## Conduction
Thermal heat transfer through contact between objects.
#### Fourier's Law
$q''_{x}=\frac{q_{x}}{A} = -k \frac{ \partial T }{ \partial x }$
$k$ = thermal conductivity $\left[ \mathrm{\frac{W}{mK}} \right]$
$q''$ = heat flux $\left[ \mathrm{\frac{W}{m^2}} \right]$
or more generally,
$\vec{q}'' = -k\nabla T$
## Convection
Mode of heat transfer due to conduction / thermal diffusion plus bulk motion of fluid.
$\mathrm{Nu} = \frac{hD}{k}=f(\mathrm{Re}, \mathrm{\mathrm{Pr}})$ $\mathrm{Pr} = \frac{\nu}{\alpha}=\frac{\text{kinematic viscosity}}{\text{thermal diffusivity}}$
1. Calculate the Reynolds number ($\frac{\rho vL}{v}$) and Prandtl number ($\frac{\nu}{\alpha}=\frac{c_{p}\mu}{k}$)
2. Choose the right correlation and calculate Nusselt number
3. Calculate $h$ from Nusselt number: $h= \frac{\mathrm{Nu}\cdot k}{d}$
4. $\dot{q}''=h\Delta T$ or $\dot{Q}=hA\Delta T$
### Newton's Law of Cooling
$\dot{Q} = hA\Delta T$
$q''_{s} = h(T_{s} - T_{\infty})$
$q =$ heat transfer $\mathrm{W}$
$q =$ heat flux $\mathrm{W/m^{2}}$
$h =$ heat transfer coefficient $\mathrm{\frac{W}{m^{2}K}}$
### External Flow
#### Thermal Boundary Layer
Region in which you have thermal gradients between $T_{s}$ and $T_{\infty}$
![[Pasted image 20250316225128.png|350]]
#### Total Heat Transfer Rate
$q = \bar{h} A_{s} (T_{s} - T_{\infty})$
where average convection coefficient $\bar{h}$ is defined as
$\bar{h} = \frac{1}{A_{s}} \int _{A_{s}} h \, dA_{s}$
which for a flat plate reduces to
$\bar{h} = \frac{1}{L} \int_{0}^{L} h \, dx$
#### Empirical Correlation
With dimensional analysis and experimentation the following relationship has been shown:
$\overline{Nu_{L}} = C \mathrm{Re}_{L}^{m}\mathrm{Pr}^{n}$
$C$, $m$, and $n$, are typically independent of the nature of the fluid and instead vary with surface geometry and type of flow.
###### Flow over Isothermal Plate
Assuming steady, incompressible, laminar flow with constant fluid properties and negigible viscous dissipation and $\frac{dp}{dx}=0$
| Flow Type | Condition | Correlation |
| --------- | -------------------------------------------------------------------------------------- | ------------------------------------------------------------------------- |
| Laminar | $\mathrm{Pr} \geq 0.6$ | $\overline{Nu}_{x} = 0.664 \mathrm{Re}_{x}^{1/2} \mathrm{Pr}^{1/3}$ |
| Turbulent | $\mathrm{Pr} \geq 0.6$ | $\overline{Nu}_{x} = 0.0296 \mathrm{Re}_{x}^{4/5} \mathrm{Pr}^{1/3}$ |
| Mixed | $0.6 \leq \mathrm{Pr} \leq 60
lt;br>$\mathrm{Re}_{x,c} \leq \mathrm{Re}_{L} \leq 10^{8}$ | $\overline{Nu}_{x} = (0.037 \mathrm{Re}_{x}^{4/5} - A) \mathrm{Pr}^{1/3}$ |
### Natural Convection
No forced / defined bulk flow velocity, so we cannot use Reynold's number
Instead use Rayleigh number:
$\mathrm{Ra} = \mathrm{Gr}\times \mathrm{Pr}=\frac{\text{Buoyancy}}{\text{viscosity}}\times\frac{\text{momentum diffusivity}}{\text{thermal diffusivity}}$
$Nu = f(\mathrm{Ra},\mathrm{Pr})$
### Internal Flows
Laminar, fully developed flow → $\mathrm{Nu}= \begin{cases}\text{constant heat flux } \mathrm{ q''}: & 4.36\\ \text{constant wall }T_{s}: & 3.66\end{cases}$
$\Delta T = T_{s}-T_{b}$, where $T_{b}$ is a chosen "bulk mean temperature" analogous to $v_{\text{avg}}$
$\dot{Q} = hA\Delta T_{mean}$
where $\Delta T_{mean}$ is the mean of the temperature difference which logarithmically decays across the length of the pipe
$\Delta T_{mean} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln(\Delta T_{1})-\ln\Delta (T_{2})}$
![[Pasted image 20250516235942.png]]
![[Pasted image 20250517000532.png]]
###### Fully developed flow through a circular tube
| Flow Type | Condition | Correlation |
| ----------------------- | ----------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------- |
| Turbulent (Internal) | $0.7 \leq \mathrm{Pr} \leq 160lt;br>$\mathrm{Re}_D > 10,\!000lt;br>$L/D \geq 10$ | $\mathrm{Nu}_D = 0.023 \mathrm{Re}_D^{4/5} \mathrm{Pr}^{n}lt;br>$n = 0.3$ for cooling, $n = 0.4$ for heating |
| Turbulent (Sieder-Tate) | $0.7 \leq \mathrm{Pr} \leq 16,\!700lt;br>$\mathrm{Re}_D > 10,\!000lt;br>$L/D \geq 10$ | $\mathrm{Nu}_D = 0.027 \mathrm{Re}_D^{4/5} \mathrm{Pr}^{1/3} \left( \frac{\mu}{\mu_s} \right)^{0.14}$ |
| Turbulent (Gnielinski) | $0.5 \leq \mathrm{Pr} \leq 2000lt;br>$3,\!000 \leq \mathrm{Re}_D \leq 5 \times 10^6$ | $\mathrm{Nu}_D = \frac{(f/8)(\mathrm{Re}_D - 1000)\mathrm{Pr}}{1 + 12.7(f/8)^{1/2}(\mathrm{Pr}^{2/3} - 1)}$ |
## Fins
### Heat Transfer from a Fin
**Maximum possible heat transfer** (if the entire fin were at base temperature):
$Q_{\text{max}} = h A_f (T_b - T_\infty)$
**Actual heat transfer** depends on geometry and tip condition.
For example, for an adiabatic tip:
$Q_f = \sqrt{hPkA_c} (T_b - T_\infty) \tanh(mL)$
**Fin efficiency** (how well a fin performs compared to ideal):
$\eta_f = \frac{Q_{\text{actual}}}{Q_{\text{max}}}$
### Fin Resistance
- Fin resistance (includes efficiency):
$R_f = \frac{1}{\eta_f h A_f}$
### Fin Parameter and Definitions
Fin parameter:
$m = \sqrt{\frac{hP}{kA_c}}$
- Fin surface area: $A_f$
Fin perimeter: $P$
Cross-sectional area: $A_c$
- Modified fin heat transfer term:
$M = h P k A_c (T_b - T_\infty)$
### Efficiency Expressions (Special Cases)
- Infinite fin:
$\eta_f = \frac{1}{mL}$
- Adiabatic tip:
$\eta_f = \frac{\tanh(mL)}{mL}$
### Governing Equation and Solution
- 1D conduction + convection (steady state):
$\frac{d^2 \theta}{dx^2} - m^2 \theta = 0$, where $\theta(x) = T(x) - T_\infty$
- General solution:
$\theta(x) = A e^{mx} + B e^{-mx}$
- With adiabatic tip boundary condition:
$\frac{d\theta}{dx}\big|_{x=L} = 0$
- With prescribed tip temperature:
$\theta(L) = \theta_L$
### Heat Transfer Rate (Adiabatic Tip)
- Heat conducted at base:
$Q_f = \sqrt{hPkA_c} (T_b - T_\infty) \tanh(mL)$
- Can also be written:
$Q_f = M \tanh(mL)$
#### Fin Tip Conditions
| Tip Condition | Temperature Distribution $\theta(x)/\theta_b$ | Fin Heat Transfer Rate $Q_f$ |
|------------------------|---------------------------------------------------------------------------|--------------------------------------------|
| Infinite fin | $\theta(x)/\theta_b = e^{-mx}$ | $Q_f = M$ |
| Adiabatic tip | $\theta(x)/\theta_b = \frac{\cosh(m(L-x))}{\cosh(mL)}$ | $Q_f = M \tanh(mL)$ |
| Prescribed temp at tip | $\theta(x)/\theta_b = \frac{\sinh(mL) + \sinh(m(L-x))}{\sinh(mL)}$ | $Q_f = M \left(\frac{\cosh(mL) - 1}{\sinh(mL)}\right)$ |
Where:
- $\theta(x) = T(x) - T_\infty$
- $\theta_b = T_b - T_\infty$
- $m = \sqrt{\frac{hP}{kA_c}}$
- $M = \sqrt{hPkA_c} (T_b - T_\infty)$
## Heat Exchangers
Chapter 11, [[incropera.pdf]]
### Heat Exchanger Variants
A heat exchanger transfers heat between two fluids without mixing them
Parallel-flow vs counterflow
![[Pasted image 20250414112154.png|300]] ![[Pasted image 20250414112225.png|325]]
Crossflow:
![[Pasted image 20250414112442.png|625]]
### 1st Law
$\dot{Q}_{H} = (\dot{m}c_{p})_{H}\Delta T_{H}$
$\dot{Q}_{C}=(\dot{m}c_{p})_{C}\Delta T_{C}$
$\dot{Q}_{H}=\dot{Q}_{C}$
### LMTD Method
- all four inlet and outlet temps are known
- you want to find required heat exchanger area
$\dot{Q} = UA \Delta T_{\text{LM}}$
$U =$ overall heat transfer coeff
$A =$ heat transfer surface area
${} \Delta T_{LM} =$ log mean temperature difference
$\Delta T_{LM} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln(\Delta T_{1}/\Delta T_{2})}$
| Type | $\Delta T_{1}$ | $\Delta T_{2}$ |
| ------------- | ----------------------- | ----------------------- |
| Parallel Flow | ${} T_{h,i}-T_{c,i} {}$ | ${} T_{h,o}-T_{c,o} {}$ |
| Counterflow | ${} T_{h,i}-T_{c,o} {}$ | $T_{h,o}-T_{c,i}$ |
For crossflow and shell-and-tube, use a correction factor $F$ from charts:
$\Delta T_{LM,eff} = F \cdot \Delta T_{LM}$
### Effectiveness-NTU Method
- Heat exchanger effectiveness is known
- Missing some fluid temperatures
- You want to solve for required area or UA
1. Find heat capacity rates $C~\mathrm{[W/K]}$
$C = \dot{m}c_{p}$, $C_\text{min}=\text{min}(C_{c},C_{h})$, $C_\text{max}=\text{max}(C_{c},C_{h})$, $C_{r}=\frac{C_\text{min}}{C_\text{max}}$
2. Determine $q_\text{max}$
$q_{\text{max}} = C_\text{min}(T_{h,i}-T_{c,i})$
3. Determine effectiveness $\epsilon$ from charts
$\varepsilon = \frac{\dot{Q}_\text{actual}}{\dot{Q}_{\text{max}}} = f\left( \mathrm{NTU}, \frac{C_\text{min}}{C_\text{max}} \right)$
4. Determine ${} Q_{\text{actual}}$ from effectiveness
$Q_{\text{actual}}=\varepsilon Q_{\text{max}}$
5. Use $\varepsilon$-NTU relations to determine Number of Transfer Units
$\mathrm{NTU} = \frac{UA}{C_\text{min}}$
6. Solve for UA or required area
High NTU = high heat transfer for a given minimum heat capacity rate
![[Pasted image 20250414121113.png]]
| Heat Exchanger Type | $\varepsilon$ Formula | NTU as a function of $\varepsilon$ |
| ------------------------------------------------------------ | ------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------- |
| Counterflow | $\varepsilon = \frac{1 - e^{-\text{NTU}(1 - C_r)}}{1 - C_r e^{-\text{NTU}(1 - C_r)}}$ | ${} \mathrm{NTU} = \frac{1}{C_{r}-1}\ln\left( \frac{\varepsilon-1}{\varepsilon C_{r}-1} \right) {}$ |
| Parallel flow | $\varepsilon = \frac{1 - e^{-\text{NTU}(1 + C_r)}}{1 + C_r}$ | $\text{NTU} = -\frac{\ln(1 - \varepsilon (1 + C_r))}{1 + C_r}$ |
| Crossflow:<br>$C_\text{max}$ mixed<br>$C_\text{min}$ unmixed | ${} \varepsilon = \frac{1}{C_{r}}(1-e^{-C_{r}(1-e^{-C_{r}(\mathrm{NTU})})}) {}$ | ${} \mathrm{NTU} = -\ln\left( 1 + \frac{1}{C_{r}}\ln(1-\varepsilon C_{r}) \right) {}$ |
| Crossflow:<br>$C_\text{max}$ unmixed<br>$C_\text{min}$ mixed | ${} \varepsilon = 1-e^{-\mathrm{NTU}} {}$ | $Q\mathrm{NTU}=-\frac{1}{C_{r}}\ln(C_{r}\ln(1-\varepsilon)+1)$ |
![[Pasted image 20250421183147.png]]