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hw7

Previousq7Nexthw6

Last updated 4 years ago

  • 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 $y$.

ΞΌdudy=gρ(Ξ΄βˆ’y)βˆ’dpdx(Ξ΄βˆ’y)0=gρvβˆ’dpdxβ†’dpdx=gρvΞΌdudy=g(Οβˆ’Οv)(Ξ΄βˆ’y)β†’u=(g(Οβˆ’Οv)ΞΌ)(Ξ΄yβˆ’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}ΞΌdydu​0ΞΌdydu​​=gρ(Ξ΄βˆ’y)βˆ’dxdp​(Ξ΄βˆ’y)=gρvβ€‹βˆ’dxdp​→dxdp​=gρv​=g(Οβˆ’Οv​)(Ξ΄βˆ’y)β†’u=(ΞΌg(Οβˆ’Οv​)​)(Ξ΄yβˆ’2y2​)​​

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

Ξ“=ρ∫0Ξ΄udy=ρ∫0Ξ΄(g(Οβˆ’Οv)ΞΌ)(Ξ΄yβˆ’y22)dy=gρ(Οβˆ’Οv)ΞΌΞ΄33\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}\\Ξ“=ρ∫0δ​udy=ρ∫0δ​(ΞΌg(Οβˆ’Οv​)​)(Ξ΄yβˆ’2y2​)dy=ΞΌgρ(Οβˆ’Οv​)​3Ξ΄3​

Take derivative to find mass change per unit length:

Ξ“=gρ(Οβˆ’Οv)ΞΌΞ΄33β†’d/dΞ΄dΞ“dΞ΄=gρ(Οβˆ’Οv)ΞΌΞ΄2\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}Ξ“=ΞΌgρ(Οβˆ’Οv​)​3Ξ΄3​d/dδ​dΞ΄dΓ​=ΞΌgρ(Οβˆ’Οv​)​δ2

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

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.

Part a: Equate mass change to obtain film growth:

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

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

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.

T(y)=Tw+Tsatβˆ’TwΞ΄yT(y) =T_{w}+\frac{T_{s a t}-T_{w}}{\delta} yT(y)=Tw​+Ξ΄Tsatβ€‹βˆ’Tw​​y
qAdx=kΞ”TΞ΄dx=havdΞ“β†’1/dxqA=kΞ”TΞ΄=havdΞ“dx=(hfg+38cΞ”T)dΞ“dxdΞ“=kΞ”TΞ΄(hfg+38cΞ”T)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 xAq​dx=kδΔT​dx=hav​dΞ“1/dx​Aq​=kδΔT​=hav​dxdΓ​=(hfg​+83​cΞ”T)dxdΓ​dΞ“=Ξ΄(hfg​+83​cΞ”T)kΞ”T​dx
dΞ“=gρ(Οβˆ’Οv)ΞΌΞ΄2dδ⏞FromΒ massΒ change=kΞ”TΞ΄(hfg+38cΞ”T)dx⏞FromΒ energyΒ transportΞ΄=4kΞΌΞ”Txgρ(Οβˆ’Οv)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}dΓ​=ΞΌgρ(Οβˆ’Οv​)​δ2dδ​FromΒ massΒ change​=Ξ΄(hfg​+83​cΞ”T)kΞ”T​dx​FromΒ energyΒ transport​δ=4gρ(Οβˆ’Οv​)hfg​4kΞΌΞ”Tx​​​​​
Ξ΄3β‹…dΞ΄=(gρ(Οβˆ’Οv)ΞΌ)βˆ’1β‹…kΞ”Thfg+38cΞ”Tdx=ΞΌgρ(Οβˆ’Οv)β‹…kΞ”TΞ΄(hfg+38cΞ”T)⏞IndependentΒ termsdxΞ΄4/4=ΞΌkΞ”Tgρ(Οβˆ’Οv)havgxΞ΄=4ΞΌkΞ”Tgρ(Οβˆ’Οv)havgx4Ξ΄3dΞ΄=kΞΌΞ”Tgρ(Οβˆ’Οv)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}Ξ΄3β‹…dδδ4/4δδ3dδ​=(ΞΌgρ(Οβˆ’Οv​)​)βˆ’1β‹…hfg​+83​cΞ”TkΞ”T​dx=gρ(Οβˆ’Οv​)μ​⋅δ(hfg​+83​cΞ”T)kΞ”T​​IndependentΒ terms​dx=gρ(Οβˆ’Οv​)havg​μkΞ”T​x=4gρ(Οβˆ’Οv​)havg​4ΞΌkΞ”T​x​=gρ(Οβˆ’Οv​)havg​kΞΌΞ”T​dx​​
∫δ3dΞ΄=ΞΌgρ(Οβˆ’Οv)β‹…kΞ”TΞ΄(hfg+38cΞ”T)⏞IndependentΒ terms∫dxΞ΄44=ΞΌkΞ”Tgρ(Οβˆ’Οv)havgxΞ΄=4ΞΌkΞ”Tgρ(Οβˆ’Οv)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}∫δ3dΞ΄4Ξ΄4​​=gρ(Οβˆ’Οv​)μ​⋅δ(hfg​+83​cΞ”T)kΞ”T​​IndependentΒ termsβ€‹βˆ«dx=gρ(Οβˆ’Οv​)havg​μkΞ”T​xΞ΄=4gρ(Οβˆ’Οv​)havg​4ΞΌkΞ”T​x​​​​
havg=1L∫0Lhdx=1L∫0L(gρ(Οβˆ’Οv)k3hfgβ€²4xΞΌΞ”T4)dx=1L(gρ(Οβˆ’Οv)k3hfgβ€²4ΞΌΞ”T4)⏞constants∫0L(1x1/4)dx=1L∫0L(1x1/4)dx⏟0.943L4gρ(Οβˆ’Οv)k3hfgβ€²ΞΌΞ”T4=0.943L4gρ(Οβˆ’Οv)k3hfgβ€²ΞΌΞ”T4=0.943gρ(Οβˆ’Οv)k3hfgβ€²LΞΌΞ”T4havg=0.943gρliq(ρliqβˆ’Οv)kliq3hfgLΞΌliq(Tsatβˆ’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}havg​​=L1β€‹βˆ«0L​hdx=L1β€‹βˆ«0L​​44xΞΌΞ”Tgρ(Οβˆ’Οv​)k3hfg′​​​​dx=L1​​44ΞΌΞ”Tgρ(Οβˆ’Οv​)k3hfg′​​​​​constantsβ€‹βˆ«0L​(x1/41​)dx=L40.943​L1β€‹βˆ«0L​(x1/41​)dx​​4ΞΌΞ”Tgρ(Οβˆ’Οv​)k3hfg′​​​=L40.943​4ΞΌΞ”Tgρ(Οβˆ’Οv​)k3hfg′​​​=0.9434LΞΌΞ”Tgρ(Οβˆ’Οv​)k3hfg′​​​havg​=0.9434LΞΌliq​(Tsatβ€‹βˆ’Tw​)gρliq​(ρliqβ€‹βˆ’Οv​)kliq3​hfg​​sinϕ​​​​