Heat Transfer Cheatsheet

Cheatsheet Content

### Introduction to Heat Transfer - **Definition:** Heat transfer is the study of thermal energy in transit due to a temperature difference. - **Three Modes:** Conduction, Convection, Radiation. - **Importance:** Fundamental to countless engineering applications (power generation, HVAC, electronics cooling, biomedical). ### Fourier's Law of Conduction - **Definition:** Describes heat conduction in materials due to molecular activity. $$ \vec{q} = -k \nabla T $$ Where: - $\vec{q}$ is the **heat flux** ($W/m^2$) - rate of heat transfer per unit area. - $k$ is the **thermal conductivity** ($W/m \cdot K$) - material's ability to conduct heat. - $\nabla T$ is the **temperature gradient** ($K/m$) - change in temperature per unit length. - For one-dimensional heat transfer in the x-direction: $$ q_x = -k \frac{dT}{dx} $$ - **Heat Transfer Rate (Q):** $$ Q = q_x A = -kA \frac{dT}{dx} $$ Where $A$ is the cross-sectional area normal to heat flow. - **Key Concept:** The negative sign ensures $Q$ is positive when heat flows in the positive x-direction (i.e., from higher to lower temperature). - **Thermal Conductivity Values (room temperature):** | Material | $k \ [W/(m \cdot K)]$ | Category | |-----------------|-----------------------|--------------------| | Diamond | 2300 | Excellent conductor | | Copper | 387.6 | Good conductor | | Aluminium | 202.4 | Good conductor | | Iron | 80.2 | Moderate conductor | | Glass | 0.78 | Poor conductor | | Water, liquid | 0.61 | Poor conductor | | Wood, oak | 0.173 | Insulator | | Air | 0.026 | Super insulator | - **Physical Insight:** High $k$ means the material conducts heat well. Poor conductors (insulators) have low $k$. - **Insulation Materials:** Work by trapping still air (low $k$) and preventing convection within air pockets. #### Example: Steady-State Conduction in a Plane Wall A plane wall of thickness $L$, area $A$, and thermal conductivity $k$ has inner surface temperature $T_1$ and outer surface temperature $T_2$. $$ Q = kA \frac{T_1 - T_2}{L} $$ _This is derived from Fourier's Law for a constant temperature gradient across the wall._ ### Convection Heat Transfer - **Definition:** Heat transfer between a surface and a moving fluid. - **Newton's Law of Cooling:** $$ q = h (T_s - T_\infty) $$ Where: - $q$ is the **heat flux** ($W/m^2$) - $h$ is the **convection heat transfer coefficient** ($W/m^2 \cdot K$) - $T_s$ is the surface temperature - $T_\infty$ is the fluid temperature far from the surface - **Heat Transfer Rate (Q):** $$ Q = h A (T_s - T_\infty) $$ Where $A$ is the surface area. - **Types of Convection:** - **Forced Convection:** Fluid motion is caused by external means (e.g., fan, pump). - **Natural/Free Convection:** Fluid motion is caused by buoyancy forces resulting from density variations due to temperature differences. - **Typical Convection Coefficient Values:** | Process | $h \ [W/(m^2 \cdot K)]$ | |---------------------------|-------------------------| | Natural convection (air) | 2-25 | | Forced convection (air) | 25-250 | | Forced convection (water) | 100-20,000 | | Boiling water | 2,500-25,000 | | Condensing steam | 5,000-100,000 | _Water-cooled systems are more effective than air-cooled systems for high-performance electronics due to water's significantly higher $h$ values._ ### Radiation Heat Transfer - **Definition:** Heat transfer via electromagnetic waves; no medium required! - **Stefan-Boltzmann Law (Blackbody):** $$ q_b'' = \sigma T^4 $$ Where: - $q_b''$ is the blackbody emissive power ($W/m^2$) - $\sigma = 5.67 \times 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant - $T$ is the absolute temperature in Kelvin - **Emissivity ($\epsilon$):** Ratio of actual emission to blackbody emission ($0 \le \epsilon \le 1$). $$ q'' = \epsilon \sigma T^4 $$ - **Net Radiation Exchange (Surface to Surroundings):** $$ Q = \epsilon A \sigma (T_s^4 - T_{surr}^4) $$ Where $T_s$ is surface temperature and $T_{surr}$ is surroundings temperature, both in Kelvin. - For a small object in a large enclosure. - If $T_{surr} \approx T_\infty$, a linearized radiation coefficient $h_r$ can be used, $h_{combined} = h_{conv} + h_r$. #### Example: Comparing Radiative Heat Flux Two surfaces, one polished ($\epsilon_1=0.1$, $T_1=1070^\circ C$) and one oxidized ($\epsilon_2=0.78$), emit the same heat flux. Determine $T_2$. 1. **Convert Temperatures to Kelvin:** $T_1 = 1070 + 273 = 1343 \ K$ 2. **Equate Heat Fluxes:** $$ q_1'' = q_2'' \implies \epsilon_1 \sigma T_1^4 = \epsilon_2 \sigma T_2^4 $$ $$ T_2 = \left( \frac{\epsilon_1}{\epsilon_2} \right)^{1/4} T_1 $$ 3. **Calculate $T_2$:** $$ T_2 = \left( \frac{0.1}{0.78} \right)^{1/4} \times 1343 \ K \approx 803.62 \ K $$ 4. **Convert back to Celsius:** $$ T_2 = 803.62 - 273 = 530.62^\circ C $$ _The oxidized surface needs to be at a lower temperature to emit the same heat flux as the polished surface due to its higher emissivity._ ### Thermal Resistance - **Concept:** Analogous to electrical resistance ($V=IR$), heat transfer occurs due to a temperature difference across a thermal resistance: $$ Q = \frac{\Delta T}{R_{th}} $$ - **Conduction Resistance ($R_{cond}$):** - Plane Wall: $$ R_{cond} = \frac{L}{kA} $$ - Cylinder: $$ R_{cyl} = \frac{\ln(r_2/r_1)}{2\pi L k} $$ - Sphere: $$ R_{sph} = \frac{r_2 - r_1}{4\pi r_1 r_2 k} $$ - **Convection Resistance ($R_{conv}$):** $$ R_{conv} = \frac{1}{hA} $$ - **Radiation Resistance ($R_{rad}$):** $$ R_{rad} = \frac{1}{h_r A} $$ Where $h_r = \epsilon \sigma (T_s^2 + T_{surr}^2)(T_s + T_{surr})$. - **Contact Resistance ($R_{contact}$):** $$ R_{contact} = \frac{1}{h_c A} $$ Where $h_c$ is the thermal contact conductance. - **Key Insight:** Contact resistance is significant for good conductors (metals) but can be disregarded for poor conductors (insulators). #### Resistance Networks - **Series:** $R_{total} = R_1 + R_2 + ...$ (Heat flows through each component sequentially) - **Parallel:** $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$ or $R_{total} = \frac{R_1 R_2}{R_1 + R_2}$ (for two parallel resistances) (Heat divides to flow through components simultaneously) #### Example: Heat Transfer Through a Window A glass window (thickness $L=6mm$, $k=0.78 W/mK$) has inner and outer convection ($h_1=10 W/m^2K$, $h_2=25 W/m^2K$). Room temperature $T_{room}=24^\circ C$, outdoor temperature $T_{out}=-5^\circ C$. Window area $A = 1.5m \times 2.4m = 3.6m^2$. 1. **Resistances:** - Inner Convection: $R_{conv,i} = \frac{1}{h_1 A} = \frac{1}{10 \times 3.6} = 0.02778 \ K/W$ - Conduction through glass: $R_{glass} = \frac{L}{kA} = \frac{0.006}{0.78 \times 3.6} = 0.00214 \ K/W$ - Outer Convection: $R_{conv,o} = \frac{1}{h_2 A} = \frac{1}{25 \times 3.6} = 0.01111 \ K/W$ 2. **Total Resistance (Series):** $$ R_{total} = R_{conv,i} + R_{glass} + R_{conv,o} = 0.02778 + 0.00214 + 0.01111 = 0.04103 \ K/W $$ 3. **Steady Rate of Heat Transfer (Q):** $$ Q = \frac{T_{room} - T_{out}}{R_{total}} = \frac{24 - (-5)}{0.04103} = \frac{29}{0.04103} \approx 706.8 \ W $$ ### Energy Balance for Systems - **General Equation:** $$ \frac{\partial E_{system}}{\partial t} = \dot{E}_{in} - \dot{E}_{out} + \dot{E}_{gen} $$ Where: - $\frac{\partial E_{system}}{\partial t}$ is the rate of change of energy stored in the system. - $\dot{E}_{in}$, $\dot{E}_{out}$ are rates of energy transfer in/out (conduction, convection, radiation, mass flow). - $\dot{E}_{gen}$ is the rate of energy generation within the system. - **Simplifications:** - **Steady-State:** $\frac{\partial E_{system}}{\partial t} = 0$ - **No Generation:** $\dot{E}_{gen} = 0$ - **Surface Energy Balance:** For a surface, the sum of all incoming and outgoing heat fluxes must be zero at steady state. $$ \sum \dot{Q}_{in} = \sum \dot{Q}_{out} $$ #### Example: Internal Energy Generation in a Plane Wall A plane wall (thickness $L$, thermal conductivity $k$) has uniform internal energy generation $\dot{q}$ ($W/m^3$). The surface at $x=0$ is at temperature $T_0$ and experiences convection ($h$, $T_\infty$). The surface at $x=L$ is well insulated (adiabatic). 1. **Differential Equation:** For steady-state, one-dimensional conduction with internal heat generation: $$ \frac{d}{dx}\left( -k \frac{dT}{dx} \right) + \dot{q} = 0 \implies k \frac{d^2T}{dx^2} + \dot{q} = 0 $$ 2. **Boundary Conditions:** - At $x=0$ (convection): $-k \frac{dT}{dx}|_{x=0} = h(T(0) - T_\infty)$ - At $x=L$ (insulated): $\frac{dT}{dx}|_{x=L} = 0$ 3. **Temperature Distribution:** Integrating and applying BCs gives a parabolic temperature profile, with maximum temperature occurring at the insulated surface. _The overall energy balance states that total heat generated must be equal to heat transferred out of the system._ $$ \dot{E}_{gen} = \dot{Q}_{out, conv} \implies \dot{q} (A \cdot L) = h A (T_0 - T_\infty) $$ $$ \dot{q} L = h (T_0 - T_\infty) $$ ### Conduction in Cylinders - **Heat Transfer Rate for a Cylindrical Layer:** $$ Q = \frac{2\pi L k (T_1 - T_2)}{\ln(r_2/r_1)} $$ - **Conduction Resistance for a Cylindrical Layer:** $$ R_{cyl} = \frac{\ln(r_2/r_1)}{2\pi L k} $$ #### Example: Multilayered Cylinders Consider a pipe with an insulation layer. Heat transfer occurs from the inner fluid ($T_{\infty,1}, h_1$) through the pipe wall ($k_p$) and insulation ($k_{ins}$) to the outer fluid ($T_{\infty,2}, h_2$). The total resistance is a series combination: $$ R_{total} = R_{conv,1} + R_{pipe} + R_{ins} + R_{conv,2} $$ $$ R_{total} = \frac{1}{h_1 A_1} + \frac{\ln(r_2/r_1)}{2\pi L k_p} + \frac{\ln(r_3/r_2)}{2\pi L k_{ins}} + \frac{1}{h_2 A_3} $$ Where $A_1 = 2\pi r_1 L$ and $A_3 = 2\pi r_3 L$. The heat transfer rate is: $$ Q = \frac{T_{\infty,1} - T_{\infty,2}}{R_{total}} $$ ### Critical Radius of Insulation - **Concept:** For a plane wall, adding insulation always decreases heat transfer. For cylindrical or spherical geometries, adding insulation can initially *increase* heat transfer if the outer surface area increases significantly while resistance to conduction is still low. - **Critical Radius for a Cylinder ($r_{cr}$):** $$ r_{cr} = \frac{k}{h} $$ Where $k$ is the thermal conductivity of the insulation and $h$ is the convection coefficient at the outer surface. - **Interpretation:** - If initial radius $r_1 < r_{cr}$: Adding insulation *increases* heat transfer until $r_{insulation} = r_{cr}$.