🔏
brainless
  • What is this?
  • Responsibility
  • Changelog
  • meta
    • Sharing
      • Inspirations
      • Workflows
      • Social Media
    • Geography
      • Life
      • Death
        • Family Death
    • Research
      • Project Index
      • 3D Printing
      • Photogrammetry
      • Drone Building
    • External Websites
    • [unreleased]
      • [Template]
      • [TEMP] CL V0.0.5 or whatever
      • Skincare
      • Travel
      • Working and Staying Busy
      • Stride
      • Funeral Playlist
      • Notes and Ideas
      • Boredom
      • Four Noble Truths of "Thermo"
      • Respect
      • Work
  • STE[A]M
    • [guide]
    • Science
      • Materials Modeling
        • Syllabus Description
        • Lecture Slides
        • Student Notes
        • Assignments
          • index
          • Carbon Nanotubes
    • Technology
      • Computer Science
        • Commands
      • Photogrammetry
      • Quantum Computing
      • Computers
      • Programs
        • Matlab and Octave
        • Audacity
        • Google Chrome
          • Websites
            • Google Suite Sites
            • Github
              • Version Control
            • Product Hunt
            • Twitter
            • Youtube
              • Channels
            • Vimeo
          • Extensions
            • Dark Reader
            • Vimium
        • [miscellaneous]
          • Octave
          • PureRef
          • git
          • gnu stow
          • mermaid.js
        • Excel
        • Blender
        • LaTeX
        • Sublime: Text Editor
        • Spotify
        • VLC Media Player
      • Android and iOS
      • Operating Systems
        • macOS
          • mackup
        • Unix
          • folder structure
        • Windows
          • App Installation
          • Meshroom
          • Corsair Utility Engine
      • 3D Printing
    • Engineering
      • Accreditation
        • Fundamentals of Engineering
        • Professional Engineering
      • Continuum Mechanics
        • Fluid Mechanics
          • Incompressible Flow
            • corona final
            • indexhw4
            • indexhw3
            • index
            • hw2
            • hw1
          • Syllabus Description
          • Lecture Slides
          • Student Notes
            • Dynamic or Kinematic Viscosity
          • Assignments
            • Homeworks
              • Homework 1
              • Homework 2
              • Homework 3
              • Homework 4
              • Homework 5
            • Vortex Project
        • Solid Mechanics
          • Syllabus Description
          • Lecture Slides
          • Student Notes
          • Assignments
        • Incompresible Flow
          • Syllabus Description
          • Lecture Slides
          • Student Notes
          • Assignments
      • Experimental Mechanics
        • Syllabus Description
        • Lecture Slides
        • Student Notes
        • Assignments
      • Finite Element Methods
        • Intro to Finite Elements
          • Syllabus Description
          • Lecture Slides
          • Student Notes
          • Assignments
        • Fundamentals of FEM
          • Syllabus Description
            • index
          • Lecture Slides
          • Student Notes
          • Assignments
            • Project
              • index
              • Untitled
            • Homework 1
            • Homework 4
            • index
      • Heat Transfer
        • Syllabus Description
        • Lecture Slides
        • Student Notes
        • Assignments
          • homework
            • hw10
            • q9
            • q8
            • q7
            • hw7
            • hw6
            • q5
            • q3
            • 1 ec
          • Discussions
            • d11
            • d10
            • d9
            • d8
            • d6
            • d4
            • d3
          • Project Notes
      • Machine Dynamics
        • Syllabus Description
        • Lecture Slides
        • Student Notes
        • Assignments
    • Art
      • Color Theory
      • Origami
        • FolderMath
          • Surveying Origami Math
          • Represent a Folded Object
          • Creating a Crease Pattern
          • Making the Folds
          • Simulating Folding Origami
          • List of Resources
            • Codes
            • Papers, Programs, and Inspirations
    • Mathematics
      • Complex Numbers
        • What is i^i?
      • Analytic Hierarchy Process
      • Probability
      • Conway's Game of Life
      • Metallic Numbers
      • Cauchy's Formula for Repeated Integration
      • Wavelet Transform
      • Laplace Tidal Equation
      • Alternating Summation of Ones
      • Constants
      • Bad Maths
      • Calculus
        • Syllabus Description
        • Miscellaneous
  • Thoughts
    • Marksmanship
      • Archery
    • Schooling
    • ...and Ideas?
      • Perceived Time and Learning
      • Content Comprehension
    • Comics and Games
      • Rubik's Cube
      • Dungeons and Dragons
      • Beyond-All-Reason
      • Sekiro: Shadows Die Twice
      • Super Smash Bros
        • Project M
        • Project +
      • League of Legends
      • Satisfactory
    • Literature and Art
      • Books
      • Reading is Hard
      • Various Words and Phrases
      • Poems
      • Interviews
      • Quotes
        • Phrases
      • Jokes
      • ASCII Art
    • Shows and Films
      • Cowboy Bebop
      • My Hero Academia
      • Sword of the Stranger
    • Working and Life Balance
  • Projects
Powered by GitBook
On this page
  1. STE[A]M
  2. Engineering
  3. Heat Transfer
  4. Assignments
  5. homework

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}μdydu​0μdydu​​=gρ(δ−y)−dxdp​(δ−y)=gρv​−dxdp​→dxdp​=gρv​=g(ρ−ρv​)(δ−y)→u=(μg(ρ−ρv​)​)(δy−2y2​)​​

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}\\Γ=ρ∫0δ​udy=ρ∫0δ​(μg(ρ−ρv​)​)(δy−2y2​)dy=μgρ(ρ−ρv​)​3δ3​

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}Γ=μgρ(ρ−ρv​)​3δ3​d/dδ​dδdΓ​=μgρ(ρ−ρv​)​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.

T(y)=Tw+Tsat−TwδyT(y) =T_{w}+\frac{T_{s a t}-T_{w}}{\delta} yT(y)=Tw​+δTsat​−Tw​​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 xAq​dx=kδΔT​dx=hav​dΓ1/dx​Aq​=kδΔT​=hav​dxdΓ​=(hfg​+83​cΔT)dxdΓ​dΓ=δ(hfg​+83​cΔT)kΔT​dx

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}dΓ​=μgρ(ρ−ρv​)​δ2dδ​From mass change​=δ(hfg​+83​cΔT)kΔT​dx​From energy transport​δ=4gρ(ρ−ρv​)hfg​4kμΔTx​​​​​

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}δ3⋅dδδ4/4δδ3dδ​=(μgρ(ρ−ρv​)​)−1⋅hfg​+83​cΔTkΔT​dx=gρ(ρ−ρv​)μ​⋅δ(hfg​+83​cΔT)kΔT​​Independent terms​dx=gρ(ρ−ρv​)havg​μkΔT​x=4gρ(ρ−ρv​)havg​4μkΔT​x​=gρ(ρ−ρv​)havg​kμΔT​dx​​

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}∫δ3dδ4δ4​​=gρ(ρ−ρv​)μ​⋅δ(hfg​+83​cΔT)kΔT​​Independent terms​∫dx=gρ(ρ−ρv​)havg​μkΔT​xδ=4gρ(ρ−ρv​)havg​4μkΔT​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

\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}havg​​=L1​∫0L​hdx=L1​∫0L​​44xμΔTgρ(ρ−ρv​)k3hfg′​​​​dx=L1​​44μΔTgρ(ρ−ρv​)k3hfg′​​​​​constants​∫0L​(x1/41​)dx=L40.943​L1​∫0L​(x1/41​)dx​​4μΔTgρ(ρ−ρv​)k3hfg′​​​=L40.943​4μΔTgρ(ρ−ρv​)k3hfg′​​​=0.9434LμΔTgρ(ρ−ρv​)k3hfg′​​​havg​=0.9434Lμliq​(Tsat​−Tw​)gρliq​(ρliq​−ρv​)kliq3​hfg​​sinϕ​​​​
Previousq7Nexthw6

Last updated 4 years ago