2.4 Thermal Physics, Linear/Volumetric Expansion & Heat Capacity

2.3 Thermal Physics, Linear/Volumetric Expansion & Heat Capacity

Quick Answer: Thermal expansion arises from the anharmonic vibration of crystal lattices as temperature rises: $\Delta L = L_0 \cdot \alpha \cdot \Delta T$. When thermal expansion is prevented by mechanical restraint ($R = 1.0$), an elastic thermal stress develops: $\sigma_{\text{th}} = -E \cdot \alpha \cdot \Delta T$. Because the yield strength of metals drops sharply at elevated temperatures, the heated weld zone undergoes compressive plastic upsetting. Upon cooling, this plastically shortened zone is pulled in tension by the colder surrounding base metal, generating tensile residual stresses equal to the room-temperature yield strength ($\sigma_{\text{residual}} \approx \sigma_y$). Melting the metal requires both sensible heat ($Q = m C_p \Delta T$) and the latent heat of fusion ($\Delta H_f \approx 270\text{ kJ/kg}$ for steel).

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1. Temperature Scales & Thermophysical Relationships

Welding engineering requires precise conversions between thermodynamic absolute scales and engineering scales:

• Celsius ($^\circ\text{C}$): Defined around the ice-water phase equilibrium ($0^\circ\text{C}$) and boiling point ($100^\circ\text{C}$) at standard atmospheric pressure ($101.325\text{ kPa}$).
• Kelvin ($\text{K}$): The SI base unit of thermodynamic temperature. Absolute zero ($0\text{ K}$) is the state of minimum molecular kinetic energy: $$T_K = T_C + 273.15$$
• Fahrenheit ($^\circ\text{F}$): Common US Customary scale: $$T_F = \frac{9}{5} T_C + 32 = 1.8 T_C + 32$$
• Rankine ($^\circ\text{R}$): The absolute thermodynamic scale corresponding to Fahrenheit: $$T_R = T_F + 459.67 = 1.8 T_K$$

Temperature Difference Conversion Note

When calculating thermal expansion, thermal strain, or heat transfer, calculations depend on a temperature increment ($\Delta T$), where:

$$\Delta T \text{ (in } ^\circ\text{C}) = \Delta T \text{ (in K)}$$ $$\Delta T \text{ (in } ^\circ\text{F}) = \Delta T \text{ (in } ^\circ\text{R}) = 1.8 \cdot \Delta T \text{ (in } ^\circ\text{C})$$

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2. Linear & Volumetric Thermal Expansion

At the atomic scale, atoms vibrate within interatomic potential wells (the Lennard-Jones or Morse potential). Because the potential energy curve is anharmonic (steeper at close repulsive distances than at extended attractive distances), increasing thermal kinetic energy increases the average equilibrium distance between atomic nuclei, manifesting macroscopically as thermal expansion.

  Potential Energy (U)
      ^
      |      /
      |     /  <-- Steeper repulsive wall
      |    /
    0 +---/-----------------------------------> Interatomic Distance (r)
      |  /   \
      | /     \
      |/   r0  \   <-- Asymmetry causes average r to expand with T
      |---------

Mathematical Formulations

• Linear Thermal Expansion: $$\Delta L = L_0 \cdot \alpha \cdot \Delta T \implies L = L_0 (1 + \alpha \cdot \Delta T)$$ Where $L_0$ is initial length, $\Delta L$ is change in length, $\Delta T$ is temperature change ($T - T_0$), and $\alpha$ is the mean coefficient of linear thermal expansion ($\text{K}^{-1}$ or $^\circ\text{C}^{-1}$).
• Volumetric Thermal Expansion: For an isotropic polycrystalline solid: $$\Delta V = V_0 \cdot \beta \cdot \Delta T \implies \beta \approx 3\alpha$$ Where $\beta$ is the volumetric expansion coefficient.

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3. Thermophysical Properties of Structural Alloys

Differences in thermal properties among engineering alloys govern distortion, cooling rates, and residual stress states:

The Austenitic Stainless vs. Carbon Steel Disparity

Notice that austenitic stainless steel (AISI 304) has:

1. An expansion coefficient ($\alpha \approx 17.5 \times 10^{-6}/\text{K}$) that is $\approx 45\% - 50\%$ higher than carbon steel ($\approx 12.0 \times 10^{-6}/\text{K}$).
2. A thermal conductivity ($k \approx 15\text{ W}/[\text{m}\cdot\text{K}]$) that is only $\approx 30\%$ of carbon steel ($52\text{ W}/[\text{m}\cdot\text{K}]$).

This combination makes stainless steel extremely susceptible to severe weld distortion and high residual stresses: heat stays localized in the joint while the metal expands drastically.

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4. Thermal Strain, Restraint & Residual Stress Generation

When a structural element is subjected to a temperature rise $\Delta T$, its free thermal strain is:

$$\varepsilon_{\text{th}} = \alpha \cdot \Delta T$$

The Four Stages of Residual Stress Generation

1. Heating with Free Expansion ($R = 0$): If the metal is completely unrestrained, it expands freely. Mechanical strain is zero ($\varepsilon_{\text{mech}} = 0$), and no thermal stress develops ($\sigma = 0$).
2. Heating with Full Restraint ($R = 1.0$): If rigid boundary conditions prevent expansion, the total strain must be zero ($\varepsilon_{\text{total}} = \varepsilon_{\text{mech}} + \varepsilon_{\text{th}} = 0$). This forces a compressive mechanical strain: $$\varepsilon_{\text{mech}} = -\varepsilon_{\text{th}} = -\alpha \cdot \Delta T$$ The resulting elastic thermal stress is: $$\sigma_{\text{th}} = E \cdot \varepsilon_{\text{mech}} = -E \cdot \alpha \cdot \Delta T$$
3. Plastic Upsetting (Elevated Temperature): As temperature increases beyond $300^\circ\text{C} - 500^\circ\text{C}$, the material's yield strength ($\sigma_y(T)$) drops precipitously. When the compressive thermal stress exceeds this reduced yield limit, the hot zone yields in compression. The hot metal flows plastically—a phenomenon termed plastic upsetting.
4. Cooling & Tensile Residual Stress Lock-In: Upon cooling back to ambient temperature, the plastically shortened hot zone attempts to contract by $\Delta L = L_0 \alpha \Delta T$. The rigid, cold surrounding parent plate restrains this contraction, pulling the cooled weldment into severe tension. Because the room-temperature yield strength is restored, the residual tensile stress in the weld fusion zone and immediate HAZ reaches the room-temperature yield strength ($\sigma_{\text{residual}} \approx \sigma_y$).

  Stress (σ)
    ^
 +σy|------------------------- Tensile Residual Stress upon Cooling
    |                        /
    |                       / (Elastic unloading)
  0 +----------------------/-------------------------> Temperature (T)
    |  \                  /
    |   \ Compressive    /
    |    \ Plastic      /
 -σy|     \ Upsetting  /
    v      -----------

Degree of Restraint Factor ($R$)

In practical weldments, restraint is rarely zero or 100% rigid. A dimensionless restraint factor $R \in [0, 1.0]$ characterizes the joint rigidity:

$$\sigma = R \cdot E \cdot \alpha \cdot \Delta T$$

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5. Specific Heat Capacity & Latent Heat Transformations

Sensible Heat ($Q_{\text{sensible}}$)

Specific heat capacity ($C_p$) is the quantity of thermal energy required to raise the temperature of a unit mass of material by one degree:

$$Q_{\text{sensible}} = \int_{T_1}^{T_2} m \cdot C_p(T) \cdot dT \approx m \cdot \bar{C}_p \cdot \Delta T$$

Volumetric Heat Capacity and Thermal Diffusivity

In Rosenthal's classic weld heat conduction analysis, the cooling rate depends on the thermal diffusivity ($a$):

$$a = \frac{k}{\rho \cdot C_p} \quad (\text{units: } \text{m}^2/\text{s})$$

Where $k$ is thermal conductivity, $\rho$ is mass density, and $\rho C_p$ is volumetric heat capacity ($\text{J}/[\text{m}^3\cdot\text{K}]$). High thermal diffusivity materials (e.g., aluminum, $a \approx 7 \times 10^{-5}\text{ m}^2/\text{s}$) dissipate heat rapidly, requiring higher preheats and high-amperage heat sources, whereas low diffusivity materials (e.g., titanium and stainless steel, $a \approx 4 \times 10^{-6}\text{ m}^2/\text{s}$) retain heat, expanding HAZ dimensions.

Latent Heat of Fusion ($\Delta H_f$) and Vaporization ($\Delta H_v$)

Melting and vaporization require breaking interatomic bonds without changing temperature:

• Latent Heat of Fusion ($\Delta H_f$): Thermal energy absorbed at the melting point to transform solid crystal lattice into amorphous liquid. For structural steel, $\Delta H_f \approx 270\text{ kJ/kg}$.
• Total Heat of Melting ($Q_{\text{melt}}$): $$Q_{\text{melt}} = m \left[ \bar{C}_{p,\text{solid}} (T_m - T_0) + \Delta H_f + \bar{C}_{p,\text{liquid}} \Delta T_{\text{superheat}} \right]$$
• Latent Heat of Vaporization ($\Delta H_v$) & Element Vaporization: In high-intensity arc columns and laser/electron beam keyholes, temperatures exceed boiling points ($\approx 2,860^\circ\text{C}$ for iron). Highly volatile alloying elements with lower boiling points preferentially vaporize (Rayleigh distillation):
  • Zinc in brass and galvanized steels ($T_{\text{boil}} = 907^\circ\text{C}$) — explodes into violent vapor plumes, causing weld porosity.
  • Magnesium in 5xxx series aluminum alloys ($T_{\text{boil}} = 1,090^\circ\text{C}$) — burns out of the pool, reducing as-welded tensile strength.
  • Manganese in structural carbon steels ($T_{\text{boil}} = 2,061^\circ\text{C}$) — vaporizes into hazardous welding fume.

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6. Worked Numerical Examples

Problem 1: Thermal Stress in a Restrained Pipe Joint

A schedule 80 seamless carbon steel pipe spool ($E = 205\text{ GPa} = 205,000\text{ MPa}$, $\alpha = 12.2 \times 10^{-6}/\text{K}$, specified minimum yield strength $\sigma_y = 350\text{ MPa}$) is welded between rigid immovable bulkhead anchors. The pipeline is tied in at an ambient temperature $T_1 = 15^\circ\text{C}$. During service with hot hydrocarbon fluids, the pipe wall reaches $T_2 = 165^\circ\text{C}$ ($\Delta T = 150^\circ\text{C}$). The rigidity of the anchor foundations yields an effective joint restraint factor $R = 0.85$.

Calculate:

1. The unconstrained free thermal expansion ($\Delta L_{\text{free}}$) of a $6.0\text{ m}$ pipe section.
2. The restrained compressive thermal stress ($\sigma_{\text{th}}$) developed in the pipe.
3. Determine whether the pipe yields in compression or remains elastic.
4. The maximum permissible temperature increase ($\Delta T_{\text{max}}$) before the onset of compressive yielding.

Solution:

Step 1: Free thermal expansion ($\Delta L_{\text{free}}$)

$$\Delta L_{\text{free}} = L_0 \cdot \alpha \cdot \Delta T = 6.0\text{ m} \times (12.2 \times 10^{-6}/\text{K}) \times 150\text{ K} = 6.0 \times 0.00183\text{ m} = 0.01098\text{ m} = 10.98\text{ mm}$$

Step 2: Restrained thermal stress ($\sigma_{\text{th}}$)

$$\sigma_{\text{th}} = R \cdot E \cdot \alpha \cdot \Delta T$$ $$\sigma_{\text{th}} = 0.85 \times 205,000\text{ MPa} \times (12.2 \times 10^{-6}/\text{K}) \times 150\text{ K}$$ $$\sigma_{\text{th}} = 0.85 \times 205,000 \times 0.00183 = 0.85 \times 375.15\text{ MPa} \approx 318.88\text{ MPa}$$

Step 3: Yield determination Comparing with the yield strength:

$$\sigma_{\text{th}} = 318.9\text{ MPa} < \sigma_y = 350.0\text{ MPa}$$

The thermal stress remains elastic with a safety margin of $350.0 - 318.9 = 31.1\text{ MPa}$.

Step 4: Maximum allowable $\Delta T$ before yielding

$$\sigma_y = R \cdot E \cdot \alpha \cdot \Delta T_{\text{max}} \implies \Delta T_{\text{max}} = \frac{\sigma_y}{R \cdot E \cdot \alpha}$$ $$\Delta T_{\text{max}} = \frac{350\text{ MPa}}{0.85 \times 205,000\text{ MPa} \times 12.2 \times 10^{-6}/\text{K}} = \frac{350}{2.12585} \approx 164.6^\circ\text{C}$$

The pipe can withstand a maximum temperature increase of $164.6^\circ\text{C}$ before yielding commences.

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Problem 2: Energy to Melt GMAW Electrode Wire

In an automated GMAW station, solid ER70S-6 wire ($1.2\text{ mm}$ diameter, density $\rho = 7,850\text{ kg/m}^3$) is delivered at a wire feed speed $WFS = 8.5\text{ m/min}$. The ambient wire temperature is $T_0 = 20^\circ\text{C}$, the melting temperature (liquidus) is $T_m = 1,530^\circ\text{C}$, and the droplet detachment superheat temperature is $T_{\text{drop}} = 1,650^\circ\text{C}$ ($\Delta T_{\text{super}} = 120^\circ\text{C}$).

Thermodynamic properties:

• Mean solid specific heat: $\bar{C}_{p,\text{solid}} = 650\text{ J}/(\text{kg}\cdot\text{K})$
• Latent heat of fusion: $\Delta H_f = 270\text{ kJ/kg} = 270,000\text{ J/kg}$
• Mean liquid specific heat: $\bar{C}_{p,\text{liquid}} = 820\text{ J}/(\text{kg}\cdot\text{K})$

Calculate:

1. The mass wire consumption rate ($\dot{m}$) in $\text{kg/h}$ and $\text{kg/s}$.
2. The specific enthalpy of melting ($h_{\text{melt}}$) in $\text{kJ/kg}$.
3. The minimum theoretical thermal power ($P_{\text{melt}}$) in $\text{kW}$ required purely to melt the electrode wire.

Solution:

Step 1: Mass wire feed rate ($\dot{m}$) Wire cross-sectional area:

$$A_{\text{wire}} = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.0012\text{ m})^2 = 1.131 \times 10^{-6}\text{ m}^2$$

Volumetric feed rate:

$$\dot{V} = A_{\text{wire}} \times \frac{WFS}{60} = 1.131 \times 10^{-6}\text{ m}^2 \times \frac{8.5\text{ m/min}}{60\text{ s/min}} = 1.602 \times 10^{-7}\text{ m}^3/\text{s}$$

Mass rate:

$$\dot{m} = \rho \cdot \dot{V} = 7,850\text{ kg/m}^3 \times 1.602 \times 10^{-7}\text{ m}^3/\text{s} = 1.258 \times 10^{-3}\text{ kg/s}$$

In kilograms per hour:

$$\dot{m} = 1.258 \times 10^{-3}\text{ kg/s} \times 3,600\text{ s/h} \approx 4.529\text{ kg/h}$$

Step 2: Specific enthalpy of melting ($h_{\text{melt}}$)

$$q_{\text{solid}} = \bar{C}_{p,\text{solid}} \cdot (T_m - T_0) = 650\text{ J}/(\text{kg}\cdot\text{K}) \times (1,530 - 20)\text{ K} = 650 \times 1,510 = 981,500\text{ J/kg} = 981.5\text{ kJ/kg}$$ $$q_{\text{fusion}} = \Delta H_f = 270.0\text{ kJ/kg}$$ $$q_{\text{superheat}} = \bar{C}_{p,\text{liquid}} \cdot \Delta T_{\text{super}} = 820\text{ J}/(\text{kg}\cdot\text{K}) \times 120\text{ K} = 98,400\text{ J/kg} = 98.4\text{ kJ/kg}$$ $$h_{\text{melt}} = q_{\text{solid}} + q_{\text{fusion}} + q_{\text{superheat}} = 981.5 + 270.0 + 98.4 = 1,349.9\text{ kJ/kg}$$

Step 3: Theoretical melting power ($P_{\text{melt}}$)

$$P_{\text{melt}} = \dot{m} \cdot h_{\text{melt}} = (1.258 \times 10^{-3}\text{ kg/s}) \times (1,349.9 \times 10^3\text{ J/kg}) \approx 1,698.2\text{ W} \approx 1.70\text{ kW}$$

The arc must transfer a minimum of $1.70\text{ kW}$ directly into the wire anode to maintain continuous wire droplet detachment.

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7. Real-World Engineering Scenario: Dissimilar Transition Joint Failure

In a petrochemical steam cracker, austenitic AISI 321 stainless steel tubes ($18\text{Cr}-10\text{Ni}-\text{Ti}$) were welded directly to $2.25\text{Cr}-1\text{Mo}$ ferritic steel superheater headers using an austenitic ER309L filler metal. The operating temperature cycled between $40^\circ\text{C}$ during shutdowns and $540^\circ\text{C}$ during production runs ($\Delta T = 500^\circ\text{C}$). After two years of service, circumferential cracks opened completely around the HAZ of the ferritic header adjacent to the fusion line.

Root Cause Analysis: The coefficient of thermal expansion of the austenitic weld metal was $\alpha_{\text{austenitic}} \approx 17.5 \times 10^{-6}/\text{K}$, whereas the ferritic base metal was $\alpha_{\text{ferritic}} \approx 13.0 \times 10^{-6}/\text{K}$. The differential expansion was:

$$\Delta \alpha = 17.5 \times 10^{-6} - 13.0 \times 10^{-6} = 4.5 \times 10^{-6}/\text{K}$$

Over the $\Delta T = 500^\circ\text{C}$ thermal cycle, this mismatch induced an interfacial cyclic thermal strain:

$$\Delta \varepsilon_{\text{thermal}} = \Delta \alpha \cdot \Delta T = (4.5 \times 10^{-6}/\text{K}) \times 500\text{ K} = 0.00225 \quad (0.225\%)$$

Because this cyclic strain was concentrated within an ultra-narrow planar boundary at the fusion line, it exceeded the elastic strain limit of the partially decarburized ferritic HAZ at $540^\circ\text{C}$, causing low-cycle thermal fatigue cracking. Furthermore, carbon migrated from the high-chemical-potential ferritic steel into the chromium-rich austenitic weld metal, creating a weak, carbon-depleted soft zone right at the fusion boundary.

Engineering Redesign: The joint was redesigned using an Inconel (nickel-base) filler metal (ERNiCr-3 / Alloy 82). Nickel-base alloys have an intermediate coefficient of expansion ($\alpha \approx 13.8 \times 10^{-6}/\text{K}$), virtually matching the ferritic steel and reducing interfacial shear strains by over $75\%$. Additionally, nickel alloys do not promote carbon migration, permanently solving the cracking problem.

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8. Common CWEng Exam Traps

1. Assuming Linear Stress Accumulation at High Temperatures: Candidates frequently calculate $\sigma = E \cdot \alpha \cdot \Delta T$ across a $1,000^\circ\text{C}$ welding cycle, obtaining fictitious stresses exceeding $2,500\text{ MPa}$. They forget that structural steel yields plastically above $350^\circ\text{C} - 450^\circ\text{C}$ because the material's yield strength drops toward zero near the melting point.
2. Neglecting the Austenite-to-Ferrite Expansion Disparity in Joint Design: Forgetting that austenitic stainless steel expands $\approx 50\%$ more than carbon steel is the most common cause of underestimating distortion and restraint cracking in dissimilar joints.
3. Ignoring Latent Heat in Melting Calculations: Treating melting solely as a sensible heat calculation ($Q = m C_p \Delta T$) misses the latent heat of fusion ($\Delta H_f = 270\text{ kJ/kg}$), which represents roughly $20\%$ of the total enthalpy required to melt steel.