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 | |

Brazing | |

Adhesive bonding | |

Spot welding |

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.

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).

80 | |

120 | |

180 | |

25 |

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}

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).

500 | |

460 | |

320 | |

520 |

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

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 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).

\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 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

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).

30.5 | |

39.25 | |

45.75 | |

55.75 |

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}

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.

29.

Value of resistance wrong given

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