hw7

  • Chris Nkinthorn, 20190716

Prompt: Derive the following expressions:

  • Boundary Layer Height: $\delta(x)=\left[\frac{4 x k{l}\left(T{s a t}-T{w}\right) v{l}}{h{f g} g\left(\rho{l}-\rho_{v}\right)}\right]^{1 / 4}$

  • Heat Transfer Coefficient: $h(x)=\left[\frac{h{f g} g\left(\rho{l}-\rho{v}\right) k{l}^{3}}{4 x\left(T{s a t}-T{w}\right) v_{l}}\right]^{1 / 4}$

  • Average Heat Transfer Coefficient: $\overline{h}{L}=0.943\left[\frac{h{f \beta} g\left(\rho{l}-\rho{v}\right) k{l}^{3}}{L\left(T{s a t}-T{w}\right) v{l}}\right]^{1 / 4}$

Solution

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

μdudy=gρ(δ−y)−dpdx(δ−y)0=gρv−dpdx→dpdx=gρvμdudy=g(ρ−ρv)(δ−y)→u=(g(ρ−ρv)μ)(δy−y22)\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

Γ=ρ∫0δudy=ρ∫0δ(g(ρ−ρv)μ)(δy−y22)dy=gρ(ρ−ρv)μδ33\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:

Γ=gρ(ρ−ρv)μδ33→d/dδdΓdδ=gρ(ρ−ρv)μδ2\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)=Tw+Tsat−TwδyT(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.

qAdx=kΔTδdx=havdΓ→1/dxqA=kΔTδ=havdΓdx=(hfg+38cΔT)dΓdxdΓ=kΔTδ(hfg+38cΔT)dx\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:

dΓ=gρ(ρ−ρv)μδ2dδ⏞From mass change=kΔTδ(hfg+38cΔT)dx⏞From energy transportδ=4kμΔTxgρ(ρ−ρv)hfg4\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.

δ3⋅dδ=(gρ(ρ−ρv)μ)−1⋅kΔThfg+38cΔTdx=μgρ(ρ−ρv)⋅kΔTδ(hfg+38cΔT)⏞Independent termsdxδ4/4=μkΔTgρ(ρ−ρv)havgxδ=4μkΔTgρ(ρ−ρv)havgx4δ3dδ=kμΔTgρ(ρ−ρv)havgdx\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 $h{fg}’ = h{fg} + \frac 3 8 c \Delta T$ for boundary layer height:

∫δ3dδ=μgρ(ρ−ρv)⋅kΔTδ(hfg+38cΔT)⏞Independent terms∫dxδ44=μkΔTgρ(ρ−ρv)havgxδ=4μkΔTgρ(ρ−ρv)havgx4\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.

havg=1L∫0Lhdx=1L∫0L(gρ(ρ−ρv)k3hfg′4xμΔT4)dx=1L(gρ(ρ−ρv)k3hfg′4μΔT4)⏞constants∫0L(1x1/4)dx=1L∫0L(1x1/4)dx⏟0.943L4gρ(ρ−ρv)k3hfg′μΔT4=0.943L4gρ(ρ−ρv)k3hfg′μΔT4=0.943gρ(ρ−ρv)k3hfg′LμΔT4havg=0.943gρliq(ρliq−ρv)kliq3hfgLμliq(Tsat−Tw)sin⁡ϕ4\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}
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