The Heat Input Formula calculates the amount of electrical energy delivered per unit length of weld during the arc welding process, expressed as \( Q = \frac{V \times I \times 60}{S} \). This formula is a key concept in thermal physics and engineering thermodynamics, covered in NCERT Class 11 Physics (Chapter 12 — Thermodynamics) and extended in Class 12 and competitive syllabi. JEE Main and NEET aspirants encounter heat transfer and energy input concepts in multiple question types each year. This article covers the full derivation, variable table, complete formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, JEE/NEET applications, and detailed FAQs.
Key Heat Input Formulas at a Glance
Quick reference for the most important formulas related to heat input and arc welding thermodynamics.
- Basic Heat Input: \ ( Q = \frac{V \times I \times 60}{S} \)
- Heat Input with Efficiency: \ ( Q_{eff} = \frac{\eta \times V \times I \times 60}{S} \)
- Arc Energy: \ ( E = V \times I \)
- Heat Transfer (Conduction): \ ( Q = mc\Delta T \)
- Thermal Power: \ ( P = \frac{Q}{t} \)
- Heat Flow Rate (Fourier): \ ( \frac{dQ}{dt} = kA\frac{\Delta T}{L} \)
- Specific Heat Capacity: \ ( c = \frac{Q}{m \Delta T} \)
What is the Heat Input Formula?
The Heat Input Formula defines the total electrical energy supplied to a weld joint per unit length of weld bead during the arc welding process. It is a critical parameter that controls weld quality, microstructure, and mechanical properties of the joint. A higher heat input leads to a wider heat-affected zone (HAZ) and slower cooling rates, while a lower heat input produces a narrower HAZ with faster cooling.
In the NCERT curriculum, the underlying principles of heat input are rooted in Class 11 Physics, Chapter 12 (Thermodynamics) and Class 11 Physics, Chapter 11 (Thermal Properties of Matter). The concept of energy transfer, specific heat, and thermal power all feed directly into understanding how heat input behaves in physical and engineering systems.
The Heat Input Formula is not only relevant to welding engineering but also forms the conceptual backbone of heat transfer problems in CBSE board exams and competitive entrance tests. It bridges electrical energy (voltage and current) with thermal energy delivered to a material, making it a cross-disciplinary formula of high importance.
In arc welding specifically, the formula accounts for welding voltage, welding current, and the travel speed of the electrode. The result is expressed in Joules per millimetre (J/mm) or kilojoules per millimetre (kJ/mm).
Heat Input Formula — Expression and Variables
The standard Heat Input Formula used in arc welding and thermal physics is:
\[ Q = \frac{V \times I \times 60}{S} \]
Where Q is the heat input in Joules per millimetre (J/mm), V is the arc voltage in Volts, I is the welding current in Amperes, and S is the travel speed in mm/min. The factor 60 converts minutes to seconds for dimensional consistency.
When thermal efficiency of the welding process is included, the formula becomes:
\[ Q_{eff} = \frac{\eta \times V \times I \times 60}{S} \]
Here, \ ( \eta \) (eta) is the thermal efficiency factor, which varies by welding process (e.g., 0.8 for SMAW, 0.6 for GTAW).
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | Heat Input per unit weld length | J/mm or kJ/mm |
| V | Arc Voltage | Volts (V) |
| I | Welding Current | Amperes (A) |
| S | Travel Speed (weld speed) | mm/min |
| η | Thermal Efficiency of Welding Process | Dimensionless (0 to 1) |
| 60 | Time conversion factor (min to sec) | s/min |
Derivation of the Heat Input Formula
The derivation starts from the definition of electrical power. The power delivered by an arc is:
\[ P = V \times I \]
This gives power in Watts (Joules per second). To find the energy delivered over time t (in seconds):
\[ E = P \times t = V \times I \times t \]
The travel speed S is given in mm/min. The time to travel 1 mm is \ ( t = \frac{60}{S} \) seconds. Substituting this into the energy equation:
\[ Q = V \times I \times \frac{60}{S} = \frac{V \times I \times 60}{S} \]
This gives heat input per mm of weld length. Including thermal efficiency \ ( \eta \) accounts for heat losses to the surrounding environment and gives the effective heat input \ ( Q_{eff} = \eta \times Q \).
Complete Thermodynamics and Heat Transfer Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat Input (Arc Welding) | \ ( Q = \frac{V \times I \times 60}{S} \) | V=Voltage, I=Current, S=Speed | J/mm | Class 11, Ch 12 |
| Heat Input with Efficiency | \ ( Q_{eff} = \frac{\eta V I \times 60}{S} \) | η=efficiency, V, I, S | J/mm | Class 11, Ch 12 |
| Specific Heat Capacity | \ ( Q = mc\Delta T \) | m=mass, c=specific heat, ΔT=temp change | J | Class 11, Ch 11 |
| Thermal Power | \ ( P = \frac{Q}{t} \) | Q=heat energy, t=time | W (J/s) | Class 11, Ch 6 |
| Fourier’s Law of Heat Conduction | \ ( \frac{dQ}{dt} = kA\frac{\Delta T}{L} \) | k=thermal conductivity, A=area, L=length | W | Class 11, Ch 11 |
| Newton’s Law of Cooling | \ ( \frac{dT}{dt} = -k(T – T_0) \) | T=temperature, T₀=ambient temp, k=constant | K/s | Class 11, Ch 11 |
| Stefan-Boltzmann Law | \ ( P = \sigma A T^4 \) | σ=5.67×10⁻&sup8; W/m²K⁴, A=area, T=temp | W | Class 11, Ch 11 |
| First Law of Thermodynamics | \ ( \Delta U = Q – W \) | ΔU=internal energy, Q=heat, W=work | J | Class 11, Ch 12 |
| Efficiency of Heat Engine | \ ( \eta = 1 – \frac{T_2}{T_1} \) | T₁=source temp, T₂=sink temp | Dimensionless | Class 11, Ch 12 |
| Latent Heat | \ ( Q = mL \) | m=mass, L=latent heat | J | Class 11, Ch 11 |
Heat Input Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: A welding process uses an arc voltage of 25 V and a welding current of 200 A. The travel speed is 300 mm/min. Calculate the heat input per mm of weld length.
Given:
- Voltage, V = 25 V
- Current, I = 200 A
- Travel Speed, S = 300 mm/min
Step 1: Write the Heat Input Formula: \ ( Q = \frac{V \times I \times 60}{S} \)
Step 2: Substitute the known values:
\[ Q = \frac{25 \times 200 \times 60}{300} \]
Step 3: Calculate the numerator: \ ( 25 \times 200 \times 60 = 300{,}000 \)
Step 4: Divide by travel speed: \ ( Q = \frac{300{,}000}{300} = 1{,}000 \) J/mm
Answer
Heat Input Q = 1,000 J/mm = 1 kJ/mm
Example 2 (Class 11-12 Level — With Thermal Efficiency)
Problem: In a Gas Tungsten Arc Welding (GTAW) process, the arc voltage is 18 V, the welding current is 150 A, and the travel speed is 180 mm/min. The thermal efficiency of GTAW is 0.60. Find the effective heat input per mm of weld.
Given:
- Voltage, V = 18 V
- Current, I = 150 A
- Travel Speed, S = 180 mm/min
- Thermal Efficiency, η = 0.60
Step 1: Write the effective Heat Input Formula: \ ( Q_{eff} = \frac{\eta \times V \times I \times 60}{S} \)
Step 2: Substitute all values:
\[ Q_{eff} = \frac{0.60 \times 18 \times 150 \times 60}{180} \]
Step 3: Calculate the numerator: \ ( 0.60 \times 18 \times 150 \times 60 = 0.60 \times 162{,}000 = 97{,}200 \)
Step 4: Divide by travel speed: \ ( Q_{eff} = \frac{97{,}200}{180} = 540 \) J/mm
Step 5: Convert to kJ/mm: \ ( Q_{eff} = 0.54 \) kJ/mm
Answer
Effective Heat Input Qeff = 540 J/mm = 0.54 kJ/mm
Example 3 (JEE/NEET Level — Concept Application with Heat Transfer)
Problem: During a Shielded Metal Arc Welding (SMAW) process, the arc voltage is 30 V, the welding current is 250 A, and the travel speed is 200 mm/min. The thermal efficiency is 0.80. If the weld bead is deposited on a steel plate of mass 0.5 kg and specific heat capacity 500 J/(kg·K), estimate the maximum possible temperature rise of the plate assuming all effective heat input goes into heating the plate over a 100 mm weld length.
Given:
- V = 30 V, I = 250 A, S = 200 mm/min, η = 0.80
- Mass of plate, m = 0.5 kg
- Specific heat, c = 500 J/(kg·K)
- Weld length = 100 mm
Step 1: Calculate effective heat input per mm:
\[ Q_{eff} = \frac{0.80 \times 30 \times 250 \times 60}{200} = \frac{360{,}000}{200} = 1{,}800 \text{ J/mm} \]
Step 2: Find total heat input for 100 mm weld:
\[ Q_{total} = 1{,}800 \times 100 = 180{,}000 \text{ J} = 180 \text{ kJ} \]
Step 3: Apply the specific heat formula \ ( Q = mc\Delta T \) to find temperature rise:
\[ \Delta T = \frac{Q_{total}}{mc} = \frac{180{,}000}{0.5 \times 500} = \frac{180{,}000}{250} = 720 \text{ K} \]
Step 4: This is the theoretical maximum. In practice, heat losses reduce the actual temperature rise.
Answer
Maximum possible temperature rise ΔT = 720 K (720°C rise above initial temperature)
CBSE Exam Tips 2025-26
- Memorise the formula structure: We recommend writing \ ( Q = \frac{V \times I \times 60}{S} \) on your formula sheet first. The factor of 60 is the most commonly forgotten element in CBSE answers.
- Unit consistency is key: Always check that travel speed is in mm/min before substituting. If it is given in m/min or cm/min, convert it first. CBSE examiners award marks for correct unit handling.
- Link to thermodynamics: In 2025-26 board papers, heat input questions may be combined with the first law of thermodynamics \ ( \Delta U = Q – W \). Practise these combined problems from NCERT exemplar exercises.
- Show all steps: Our experts suggest writing the formula, then substituting values, then simplifying. Partial marks are awarded at each step in CBSE marking schemes.
- Know the efficiency values: For SMAW, η = 0.8; for GTAW, η = 0.6; for GMAW, η = 0.75. These values are standard and may be given or expected in application questions.
- Practice dimensional analysis: Verify your answer unit (J/mm) by checking dimensions on both sides. This habit prevents unit-related errors in 3-mark and 5-mark questions.
Common Mistakes to Avoid
- Forgetting the factor of 60: Many students write \ ( Q = \frac{VI}{S} \) without the 60. This gives a wrong answer because speed is in mm/min, not mm/s. Always include 60 to convert minutes to seconds.
- Confusing arc energy with heat input: Arc energy is simply \ ( E = V \times I \) in Watts. Heat input is arc energy divided by travel speed, adjusted for time. These are different quantities with different units.
- Ignoring thermal efficiency: When a problem states the welding process type (SMAW, GTAW, GMAW), always apply the efficiency factor \ ( \eta \). Using Q instead of Qeff overestimates the actual heat delivered.
- Wrong unit for travel speed: Travel speed must be in mm/min in the standard formula. If the problem gives speed in mm/s, multiply by 60 before substituting, or adjust the formula to \ ( Q = \frac{V \times I}{S_{mm/s}} \).
- Mixing up Q (heat input) with Q (heat in thermodynamics): In NCERT thermodynamics, Q represents heat absorbed by a system in Joules. In the welding formula, Q is heat per unit length in J/mm. Read the context carefully to avoid confusion.
JEE/NEET Application of the Heat Input Formula
In our experience, JEE aspirants encounter the Heat Input Formula and its underlying principles in multiple forms across thermodynamics and heat transfer chapters. Understanding this formula builds a strong conceptual foundation for tackling high-difficulty problems.
Application Pattern 1: Combined Energy and Temperature Problems
JEE Main frequently combines the heat input concept with the specific heat formula \ ( Q = mc\Delta T \). A problem may give welding parameters and ask for the temperature rise of a component. Students must first calculate total heat input using the welding formula, then apply the calorimetry equation. This two-step approach is a classic JEE pattern.
Application Pattern 2: Thermodynamic Efficiency Problems
JEE Advanced problems on heat engines and Carnot cycles use the concept of “heat input” as the energy supplied to the hot reservoir. The efficiency formula \ ( \eta = 1 – \frac{T_2}{T_1} \) is directly related to how much of the heat input is converted to useful work. Understanding that \ ( W = \eta \times Q_{input} \) is critical for solving multi-step thermodynamics problems in JEE Advanced.
Application Pattern 3: NEET Heat Transfer Conceptual Questions
NEET Biology and Physics papers include questions on heat transfer in biological and physical systems. The principle that heat input per unit time equals thermal power \ ( P = \frac{Q}{t} \) is used in questions about metabolic heat production, body temperature regulation, and industrial heating. Our experts recommend practising at least 10 NEET-level MCQs on thermal power and specific heat annually to consolidate this concept.
In our experience, JEE aspirants who master the dimensional analysis of the Heat Input Formula — understanding why the unit is J/mm and not just J — perform significantly better on unfamiliar thermodynamics problems in both JEE Main and Advanced.
FAQs on Heat Input Formula
For more related concepts, explore our detailed guide on the Electric Current Formula and the Current Density Formula, which provide the electrical foundation for understanding heat input calculations. You can also visit our complete Physics Formulas hub for a full list of NCERT-aligned formula articles. For official NCERT textbook references, visit ncert.nic.in.