Question 1
Two surfaces P and Q are to be joined together. In which of the given joining operation(s), there is no melting of the two surfaces P and Q for creating the joint?
Arc welding
Adhesive bonding
Spot welding
GATE ME 2023   Manufacturing Engineering
Question 1 Explanation: 
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.
Question 2
In a direct current arc welding process, the power source has an open circuit voltage of 100 V and short circuit current of 1000 A. Assume a linear relationship between voltage and current. The arc voltage (V) varies with the arc length (l) as V = 10 + 5l, where V is in volts and l is in mm. The maximum available arc power during the process is _________ kVA (in integer).
GATE ME 2022 SET-2   Manufacturing Engineering
Question 2 Explanation: 
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}

Question 3
An assignment problem is solved to minimize the total processing time of four jobs (1, 2, 3 and 4) on four different machines such that each job is processed exactly by one machine and each machine processes exactly one job. The minimum total processing time is found to be 500 minutes. Due to a change in design, the processing time of Job 4 on each machine has increased by 20 minutes. The revised minimum total processing time will be _____________ minutes (in integer).
GATE ME 2022 SET-1   Manufacturing Engineering
Question 3 Explanation: 
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
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
Minimum total processing time = 80+160+80+200 = 520 mins
Question 4
Two mild steel plates of similar thickness, in buttjoint configuration, are welded by gas tungsten arc welding process using the following welding parameters.
\begin{array}{|c|c|}\hline \text{Welding voltage}&20V \\ \hline \text{Welding current} & 150A\\ \hline \text{Welding speed}&5mm/s \\ \hline \end{array}
A filler wire of the same mild steel material having 3 mm diameter is used in this welding process. The filler wire feed rate is selected such that the final weld bead is composed of 60% volume of filler and 40% volume of plate material. The heat required to melt the mild steel material is 10 J/mm^3. The heat transfer factor is 0.7 and melting factor is 0.6. The feed rate of the filler wire is __________ mm/s (round off to one decimal place).
GATE ME 2022 SET-1   Manufacturing Engineering
Question 4 Explanation: 
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
Question 5
A spot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5 mm and a thickness of 1 mm. The welding time was 0.1 s. The melting energy for the steel is 20 J/mm^3. Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding operation is _________kW (round off to two decimal places).
GATE ME 2021 SET-2   Manufacturing Engineering
Question 5 Explanation: 
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}

There are 5 questions to complete.

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