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 $y$.
μ d u d y = g ρ ( δ − y ) − d p d x ( δ − y ) 0 = g ρ v − d p d x → d p d x = g ρ v μ d u d y = g ( ρ − ρ v ) ( δ − y ) → u = ( g ( ρ − ρ v ) μ ) ( δ y − y 2 2 ) \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} μ d y d u 0 μ d y d u = g ρ ( δ − y ) − d x d p ( δ − y ) = g ρ v − d x d p → d x d p = g ρ v = g ( ρ − ρ v ) ( δ − y ) → u = ( μ g ( ρ − ρ v ) ) ( δy − 2 y 2 ) Using this expression for the velocity, as a function of height and the pressure difference relative to vapor pressure
Γ = ρ ∫ 0 δ u d y = ρ ∫ 0 δ ( g ( ρ − ρ v ) μ ) ( δ y − y 2 2 ) d y = g ρ ( ρ − ρ v ) μ δ 3 3 \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}\\ Γ = ρ ∫ 0 δ u d y = ρ ∫ 0 δ ( μ g ( ρ − ρ v ) ) ( δy − 2 y 2 ) d y = μ g ρ ( ρ − ρ v ) 3 δ 3 Take derivative to find mass change per unit length:
Γ = g ρ ( ρ − ρ v ) μ δ 3 3 → d / d δ d Γ d δ = g ρ ( ρ − ρ v ) μ δ 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} Γ = μ g ρ ( ρ − ρ v ) 3 δ 3 d / d δ d δ d Γ = μ g ρ ( ρ − ρ v ) δ 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 + T s a t − T w δ y T(y) =T_{w}+\frac{T_{s a t}-T_{w}}{\delta} y T ( y ) = T w + δ T s a t − T w 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 $\delta$.
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.
q A d x = k Δ T δ d x = h a v d Γ → 1 / d x q A = k Δ T δ = h a v d Γ d x = ( h f g + 3 8 c Δ T ) d Γ d x d Γ = k Δ T δ ( h f g + 3 8 c Δ T ) d x \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 A q d x = k δ Δ T d x = h a v d Γ 1/ d x A q = k δ Δ T = h a v d x d Γ = ( h f g + 8 3 c Δ T ) d x d Γ d Γ = δ ( h f g + 8 3 c Δ T ) k Δ T d x Part a: Equate mass change to obtain film growth:
d Γ = g ρ ( ρ − ρ v ) μ δ 2 d δ ⏞ From mass change = k Δ T δ ( h f g + 3 8 c Δ T ) d x ⏞ From energy transport δ = 4 k μ Δ T x g ρ ( ρ − ρ v ) h f g 4 \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} d Γ = μ g ρ ( ρ − ρ v ) δ 2 d δ From mass change = δ ( h f g + 8 3 c Δ T ) k Δ T d x From energy transport δ = 4 g ρ ( ρ − ρ v ) h f g 4 k μ Δ T x Solving the equation, group differentials in mass and linear position on either side, to collect mass coefficients and energy transport together.
δ 3 ⋅ d δ = ( g ρ ( ρ − ρ v ) μ ) − 1 ⋅ k Δ T h f g + 3 8 c Δ T d x = μ g ρ ( ρ − ρ v ) ⋅ k Δ T δ ( h f g + 3 8 c Δ T ) ⏞ Independent terms d x δ 4 / 4 = μ k Δ T g ρ ( ρ − ρ v ) h a v g x δ = 4 μ k Δ T g ρ ( ρ − ρ v ) h a v g x 4 δ 3 d δ = k μ Δ T g ρ ( ρ − ρ v ) h a v g d x \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} δ 3 ⋅ d δ δ 4 /4 δ δ 3 d δ = ( μ g ρ ( ρ − ρ v ) ) − 1 ⋅ h f g + 8 3 c Δ T k Δ T d x = g ρ ( ρ − ρ v ) μ ⋅ δ ( h f g + 8 3 c Δ T ) k Δ T Independent terms d x = g ρ ( ρ − ρ v ) h a vg μ k Δ T x = 4 g ρ ( ρ − ρ v ) h a vg 4 μ k Δ T x = g ρ ( ρ − ρ v ) h a vg k μ Δ T d x Integrate and conduct algebra for expression, where $h{fg}’ = h {fg} + \frac 3 8 c \Delta T$ for boundary layer height:
∫ δ 3 d δ = μ g ρ ( ρ − ρ v ) ⋅ k Δ T δ ( h f g + 3 8 c Δ T ) ⏞ Independent terms ∫ d x δ 4 4 = μ k Δ T g ρ ( ρ − ρ v ) h a v g x δ = 4 μ k Δ T g ρ ( ρ − ρ v ) h a v g x 4 \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} ∫ δ 3 d δ 4 δ 4 = g ρ ( ρ − ρ v ) μ ⋅ δ ( h f g + 8 3 c Δ T ) k Δ T Independent terms ∫ d x = g ρ ( ρ − ρ v ) h a vg μ k Δ T x δ = 4 g ρ ( ρ − ρ v ) h a vg 4 μ k Δ T x 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
Part c: Average HTC comes from the average as a function of position.
h a v g = 1 L ∫ 0 L h d x = 1 L ∫ 0 L ( g ρ ( ρ − ρ v ) k 3 h f g ′ 4 x μ Δ T 4 ) d x = 1 L ( g ρ ( ρ − ρ v ) k 3 h f g ′ 4 μ Δ T 4 ) ⏞ constants ∫ 0 L ( 1 x 1 / 4 ) d x = 1 L ∫ 0 L ( 1 x 1 / 4 ) d x ⏟ 0.943 L 4 g ρ ( ρ − ρ v ) k 3 h f g ′ μ Δ T 4 = 0.943 L 4 g ρ ( ρ − ρ v ) k 3 h f g ′ μ Δ T 4 = 0.943 g ρ ( ρ − ρ v ) k 3 h f g ′ L μ Δ T 4 h a v g = 0.943 g ρ l i q ( ρ l i q − ρ v ) k l i q 3 h f g L μ l i q ( T s a t − T w ) sin ϕ 4 \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} h a vg = L 1 ∫ 0 L h d x = L 1 ∫ 0 L 4 4 xμ Δ T g ρ ( ρ − ρ v ) k 3 h f g ′ d x = L 1 4 4 μ Δ T g ρ ( ρ − ρ v ) k 3 h f g ′ constants ∫ 0 L ( x 1/4 1 ) d x = L 4 0.943 L 1 ∫ 0 L ( x 1/4 1 ) d x 4 μ Δ T g ρ ( ρ − ρ v ) k 3 h f g ′ = L 4 0.943 4 μ Δ T g ρ ( ρ − ρ v ) k 3 h f g ′ = 0.943 4 Lμ Δ T g ρ ( ρ − ρ v ) k 3 h f g ′ h a vg = 0.943 4 L μ l i q ( T s a t − T w ) g ρ l i q ( ρ l i q − ρ v ) k l i q 3 h f g sin ϕ