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).

80 | |

120 | |

180 | |

25 |

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}

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).

500 | |

460 | |

320 | |

520 |

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

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).

\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).

18.25 | |

32.12 | |

10.69 | |

8.45 |

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

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).

30.5 | |

39.25 | |

45.75 | |

55.75 |

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}

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).

33.6 | |

14.2 | |

36.2 | |

48.6 |

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}

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

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 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).

0.2 | |

0.5 | |

0.9 | |

0.7 |

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}

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)?

Oxy-acetylene gas welding | |

Tungsten inert gas welding | |

Plasma arc welding | |

Laser beam welding |

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).

6.5 | |

7.5 | |

8.6 | |

9.2 |

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

groove | |

spot | |

fillet | |

plug |

Question 10 Explanation:

There are 10 questions to complete.

29.

Value of resistance wrong given

Thank you for your suggestion. We have updated it.