1. Energy Balance to solve Convective Heat Transfer

Velocity Averaged Temperature, $T_m$

Mass Flow Rate, $\dot m$

$$ \dot m = \int_A \rho u dA\\ \dot m = \rho \bar u \pi r_0^2 $$

Velocity Averaged Temperature, $T_m$

$$ \dot m c_p T_m = \int_A \rho u c_p T dA\\ T_m = \frac{1}{\rho \bar u \pi r_0^2 c_p}\int_0^{r_0} \rho u T c_p (2\pi r dr)\\ T_m = \frac{2}{\bar ur_0^2}\int_0^{r_0}uTrdr $$

Heat Convection, $q_s$

$$ q_s = h(T_s-T_m) $$

Recall Integral form of Heat Transport Equation:

$$ \cancel{\frac{\partial}{\partial t}\int_\text{CV}\rho c_p T dV} = \cancel{\int_\text{CV} \dot{S}V dV} -\oint\text{CS}(\vec{q} \cdot \hat{n})dA - \oint_\text{CS}\rho c_p T (\vec{v}\cdot\hat{n}) dA $$

• The first term is cancelled as Steady State
• The second term is cancelled as NO Heat Generation

Therefore,

$$ \oint_\text{CS}(\vec{q} \cdot \hat{n})dA

\underbrace{- \oint_\text{CS}\rho c_p T (\vec{v}\cdot\hat{n}) dA}\text{definition of $T_m$} \\ \oint\text{CS}(\vec{q} \cdot \hat{n})dA

(\dot m c_p T_m)_2-(\dot m c_p T_m)_1 \\ q_s (\pi d) dx = \dot m c_p(T_m + dT_m)-\dot m c_p T_m\\ q_s (\pi d) dx = \dot m c_p dT_m \\ \frac{\partial T_m}{\partial x} = \frac{q_s (\pi d)}{\dot m c_p} $$

where $q_s = h(T_s-T_m)$, hence:

$$ \frac{\partial T_m}{\partial x} = \frac{\pi d h}{\dot m c_p}(T_s-T_m) $$

Isothermal Surface

Temperature, $T$

For an isothermal surface, $T_s=const.$

$$ \frac{\partial T_m}{\partial x}

\frac{\pi d h}{\dot m c_p}(T_s-T_m)\\ -\frac{\partial \Delta T}{\partial x} = \frac{\pi d h}{\dot m c_p}\Delta T, \qquad \Delta T = T_s-T_m\\ \int_{T_{in}}^{T_{out}}\frac{1}{\Delta T}\partial \Delta T

-\frac{\pi d}{\dot m c_p}\int_0^L h dx\\ \ln(\frac{\Delta T_{out}}{\Delta T_{in}}) = -\frac{\pi d L}{\dot m c_p}\bar h\\ \therefore \frac{\Delta T_{out}}{\Delta T_{in}} = \exp(- \frac{A}{\dot m c_p}\bar h) $$

Convective Heat Flux, $q$

From Energy Balance:

$$ q_{conv} = \dot m c_p (T_{m, out} - T_{m, in}) = \dot m c_p [(T_s - T_{m, out}) - (T_s - T_{m, in})] \\ q_{conv} = \dot m c_p (\Delta T_{in} - \Delta T_{out}) $$