Design a 150 KVA, 33KV/11kV, 50 Hz, 3 Phase star/delta core type oil immersed natural cooled distribution transformer, maximum temperature rise not to exceed 40°c.

Let, maximum flux density (Bm) =1.2 wb/m2 

current density (δ) =2.3A/mm2 

Let, k=0.45 for three phase core type distribution transformer

 Overall Dimension Core Design: 

Voltage per turn (Et) = K√Q= 0.45*√150= 5.5 V

 Voltage per turn (Et) = 4.44*f*Bm*Ai Ai = 5.5/(4.44*50*1.1) = 20.6*103 mm2

 Using a cruciform core, Ai = 0.56*d2 

Diameter of circumscribing circle (d) = √((20.6*〖10〗^3)/0.56) = 192.2 mm 

Length of largest core stamping (a) =0.85*d = 0.85*191.8 = 163.37 mm

 Length of second largest core stamping (b) =0.53*d = 0.53*191.8 = 101.87 mm

 Window Dimension:

Window space factor (kw) = 10/(30+kv) =10/(30+33/√3) = 0.2

 Output of transformer (Q) = 3.33*f*Bm*Ai*kw*Aw* δ*10-3 KVA

 Aw =150/(3.33*50*1.2*20.6*〖10〗^(-3)*0.2*2.3*〖10〗^6*〖10〗^(-3) ) = 79.22*103 mm2 Taking the ratio of height to width of window as 3,

 Hw* Ww = 79.22*103 3Ww2 =79.22*103

 Ww = 160.5mm

 Hw = 487.5 mm

 Distance between adjacent core centres (D) = Ww + d =160.5 + = 333.321 mm 

Yoke Design 

The area of yoke is taken as 1.2 times that of limb. 

∴ Flux density in yoke = 1.2/1.2 =1.0 wb/m2 

Net area of yoke = 1.2*20.6*103 = 24.72*103 mm2

 Gross area of yoke = (24.72*〖10〗^3)/0.9 = 27.47*103 mm2 

Taking the section of the yoke as rectangle.

 Depth of yoke (Dy) = a = 163.37 mm 

Height of yoke (Hy) = (27.467*〖10〗^3)/163.37 =168.13 mm

 Overall Dimension of frame:

Height of frame (H) = Hw + 2*Hy = 823.76mm 

Width of frame (W) = 2*D + a = 911.17mm 

Depth of frame (Dy) = a = 163.37 mm

 Lower Voltage side winding:

Secondary line voltage = 11000 v ; 

 connection = delta

 Secondary phase voltage = 11000 v

 Number of turns per phase (Ts) = 11000/5.5 = 2000

 Secondary phase current (Is) = (150*〖10〗^3)/(3*11000) = 4.54 A 

LV winding line current = 7.86 A

 A current density of 2.3 A/mm2 is used. 

Area of secondary conductor (as) = (4.54 )/2.31 = 1.978 mm2

 Diameter of bare conductor, d=√(4/π*1.978) =1.58mm

 From table 23.1 (IS:1897-1962), 

using a bare conductor of diameter of 1.6 mm and insulated diameter of 1.775 mm 

Modified area of conductor=π/4*1.62 =2.01mm2

 .∙.Actual value of current density δp=4.54/2.01=2.26 A/mm2 

Using 8 coils of 250 turns each Voltage per turn=11000/8=1375 V

 which is within the limit.

 Using normal coil of 250 turns Taking 10 layers per coil, turn per coil =250/10 =25

 Maximum voltage between layers = 2*25*5.5 =275 V, which is below the allowable limit.

 Axial depth of one coil ofL.V winding = 25*1.775 = 44.375 mm

 The spacers used between adjacent coils are 5 mm in height

. Axial length of LV winding ,Lcs=8*44.375 + 8*5=395 mm 

The height of window is 487.5 mm. This leaves a clearance of (487.5 – 395) = 46.25mm on each side of the winding. 

using 0.5 mm pressboard cylinders between layers,

 Radial depth of L.v winding (bs) = number of layers * radial depth of conductor + insulation between layers = 10*1.775 + 9*0.5 = 22.25 mm

 Diameter of circumscribing circle (d) = 192.2 mm 

Using pressboard wraps 1.5 mm thick as insulation between l.v winding and core.

 Inside diameter of l.v winding = 192.2+2*1.5 = 195.2 mm 

Outside diameter of l.v winding = 195.2+2*22.25 = 240.2 mm 

Higher voltage side winding

Primary line voltage=33000 v 

 connection type: Star 

Primary phase voltage=19052 v

 Number of turn per phase, Tp=2000* 19052/11000 = 3464

 Using 13 coils, Voltage per coil=19052/13=1465.53 V , which is within the limit.

 Turns per coil=( 3464)/13=266.46 

Using 12 coils of 250 turns each and a reinforced coil of 464 turns, 

 Total hv turn Tp=12*250+464=3464 

Taking 10 layers per coil, turn per coil =280/10 =25

 maximum voltage between layer= 2*25*5.5=275 v 

HV winding phase current=150* 1000/(3*19052) =2.624 A

 Taking current density as 2.3A/mm2 , 

 Area of high voltage conductor ,as=2.624/2.3=1.14 mm2

 Diameter of bare conductor, d=√(4/π*1.14) =1.20mm

 From the table(BIS: 3454-1966), 

nearest standard conductor size has Bare diameter= 1.25mm Insulated diameter= 1.45 mm with fine covering 

Modified area of conductor=π/4*1.52 =1.22 mm2 

.∙.Actual value of current density δp=2.624/1.22=2.13 A/mm2

 Axial depth of 1 coil=25*1.45=36.25mm 

The spacers used between adjacent coils are 3 mm in height. 

Axial length of HV winding ,Lcp=12*36.25+12*3=471 mm 

Height of window is 487.5mm, Space left between winding and window is 487.5-471=16.5mm Clearance left on each side=8.25mm which is within the limit for 33 Kv.

 Let the insulation between the layer is 0.5mm thick paper. 

∴ The radial depth of HV coil, bP=10*1.45+0.5*9=19mm

 The thickness of insulation between HV and LV =5+0.9*19.052=24mm 

The insulation between HV and LV winding is 24 mm thick 0.8mm thick paper insulated LV winding, 8mm is left of oil duct and again 8mm insulation is warped.

 ∴ Inside diameter of HV winding, DHVI=DLVO+2*thickness of insulation =240.2+2*24 mm =288.2mm 

∴Outside diameter of HV winding,DHVO= DHVI+2*bp =288.2+2*19 =326.2mm 

Clearance between winding=333.321-326.2=7.121 mm  

Resistance: 

Mean diameter of primary=(288.2+326.2)/2=307.2

 ∴ Length of mean turns of primary L_mtp=π*307.2 =0.965m

 Let ρ =0.021Ω〖mm〗^2/m 

∴Resistance of primary r_p=(ρ*l)/A=(ρ*L_mtp*T_p)/A_p

 ∴r_p=(0.021*0.965*3464)/1.22=57.53 Ω

 Now, Mean diameter of secondary =(240.2+195.2)/2=217.7 mm

 ∴Length of mean turn of secondary L_mts=π*217.7 =0.683m 

∴r_(s=) (0.021*0.683*2000)/2.01=14.27Ω

 Hence,

 ∴Resistance referred to primary =r_p+(〖3464/2000)〗^2 r_s = 100Ω 

∴ P.U. resistance of transformer =〖I_p R〗_p/V_p =(2.624*100)/19052=0.013

P.U Leakage Reactance:

Mean diameter of winding = (326.2+195.2)/2=260.7mm

 ∴Length of mean turns,L_mt=π*250.75= 0.819 m 

Height of winding L_c=L_(cs+L_cp )/2=(471+395)/2=0.433 mm 

∴ Leakage reactance of transformer referred to primary side, 〖 X〗_p =2*π*f*μ_0*〖T_p〗^2*L_mt/lc(a+b_(p+b_s )/3)*〖10〗^(-3)

 Where, a=24 mm 〖 b〗_p= 19 mm bs= 22.25 mm 

∴X_p =2*π*50*μ_0*〖2142〗^2*0.788/0.358(15+(20.85+12.1)/3)*〖10〗^(-3) = 338.24 Ω 

 ∴ P.U leakage reactance E_x=〖I_p x〗_p/V_p =(2.624*338.24)/19052=0.046 P.U

 ∴ P.U. impedance =√(〖E_x〗^2+〖E_r〗^2 ) =0.048 P.U.

 Regulation:

∴P.U. regulation ε=E_x sin⁡ϕ +E_r cos⁡ϕ

s.n	cos⁡ϕ	ε
1.	1	0.013
2.	0	0.046
3.	0.8(lag)	0.038

 Losses: 

1)〖 I〗^2R loss〖(P〗_c): 

At 〖75〗^oC, ∴I^2R loss = 3〖I_p〗^2 R_p=3*〖2.624〗^(2*) 100=2065.6watt

 Taking stray load loss 15% of above,

 ∴Total I^2R loss (P_c)=1.15*2065.6=2375.45 watt

 2) Core loss =P_i

 Let, density of lamination be 〖7.6*10〗^3 〖kg/m〗^3

 We use the grade 92 HRS so, from loss curve of Electrical transformer steel at B_m=1.2 Wb/m^2(for limb), specific iron loss=1.8watt/Kg Also, 

Weight of limb=3*H_w*A_i*〖7.6*10〗^3 =3*0.487*0.0206〖*7.6*10〗^3 =228.73 kg

 ∴Limb loss =1.8*228.73=411.72 watt

 Weight of two yoke =2*W*A_w 〖*7.6*10〗^3 =2*0.911*0.079*7.6*1000 =1093.92 kg Corresponding to flux density 1 Wb/m^2 specific core loss is 1.2watt/kg

 ∴core loss in yoke =1.2*1093.92 =1312.71 watt Hence,

 ∴Total core loss=411.72+1312.71 =1724.43 watt

 Efficiency: 

∴ɳ at full load and unity power factor =(output power)/(input power)*100% =150000/(150000+1724.43+2375.45)*100% =97.39% 

 For maximum efficiency , x2Pc=Pi ∴x=0.85 

Thus, the maximum efficiency occurs at 85% of full load. 

No load current: Flux density of core= 1.2Wb/mm2 

 Flux density of yoke=1.0Wb/ mm2

 From B-H curve for electrical transformer steel grade 92, atc=400A/m and aty=120A/m

 ∴Total magnetizing mmf=3*400*0.487+2*120*0.9117=1419.3 AT

 Now,

 ∴Magnetizing mmf per phase,AT0=1419.3/3=473.1 AT

 ∴Magnetizing current,Im=AT0/(√2 Tp)=0.0966A 

 ∴Loss component of no load current ,IL= core loss/3vp=1724.43/57156=0.03A

 ∴No load current I0=0.101A Hence,

 ∴No load current as percentage of full load current =0.101/2.624*100% =3.85% 

 Tank: 

Height of frame, H=823.76mmso,

 now allowing 150mm for base and 250mm for oil .

 Height of oil level=0.8237+0.15+0.25=1.22m 

Allowing 250 mm for lead,

 Hence, height of tank,Ht=1.061+0.25=1.47m 

Assuming clearance of 50 mm on each side,  

Width of tank=2D+DHVO+2*50 mm =2*0.3333+0.326+0.1 =1.086 m 

Assuming,50mm clearance on both sides 

∴length of tank,Lt=DHVO+2*50 mm =0.326+0.1 =0.426 m 

 Tubes:

Total loss dissipating surface of tank,St=2*(1.086+0.426) =3.024 m2

 Total specific loss due to radiation and convection is 12.5 w/m2-°c

 Temperature rise =(2375.45+1724.4)/(12.5*3.024) =108.42 °c 

Since temperature rise exceed the maximum limitation given i.e 40 °c,so plain tank is not sufficient for cooling and tubes are required. 

Heat dissipation by plain tank = 12.5 w/°c 

Heat dissipation by tubes by convection =6.5*1.4*xst = 9.1 xst 

Total heat dissipation = st (12.5+9.1x) 

Total area of tank wall and tubes =st(1+x) xst = 1/9.1 [(total loss)/θ - 12.5st ] = 7.1095 m2 

Area of each tubes = π *dt *lt = 0.19 m2

 Where dt = diameter of tubes =50 mm lt = length of tubes = 1.25 m 

Numbers of tubes = (Area of tubes)/(Area of each tubes) = 7.1095/0.19 = 36 each spaced at 75 mm

Design sheet

KVA  150                                phase 3                 frequency 50 Hz    Star/delta

Line voltage h.v. 33000 v         phase voltage h.v 19052 v

                       l.v.11000 v                                        l.v.11000 v

Line current  h.v. 2.624A                    phase current h.v 2.624A

                        l.v.  7.86 A                                        l.v. 4.54 A

Type – Core                                                 Type of cooling – ON

Core

1	Materials		0.35 mm thick 92 grade
2	Output constant	K	0.45
3	Voltage per turn	Et	5.5v
4	Circumscribing circle diameter	d	192.2 mm
5	Number of steps		2
6	Dimensions
		a	163.37 mm
		b	101.87mm
7	Net iron area	Ai	20.6*103 mm
8	Flux density	Bm	1.2Wb/m2
9	Flux	m	0.02472 Wb
10	Weight		228.73Kg
11	Specific iron loss		1.8Watt/Kg
12	Iron loss		411.72Watt

Yoke

1	Depth of yoke	Dy	163.37mm
2	Height of yoke	Hy	168.13mm
3	Net yoke area		24.72*103 mm2
4	Flux density		1 Wb/m2
5	Flux		0.02472 Wb
6	weight		1093.92 Kg
7	Specific iron loss		1.2 Watt/Kg
8	Iron loss		1372.71 Watt

Windows

1	Number		2
2	Window space factor	Kw	0.2
3	Height of window	Hw	487.5mm
4	Width of window	Ww	160.5 mm
5	Window area	Aw	79.22*103 mm2

Frame

1	Distance between adjacent limbs	D	333.32 mm
2	Height of frame	H	823.76 mm
3	Width of frame	W	911.17 mm
4	Depth of frame	Dy	163.37 mm

S.N	Windings	L.v	H.v
1	Type of winding	Cross over	Cross over
2	connection	delta	star
3	Conductor
	Dimension- Bare	1.6 mm	1.25 mm
	Insulated	1.775 mm	1.45 mm
	Area	2.01mm2	1.22 mm2
	Number in parallel	None	None
4	Current density	2.26A/mm2	2.13 A/mm2
5	Turns per phase	2000	3464
6	Coils total number	24	39
	Per core leg	8	13
7	Turns per coil	250	12of 250, 1 of 464 turns
	per layers	25	25
8	Number of layers	10	10
9	Height of winding	395mm	471 mm
10	Depth of winding	22.25mm	19 mm
11	Insulation between layers	1.5 mm pressboard	0.8 mm
	Between coils	5.0 mm spacers	3.0 mm spacers
12	Coils diameters inside	195.2 mm	288.2 mm
	outside	240.2mm	326.2 mm
13	Length of mean turn	0.683 m	0.965 m
14	Resistance at 75°c	14.27	57.53

Insulation

1	Between l.v. winding and core	Press board wraps 1.5 mm
2	Between l.v. winding and h.v. winding	Bakelized paper 8.0 mm
3	Width of duct between l.v. and h.v.	8 mm

Tank

1	Dimensions:height	Ht	1.47 m
	Length	Lt	0.426 m
	Width	Wt	1.086 m
2	Oil level		1.221m
3	Tubes		7.1095m2, 36 tubes each of area 0.19 m2
4	Temperature rise		40°C

Impedance

1	p.u resistance	0.013 p.u
2	p.u reactance	0.046 p.u
3	p.u impedance	0.048 p.u

Losses

1	Total core loss	1724.43watt
2	Total copper loss	2375.45 watt
3	Total losses at full load	4099.88 watt
4	Efficiency at full load and unity p.f	97.39%