ANALYSIS OF AN ALTERNATING CURRENT SINGLE PHASE ELECTRIC ARC WELDING MACHINE USED IN MOBILE ENGINEERING WORKSHOPS IN NIGERIA

ANALYSIS OF AN ALTERNATING CURRENT SINGLE PHASE ELECTRIC ARC WELDING MACHINE USED IN MOBILE ENGINEERING WORKSHOPS IN NIGERIA

ABSTRACT

Arc welding is one of the activities that are very common in automobile, dam, bore hole, highway, drilling of oil well and other engineering practices. This article will examine into details the design specifications of an alternating current arc welding machine. The paper will also look into the design using the core type single phase concentric coil winding step down transformer. The machine is of 50Hz 220V input, 62V output and 5KVA rating. The machine is design locally using the available local materials in Nigeria.

Keywords: Design, concentric winding, step down transformer, specifications rating and arc welding

INTRODUCTION

A well developed transport, mining oil well drilling, dam and borehole engineering services company must have a mobile workshop equipped with all available maintenance and repair tools, devices and machines for their company to function  well.

Consider a breakdown of an engineering implement “already in a site for execution of a project” that involves joining two metals together. The maintenance engineer needs not to go back to the base station for the fabrication of the component (Althouse and Turnquist, 2005). The broken components can be fasten or welded together in the site with the same implement running to drive a single phase 220v alternator that can power single phase arc welding machine.

DESIGN SPECIFICATION

Power rating	5KVA
Primary (input) voltage	220 volts
Secondary (output) voltage	62 volts
Primary current	20 amps
Secondary current	8.2amps
Frequency	50Hz
Current density	1.7 – 30A/mm2
Window space factor	0.6
Maximum flux density	1.5wb/m2
Stacking factor	0.9
Assumed efficiency	95%
Cooling medium	Forced air (artificial)
Material used for core	Silicon steel
Shape of core cross section	Square
Number of windings	Concentric coil
Number of phase	One
Clamping	Bolts and nuts

DESIGN PROCEDURES AND CALCULATIONS

NUMBER OF TURNS

 To ascertain the number of turns, it is necessary to find the voltage per turn first (Theraja, 2006). By applying the equation from transformer principle an equation is got and could be used to ascertain the number of turns.

Analysis of an Alternating Current Single Phase Electric Arc Welding Machine Used in Mobile Engineering Workshops in Nigeria

i.e.

where V_t = voltage per turns

          E = primary and secondary voltage

         f = frequency = 50Hz

            Φ = flux weber = B x Ai …………………2.0

Where B = flux density in wb/m²

              A = cross-sectional area of the core in m²

Putting the value of the flux (Ø) in form of KVA (kilovolt amperes) gives:

KVA = E x I x 10^-3 …………………………….3.0

            = x I x 10^-3

         = 4.44 ²/r x 10^-3

Where, r = Φ/IN = magnetic load /(electric loading)²

            Φ² = [(r/4.44f x 10^-3) x KVA]

            Φ = [r/4.44f x 10^-3) x KVA]^1/2

Substituting the values of Φ in equation 1.0

V_t =

V_t = (4.44f x 10³ x r x KVA)^1/2

Therefore Vt = C (KVA)^1/2 …………………………4.0

Where C = (4.44f x 10³ x r)^1/2

It should be noted that the value C is a given range of constant depending on the type of transformer construction and use.

For a core type that was designed in this paper the following range of values may be adopted 0.55 to 0.65 (Slemon and Straughen, 2008). The volt/turn is gotten thus:

Vt = C(KVA)^1/2

Vt = 0.55(5)^1/2

    = 1.229 volt per turn

PRIMARY AND SECONDARY NUMBER OF TURNS

For the primary number of turns

Where E_I = primary voltage

              N_I = primary number of turns

Substituting values,

N_I

Also, the secondary number of turns is gotten as:

Where E₂ = secondary voltage

           N₂ = secondary number of turns

Substituting values,

Considering a secondary compensation

PRIMARY AND SECONDARY CURRENT

The primary current is the maximum root mean square (r.m.s) value or nominal value of current that can be drawn by the transformer from the supply without the equivalent being adversely affected as a result of overheating. Therefore the primary current is calculated from the VA rating of the machine and the (r.m.s). value of the applied voltage.

That is. VA = V_II_I = maximum power ………………………….5.0

Therefore I_I =

Primary current ≈ 22A from the power relation

V_II_I = V₂I₂

I₂ =

 I₂ =  = 80.6 ≈ 81 Amp.

CORE DESIGN CALCULATION

The core area is of square cross section. The gross area, A g  = 0.5d²

The effect core area,

  …………………………6.0

Where 0.9 is the stacking factor.

The effective core area is calculated thus:

Taking flux density B as 1.5wb/m²

Where

 1.229 = 4.44 x 50 x 1.5 x

 =

    =

     =

     =

Now,  =

Therefore d² =

      =

Therefore d =

The window area comprises of length and width of the lamination (Balbin, 2010). Since the core is of square section; the width of lamination is given as:

W_c = 0.71d

      =

     =  6cm

The window area is calculated from KVA –Ph

KVA –Ph = 10^-3 ……………………………….7.0

Where KVA = total kilovolt ampere of a 1 phase transformer = 5KVA

frequency, 50Hz

Φ =  ………………………………………………………………….8.0

Where B = flux density in Weber/m² = 1.5wb/m²

Cross sectional area = 36.91cm² = 0.003691mm

J = current density = 3.0A/m²

Kw = window space factor = 0.6

Note the value of J has been given to be 3A/mm²

Inserting the values in equation 7.0 above

Taking the ratio of the height (length) of the window of the width of the as

Therefore

But Aw = L x W = 3.2W x W = 3.2W² ……………………….9.0

W =

The length of window is given as:

   =

L = 12.03cm

The height of the stack as shown is equal to 0.71d for a square cross sectional area (Harry and John 2001).

Therefore Hs =

The number of lamination sheet is given by:

Ns =

The power equation for an 100% efficiency system is V₁I₁ = V₂I₂………..10.0

Assuming an efficiency (η) of 95% for this design,

The power rating at this efficiency = 5KVA

Therefore, input,

At this output, the primary current,

I₁ =

From equation 10.0, secondary current is gotten as:

230 x 22.86 = 62 x I₂

I₂ =

The mean value of current between the output power 5KVA and that of 5.26KVA is gotten as:

I₁ = and I₂ =

Gauge of wire = 105wG = 8300mm² = 8.3mm²

Current carrying capacity = 25A

²

For the secondary winding, resistance =

Area of wire = 32.2mm²

Diameter = 6.4mm²

Current capacity = 97A

The current density (J) = ²

Table 1.0:        Gauges, diameters, areas, resistance and current carrying capacity of copper wires.

Gauge No (SWG)	Diameter (mm)	Area (mm2)	Resistance (Ω/m)	Capacity in Amp
25	0.508	0.203	0.0832	1.02
24	0.558	0.245	0.0691	1.32
23	0.609	0.292	0.0579	1.46
22	0.711	0.397	0.0427	2.0
21	0.812	0.519	0.327	2.6
20	0.914	0.637	0.0258	2.63
19	1.016	0.811	0.0209	3.24
18	1.219	1.17	0.0145	4.68
17	1.422	1.59	0.0107	4.77
16	1.625	2.07	0.0082	6.21
15	1.828	2.63	0.00645	7.87
14	2.032	3.24	0.00523	9.72
13	2.336	4.29	0.00395	12.87
12	2.640	5.48	0.00309	16.44
11	2.946	6.82	0.00248	20.5
10	3.250	8.30	0.00204	25
9	3.657	10.5	0.00161	31.5
8	4.064	13.0	0.00130	39
7	4.470	15.7	0.00108	47
6	4.876	22.8	0.00094	56
5	5.384	27.3	0.000740	68
4	5.892	32.2	0.000649	82
3	6.400	38.6	0.000525	94
2	7.010	45.6	0.000438	116
1	7.620	53	0.000371	137
0	8.229	193	0.000318	160
00	8.839	78.131	0.000216	214

Standard wire gauge and sizes (Balbin, 2010).

THE BOBBIN OR FORMER

The bobbin width for each limb is given by:

Number of each windings are 4 with 2 units on each limb as shown

Space factor = K w = 0.6

Perimeter of window = (L + W) ……………………………….13.0

= (12.03 + 3.758) cm

 Width of bobbin = 2

= 4.736cm = 47.36m

NUMBER OF TURNS PER LAYER

This is given as the ratio of the width of the bobbin (former) to the conductors’ diameter, applying a factor of 0.6 as allowance for insulation on the conductors (Slemon and Straughen, 2007).

i.e. primary number of turn per layer =

Analysis of an Alternating Current Single Phase Electric Arc Welding Machine Used in Mobile Engineering Workshop in Nigeria

For the primary side, Turns per layer =

Total number of layer of the primary layer scale =

This means 8 layers for limb.

For the secondary, the secondary number of  

=

This means 4 layers per limb.

THE COIL DEPTH AND LENGTH

The radial depth is given as follows:

(height of each limb x number of layers) + (total number of thickness of interlayers wraps) + (winding allowance of 0.75mm).

Where, for the primary,

Height of turn = diameter of coil = 3.25mm

Thickness of interlayer wrap = 0.15mm

Radial depth = (3.25 x 16) + (12 x 0.15) + 0.75mm = 54.55mm

For each limb, radial depth will be: for the primary

For the secondary,

Height of turn = diameter of coil = 6.40mm

Radial depth = (6.40 x 8) + (0.15 x 6) + 0.75mm = 52.85mm  53mm

Denoting the length of conductor of the primary as Lp and that of the secondary by Ls.

The factor 8 is gotten from the fact that stalk height is in four different directions and the radial depth is in four sides at the core (Richard 2009).

Therefore, Bt – bobbin thickness = 0.2 = 2mm

                 Hs – number of turns in primary side

                 Hr – radial height = 27.2

                   LP = 8(27.2 + 64 + 2) = 69713.6mm = 69.71m

         For the secondary Ls = 8(26.5 + 64 + 2)

Thus, for wire gauge of 10 copper resistance = 0.00204Ω/m

And a wire gauge of 3 (copper) resistance = 0.000525Ω/m

Therefore the resistance winding of the primary can be gotten by multiplying the length of the primary winding and its resistance per metre (Theraja 2006).

Rp = Lp x 0.00204Ω/m

Rp = 69.71 x 0.00204Ω/m = 0.1422Ω

Similarly, for the secondary resistance

Rs = Ls x 0.00525Ω/m

Rs = 18.5 x 0.000525 = 0.0097Ω

COPPER LOSS

If Pcu loss is the copper loss in the primary and equal to

Pcu loss = R₁………………………………………………..16.0

Using the properties of copper and relating with the current gives:

Pcu loss = R₁ = (21.74)² x 0.1422 = 67.21watts

For secondary, copper loss.

            Pcu loss = R₂ = (80.60)² x 0.0097 = 63.02 watts

Therefore total copper loss:

            PcuL = PcuL₁ + PcuL₂

                        = 67.21 + 63.02 = 130.23 watts

VOLTAGE DROP

From ohms law, total voltage drop is given as voltage drop = V = IR

For the primary, Vd₁ = I₁R₁

                                       = 21.74 x 0.1422 = 3.09 volts

For the secondary, Vd₂ = I₂R₂ = 80.60 x 0.0097 = 0.7818 volts

Therefore, the voltage drop is:

Vd = Vd₁­ + Vd₂

      = 3.09 + 0.7818

     = 3.878 volts

CONCLUSION

The basic design specification, procedures and calculation for the fabrication of single phase (1Φ) arc welding machine has been fully investigated. All other equipment that will make the machine to function effectively was equally briefed in the recommendation of the research work.

RECOMMENDATION

Nigeria is still a developing country that seriously needs a standard and highly facilitated mobile engineering workshop. Nigerian investors and government should encourage local production of single phase arc welding machine using the parameters in the above design and analysis.

            Normally, transport engineering company, engineering market company, dam bore hole and oil well drilling company must have the following equipment:

·         Single phase and three phase generator

·         Heavy duty fork lift

·         Heavy duty clamping device

·         Drilling machine

·         Grinding machine

·         Shaping machine

·         Compressor and

·         Pneumatic hydraulic hard core breaker in their mobile container for arc welding machine to function effectively.

REFERENCES

Althouse A.D. Turnquist, Colt and Bowditch W.A. (2007): “Modern welding” Chapman and hall limited. London. Pp. 130-136.

Theraja B.L. and Theraja A.K. (2006): “Electrical Technology”, Nirja Construction and development co. Ltd. New Delhi, pp 773-1278.

Slemon G.R. and Straughen A. (2008): “Electric Machines”, Addison – Wesley Publishing Company, Massachusetts, USA PP 114-121.

Richard L.L. (2009): “Welding and Welding Technology”, McGraw-Hill, New York, pp 11-15.

Balbin Singh (2010): “Electrical Machines Design” Lansome and Hal Ltd. London, pp 104-109.

Harry U. Edward and John W. (2001): “Welding for Engineers”, Chapman and Hall Ltd, London, pp. 76-84.