choice of cooling method depends upon the size, type of application and type of conditions obtaining at the site where the transformer in installed. • Air natural is used for transformers up to 1.5 MV A. Since cooling by air is not so effective and proves insufficient for transformers of medium sizes, oil is used as a coolant. Oil is used for almost all transformers except for the transformers used for special applications. Both plain walled and corrugated walled tanks are used in oil cooled transformer. 3.11Transformer oil as a cooling medium: For the transformer oil, the specific heat dissipation due to convection of oil is given by λ convection = 4.03*( θ/ H )1/4 where θ = Temperature difference of the surface relative to oil, H = Height of dissipating surface in meters 3.12 Temperature rise in plain walled tanks: • The transformer core and winding assembly is placed inside a container called tank. The walls of the tank dissipate heat by both radiation and convection. • For a temperature rise of 40°C above the ambient temperature of 20 °C, the specific heat dissipation are as follows, 1. Specific heat dissipating due to radiation, λ rad = 6 W/m2 0C 2. Specific heat dissipation due to convection, λ con = 6.5 W/m2 0C . The total specific heat dissipation in plain walled tanks is 12.5 W/ W/m2 0C • Totalloss Specificheat * Heatdissipationsurfaceof tan k = (Pi + Pc )/ 12.5 Si The temperature rise,θ = where, Pi = Iron loss Pc = Copper loss Si = Heat dissipating surface of the tank 3.13 Design of tank with cooling tubes: • The transformers are provided with cooling tubes to increase the heat dissipating area. • The tubes are mounted on the vertical sides of the 'transformer tank. ' But the increase in dissipation of heat is not proportional to increase in area, because the tubes would screen (conceals) some of the tank surface preventing radiations from the screened surface. • On the other-hand the tubes will improve the circulation of oil. This improves the dissipation of loss by convection. The circulation of oil is due to more effective pressure heads produced by columns of oil in tubes. • The improvement in loss dissipation by convection is equivalent to loss dissipated by 35% of tube surface area. Hence to account for this improvement in dissipation of loss by convection an additional 35 % tube area is added to actual tube surface area or the specific heat dissipation due to convection is taken as 35% more than that without tubes. Let, the dissipating surface of the tank = Si The dissipating surface of the tubes = x * Si Loss dissipated by surface of the tank by radiation and convection = (6 + 6.5) St = 12.5 St 135 Loss dissipated by tubes by convection = 6.5 * *X*St 100 • Total loss dissipated by walls and tubes = 12.5 St + 8.8*X*St = (12.5 + 8.8 X)* St ---- 1 Actual total area of tank walls and tubes = St + X* St = (1+X) * St Total loss dissipated Loss dissipated per m2 of dissipating surface = Total area = St(12.5 + 8.8X) / St (1 + X) = (12.5 + 8.8X) / (1 + X) ---- 2 Total loss Temperature rise in transformer with cooling tubes θ = Loss dissipated Total loss, Ploss = Pi + Pc ---- 3 Where Pi = Iron loss Pc = Copper loss From equation 1 & 2 we get θ = (Pi + Pc)/ (12.5 + 8.8 X)* St Or (12.5 + 8.8 X) = (Pi + Pc)/ θ * St ----- 4 1 X = ((Pi + Pc)/ θ * St - 12.5 ) * ---- 4 8 .8 Total area of cooling tubes = X*St On substituting for X from equation 4 we get, 1 [(Pi + Pc)/ θ - 12.5 )] *St --- 5 Total area of cooling tubes = 8 .8 Calculation of number of tubes: Let, lt = Length of the tube dt = Diameter of the tube Surface area of each tube = Π dt lt ( surface area of a cylinder) Total area of tubes Total number of tubes = Area of each tube 1 [(Pi + Pc)/ θ - 12.5 )] *St ---- 6 8 .8 The standard diameter of the cooling tubes is 50 mm and the length of the tube depends on the height of the tank. The tubes are arranged with a centre to centre spacing of75 mm. The dimensions of the tank are decided by the dimensions of the transformer frame and clearance required on all the sides. The dimensions of the tank are shown in following figure. = (1/ Π dt lt ) * • • • • Figure 3.7 Dimensions of the tank With reference to above figure it can be written, Width of the tank, WT = 2D + Doc + 2 C1 (for 3-phase) --- 7 = D + Doc + 2C1 (for I-phase) --- 8 Length of the tank, LT = Doc + 2 C2 ---- 9 Height of the tank, HT = H + C3 + C4 ---- 10 Where, C1 - Clearance between winding and tank along the width. C2 - Clearance between the winding and tank along the length. C3 - Clearance between the transformer frame and the tank at the bottom. C4 - Clearance between the transformer frame and the tank at the top. Doc - Outer diameter of coil EXAMPLE3.5: A 250 KVA, 6600V/400V, 3-phase core type transformer has a total loss of 4800W at full load. The transformer tank is 1.25m in height and 1m * 0.5m in plan. Design suitable scheme of tubes if the average temperature rise is limited to 350 C. The diameter of tube is 50mm and is spaced 75mm from each other. The avarege height of the tubes is 1.05m. Specific heat dissipation due to radiation and convection is respectively 6 and 6.5 W/m2 0 C. Assume that the convection is improved by 35% due to provision of tubes. (AnnaUni – Ap/May 2005) Solution: Step 1: Calculation of Heat dissipating surface of the tank: Given that Tank dimension = 0.5 x 1 x 1.25 m Figure 3.8 Dimensions of the tank So it is given that LT = Length = 0.5 m, HT = Height = 1.25 m WT = Width = 1 m Now the heat dissipating surface of the tank = St = Total area of vertical sides = 2(LT HT + WT HT) = 2 HT (LT + WT) St = 2 x 1.25 x (0.5 + I) =3.75 m2 Loss dissipated by surface of the tank by radiation and convection = (6 + 6.5) St = 12.5 St Let the dissipating surface of the tubes = x * Si 135 Loss dissipated by tubes by convection = 6.5 * *X*St 100 Total loss dissipated by walls and tubes = 12.5 St + 8.8*X*St = (12.5 + 8.8 X)* St Temperature rise in transformer with cooling tubes θ = Total loss Loss dissipated Given that, Total loss, P loss = 4800 W θ = 4800 / (12.5 + 8.8 X)* St 1 X = (4800/ θ * St - 12.5) * 8 .8 1 = ( 4800/ 35 * 3.75 - 12.5 ) * 8 .8 = 2.7354 Total area of cooling tubes = X*St = =2.7354 x 3.75 = 10.2578 m2 Area of each cooling tube = Π dt lt = Π X 50 x 10-3 x 1.05 = 0.1649 m2 Total area of tubes Total number of tubes = n i = Area of each tube 10.2578 = = 62.206 = 62 tubes 0.1649 • The width of the tank is 1000 mm. If we leave an edge spacing of 87.5 mm on either sides, then we can arrange 12 tubes widthwise with a spacing of 75mm between the centres of tubes. • The length of the tank is 500 mm. If we leave an edge spacing of 100mm on either sides, then we can arrange 5 tubes lengthwise with a spacing of75mm between the centres of tubes. • But one row is not sufficient to accommodate the required 62 cooling tubes. Hence 2 rows of cooling tubes are provided on both lengthwise and widthwise. The plan of the cooling tubes is shown the following figure Result: Figure 3.9 The plan of the cooling tubes Total number of cooling tubes provided = 64 They are arranged as 2 rows on widthwise with each row consisting of 12 & 11 tubes and 2 rows on lengthwise with each row consisting of 5 & 4 tubes. 3.14 Estimation of no load current of transformer: • The no-load current of a transformer has two components. They are magnetizing component and loss component. • The magnetizing current depends on the mmf required to establish the desired flux. • The loss component of no-load current depends on the iron loss. 3.15 No-Load current of single phase transformer: Figure 3.10 Dimension of core - 1Ф transformer No-load current, I0 = √ Im2 + Il2 --- 1 I m = magnetizing current I l = loss component of no-load current Magnetizing current: In above figure, Total length of core = 2 l c Total length of yoke = 2 l y Here, l c = H w = Height of window, l y = Ww = Width of window Mmf for core = mmf per meter for maximum flux density in core * Total length of core = at c * 2 l c = 2 at c l c Mmf for yoke = mmf per meter for maximum flux density in yoke * Total length of yoke = = at y * 2 l y = 2 at y l y Total magnetizing mmf, AT 0 = mmf for core + mmf for yoke + mmf for joints = 2 at c l c + 2 at y l y + mmf for joints ---- 2 Maximum value of magnetizing current = ATo / Tp ---- 3 If the magnetizing current is sinusoidal then, rms value of magnetizing current = Im = ATo / √2 Tp ---- 4 The values of at c and at y are taken from B-H curves for transformer steel. Loss component of no-load current: • The iron losses are calculated by finding the weight of cores and yokes. • The loss per kg of iron is taken from the loss curves given by the manufacturer of transformer laminations. The loss component of no-load current, I0 = Pi / Vp ----- 5 Pi - Iron loss in watts Vp =Terminal voltage of primary winding 3.16 No-Load current of three phase transformer: No-load current/ phase I0 = √ Im2 + Il2 --- 6 I m = magnetizing current I l = loss component of no-load current Magnetizing current: Figure 3.11 Dimension of core - 3Ф transformer In above figure, Total length of core = 3 l c Total length of yoke = 2 l y Here, l c = H w = Height of window, l y = 2Ww + d Where Ww = = Width of window d = diameter of circumscribing circle Mmf for core = mmf per meter for maximum flux density in core * Total length of core = at c * 3 l c = 3 at c l c Mmf for yoke = mmf per meter for maximum flux density in yoke * Total length of yoke = = at y * 2 l y = 2 at y l y Total magnetizing mmf, = mmf for core + mmf for yoke + mmf for joints = 3 at c l c + 2 at y l y + mmf for joints ---- 7 Total magnetizing mmf, AT 0 per phase = ( 3 at c l c + 2 at y l y + mmf for joints )/3 --8 Maximum value of magnetizing current/ phase = ATo / Tp If the magnetizing current is sinusoidal then, rms value of magnetizing current = Im = ATo / √2 Tp ---- 4 The values of at c and at y are taken from B-H curves for transformer steel. Loss component of no-load current: • The iron losses are calculated by finding the weight of cores and yokes. • The loss per kg of iron is taken from the loss curves given by the manufacturer of transformer laminations. Let, Pi = Total iron loss for the three phases Pi = 3 3 Vp * Il Loss component of no-load current, I l = P i /3 Vp EXAMPLE 3.6: A 15,000 KVA, 33/6.6KV,star/delta 3-phase core type transformer has the following data.Net iron area of each limb = 1.5* 10 -3 m2. Net area of yoke = 1.8* 10 -3 m2. mean length of magnetic flux path in each limb = 2.3 m, mean length of magnetic flux path in each yoke = 1.6 m, number of turns per phase of HV windings = 450. Calculate the no load current. Density of core steel =7.8*103 Kg/m3 Use the following data: Flux density Bm 0.9 1.0 1.2 1.3 1.4 Mmf/m 130 210 420 660 1300 Iron loss/Kg 0.8 1.3 1.9 2.4 2.9 (AnnaUni – Ap/May 2005) Solution: H.V winding voltage per phase = 33000/ √3 = 19100 V E Emf per turn Et = = 4.44 f Фm volts T Фm = 19100/ 4.44*50*450 = 0.191 Wb Flux density in the limbs = Bm (limb) = 0.191/0.15 = 1.27 Wb/m2 Referring to the given data corresponding to this flux density At c = 560 A/m P i = 2.25W/Kg Flux density in the yokes = Bm (yoke)= 0.191/0.18 = 1.06 Wb/m2 Referring to the given data corresponding to this flux density At y = 260 A/m P i = 1.4W/Kg Total mmf for three limbs = 3*560*2.3 = 3860A Total mmf for two yokes = 2*260*1.6 = 832A Total mmf limbs and yokes = 3860+832 = 4692A Total magnetizing mmf, AT 0 per phase = 4692/3 = 1564 A rms value of magnetizing current = Im = ATo / √2 Tp = 1564 / √2 *450 = 2.6 A Volume of three limbs = 3*2.3*0.15 = 1.035 m3 weight of three limbs = 1.035*7.8*103 = 8.08*103 Kg Volume of two yokes = 3*1.6*0.18 = 0.576 m3 weight of two yokes = 0.576*7.8*103 = 4.49*103 Kg Loss in limbs = 8.08*103 *2.25 = 18.2*103 W Loss in yokes = 4.49*103 *1.4 = 6.3*103 W Total iron loss = 18.2*103 +6.3*103 = 24.5*103 W Loss per phase = 24.5*103 /3 = 8.16 *103 W Loss component of no-load current, I l = P i / Vp = 8.16 *103/ 19100 = 0.427A No-load current/ phase I0 = √ Im2 + Il2 = √ 2.462 + 0.4272 = 2.5A Questions Unit – 3 – Design of Transformers Part –A 1. State various methods of cooling of large power transformers. 2. State the various types of limb sections of core type transformer. 3. The voltage per turn of a 500 KVA, 11 KV/415V, ∆/Υ, 3-phase transformer is 8.7 V. Calculate number of turns per phase of LV and HV windings. 4. What is conservator? 5. Define voltage regulation of a transformer and state its importance. 6. State the factors on which the thermal time constant of a transformer depends. 7. State the reasons for preferring circular coils in comparison to rectangular coil in transformer winding. 8. Top and bottom surfaces of the transformer tank are not considered for the design of cooling tubes for transformer. Why? 9. Define window space factor the design of a transformer. 10. What is the functional difference between CT and PT? 11. What are the different losses in a transformer? 12. State the merits of three phase transformers over single phase transformers. 13. What is the significance of the ratio of magnetic loading to electric loading in the design of transformer. 14. What are the steps followed for the calculation of magnetizing current in a single phase transformer? 15. Draw the cruciform section of the transformer core and give the optimum dimensions in terms of circumscribing circle diameter d. Part –B 1. Derive the voltage per turn equation for a single phase transformer (8) 0 2. A 250 KVA transformer give the temperature rise of 20 C after 1 hour of full load and 33.50 C after 2 hours.Find out the percentage overload to which the transformer can be subjected safely for 1 hour if it has maximum efficiency on full load. (8) 3. Explain the design procedure of cooling tubes for a transformer. (8) 4. Estimate the no load current of a 400 volt, 50 Hz, single phase transformer With the following particulars: length of the mean magnetic flux path – 200 cm, gross cross section 100 cm2, joints are equivalent to 0.1mm air gap. Maximum flux density – 0.7 Wb/m2, magntising force corresponding to 0.7 Wb/m2 is 2.2 AT/cm and specific loss corresponding to0.7 Wb/m2 is 0.5 Watt/kg. Assume a stacking factor of 0.9. (8) 5. Derive the output equation of a 3-phase transformer. (6) 6. A 250 KVA, 6600V/400V, 3-phase core type transformer has a total loss of 4800W at full load. The transformer tank is 1.25m in height and 1m * 0.5m in plan. Design suitable scheme of tubes if the average temperature rise is limited to 350 C. The diameter of tube is 50mm and is spaced 75mm from each other. The avarege height of the tubes is 1.05m. Specific heat dissipation due to radiation and convection is respectively 6 and 6.5 W/m2-0 C. Assume that the convection is improved by 35% due to provision of tubes. (10) 7. How will you estimate no-load current in single and 3-phase transformers. (6) 8. A 15,000 KVA, 33/6.6KV,star/delta 3-phase core type transformer has the following data.Net iron area of each limb = 1.5* 10 -3 m2. Net area of yoke limb = 1.8* 10 -3 m2. mean length of magnetic flux path in each limb = 2.3 m, mean length of magnetic flux path in each yoke = 1.6 m, number of turns per phase of HV windings = 450. Calculate the no load current. Use the following data: (10) Mmf/m 130 210 420 660 1300 Iron loss/Kg 0.8 1.3 1.9 2.4 2.9 9. Derive the output equation of a single phase transformer and point out salient features of this equation. (4) 10. Explain different methods of cooling of a transformers with relevant sketches. State relative merits and limitations of these methods. (4) 11. Compute the main dimensions of the core of a 5 KVA, 11000/400 volts 50 Hz single phase core type transformer. Window space factor = 0.2. The height of the window is 3 times its width. Current density = 1.4 A/mm2. Bmax = 1.0 Tesla. stacking factor = 0.9. Net conductor area in the window = 0.6 times the net cross sectional area of iron in the core.Also find no. of primary and secondary terms. (8 12. Calculate the approximate overall dimensions for a 200 KVA, 6600/400 volts 50 Hz three phase core type transformer. Data: Emf/turn = 9 volts, Bmax = 1.0 Tesla, Current density δ = 2.5 A/mm2 Window space factor Kw= 0.3. Overall height = Overall width. Core is 4 stepped. Assume the stacking factor as 0.91 (16) 13. Determine the dimensions of core and yoke for a 200 KVA, 50 Hz single phase core type transformer. A cruciform core is to be used with distance between adjacent limbs equal to 1.6 times the width of core laminations. Assume voltage per turn is 14 volts, Bmax = 1.1 Tesla,Window space factor is 0.32. Current density is 3 A/mm2 and stacking factor is 0.9. Net iron area is 0.56 m2 where d is the diameter of the circumscribing circle of cruciform core. Width of largest stamping is 0.85d. (16) 14. A 100 kVA,6600/440 volts ,50Hz three phase delta/star core type oil immersed natural cooled transformer has the following data :Distance between centers of adjacent limbs =0.47 metre,Outer diameter of high voltage winding =0.44 metre, Height of frame =1.24 metre, Core loss =3.7 kw and rated coper loss =10.5 kw. Design a suitable tank with tubes for this transformer. The average temperature rise of oil should not exceed 350 C. Specific heat Dissipation from tank walls due to radiation and convection is respectively 6 and 6.5 W/m2-0 C. Assume that the convection is improved by 35% due to provision of tubes. (16 ) 15. Calculate the core and window area and make an estimate of the weights of copper and iron required for a 125 KVA,2000/400 V,50Hz,single phase shell type transformer.Data: Bm =1.1 wb/m2 , δ = 2.2 amp/mm2, Voltage/turn =11.2 V Window area constant =0.33. Core is a rectangular and the stampings are all 7cm wide. Sketch the core inserting the dimensions. (16) 16. The tank of a 150 KVA transformer has the dimensions 100cm *55cm*120cm height. Design a suitable arrangement of cooling tubes of mean length 100com,and diameter of 5cm to limit the temperature rise to 350 C if the full load losses to be dissipated are 5KW.Assume other values suitably. (16)