Production Engineering Test 4

Both brazing and soldering are the metal joining processes in which parent metal does not melt but only filler metal melts filling the joint with capillary action.

• If the filler metal is having melting temperature more than 450°C but lower than the melting temperature of components, then it is termed as a process of brazing or hard soldering
• However, if the melting temperature of filler metal is lower than 450°C and lower than the melting point of the material of components then it is known as soldering or soft soldering

Concept:

\(\text{Melting}\text{ efficiency }\!\!~\!\!\text{ }=\frac{Actual~amount~of~heat~utilised}{Total~amount~of~heat~supplied}\)

Actual utilised = specific melting energy × volume

Total supplied = η VI

Calculation:

Total supplied = 0.91 × 25 × 300 = 6825 J

\(\text{Actual}=1\times \frac{4}{3}\pi \times {{6.5}^{3}}=1150.34~J\)

\(\therefore {{\eta }_{m}}=\frac{1150.34}{6825}\)

Concept:

The portion of heat generated was used to form the weld is given as, \(\frac{Heat ~required ~to ~melt }{Heat ~generated ~in ~welding}\times100=\frac{H_m}{H_g}\times 100\)

Heat generated in welding (H_g) = I2Rt and

Heat required to melt required volume (H_m) = volume of nugget (V) × energy required to melt unit volume (H)

Volume of nugget = \(\frac{\pi }{4}{d^2}h\)

Calculation:

Given:

I = 12000 Amp, R = 0.0001 ohms, t = 0.2 sec, d = 6 mm, h = 2.5 mm, H = 12 J/mm³

where, I = current, R = resistance, t = time duration, d = diameter of nugget, h = height of nugget, H = unit melting energy 

Volume of nugget (V) = \(\frac{\pi }{4} \times {6^2} \times 2.5 = 70.68\) mm³

Heat required to melt required volume (H_m) = volume of nugget × energy required to melt unit volume

= 70.68 × 12 = 848.16 J

Heat generated in welding (H_g) = 12000² × 0.0001 × 0.2 = 2880 J

Portion of heat used = \(\frac{{848.16}}{{2880}}\times100=0.294\) = 29.4%

Concept:

Melting efficiency \({\eta _m} = \frac{{{H_m}}}{{{H_s}}}\) 

Where H_m is heat required for melting & H_s is heat supplied.

We know that, \({H_s} = \frac{{VI}}{{{A_b} \times v}} \times {\eta _h}\;\;J/m{m^3}\)

Where, H_S = Heat input per unit Volume, V = Arc voltage, I = Current supplied, η_h = Arc or thermal efficiency, v = welding speed, A_b is weld bead area

For heat input in per unit length:  \({H_s} = \frac{{VI}}{v} \times {\eta _h}\;\;J/mm\)

\(\Rightarrow {\eta _m} = \frac{{{H_m}}}{{{H_s}}} = \frac{{{H_m}}}{{\frac{{VI{\eta _h}}}{{{A_b} \times v}}}}\)

Calculation:

H’_s = 1000 J/mm, η_m = 0.45, H_m = 20 J/mm³

\({\rm{\;}}\frac{{VI}}{v} \times {\eta _h} = 1000\;J/mm\)

\(\therefore {\eta _m} = \frac{{{H_m}}}{{\frac{{1000}}{{{A_b}}}}}\)

\(\Rightarrow 0.45 = \frac{{20}}{{\frac{{1000}}{{{A_b}}}}} \Rightarrow {A_b} = 22.5\;m{m^2}\)

Points to remember:

• If heat input is asked in per unit length, use \({H_S} = \frac{{VI}}{v} \times {\eta _h}\)
• If heat input is asked in per unit volume, use \({H_S} = \frac{{VI}}{{{A_b} \times v}} \times {\eta _h}\) where A_b is weld bead area

Concept:

Heat supplied = I²Rt

I = current, R = resistance, t = time

Height of weld nugget = 2 (thickness – indentation)

Volume of nugget \(= \frac{\pi }{4}{d^2}h\)

\(d = 6\sqrt t\)  (Using Unwin’s formula)

Heat required = u(v) × P

u = specific melting energy

\({\rm{Efficiency\;}}\left( \eta \right) = \frac{{Heat\;required}}{{Heat\;supplied}} \times 100\;\% \)

Calculation:

Given:

I = 11000 A, t = 0.1 sec, P = 10 gm/cc, u = 1390 J/gm, R = 300 × 10^-6 Ω.

H = I² Rt = 121 × 10⁶ × 300 × 10^-6 × 0.1

∴ H = 3630 J

Now,

Height (h) = 2 (2 – 0.15 × 2) = 3.4 mm

\(Volume\;\left( V \right) = \frac{\pi }{4} \times 36 \times 2 \times 3.4 = 192.265 \times {10^{ - 3}}\;c{m^3}\)

\({\rm{Mass}} = \rho V = 192.265 \times {10^{ - 2}}\;gm = 1.922\;gm\) 

Now,

Heat required = 1390 × 1.922

∴ Heat required = 2672.49 J

\(\eta = \frac{{2672.49}}{{3630}} \times 100\)

Concept:

Using the relations, calculate the relation between I and ℓ

I = f(ℓ)

Calculation:

I² = - 400 (V – 80)

V = 30 + 10 ℓ

I² = -400 (30 + 10 ℓ - 80)

I² = 400 (50 – 50 L)

Now,

When L = 4 mm

I² = 400 × (50 – 40) = 4000

\(I=\sqrt{4000}=63.245~A\)

Now,

L = 3.8 mm

I² = 400 × (50 – 38) = 4800

\(I=\sqrt{4800}=69.282~A\)

Now,

Concept:

During operation, V_p = V_a

Power = V_aI

\({\rm{Maximum\;Power}},{\rm{\;}}\left( {\frac{{dP}}{{d{l_a}}}} \right) = 0\)

Calculation:

V_p = V_a

\(40 - \frac{I}{{64}} = 3{L_a} + 27\)

\(\frac{I}{{64}} = - 3{L_a} + 13\)

I = -192 L_a + 832

Now,

Power (P) = V_aI

P = (3L_a + 27) (832 – 192 L_a)

P = 2496 L_a + 22464 – 5184 L_a + \(\left( { - 576\;L_a^2} \right)\) 

\(P = - 576\;L_a^2 - 2688\;{l_a} + 22464\)

\(\frac{{dP}}{{d{l_a}}} = 0 \Rightarrow 2688 = - 2 \times 576\;{L_a}\)

L_a is L₀ (Ignore)

Power at L_a = 3 mm, P = 9216 W

L_a = 4 mm, P = 2496 W

ΔP = 6720 W

When,

I = 380 A

380 = 832 – 192 L_a

∴ L_a = 2.35 mm 

Concept:

Heat required to form the weld will be \(H = {U_m}V\)

Also, as the current varies with time, the heat generated will be:

\(H = \eta \mathop \smallint \nolimits_0^t {I^2}Rdt\)

Calculation:

\(H = 10 \times \frac{\pi }{4} \times {4^2} \times 2 = 251.327\;J\)

Also,

\(H = 0.25 \times \mathop \smallint \nolimits_0^t {\left( {100000{\rm{t}}} \right)^2} \times 85 \times {10^{ - 6}}\;dt = 212500 \times \mathop \smallint \nolimits_0^t {{\rm{t}}^2}dt = \;\frac{{212500}}{3}{t^3}\)

\(H = \;\frac{{21250000}}{3}{t^3} = 251.324 \Rightarrow t = 0.152\;seconds\)

Concept:

The duty cycle is the percentage of time a welding machine can be used continuously. A 60% duty

cycle means that out of 10 minutes, the machine can be used for a total of six minutes at the

 maximum rated current. When providing power at this level, it must be cooled off four minutes out of every 10 minutes. The duty cycle increases as the amperage is lowered & decreased for higher amperage. The relation is given by:

I²D = constant

Where, D: duty cycle

(I)² × 1 = (600)² × 0.6

I = 464.7580 A

\(\% \;decrease = \frac{{600 - 464.7580}}{{600}} \times 100\)