# Joining-Welding

 Question 1
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).
 A 80 B 120 C 180 D 25
GATE ME 2022 SET-2   Manufacturing Engineering
Question 1 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 2
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).
 A 500 B 460 C 320 D 520
GATE ME 2022 SET-1   Manufacturing Engineering
Question 2 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
$J_1-M_1=80$
$J_2-M_4=160$
$J_3-M_2=80$
$J_4-M_3=180$
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
 Question 3
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).
 A 18.25 B 32.12 C 10.69 D 8.45
GATE ME 2022 SET-1   Manufacturing Engineering
Question 3 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 4
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).
 A 30.5 B 39.25 C 45.75 D 55.75
GATE ME 2021 SET-2   Manufacturing Engineering
Question 4 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}
 Question 5
The resistance spot welding of two 1.55 mm thick metal sheets is performed using welding current of 10000 A for 0.25 s. The contact resistance at the interface of the metal sheets is 0.0001 $\Omega$. The volume of weld nugget formed after welding is 70 $mm^3$. Considering the heat required to melt unit volume of metal is 12 $J/mm^3$, the thermal efficiency of the welding process is ______% (round off to one decimal place).
 A 33.6 B 14.2 C 36.2 D 48.6
GATE ME 2021 SET-1   Manufacturing Engineering
Question 5 Explanation:
\begin{aligned} \text { Thickness } &=1.55 \mathrm{~mm} \\ I &=10000 \mathrm{~A} \\ t &=0.25 \mathrm{~s} \\ R &=0.0001 \Omega \\ V_{n} &=70 \mathrm{~mm}^{3} \\ H_{m} &=12 \mathrm{~J} / \mathrm{mm}^{3} \\ \eta_{m} &=? \\ \eta_{m} &=\frac{H_{m}}{H_{s}} \end{aligned}
Melting efficiency = Thermal efficiency
\begin{aligned} \eta_{m} &=\frac{H_{m}}{I^{2} R t}=\frac{12 \times 70}{(10000)^{2} \times 0.0001 \times 0.25} \\ &=0.336 \\ \eta_{m} &=33.6 \% \end{aligned}
 Question 6
Two plates, each of 6 mm thickness, are to be butt-welded. Consider the following processes and select the correct sequence in increasing order of size of the heat affected zone

1. Arc welding
2. MIG welding
3. Laser beam welding
4. Submerged arc welding
 A 1-4-2-3 B 3-4-2-1 C 4-3-2-1 D 3-2-4-1
GATE ME 2020 SET-2   Manufacturing Engineering
Question 6 Explanation:
Processes with low rate of heat input (slow heating) tend to produce high total heat constant within the metal, slow cooling rates, and large heat-affected zones. high heat input process, have low total heats, fast cooling rates and small heat affected zones
 Question 7
A gas tungsten arc welding operation is performed using a current of 250A and an arc voltage of 20V at a welding speed of 5mm/s. Assuming that the arc efficiency is 70%, the net heat input per unit length of the weld will be ______kJ/mm(round off to one decimal place).
 A 0.2 B 0.5 C 0.9 D 0.7
GATE ME 2019 SET-2   Manufacturing Engineering
Question 7 Explanation:
GTAW, current $\mathrm{I}=250 \mathrm{A}$
Voltage, $\mathrm{V}=20 \mathrm{V},$
Speed, $v=5 \mathrm{mm} / \mathrm{sec}$
Arc efficiency, $\eta=0.7$
Duty cycle, D=1 (not given)
Heat input per unit length, $\frac{\mathrm{H}}{\ell}=\frac{\eta \mathrm{D} \mathrm{VI}}{v}$
$\begin{array}{l} =\frac{0.7 \times 1 \times 250 \times 20}{5} \\ =700 \mathrm{J} / \mathrm{mm}=0.7 \mathrm{kJ} / \mathrm{mm} \end{array}$
 Question 8
Which one of the following welding methods provides the highest heat flux ($W/mm^2$)?
 A Oxy-acetylene gas welding B Tungsten inert gas welding C Plasma arc welding D Laser beam welding
GATE ME 2019 SET-1   Manufacturing Engineering
Question 8 Explanation:
Heat flux means power input in terms of heat per unit area. It will be maximum in Laser beam welding.
 Question 9
A welding operation is being performed with voltage = 30 V and current = 100 A. The cross-sectional area of the weld bead is 20 $mm^{2}$. The work-piece and filler are of titanium for which the specific energy of melting is 14 J/$mm^{3}$. Assuming a thermal efficiency of the welding process 70%, the welding speed (in mm/s) is __________ (correct to two decimal places).
 A 6.5 B 7.5 C 8.6 D 9.2
GATE ME 2018 SET-2   Manufacturing Engineering
Question 9 Explanation:
\begin{aligned} \text{Welding speed } (\mathrm{mm} / \mathrm{s})&=?\\ \text { Effective power } &=\eta_{\mathrm{th}} \times(\mathrm{Vl}) \\ &=0.7 \times 30 \times 100=2100 \mathrm{J} / \mathrm{S} \\ \text { Sp.engery } &=\frac{\text { Power }}{A \times V} \\ V &=\frac{\text { Power }}{\text { sp.energy } \times \mathrm{A}} \\ &=\frac{2100}{14 \times 20}=7.5 \mathrm{mm} / \mathrm{s} \end{aligned}
 Question 10
The type of weld represented by the shaded region in the figure is
 A groove B spot C fillet D plug
GATE ME 2018 SET-1   Manufacturing Engineering
Question 10 Explanation:

There are 10 questions to complete.

### 2 thoughts on “Joining-Welding”

1. 29.
Value of resistance wrong given