Welded Joints MCQ Quiz in मराठी - Objective Question with Answer for Welded Joints - मोफत PDF डाउनलोड करा
Last updated on Mar 21, 2025
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Welded Joints Question 1:
Assuming the voltage current characteristics for a DC welding generator short-circuit current as 300 A & open circuit voltage as 60 V. Determine the optimum voltage & current setting for maximum power output. Where, the voltage arc length characteristic is given as V = 15 + 25L
Answer (Detailed Solution Below)
Welded Joints Question 1 Detailed Solution
Vo = 60 V, ISC = 300A & V = 15 + 25L
\(\begin{array}{l} \Rightarrow \;15 + 25L = {V_o} - \left( {\frac{{{V_o}}}{{{I_{sc}}}}} \right) \times I\\ \Rightarrow 15 + 25L = 60 - \frac{I}{5} \Rightarrow I = 5\left( {60 - 15 - 25L} \right)\\ \Rightarrow I = 5\left( {45 - 25L} \right) \end{array}\)
\(\Rightarrow I = 25\left( {9 - 5L} \right)\) ----(1)
Power (P) = VI = (15 + 25L) (9 – 5L) 25
⇒ P = 125 (3 + 5L) (9 – 5L)
⇒ P = 125 (27 + 30L – 25L2)
for maximization of power:
\(\frac{{dP}}{{dL}} = 125\left( {30 - 50L} \right) = 0 \Rightarrow L = \frac{3}{5}\left( {critical\;point} \right)\) -----(2)
\(\begin{array}{l} \frac{{{d^2}P}}{{d{L^2}}} = 125 \times \left( { - 50} \right)\\ \Rightarrow \left. {\frac{{{d^2}P}}{{d{L^2}}}} \right| = - 6250 < 0\\ @L = \frac{3}{5} \end{array}\)
This Indicates that power will be maximized at L = 3/5
\(\begin{array}{l} V = 15 + 25L = 15 + 25 \times \left( {\frac{3}{5}} \right) = 15 + 15 = 30V\\ I = 25\left( {9 - 5L} \right) = 25\left( {9 - 5 \times \left( {\frac{3}{5}} \right)} \right) = 25\left( 6 \right) = 150A\\ \left. {\begin{array}{*{20}{c}} {V = 30V}\\ {I = 150A} \end{array}} \right\}\; \end{array}\)
Welded Joints Question 2:
Tungsten arc welding of a steel plate is carried out with weld speed of 20 mm/min, welding current of 600 A and voltage of 40 V. Consider the heat transfer efficiency from the arc to the weld pool as 80%. The heat input per unit length (in kJ/mm) is
Answer (Detailed Solution Below)
Welded Joints Question 2 Detailed Solution
Heat input, \(\frac{H}{l} = \eta \frac{{VI}}{v}\)
\(= \frac{{0.8 \times 40 \times 600 \times 60}}{{20}}\)
= 57600 J/mm or 57.6 kJ/mmWelded Joints Question 3:
Tungsten arc welding of a steel plate is carried out with welding current of 500 A. Voltage of 20 V and weld speed of 20 mm/min. Consider the heat transfer efficiency from the arc to the weld pool as 90%. The heat input per unit length (in kJ/mm) is
Answer (Detailed Solution Below)
Welded Joints Question 3 Detailed Solution
\(= \eta \frac{{VI}}{v}\)Heat input
\(= \frac{{0.9 \times 20 \times 500 \times 60}}{{20}}\)
= 27000 J/mm or 27 kJ/mmWelded Joints Question 4:
For straight line V-I characteristics of a DC welding generator, short circuit current as 400 A and open circuit voltage as 100 V, which one of the following in the correct voltage and current setting for maximum power.
Answer (Detailed Solution Below)
Welded Joints Question 4 Detailed Solution
Concept:
The relation among, open-circuit voltage, short-circuit current, arc voltage and arc current is given by,
\(V = {V_0} - \frac{{{V_0}}}{{{I_S}}}\;I\)
where, V0 = open-circuit voltage, Is = short circuit current
Power = V × I
\(P = \left( {{V_0} - \frac{{{V_0}}}{{{I_S}}}\;I} \right)I\)
\(P = {V_0}I - \frac{{{V_0}}}{{{I_S}}}\;{I^2}\)
Calculation:
Given
Short-circuit current = SSC = 400 A
Open circuit voltage = OCV = 100 V
Now \(V = OCV - \left( {\frac{{OCV}}{{SSC}}} \right)I\)
\(= 100 - \left( {\frac{{100}}{{400}}} \right)I\)
Power P = VI
\(= \left( {100 - \frac{{1I}}{4}} \right)I = 100I - \frac{{{I^2}}}{4}\)
For maximum power
\(\frac{{dp}}{{dI}} = 0\)
\(= 100 - \frac{{2I}}{4} = 0\)
⇒ I = 200 A
V = 50 VWelded Joints Question 5:
In arc welding DC power source has a linear characteristic, with open circuit voltage = 70 V and short circuit current = 6000 Amp. The voltage length characteristic of the arc is given by Va = 30 + 5L, where L is arc length in mm. Calculate maximum arc power in (kW) kilowatt.
Answer (Detailed Solution Below)
105.00
Welded Joints Question 5 Detailed Solution
V0 = 70 V
Is = 6000 A
V = V0 \(-{({V_0}) \over(I_ s )}I\)
∴ 30 + 5L = 70 – (70/6000)I
∴ I = [40 – 5L] (600/7) ………. (1)
P = VI = (40 – 5L) (600/7) (30 + 5L)
For maximum arc power, dP/dL = 0
∴ (40 – 5L)(5) + (–5)(30 + 5L) = 0
∴ 200 – 25L – 150 – 25L = 0
∴Lmax = 1
∴from (1), Imax = (40 – 5 × 1) (600/7) = 3000 Amp.
Vmax = 30 + 5 × 1 = 35 V
Now maximum arc power Pmax = 35 × 3000 = 105 kW