Joining-Welding

Soldering is define as a metal joining process wherein coalescence is produced by heating the surface to be joint to a suitable temperature and melting the filler metal which is a fusible alloy called solder (melting point usually less than 427^{\circ}C ).
Principle underlying soldering is that when the surface to be jointed are cleaned off well from oxides they can be joined together using molten solder that may adhere easily with the workpiece.
Brazing is also one of the type of welding process which is used for joining the two similar or dissimilar metal piece together by heating the surface and by using a non-ferrous filler metal having its melting point above 427^{\circ}C but below the melting point of metal to be brazed. The molten filler metal is distributed between the joint surface by the capillary action which on cooling results in a sound joint.

Given, OCV = 100 V and SCC = 1000 A
Voltage Arc length characteristic V_{arc}=10+5l
We have,
\begin{aligned} V_P&=OCV-\frac{OCV}{SCC}I\\ V_P&=100-\frac{100}{1000}I\\ _P&=100-\frac{I}{10} \end{aligned}
For stable arc
\begin{aligned} V_{arc}&=V_P\\ 10+5l&=100-\frac{I}{10}\\ I&=900-50l\\ \text{Arc Power} (P)&=V \times I\\ P&=(10+5l)(900-50l)\\ P&=9000+4000l-250l^2 \end{aligned}
For Maximum Arc Power
\begin{aligned} \frac{dP}{dl}&=0\\ \frac{d}{dl}(9000+400l-250l^2)&=0\\ l=4000/500&=8mm \end{aligned}
Hence, for Maximum arc power, arc length is 8 mm
Voltage at l=8mm
\begin{aligned} V&=10+5l\\ V&=10+( 5 \times 8)\\ V&=50 \;Volt \end{aligned}
Current at l=8mm
\begin{aligned} I&=900-50l\\ I&=900-( 50 \times 8)\\ I&=500 \;A \end{aligned}
Maximum Arc Power
\begin{aligned} P_{max}&= V \times I\\ P_{max}&=50 \times 500\\ &=25000VA\\ &=25 kVA \end{aligned}

Minimum total processing time = 500 min
If the processing time of job 4 each machine is increased by 20 minutes then the revised minimum total processing time will be 520 minutes

Proof: Assume the following data
\begin{array}{|c|c|c|c|c|c|} \hline &M_1&M_2&M_3&M_4& \text{Row min}\\ \hline J_1& 80 & 150 & 100 & 130 & 80\\ \hline J_2& 240 & 250 & 300 & 160 & 160\\ \hline J_3& 130 & 80 & 170 & 190 & 80\\ \hline J_4& 190 & 220 & 180 & 180 & 180\\ \hline \end{array}
\begin{array}{|c|c|c|c|} \hline \underline{0}&70&20&50\\ \hline 80&90& 140&\underline{0}\\ \hline 50&\underline{0}&90&110\\ \hline 10&40&\underline{0}&0\\ \hline \end{array}
The assignments are
J_1-M_1=80
J_2-M_4=160
J_3-M_2=80
J_4-M_3=180
Minimum total processing time = 80+160+80+180 = 500 mins
If Job '4' times are increased by 20 mins then
\begin{array}{|c|c|c|c|c|c|} \hline &M_1&M_2&M_3&M_4& \text{Row min}\\ \hline J_1& 80 & 150 & 100 & 130 & 80\\ \hline J_2& 240 & 250 & 300 & 160 & 160\\ \hline J_3& 130 & 80 & 170 & 190 & 80\\ \hline J_4& 210 & 240 & 200 & 250 & 200\\ \hline \end{array}

\begin{array}{|c|c|c|c|} \hline \underline{0}&70&20&50\\ \hline 80&90& 140&\underline{0}\\ \hline 50&\underline{0}&90&110\\ \hline 10&40&\underline{0}&0\\ \hline \end{array}
The assignments are
J_1-M_1=80
J_2-M_4=160
J_3-M_2=80
J_4-M_3=200
Minimum total processing time = 80+160+80+200 = 520 mins

Welding current = 150A
Welding voltage = 20V
Heat Generated Q_1 = V \times I=150 \times 20=3000 Watt
Net Heat supplied to the Work Piece (Q_2)=Q_1 \times \eta _{HT} =300 \times 0.7=2100 Watt
Heat Required to Melt the material (Q_3)=Q_2 \times \eta _{m} =2100 \times 0.6=1260 Watt
Since Heat Required to melt the material is given directly in the question 10^5 \; J/mm^3
Hence we can write
1260=10 \times Volumetric metal Deposition Rate (VMDR)
(VMDR) =\frac{1260}{10}=126 mm^3/s
It means total 126 mm^3/s metal need to be deposited in the gap between Work Piece
It is given in the question that 60% of the total volume is to be filled by electrode.
Hence, effective volumetric metal deposition by electrode=126 \times 0.6=75.6 mm^3/s
Volumetric metal deposition rate = Cross section Area of electrode x Electrode feed rate
75.6=\frac{\pi}{4}d_e^2 \times \text{feed rate}
\text{feed rate}=\frac{75.6}{\frac{\pi}{4}(3)^2}=10.69mm/s

Heat transfer efficiency =10\%
Energy required to melt =20 \mathrm{~J} / \mathrm{mm}^{3}
Diameter of nugget =5 \mathrm{~mm}
Thickness =1 \mathrm{~mm}
Time =0.1 \mathrm{~second}
Heat required to melt, Q=20 \mathrm{~J} / \mathrm{mm}^{3} \times \frac{\pi}{4}(5)^{2} \times 1=392.5 \mathrm{~J}
\begin{aligned} \text { Power } &=\frac{\text { Heat required to melt }}{\text { Time }} \\ P &=\frac{392.5}{0.1} \mathrm{~J} / \mathrm{s} \\ P &=3925 \mathrm{~J} / \mathrm{s}=3925 \mathrm{~W} \\ \text { Actual power supplied } &=\frac{3925}{\eta_{\text {th }}} \\ \eta_{\text {th }} &=10 \% \\ P &=\frac{3925}{0.1}=39250 \mathrm{~W}=39.25 \mathrm{~kW} \end{aligned}