## CONTACT-RESISTANCE HEAT SOURCE

The electrical resistance could be used as a source of heat. It could be

(a) contact resistance of interfaces or

(b) Resistance of molten flux and slag Resistance of each hemispherical constriction

R = p(r2 - r1)/S where p = resistivity of material

(r2 — rx) = length of current path

S = geometric mean area of the two hemispheres of radii r1 and r2 respectively.

^(2nr22)(2nr12) = 2n rir2

Now R = p(r2 - Гі) = -2 as r2 >> r1 2nr1r2 2nr1 2 1

Total constriction resistance Rc of n such spheres/unit area

R. = 1J = - P-

n nr1 nnr1

This approximation does not cause an error of more than 15%

Thus Rc = 0.85 p/nnr1

Heat generation rate by this contact resistance with an applied voltage of V is Q = V2/RC per unit area.

However after a very short time (~ .001 sec) the contact resistance drops to — th of its

original value. Due to softening of material due to increase in temperature.

Example. In a resistance welding process applied voltage = 5 V Bridges formed n = 25/cm2 Bridge radius r, = 0.1 mm. = 0.01 cm

resistivity of material p = 2 x 10-5 ohm-cm.

-5

 _ 0.00022 ohm-cm2

0. 85 p _ 0.85 x 2 x 10

nnrL 25 x 3.14 x 0.01

Rate of heat generated/unit area

 5 x 5
 W/cm2

V2 Q _ n _

RC.00022

_ 1.136 x 105 W/cm2.

Examples for Revision

Example 1. Two different pairs of sheets of the same material have to be spot welded. In one pair, there are 25 bridges/cm2 and the average radius of each bridge is 0.1 mm. The other pair of sheets contains 50 bridges/cm2 with the same average radius of each bridge. Determine the ratio of the voltages to be applied in these two cases to generate the same rate of heating/unit area.

The rate of heat generated by contact resist-

V2

 p _ resistivity of the material V _ applied voltage Rc _ constriction resistance n _ number of bridges/cm2 r _ radius of bridge (average)

ance with an applied voltage V is

Rc

R _ 0.85 p

.Rс —

C nnrL

Case 1. Rate of heat generated/unit area

Vl2 Vl2 x 25 x n x r

_ RT _ 0.85 p

Case 2. Rate of heat generated/unit area

V22 V22 x 50 x n x r

 0.85 p

RC

C2

For equal heat to be generated

V,2 x 25 xnx r V22 x 50xnx r

 0.85 p

0. 85 p

 2

 Vl V2

 Vl V2 J

 =2

 1.414

Example 2. The voltage-arc length characteristic of a dc arc is given by :

V = (20 + 4l) volts.

where l is the arc-length in mm. During a welding operation it is expected that the arc length will vary between 4 mm and 6 mm. It is desired that the welding current be limited to the range 450-550 A. Assuming a linear power source characteristic, determine the open circuit voltage and short circuit current of the power source.

D. C. Arc voltage V = 20 + 41

Arc length varies between 4 mm and 6 mm

It is desired that welding current should be between 450 to 550 A (difference 100 A) Assume a linear power source characteristics

Find open circuit voltage and short circuit current voltage variation range :

V = 20 + 4 x 4 = 36 V to 20 + 4 X 6 = 44 V

8 V

 Fig. 6.5

current range (450 — 550) ;r 100 Amp.

 0.08

 80 100

 x 550

 100 V = C — ml = C — 80

 36 = C -

 Slope

 100

C = 80 Thus V = 80 — 0.08 I

V = C — .08 I

V = 80 — 0.08 I

When V = 0

80

1 = Ш = 1000 A

Short circuit current = 1000 A Open circuit voltage = 80 V

Example 3. During an experimental investigation the arc-voltage has been found to be related with arc-length as V = (22 + 4l) volts. The power source characteristics is as follows

f тЛ

 = 1

 + 2

 H Io K

 H Vo K

where V0 = open circuit voltage and I0 = open circuit current. In one of the observations V0 = 90 volts and I0 = 1000 Amp. What will be the values of welding currents for arc lengths of 3 mm and 5 mm with corresponding arc voltage of 30 volts and 40 volts.

Solution. Using the data given

 30 90

 = 1

 + 2

 1000

 I
 I =

x 1000 = 444.44 Amp

9 x 2

2

I

- I +2,

90 K H 1000

16 I 1

I2 = I 1 - — x - x 1000 = 400.61 Amp 81 у 2

The values of welding currents are 444.44 Amp and 400.61 Amp corresponding to arc - voltages of 30 and 40 volts respectively.