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Weld Heat Input worksheet

Grades 7–8

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Arc voltage: 18 V
- Welding current: 200 A
- Travel speed: 20 in/min
- Process (thermal efficiency k): Flux-cored, FCAW (k = 0.8)
Find: Heat input
Answer
2.
Problem 2.
- Arc voltage: 22 V
- Welding current: 200 A
- Travel speed: 18 in/min
- Process (thermal efficiency k): Plasma arc, PAW (k = 0.6)
Find: Heat input
Answer
3.
Problem 3.
- Arc voltage: 32 V
- Welding current: 180 A
- Travel speed: 12 in/min
- Process (thermal efficiency k): Stick, SMAW / MMA (k = 0.8)
Find: Heat input
Answer
4.
Problem 4.
- Arc voltage: 24 V
- Welding current: 125 A
- Travel speed: 8 in/min
- Process (thermal efficiency k): None: arc energy only (k = 1)
Find: Heat input
Answer
5.
Problem 5.
- Arc voltage: 18 V
- Welding current: 180 A
- Travel speed: 18 in/min
- Process (thermal efficiency k): MIG/MAG, GMAW (k = 0.8)
Find: Heat input
Answer
6.
Problem 6.
- Arc voltage: 20 V
- Welding current: 100 A
- Travel speed: 10 in/min
- Process (thermal efficiency k): Stick, SMAW / MMA (k = 0.8)
Find: Heat input
Answer
7.
Problem 7.
- Arc voltage: 30 V
- Welding current: 150 A
- Travel speed: 8 in/min
- Process (thermal efficiency k): Flux-cored, FCAW (k = 0.8)
Find: Heat input
Answer
8.
Problem 8.
- Arc voltage: 26 V
- Welding current: 250 A
- Travel speed: 8 in/min
- Process (thermal efficiency k): Flux-cored, FCAW (k = 0.8)
Find: Heat input
Answer
9.
Problem 9.
- Arc voltage: 24 V
- Welding current: 250 A
- Travel speed: 20 in/min
- Process (thermal efficiency k): TIG, GTAW (k = 0.6)
Find: Heat input
Answer
10.
Problem 10.
- Arc voltage: 30 V
- Welding current: 125 A
- Travel speed: 10 in/min
- Process (thermal efficiency k): Submerged arc, SAW (k = 1.0)
Find: Heat input
Answer

Answer key: Weld Heat Input worksheet

1. Heat input: 8.64 kJ/in
1. Find the arc's power in watts: volts × amps
  $18 \times 200$ $= 3,600 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{18 \times 200 \times 60}{1,000 \times 20}$ $= 10.8 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 10.8$ $= 8.64 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{8.64}{25.4}$ $= 0.340157 kJ/mm$
2. Heat input: 8.8 kJ/in
1. Find the arc's power in watts: volts × amps
  $22 \times 200$ $= 4,400 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{22 \times 200 \times 60}{1,000 \times 18}$ $= 14.666667 kJ/in$
3. Multiply by the process factor k = 0.6, the share of heat that reaches the work, to get heat input
  $0.6 \times 14.666667$ $= 8.8 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{8.8}{25.4}$ $= 0.346457 kJ/mm$
3. Heat input: 23.04 kJ/in
1. Find the arc's power in watts: volts × amps
  $32 \times 180$ $= 5,760 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{32 \times 180 \times 60}{1,000 \times 12}$ $= 28.8 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 28.8$ $= 23.04 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{23.04}{25.4}$ $= 0.907087 kJ/mm$
4. Heat input: 22.5 kJ/in
1. Find the arc's power in watts: volts × amps
  $24 \times 125$ $= 3,000 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{24 \times 125 \times 60}{1,000 \times 8}$ $= 22.5 kJ/in$
3. Multiply by the process factor k = 1, the share of heat that reaches the work, to get heat input
  $1 \times 22.5$ $= 22.5 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{22.5}{25.4}$ $= 0.885827 kJ/mm$
5. Heat input: 8.64 kJ/in
1. Find the arc's power in watts: volts × amps
  $18 \times 180$ $= 3,240 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{18 \times 180 \times 60}{1,000 \times 18}$ $= 10.8 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 10.8$ $= 8.64 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{8.64}{25.4}$ $= 0.340157 kJ/mm$
6. Heat input: 9.6 kJ/in
1. Find the arc's power in watts: volts × amps
  $20 \times 100$ $= 2,000 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{20 \times 100 \times 60}{1,000 \times 10}$ $= 12 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 12$ $= 9.6 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{9.6}{25.4}$ $= 0.377953 kJ/mm$
7. Heat input: 27 kJ/in
1. Find the arc's power in watts: volts × amps
  $30 \times 150$ $= 4,500 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{30 \times 150 \times 60}{1,000 \times 8}$ $= 33.75 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 33.75$ $= 27 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{27}{25.4}$ $= 1.062992 kJ/mm$
8. Heat input: 39 kJ/in
1. Find the arc's power in watts: volts × amps
  $26 \times 250$ $= 6,500 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{26 \times 250 \times 60}{1,000 \times 8}$ $= 48.75 kJ/in$
3. Multiply by the process factor k = 0.8, the share of heat that reaches the work, to get heat input
  $0.8 \times 48.75$ $= 39 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{39}{25.4}$ $= 1.535433 kJ/mm$
9. Heat input: 10.8 kJ/in
1. Find the arc's power in watts: volts × amps
  $24 \times 250$ $= 6,000 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{24 \times 250 \times 60}{1,000 \times 20}$ $= 18 kJ/in$
3. Multiply by the process factor k = 0.6, the share of heat that reaches the work, to get heat input
  $0.6 \times 18$ $= 10.8 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{10.8}{25.4}$ $= 0.425197 kJ/mm$
10. Heat input: 22.5 kJ/in
1. Find the arc's power in watts: volts × amps
  $30 \times 125$ $= 3,750 W$
2. Find the arc energy per inch: power × 60 gives energy per minute, then divide by the inches welded per minute (and by 1,000 for kJ)
  $\frac{30 \times 125 \times 60}{1,000 \times 10}$ $= 22.5 kJ/in$
3. Multiply by the process factor k = 1, the share of heat that reaches the work, to get heat input
  $1 \times 22.5$ $= 22.5 kJ/in$
4. For kJ/mm, divide by 25.4, the millimeters in an inch
  $\frac{22.5}{25.4}$ $= 0.885827 kJ/mm$