# hw7

* Chris Nkinthorn, 20190716

**Prompt:** Derive the following expressions:

* Boundary Layer Height: $\delta(x)=\left\[\frac{4 x &#x6B;*{l}\left(T*{s a t}-&#x54;*{w}\right) v*{l}}{&#x68;*{f g} g\left(\rho*{l}-\rho\_{v}\right)}\right]^{1 / 4}$
* Heat Transfer Coefficient: $h(x)=\left\[\frac{&#x68;*{f g} g\left(\rho*{l}-\rh&#x6F;*{v}\right) k*{l}^{3}}{4 x\left(&#x54;*{s a t}-T*{w}\right) v\_{l}}\right]^{1 / 4}$&#x20;
* Average Heat Transfer Coefficient: $\overline{h}*{L}=0.943\left\[\frac{h*{f \beta} g\left(\rh&#x6F;*{l}-\rho*{v}\right) &#x6B;*{l}^{3}}{L\left(T*{s a t}-&#x54;*{w}\right) v*{l}}\right]^{1 / 4}$

## Solution

Begin with a conservation on momentum balance, with no pressure gradient applied, to find the velocity as a function of elevation $y$.

$$
\begin{align}
\mu \frac{d u}{d y} &=g \rho(\delta-y)-\frac{d p}{d x}(\delta-y)\\
0 &=g \rho\_{v}-\frac{d p}{d x}\rightarrow \frac{d p}{d x}=g \rho\_{v}\\
\mu \frac{d u}{d y}&=g\left(\rho-\rho\_{v}\right)(\delta-y) \rightarrow u=\left(\frac{g\left(\rho-\rho\_{v}\right)}{\mu}\right)\left(\delta y-\frac{y^{2}}{2}\right)\\
\end{align}
$$

Using this expression for the velocity, as a function of height and the pressure difference relative to vapor pressure

$$
\Gamma=\rho \int\_{0}^{\delta} u d y=
\rho \int\_0^\delta \left(\frac{g\left(\rho-\rho\_{v}\right)}{\mu}\right)\left(\delta y-\frac{y^{2}}{2}\right)dy=\frac{g \rho\left(\rho-\rho\_{v}\right)}{\mu} \frac{\delta^{3}}{3}\\
$$

Take derivative to find mass change per unit length:

$$
\Gamma = \frac{g \rho\left(\rho-\rho\_{v}\right)}{\mu} \frac{\delta^{3}}{3} \xrightarrow{d/d\delta} \frac{d \Gamma}{d \delta}=\frac{g \rho\left(\rho-\rho\_{v}\right)}{\mu} \delta^{2}
$$

Need an expression for the temperature variation along the boundary layer, which we assume from the wall boundary to saturation temperature in the fluid.

$$
T(y) =T\_{w}+\frac{T\_{s a t}-T\_{w}}{\delta} y
$$

The temperature distribution is important because we can relate temperature to energy. The transfer of energy in this convective mode is characterized by the heat transfer coefficient $h$. We take another average, across the temperature variation through the region of interest in $\delta$.

$$
h\_{a v}=h\_{f g}+\frac{1}{\Gamma} \int\_{0}^{\delta} \rho u c\left(T\_{s a t}-T\right) d y\\
\begin{equation}
\label{eqn:avgEnth}
\begin{split}
h\_{a v} &=h\_{f g}+\frac{\mu}{g \rho\left(\rho-\rho\_{v}\right)} \frac{3}{\delta^{3}} g \rho \frac{\left(\rho-\rho\_{v}\right)}{\mu} c\left(T\_{s a t}-T\_{w}\right) \int\_{0}^{\delta}\left\[\delta y-\frac{y^{2}}{2}\right]\left\[1-\frac{y}{\delta}\right] d y\\
&=h\_{f g}+\frac{3}{8} c\left(T\_{s a t}-T\_{w}\right)\\
\end{split}\
\end{equation}
$$

From the temperature variation, through the heat transfer coefficient, we have an expression for energy as a function of both temperature variation and position in the boundary layer. Algebra on the differential in linear position.

$$
\frac{q}{A} d x=k \frac{\Delta T}{\delta} d x=h\_{a v} d \Gamma \xrightarrow{1/dx} \frac{q}{A}=k \frac{\Delta T}{\delta}=h\_{a v} \frac{d \Gamma}{d x}=\left(h\_{f g}+\frac{3}{8} c \Delta T\right) \frac{d \Gamma}{d x}\\
d\Gamma = \frac{k \Delta T}{\delta\left(h\_{f g}+\frac{3}{8} c \Delta T\right)} d x
$$

**Part a:** Equate mass change to obtain film growth:

$$
\begin{align}
d \Gamma &=
\overbrace{\frac{g \rho\left(\rho-\rho\_{v}\right)}{\mu} \delta^{2} d \delta}^\texttt{From mass change} =
\overbrace{\frac{k \Delta T}{\delta\left(h\_{f g}+\frac{3}{8} c \Delta T\right)} d x}^\texttt{From energy transport}\\
&\boxed{\delta=\sqrt\[4]{\frac{4 k \mu \Delta T x}{g \rho\left(\rho-\rho\_{v}\right) h\_{f g}}}}
\end{align}
$$

Solving the equation, group differentials in mass and linear position on either side, to collect mass coefficients and energy transport together.

$$
\begin{align}
\delta^3 \cdot d\delta&= \left(\frac{g \rho\left(\rho-\rho\_{v}\right)}{\mu}\right)^{-1} \cdot \frac{k \Delta T}{h\_{f g}+\frac{3}{8} c \Delta T} d x\\
&= \overbrace{\frac{\mu}{g \rho\left(\rho-\rho\_{v}\right)} \cdot \frac{k \Delta T}{\delta\left(h\_{f g}+\frac{3}{8} c \Delta T\right)}}^\texttt{Independent terms} d x\\
\delta^4/4 &= \frac{\mu k\Delta T}{g\rho\left(\rho-\rho\_v\right)h\_{avg}} x\\
\delta &= \sqrt\[4]{\frac{4\mu k\Delta T}{g\rho\left(\rho-\rho\_v\right)h\_{avg}} x}\\
\delta^{3} d \delta&=\frac{k \mu \Delta T}{g \rho\left(\rho-\rho\_{v}\right) h\_{avg}} d x\\
\end{align}
$$

Integrate and conduct algebra for expression, where $&#x68;*{fg}’ = h*{fg} + \frac 3 8 c \Delta T$ for boundary layer height:

$$
\begin{align}
\int \delta^3  d\delta&= \overbrace{\frac{\mu}{g \rho\left(\rho-\rho\_{v}\right)} \cdot \frac{k \Delta T}{\delta\left(h\_{f g}+\frac{3}{8} c \Delta T\right)}}^\texttt{Independent terms} \int d x\\
\frac {\delta^4} {4} &= \frac{\mu k\Delta T}{g\rho\left(\rho-\rho\_v\right)h\_{avg}} x\\
& \boxed{\delta = \sqrt\[4]{\frac{4\mu k\Delta T}{g\rho\left(\rho-\rho\_v\right)h\_{avg}} x}}\\

\end{align}
$$

**Part c:** Obtain Heat transfer coefficient (HTC), by using the two definitional expressions for $h$, Newton’s Law of Convection and equating the heat flux at the surface for conduction and convection

$$
\label{eqn:htc}
\begin{align}
& h =\overbrace{\left(\frac{q}{A}\right) \frac{1}{\left(T\_{s a t}-T\_{w}\right)}}^\texttt{Newton's Law}=\left.\frac{k}{\delta}\right}\texttt{Energy Balance} \rightarrow  \overbrace{\frac{k}{\sqrt\[4]{\frac{4\mu k\Delta T}{g\rho\left(\rho-\rho\_v\right)h\_{avg}} x}}}^\texttt{substitution}\\
& \boxed{h =\sqrt\[4]{\frac{g \rho\left(\rho-\rho\_{v}\right) k^{3} h\_{f g}^{\prime}}{4 x \mu \Delta T}}}
\end{align}
$$

**Part c:** Average HTC comes from the average as a function of position.

$$
\begin{align}
h\_{avg}&=\frac{1}{L} \int\_{0}^{L} h d x =\frac{1}{L} \int\_{0}^{L} \left(\sqrt\[4]{\frac{g \rho\left(\rho-\rho\_{v}\right) k^{3} h\_{f g}^{\prime}}{4 x \mu \Delta T}}\right) d x \\
&= \overbrace{\frac{1}{L} \left( \sqrt\[4]{\frac{g \rho\left(\rho-\rho\_{v}\right) k^{3} h\_{f g}^{\prime}}{4 \mu \Delta T}} \right) }^\texttt{constants} \int\_{0}^{L} \left( \frac{1}{x^{1/4}} \right) d x\\
&= \underbrace{\frac{1}{L} \int\_{0}^{L} \left( \frac{1}{x^{1/4}} \right) d x}*{\frac{0.943}{L^4}} \sqrt\[4]{\frac{g \rho\left(\rho-\rho*{v}\right) k^{3} h\_{f g}^{\prime}}{\mu \Delta T}}\\
&= \frac{0.943}{L^4}\sqrt\[4]{\frac{g \rho\left(\rho-\rho\_{v}\right) k^{3} h\_{f g}^{\prime}}{\mu \Delta T}}=0.943 \sqrt\[4]{\frac{g \rho\left(\rho-\rho\_{v}\right) k^{3} h\_{f g}^{\prime}}{L\mu \Delta T}}\\
& \boxed{h\_{a v g}=0.943 \sqrt\[4]{\frac{g \rho\_{l i q}\left(\rho\_{l i q}-\rho\_{v}\right) k\_{l i q}^{3} h\_{f g}}{L \mu\_{l i q}\left(T\_{s a t}-T\_{w}\right)} \sin \phi}}
\end{align}
$$
