Let's break down the problem and solve for the welding speed step by step: ### **Given:** - Length of plate, \( L = 200 \) mm (not needed for speed, just context) - Voltage, \( V = 20 \) V - Current, \( I = 150 \) A - Cross-section of weld bead: triangular, base \( b = 10 \) mm, height \( h = 5 \) mm - Electrode (Aluminum): energy needed to melt = \( 2.91 \) J/mm³ - Melting efficiency = 70% = 0.7 ### **To Find:** - Welding speed (\( v \)), in mm/s --- ### **Step 1: Calculate Cross-sectional Area of Weld Bead** Since it's a triangle: \[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 5 = 25 \text{ mm}^2 \] --- ### **Step 2: Calculate Power Supplied** \[ \text{Power supplied} = V \times I = 20 \times 150 = 3000 \text{ W} = 3000 \text{ J/s} \] --- ### **Step 3: Power Used for Melting** \[ \text{Power used for melting} = \text{Melting efficiency} \times \text{Power supplied} = 0.7 \times 3000 = 2100 \text{ J/s} \] --- ### **Step 4: Energy Needed to Melt Weld Metal per Second** \[ \text{Volume melted per second} = \text{Cross-sectional area} \times \text{welding speed} = 25 \times v \text{ mm}^3/\text{s} \] \[ \text{Energy required per second} = \text{Volume melted per second} \times \text{specific energy} = (25 \times v) \times 2.91 \text{ J/s} = 72.75v \text{ J/s} \] Set equal to the power used for melting: \[ 72.75v = 2100 \] \[ v = \frac{2100}{72.75} \] \[ v = 28.86 \text{ mm/s} \] --- ### **Check the Closest Option:** - 40 mm/s - 20 mm/s - 30.58 mm/s - **28.96 mm/s** (closest to our calculation) --- ### **Final Answer:** **28.96 mm/s** --- **So, the correct answer is:** > **28.96 mm/s**