Question 1 |
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).
30.5 | |
39.25 | |
45.75 | |
55.75 |
Question 1 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}
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 2 |
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).
33.6 | |
14.2 | |
36.2 | |
48.6 |
Question 2 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}
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 3 |
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
1. Arc welding
2. MIG welding
3. Laser beam welding
4. Submerged arc welding
1-4-2-3 | |
3-4-2-1 | |
4-3-2-1 | |
3-2-4-1 |
Question 3 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 4 |
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).
0.2 | |
0.5 | |
0.9 | |
0.7 |
Question 4 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}
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 5 |
Which one of the following welding methods provides the highest heat flux (W/mm^2)?
Oxy-acetylene gas welding | |
Tungsten inert gas welding | |
Plasma arc welding | |
Laser beam welding |
Question 5 Explanation:
Heat flux means power input in terms of heat per unit area. It will be maximum in Laser beam
welding.
Question 6 |
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).
6.5 | |
7.5 | |
8.6 | |
9.2 |
Question 6 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 7 |
The type of weld represented by the shaded region in the figure is
groove | |
spot | |
fillet | |
plug |
Question 7 Explanation:
Question 8 |
In an arc welding process, welding speed is doubled. Assuming all other process parameters to be constant, the cross sectional area of the weld bead will
Increase by 25% | |
Increase by 50% | |
Reduce by 25% | |
Reduce by 50% |
Question 8 Explanation:
H_{s}=\frac{V I}{v A_{b}}
If v is doubled (2 v) to make H_{s} constant A_{b} will be reduced by 50%
H_{s}=\frac{V I}{2 v \frac{A_{b}}{2}}=\frac{V I}{v A_{b}}
If v is doubled (2 v) to make H_{s} constant A_{b} will be reduced by 50%
H_{s}=\frac{V I}{2 v \frac{A_{b}}{2}}=\frac{V I}{v A_{b}}
Question 9 |
Spot welding of two steel sheets each 2 mm thick is carried out successfully by passing 4 kA of current for 0.2 seconds through the electrodes. The resulting weld nugget formed between the sheets is 5 mm in diameter. Assuming cylindrical shape for the nugget, the thickness of the nugget is________mm.
4.56mm | |
2.91mm | |
1.67mm | |
4.89mm |
Question 9 Explanation:
\begin{aligned} t &=2 \mathrm{mm} \\ I &=4 \mathrm{kA} \\ \mathrm{Time} &=0.25 \\ D_{N} &=5 \mathrm{mm} \\ I^{2} R t &=\text { Volume } \times \rho \times \mathrm{H.R} / \mathrm{kg} \\ 4000^{2} \times 200 \times 10^{-6} \times 0.2&=\mathrm{Vol} . \times 8000 \times 1400 \times 10^{3} \\ h&=2.91 \mathrm{mm} \end{aligned}
Question 10 |
The voltage-length characteristic of a direct current arc in an arc welding process is v=(100+40l) , where l is the length of the arc in mm and V is arc voltage in volts. During a welding operation, the arc length varies between 1 and 2 mm and the welding current is in the range 200-250 A. Assuming a linear power source, the short circuit current is_________ A.
658A | |
425A | |
158A | |
128A |
Question 10 Explanation:
\begin{aligned} V &=100+40 \mathrm{L} \\ \text{When}\quad L &=1 \quad I=250 \\ V &=(100+40 \times 1)=140 \\ &=V_{0}-\frac{V_{0}}{I_{0}} \times 250 \\ L &=2 \quad I=200 \\ V &=(100+40 \times 2) \\ 180 &=V_{0}-\frac{V_{0}}{I_{0}} \times 200 \\ V_{0} &=140+\frac{4}{5} \times 250=340 \\ I &=\frac{V_{0} \times 5}{4}=\frac{340 \times 5}{4} \\ &=425 \mathrm{A} \end{aligned}
There are 10 questions to complete.
29.
Value of resistance wrong given
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