Heat Transfer Essentials

Cheatsheet Content

Introduction to Heat Transfer Definition: Flow of heat (thermal energy) due to temperature differences. Always from high to low temperature until thermal equilibrium. Modes: Conduction Convection Radiation Modes of Heat Transfer 1. Conduction Definition: Transmission of heat from higher kinetic energy particles to lower kinetic energy particles through direct contact. Examples: Ironing clothes, melting ice cube in hands. 2. Convection Definition: Heat transfer due to bulk flow of a fluid (gas or liquid) carrying heat. Examples: Cooling warm water (natural convection), cooling human body with a fan (forced convection). 3. Radiation Definition: Transferring heat energy from a hotter body to a colder body through electromagnetic waves (photons). Occurs through vacuum or any transparent medium. Examples: Microwave radiation, UV rays from the sun. Units of Heat Transfer Heat (Q): Amount of heat transfer. SI unit: Joule (J). Heat Transfer Rate ($\dot{Q}$): Amount of heat transfer per unit time. Measured in J/s or Watt (W). Heat Flux ($q$): Rate of heat transfer per unit area normal to the direction of heat transfer. $$q = \frac{\dot{Q}}{A} \quad \text{(W/m}^2\text{)}$$ British Thermal Unit (Btu): Heat required to raise 1 lb of liquid water by $1^\circ \text{F}$. $1 \text{ Btu} = 1.055 \text{ kJ}$. Calorie (cal): Energy needed to raise 1 gram of water by $1^\circ \text{C}$. $1 \text{ cal} = 4.2 \text{ J}$. Fourier's Law of Heat Conduction Description: Rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, and inversely proportional to the thickness of the layer. $$\dot{Q}_{cond} = -k \cdot A \frac{\Delta T}{\Delta x} \quad \text{(W)}$$ Where: $k$: Thermal conductivity of the material. $A$: Heat transfer area. $\Delta T$: Temperature difference. $\Delta x$: Thickness of the layer. Differential Form: $$q = -k \cdot \nabla T$$ Where: $q$: Local heat flux density (W/m$^2$). $k$: Thermal conductivity (W/(m$\cdot$K)). $\nabla T$: Temperature gradient (K/m). Thermal Conductivity ($k$): Material's ability to conduct heat. High $k$ = good conductor, low $k$ = insulator. Thermal Resistance ($R_{th}$) Analogy with electrical resistance: $$\text{Current (I)} = \frac{\text{Potential difference (dV)}}{\text{Electrical resistance (R)}}$$ $$\text{Heat flow rate } (\dot{Q}) = \frac{\text{Temperature difference (dT)}}{dx/(k \cdot A)}$$ Thermal conduction resistance ($R_{th,cond}$) is $\frac{\Delta x}{k \cdot A}$. Reciprocal of thermal resistance is thermal conductance. Rules for combining electrical resistances (series/parallel) apply to thermal resistances. Heat Transfer by Convection (Newton's Law of Cooling) Description: Rate of convective heat transfer between a surface and an adjacent fluid in motion. $$\dot{Q}_{conv} = h \cdot A \cdot (T_s - T_f) \quad \text{(W)}$$ Where: $h$: Convective heat transfer coefficient (W/m$^2 \cdot$K). $A$: Surface area exposed to heat transfer. $T_s$: Surface temperature. $T_f$: Fluid temperature. Heat Transfer by Radiation (Stefan-Boltzmann Law) Description: Radiation emitted by a black body (at absolute temperature) per unit area is directly proportional to the fourth power of the temperature. $$\dot{Q}_{emit} = \sigma \cdot A \cdot T^4 \quad \text{(W)}$$ Where: $\sigma$: Stefan-Boltzmann constant ($5.67 \times 10^{-8} \text{ W/(m}^2 \cdot \text{K}^4)$). $A$: Surface area emitting radiation (m$^2$). $T$: Surface temperature (K). Real Surfaces: Radiation emitted by real surfaces is less than that of a blackbody. $$\dot{Q}_{emit} = \varepsilon \cdot \sigma \cdot A \cdot T^4 \quad \text{(W)}$$ Where $\varepsilon$ is the emissivity of the surface ($0 \le \varepsilon \le 1$). Radiation between two bodies: $$\dot{Q}_{rad} = \varepsilon \cdot \sigma \cdot A \cdot (T_1^4 - T_2^4) \quad \text{(W)}$$ Heat Transfer Through Composite Wall For a composite wall with slabs A, B, C in series: $$\dot{Q} = \frac{k_A \cdot A \cdot (t_1 - t_2)}{L_A} = \frac{k_B \cdot A \cdot (t_2 - t_3)}{L_B} = \frac{k_C \cdot A \cdot (t_3 - t_4)}{L_C}$$ Where $L$ are thicknesses, $k$ are thermal conductivities, $A$ is area, and $t$ are temperatures. Thermal resistances in series: $$R_{th,A} = \frac{L_A}{k_A \cdot A}, \quad R_{th,B} = \frac{L_B}{k_B \cdot A}, \quad R_{th,C} = \frac{L_C}{k_C \cdot A}$$ Total resistance $R_{total} = R_{th,A} + R_{th,B} + R_{th,C}$. $$\dot{Q} = \frac{T_{overall,in} - T_{overall,out}}{R_{total}}$$ Heat Exchangers Definition: Devices for efficient heat transfer between two fluids without mixing. Used in HVAC, power generation, chemical processing. Consist of hot and cold fluid circuits separated by a conductive barrier. Classification of Heat Exchangers 1. Design and Construction Shell and Tube: One fluid flows through tubes, the other around them in a shell. Versatile and robust. Plate and Frame: Stacked plates with alternating fluid channels. Large surface area, compact, efficient. 2. Relative Direction of Fluid Motion Parallel Flow: Both fluids enter from the same end, flow parallel, and exit at the other end. Temperature difference decreases from inlet to outlet. Counter Flow: Fluids enter from opposite ends, flow in opposite directions. Temperature difference remains nearly constant. Cross Flow: One fluid flows across the other, resulting in a perpendicular flow pattern. Heat Exchanger Analysis Parameters $U$: Overall heat transfer coefficient. $A$: Total surface area of heat transfer. $t_1, t_2$: Inlet and outlet fluid temperatures. Logarithmic Mean Temperature Difference (LMTD) Used for calculating mean temperature difference in heat exchangers. $$\theta_m = \frac{\theta_1 - \theta_2}{\ln(\theta_1/\theta_2)}$$ Overall Energy Balance: $$\dot{Q} = \dot{m}_h C_{ph} (T_{h1} - T_{h2}) = \dot{m}_c C_{pc} (T_{c2} - T_{c1})$$ $$\dot{Q} = U A \theta_m$$ Where: $\dot{m}$: Mass flow rate. $C_p$: Specific heat at constant pressure. $U$: Overall heat transfer coefficient. Effectiveness ($\varepsilon$) of Heat Exchanger Ratio of actual heat transfer rate to the maximum possible heat transfer rate. $$\varepsilon = \frac{\text{Actual heat transfer}}{\text{Maximum possible heat transfer}} = \frac{\dot{Q}}{\dot{Q}_{max}}$$ Fluid Capacity Rate ($C$): $$C_h = \dot{m}_h C_{ph} \quad (\text{Hot fluid})$$ $$C_c = \dot{m}_c C_{pc} \quad (\text{Cold fluid})$$ $C_{min}$ is the minimum of $C_h$ and $C_c$. $C_{max}$ is the maximum of $C_h$ and $C_c$. Maximum Possible Heat Transfer: $$\dot{Q}_{max} = C_{min} (T_{h,in} - T_{c,in})$$ Actual Heat Transfer using Effectiveness: $$\dot{Q} = \varepsilon C_{min} (T_{h,in} - T_{c,in})$$ Fouling Definition: Deposition of ash, dirt, scale, and fluid impurities on heat exchanger tube surfaces during operation, leading to reduced heat transfer efficiency. Numerical Examples (Practice Problems) 1. Roof Heat Loss Calculation Roof dimensions: 6m long, 8m wide, 0.25m thick. Material: Concrete, $k = 0.8 \text{ W/(m}\cdot\text{K)}$. Temperatures: Inner $15^\circ \text{C}$, Outer $4^\circ \text{C}$. Calculate: Rate of heat loss through the roof. Cost of heat lost for 10 hours if electricity costs Rs 5/kWh. 2. Carbon Steel Plate Heat Transfer Plate: $600\text{mm} \times 900\text{mm} \times 25\text{mm}$. $k = 45 \text{ W/(m}\cdot^\circ\text{C)}$. Temperature: Maintained at $310^\circ \text{C}$. Air at $15^\circ \text{C}$ blows over. Convection coefficient: $h = 22 \text{ W/(m}^2\cdot^\circ\text{C)}$. Radiation heat loss: $250 \text{ W}$. Calculate: Inside plate temperature. (Answer: $313.86^\circ \text{C}$). 3. Surface Heat Transfer to Surroundings Surface temperature: $250^\circ \text{C}$. Surroundings: $110^\circ \text{C}$. Convection coefficient: $h = 75 \text{ W/(m}^2\cdot^\circ\text{C)}$. Emissivity: $\varepsilon = 1$. Heat conducted to surface through solid: $k = 10 \text{ W/(m}\cdot^\circ\text{C)}$. Calculate: Temperature gradient at the surface in the solid. (Answer: $dt/dx = -1352.21^\circ \text{C/m}$). 4. Composite Wall Interface Temperature Two sections with equal wall thickness ($L$). Heat lost from $600\text{K}$ to ambient $300\text{K}$. Conductivity: Section 1 is $k$, Section 2 is $2k$. Calculate: Interface temperature ($T_1$). (Answer: $T_1 = 400\text{K}$). 5. Reactor Wall Heat Loss Wall: $320\text{mm}$ thick. Inner layer fire brick ($k = 0.84 \text{ W/(m}\cdot^\circ\text{C)}$), outer layer insulation ($k = 0.16 \text{ W/(m}\cdot^\circ\text{C)}$). Reactor operates at $1325^\circ \text{C}$, ambient $25^\circ \text{C}$. Calculate: Thickness of fire brick and insulation for minimum heat loss. Heat loss assuming insulation max temp is $1200^\circ \text{C}$. 6. Exterior Wall Insulation Wall: $0.1\text{m}$ common brick ($k = 0.7 \text{ W/(m}\cdot^\circ\text{C)}$), $0.04\text{m}$ gypsum plaster ($k = 0.48 \text{ W/(m}\cdot^\circ\text{C)}$). Calculate: Thickness of rock wool insulation ($k = 0.065 \text{ W/(m}\cdot^\circ\text{C)}$) to reduce heat loss/gain by 80%. (Answer: $58.8\text{mm}$). 7. Composite Wall Heat Flow Rate Given composite wall with specific dimensions and thermal conductivities: $k_A = 150 \text{ W/(m}\cdot^\circ\text{C)}$, $k_B = 30 \text{ W/(m}\cdot^\circ\text{C)}$, $k_C = 65 \text{ W/(m}\cdot^\circ\text{C)}$, $k_D = 50 \text{ W/(m}\cdot^\circ\text{C)}$. Assume one-dimensional flow. Calculate: Heat flow rate. (Answer: $1273.4 \text{ W}$). 8. Heat Exchanger Calculation Working fluid: enters $90^\circ \text{C}$, leaves $60^\circ \text{C}$. Cooling fluid: enters $15^\circ \text{C}$, leaves $30^\circ \text{C}$. Surface area: $50 \text{ m}^2$. Overall heat transfer coefficient: $100 \text{ W/(m}^2\cdot^\circ\text{C)}$. Calculate heat transfer rate for: Parallel flow. Counter flow.