Chapter 2 - Molecular Transport

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Instructor: Jan Haelssig Department of Chemical and Biological Engineering Fall 2023 CHG 8116 – Advanced Transport Phenomena Chapter 2: Molecular Transport
/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 2
/77 CHG 8116 – Chapter 2 | Jan Haelssig Learning Outcomes Students should achieve the following learning outcomes: ○ State the general rate equation. ○ Describe the difference between rate/flow and flux. ○ Define and describe the concept of gradients and driving force. ○ Describe and apply the equations for molecular transport: • Fourier’s law of heat conduction • Newton’s law of viscosity • Fick’s law of diffusion ○ State and apply the equations for molecular transport in their mathematically analogous forms. ○ Apply the multidimensional forms of the equations for molecular transport. ○ Discuss primary and secondary fluxes. 3
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 4
/77 CHG 8116 – Chapter 2 | Jan Haelssig Introduction When we disturb a system from its equilibrium state , a transport process will occur to try to ○ restore the equilibrium state or ○ bring the system to a new equilibrium state. This can be tracked by measuring a change in one of the state variables of the system: ○ Temperature, pressure, concentration, etc. Due to the disturbance, there will be spatial variation of this state variable within the system. The difference between the current state and the equilibrium state is the driving force for transport process. 5 https://www.peakpx.com/556527/gray-steel-electric-kettle https://www.pexels.com/photo/abstract-ball-jar-blue-food-coloring-327165/ https://pixabay.com/photos/floating-tire-summer-water-wave-72963/
/77 CHG 8116 – Chapter 2 | Jan Haelssig Introduction One mechanism of transport is by molecular motion . It can occur in both solids and fluids. This is also referred to as diffusion and can occur for transport of ○ electric charge ○ heat ○ mass ○ momentum We can formulate a general rate equation for this mode of transport: 6 differences in temperature, pressure, velocity, or concentration material/fluid properties and geometry of system Rate = Conductance × Driving Force = Driving Force Resistance
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 7
/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient The terms rate , flow , and flow rate are often used interchangeably. This refers to the amount of the property (heat/momentum/mass) being transported per unit time, and it is a scalar property. The flow or rate is related to the area of transfer as follows: Flow = න ? Flux ⋅ ?𝐴 = Average Flux ⋅ Area The flux , which is a vector property, is the amount of the property that passes through a surface per unit time and per unit area. 8 J/s A = 2 m 2 ʹʹ = 50 J/(s·m 2 ) A = 2 m 2 Flow Average Flux Same average flux (or flow) but different flux distribution
/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient In the three-dimensional Cartesian coordinate system, the gradient of a scalar function 𝑓 ?, ?, ? is given by the following expression: ∇𝑓 = 𝜕𝑓 𝜕? Ԧ ? + 𝜕𝑓 𝜕? Ԧ ? + 𝜕𝑓 𝜕? ? where Ԧ ? , Ԧ ? , and ? are the standard unit vectors in the directions of the ? , ? , and ? coordinates. The gradient vector can be interpreted as the direction of fastest increase as well as the magnitude of the increase . 9
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient We can obtain a numerical value for the gradient at any point within the three-dimensional system. For example, we obtain the following at the circled location: 𝜕𝑇 𝜕? = −144.79 K m 𝜕𝑇 𝜕? = −17.00 K m 𝜕𝑇 𝜕? = −17.037 K m 10
/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient Consider the green two-dimensional plane in the cube. 11
/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient Consider the blue one-dimensional line in the cube and on the green plane. 12 𝜕𝑇 𝜕?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Rate, Flow, Flux, and Gradient For most engineering purposes, it is sufficient to assume that the flux is linearly proportional to the gradient of the pertinent state variable: Flux ∝ Gradient A constant of proportionality is used to develop an equation relating flux to the gradient. These constants represent a transport property of the material. Flux = Proportionality Constant ⋅ Gradient 13 Low temperature heating element High temperature heating element Stainless steel pan Cast iron pan
/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 14
/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction • Conduction is the transfer of energy as heat by molecular motion. • In solids , heat is transferred due to excitation of vibrational energy levels for inter-atomic bonds. • In fluids , heat is transferred from molecule to molecule through collisions. 15 https://www.bbc.co.uk/bitesize/guides/z2gjtv4/revision/1 (Cengel, 2003)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction General rate-flux relationship: Heat Flow = න ? Heat Flux ⋅ ?𝐴 = Average Heat Flux ⋅ Area Energy Time Energy Area⋅Time Flux-gradient relation: Heat Flux = Proportionality Constant ⋅ Gradient Energy Area⋅Time Thermal conductivity Temperature gradient 16
/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction One-dimensional heat transfer in the x -direction : ሶ? ? ′′ W m 2 = −? W m ⋅ K ⋅ ?𝑇 ?? K m One-dimensional heat transfer in the x -direction : ሶ ? ? W = න ? ሶ? ? ′′ W m 2 ⋅ ?𝐴 m 2 = ത ሶ? ? ′′ W m 2 ⋅ 𝐴 m 2 17 Jean-Baptiste Joseph Fourier (1768 – 1830) https://en.wikipedia.org/wiki/ Joseph_Fourier
/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction • Consider a slab of solid material initially at some temperature T 1 . The entire slab is at a constant temperature of T 1 and is in thermal equilibrium . • The temperature on the left is suddenly increased to T 2 . Both temperatures T 1 and T 2 are maintained constant. • The molecules in the slab on the left in direct contact with the increased temperature T 2 will now also obtain a temperature of T 2 because of their close contact. The slab is no longer in thermal equilibrium. • The higher temperature molecules in the left layer will come in contact with the lower temperature molecules in the layer to the right. Heat is continuously transferred from left to right through the slab, causing the temperature along the x -axis to increase. Note that there will be a continuous heat input at the left and a continuous heat output at the right to maintain the temperatures at T 2 and T 1 . These input and output heat transfer rates will not be equal until steady state is reached. • The temperature profile eventually becomes linear if k is constant, the area is constant, and there is no heat generation within the material. The system is now at steady state . 18 Temperature (°C) Position in x -direction (m) time y x Constant Temperature, T y Constant Temperature, T 2 Insulation Insulation Constant Temperature, T 1 y x
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction The heat transfer rate in the x -direction is given by Fourier’s law : ሶ ? ? = −?𝐴 ?𝑇 ?? where ○ ሶ ? ? is the heat transfer rate in the x -direction [J/s = W] ○ k is the thermal conductivity of the material [W m -1 K -1 ] ○ A is the surface area normal to the heat flow direction [m 2 ] ○ dT / dx is the temperature gradient in the x -direction [K/m] When steady state is reached, this heat transfer rate is constant at any point in the material. 19 Constant Temperature, T 2 Insulation Insulation Constant Temperature, T 1 y x y x Temperature (°C) Position in x -direction (m) ሶ ? ? W ሶ ? ? W ?𝑇 ?? = Δ𝑇 Δ? = slope time
/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction The heat flux is given by the following expression: ത ሶ? ? ′′ = ሶ ? ? 𝐴 and ሶ? ? ′′ = −? ?𝑇 ?? where ○ ሶ? ? ′′ is the heat flux in the x -direction [J s -1 m -2 = W/m 2 ] 20 Constant Temperature, T 2 Insulation Insulation Constant Temperature, T 1 y x y x Temperature (°C) Position in x -direction (m) ሶ? ? ′′ W/m 2 ሶ? ? ′′ W/m 2 time ?𝑇 ?? = Δ𝑇 Δ? = slope
/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction In the previous example, the heat flux will initially be large at the hot surface (at the source of the heat) and zero at the cold surface. As the transfer process proceeds and the temperature profile develops, the system will eventually reach steady state . The heat flux will be uniform in the x-direction and the temperature profile will be linear only if ○ the thermal conductivity k is constant, ○ the area perpendicular to the direction of the flux is constant, and ○ there is no internal source of heat generation within the volume of the material. 21 Temperature (°C) Position in x -direction (m) ሶ? ? ′′ Heat flux in x -direction, (W/m 2 ) Position in x -direction (m) Rate of heat transfer in x -direction, (W) Position in x -direction (m) ሶ ? T 2 T 1
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction The thermal conductivity k is a material property. It is generally a function of temperature. It is a measure of the ability of the material to conduct heat. 22 insulators, poor conductors of heat conductor, good conductors of heat (Bergman et al, 2011) (Welty et al, 2015)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction 23 (Welty et al, 2015)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fourier’s Law of Heat Conduction Fourier’s law can be recast in terms of the gradient of the concentration of energy by multiplying both sides by ρ c p : ሶ? ? ′′ = −? ?𝑇 ?? → ሶ? ? ′′ = − ? 𝜌? 𝑝 ? 𝜌? 𝑝 𝑇 ?? The units of the gradient are as follows: kg m 3 ⋅ J kg ⋅ K ⋅ K ⋅ 1 m = J m 3 ⋅ m The thermal diffusivity α is defined as follows: 𝛼 = ? 𝜌? 𝑝 J m ⋅ K ⋅ s ⋅ m 3 kg ⋅ kg ⋅ K J = m 2 s 24 gradient of temperature gradient of energy concentration energy concentration
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 25
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity General rate-flux relationship: Momentum Flow = න ? Momentum Flux ⋅ ?𝐴 = Average Momentum Flux ⋅ Area Momentum kg⋅ m s Time s Momentum Area⋅Time Flux-gradient relation: Momentum Flux = Proportionality Constant ⋅ Gradient Momentum Area⋅Time Viscosity Velocity Gradient 26
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity General rate-flux relationship: Force = න ? Shear Stress ⋅ ?𝐴 = Average Shear Stress ⋅ Area Momentum kg⋅ m s Time s Force Area⋅Time Flux-gradient relation: Shear Stress = Proportionality Constant ⋅ Gradient Force Area⋅Time Viscosity Velocity Gradient 27
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity One-dimensional momentum transfer in the y -direction caused by flow in the x -direction : 𝜏 ?? N m 2 = −? N m 2 ⋅ s ⋅ ?𝑢 ? ?? m/s m One-dimensional momentum transfer in the y -direction caused by flow in the x -direction : 𝐹 ? N = න ? 𝜏 ?? N m 2 ⋅ ?𝐴 m 2 = ҧ 𝜏 ?? N m 2 ⋅ 𝐴 m 2 28 Sir Isaac Newton (1643 – 1727) https://en.wikipedia.org/wiki/Isa ac_Newton
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity • Consider a fluid contained between two large parallel plates separated by a small distance. The fluid is not moving and is in static equilibrium . The fluid layer is so thin that we can neglect the effect of gravity on the behaviour of the fluid. • The bottom plate is suddenly moved at a constant velocity u plate . The top plate is maintained stationary ( u = 0). • The fluid layer in direct contact with the bottom plate will also move at a velocity of u plate . This is referred to as the no-slip condition . The fluid is no longer in static equilibrium. • The faster moving particles in the bottom layer come in contact with the slower moving particles in the fluid layer above. Momentum is continuously transferred up and through the fluid, causing the x-direction velocity along the y -axis to increase. Note that to maintain this system, there needs to be a constant input of fluid from the left and a constant output of fluid on the right. There must also be a continuous force applied to the bottom plate to maintain its motion and a force applied to the top plate to keep it from moving. These forces will not be equal until steady state is reached. • The velocity profile eventually becomes linear if μ is constant, the area is constant, and there is no momentum generation within the material. The system is now at steady state and the flow is fully developed . • Note that the fluid layer in direct contact with the top plate maintains a velocity of 0 ( no-slip condition ). 29 Velocity in x -direction (m/s) Position in y -direction (m) time y x y x
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity The force needed to keep the lower plate moving at constant velocity is given by Newton’s law of viscosity : 𝐹 ? = −?𝐴 ?𝑢 ? ?? where ○ F x is the force in the x -direction [N = kg m/s 2 ] ○ μ is the viscosity of the fluid [kg m -1 s -1 = N m -2 s = Pa s] ○ A is the surface area normal to the momentum transfer [m 2 ] ○ du x / dy is the velocity gradient in the y -direction [s -1 ] When steady motion has been achieved, this force is constant . 30 y x y x Velocity in x -direction (m/s) Position in y -direction (m) F x [N] F x [N] time ?𝑢 ? ?? = Δ𝑢 ? Δ? = slope Each fluid layer will experience F x in the positive x -direction pulling it to the right and an equal force F x in the negative x - direction resisting this motion . Momentum is transferred to the top plate, where a force of F x is required in the negative x - direction to keep the top plate stationary .
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity The shear stress or momentum flux are given by the following expression: ҧ 𝜏 ?? ′′ = 𝐹 ? 𝐴 and 𝜏 ?? = −? ?𝑢 ? ?? where ○ τ yx is the shear stress [N/m 2 = kg m -1 s -2 ] 31 y x y x Velocity in x -direction (m/s) Position in y -direction (m) ?𝑢 ? ?? = Δ𝑢 ? Δ? = slope F x [N] τ yx [N/m 2 ] τ yx [N/m 2 ] F x [N] time
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity Initially, the momentum flux will be large at the moving plate (the source of the momentum) and zero at the stationary plate. As the transfer process proceeds and the velocity profile develops, the system will eventually reach a state of fully- developed flow or steady state . The momentum flux will be uniform in the y - direction and the velocity profile will be linear only if ○ the viscosity μ is constant, ○ the area perpendicular to the direction of the flux is constant, and ○ there is no internal source of momentum generation within the volume of the fluid. 32 Velocity in x -direction (m/s) Position in y -direction (m) Momentum flux in y -direction, τ yx (N/m 2 ) Position in y -direction (m) Force acting in x -direction, F x (N) Position in y -direction (m)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity Since momentum transfer is normal to the direction of fluid flow , two directions must be identified when writing Newton’s law for momentum transfer: 𝜏 ?? = −? ?𝑢 ? ?? Therefore, τ yx is the flux of x -momentum that is transferred through the fluid in the y -direction . kg m ⋅ s 2 = kg ⋅ m/s m 2 ⋅ s = momentum in ?−direction area of the ?? plane ⋅ time 33 direction of momentum transfer direction of velocity direction of velocity direction of velocity gradient (and momentum transfer)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity • The physical interpretation of τ yx can be confusing, since the it represents both a momentum flux and a shear stress ( F / A ). • Movement of a plate in the xz plane in the x -direction will cause a velocity gradient such that u x is a function of y . • Thus, there is a momentum flux τ yx that acts in the y -direction through the xz plane. • However, the shear stress acts in the x -direction because the force F acts in the x -direction to maintain the plate at velocity u plate . 34 u x = u x (y) τ yx as a momentum flux x z y F
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity The viscosity μ is often referred to as the absolute or dynamic viscosity and is a function of temperature. ○ kg/(m·s) = N s/m 2 = Pa s ○ 1 poise = 0.1 Pa s ○ 1 cP (centipoise) = 1 × 10 -3 Pa s The viscosity is a measure of resistance to flow of the fluid. Comparing two fluids flowing with the same value of du x / dy : 35 Low viscosity • Less force ( F = τ A ) needed to maintain the same average velocity • Less momentum ( τ ) is transferred in the y - direction • Less fluid friction experienced at the walls High viscosity • More force ( F = τ A ) needed to maintain the same average velocity • More momentum ( τ ) is transferred in the y - direction • More fluid friction experienced at the walls
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity • For many fluids, the viscosity is independent of the shear stress applied ( τ yx ) and independent of the shear rate ( γ̇ = du x / dy ). • These fluids obey Newton ’s law of viscosity and are called Newtonian fluids . Gases, water, and most organic liquids are Newtonian. 𝜏 ?? = −? ?𝑢 ? ?? 36 N/m 2 applied to fluid rate of deformation (shear rate, γ̇ ) Shear stress, τ [N/m 2 ] Shear rate, γ̇ [s -1 ] yield stress Slope of line is viscosity
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity • Shear thickening (dilatant) fluids: ○ Cornstarch in water ○ Fluid used in viscous coupling units • Shear thinning (pseudoplastic) fluids: ○ Ketchup ○ Molasses ○ Syrups ○ Honey • Bingham plastics: ○ Toothpaste ○ Mayonnaise 37 Shear stress, τ [N/m 2 ] Shear rate, γ̇ [s -1 ] yield stress ○ Paint ○ Blood ○ Polymers Slope of line is viscosity http://www.dandelionsonthewall.com/2012/01/liquid-or-solid-make-and-oobleck.html https://en.wikipedia.org/wiki/Mayonnaise https://en.wikipedia.org/wiki/Honey https://en.wikipedia.org/wiki/Viscous_coupling_unit https://en.wikipedia.org/wiki/Toothpaste https://en.wikipedia.org/wiki/Paint
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity The viscosity is normally a function of temperature. 38 (Welty et al, 2015)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Newton’s Law of Viscosity Newton’s law of viscosity can be recast in terms of the gradient of the concentration of momentum by multiplying both sides by ρ : 𝜏 ?? = −? ?𝑢 ? ?? → 𝜏 ?? = − ? 𝜌 ? 𝜌𝑢 ? ?y The units of the gradient are as follows: kg m 3 ⋅ m s ⋅ 1 m = kg m/s m 3 ⋅ m The kinematic viscosity ν is defined as follows: ? = ? 𝜌 Pa ⋅ s kg m 3 ⋅ N m 2 Pa ⋅ kg ⋅ m s 2 N = m 2 s 39 gradient of velocity gradient of momentum concentration momentum concentration
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 40
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion General rate-flux relationship: Species Flow = න ? Species Flux ⋅ ?𝐴 = Average Species Flux ⋅ Area Mass Time Mass Area⋅Time Flux-gradient relation: Species Flux = Proportionality Constant ⋅ Gradient Mass Area⋅Time Diffusion Coefficient Concentration Gradient 41
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion One-dimensional mass transfer in the y -direction : 𝐽 ? 𝑦 mol A m 2 ⋅ s = −? ?? m 2 s ⋅ ?? ? ?? mol A/m 3 m One-dimensional mass transfer in the y -direction : ሶ ? ? 𝑦 mol A s = න ? 𝐽 ? 𝑦 mol A m 2 ⋅ s ⋅ ?𝐴 m 2 = ҧ 𝐽 ? 𝑦 mol A m 2 ⋅ s ⋅ 𝐴 m 2 42 Adolf Eugen Fick (1829–1901) https://en.wikipedia.org/wiki/Adolf_Eu gen_Fick
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion 43 Concentration, C A (mol A/m 3 ) Position in y -direction (m) time y x L A [m 2 ] y x Air C A = C A * C A = 0 • Consider a liquid of pure component A contained in a reservoir attached to a long tube with cross-sectional area A . The tube is initially filled with air. • The layer of air directly above the liquid becomes saturated with A ( C A = C A * ). As a result, the concentration at y = 0 is C A *. Air is blown across the outlet to maintain a zero concentration of component A ( C A = 0). The fluid is no longer in chemical equilibrium . • The A molecules at y = 0 move up by random molecular motion. Molecules of A (mass) are continuously transferred up and through the air, causing the concentration of A along the y -axis to increase. Note that there will be a constant input of A molecules from the reservoir and a constant output of A molecules from the outlet at the top. • The concentration profile eventually becomes linear if D AB is constant, the area is constant, and there is no species A mass generation within the fluid. The system is now at steady state and the concentration profile is fully developed .
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion 44 Concentration, C A (mol A/m 3 ) Position in y -direction (m) y x L A [m 2 ] y x Air C A = C A * C A = 0 ?? ? ?? = Δ? ? Δ? = slope ሶ ? ? 𝑦 mol A/s ሶ ? ? 𝑦 mol A/s time The rate of mole transfer of A in the y -direction is given by Fick’s law : ሶ ? ? 𝑦 = −? ?? 𝐴 ?? ? ?? where ○ ሶ ? ? 𝑦 is the number of moles of component A that are transferred in the y -direction per unit time [mol A/s] ○ D AB is the diffusion coefficient of component A in B [m 2 /s] ○ A is the cross-sectional area normal to the flow direction [m 2 ] ○ dC A / dy is the concentration gradient of component A in the y -direction [mol A m -3 m -1 ] When steady state is reached, the rate of mass transfer is constant at any point in the material.
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion 45 time Concentration, C A (mol A/m 3 ) Position in y -direction (m) y x L A [m 2 ] y x Air C A = C A * C A = 0 ?? ? ?? = Δ? ? Δ? = slope 𝐽 ? 𝑦 mol A m 2 ⋅ s 𝐽 ? 𝑦 mol A m 2 ⋅ s The molar flux is given by the following expression: ҧ 𝐽 ? 𝑦 = ሶ ? ? ? 𝐴 and 𝐽 ? 𝑦 = −? ?? ?? ? ?? where ○ J Ay is the molar flux in the y -direction [mol A s -1 m -2 ]
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion In the previous example, the mass flux close to the liquid reservoir will initially be large (the source of the mass) and zero at the outlet near the air flow. As the transfer process proceeds and the concentration profile develops, the system will eventually reach steady state . The mass flux will be uniform in the y -direction and the concentration profile will be linear in the system only if ○ the diffusion coefficient D AB is constant, ○ the area perpendicular to the direction of the flux is constant, and ○ there is no internal source of species mass generation within the volume. 46 Concentration (mol A/m 3 ) Position in y -direction (m) Mass flux in y -direction, J Ay (mol A s -1 m -2 ) Position in y -direction (m) Rate of mass transfer in y -direction, (mol A/s) Position in y -direction (m) ሶ ? ? ?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion • The diffusion coefficient (or mass diffusivity) is generally known for a specific pair of chemical species (A and B) and is a measure of the speed at which they diffuse into each other. • The diffusion coefficient depends on the pressure, temperature, and composition of the system. • It is closely related to the density of the system, which greatly affects the mobility of molecules . • The typical ranges for diffusion coefficients are as follows: ○ Gases 10 -6 to 10 -5 m 2 /s (diffusion is quickest) ○ Liquids 10 -10 to 10 -9 m 2 /s ○ Solids 10 -14 to 10 -10 m 2 /s (diffusion is slowest) • For a binary system ? ?? = ? ?? . Multicomponent systems are often significantly more complex . 47
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion Diffusion coefficient D AB values for various species pairs can be found in published data tables. 48 (Welty et al, 2015)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion Fick’s law can also be written in terms of mole fraction , by substituting the following into the flux equation: ? ? mol A m 3 mixture = ? ? mol A mol mixture ⋅ ? mol mixture m 3 mixture 𝐽 ? 𝑦 = −? ?? ?? ? ?? = −?? ?? ?? ? ?? where ○ y A is the mole fraction of A [mol A/mol] ○ C is the total concentration of the mixture including all components [mol/m 3 ] 49
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion • Although Fick’s law worked for some systems, early investigations found that it did not always accurately describe diffusion in all systems. • One main reason for the discrepancy is because the diffusion of mass creates a bulk motion of mass (advection). • Stefan and Maxwell, using the kinetic theory of gases, proved that the mass flux relative to a stationary coordinate system is a result of two contributions: ด ? ? flux of total mass transported = ณ 𝐽 ? flux of mass transported by diffusion + ถ ? ? 𝑢 ∗ flux of mass transported by bulk motion of fluid where ○ ? ? is the molar flux relative to a set of stationary axes [mol A s -1 m -2 ] ○ 𝐽 ? is the molar flux relative to the molar-average velocity [mol A s -1 m -2 ] ○ C A is the concentration of A [mol A/m 3 ] ○ u* is the average molar velocity [m/s] 50 Jožef Štefan (1835 – 1893) James Clerk Maxwell (1831–1879) https://en.wikipedia.org/wiki/Ja mes_Clerk_Maxwell https://en.wikipedia.org/wiki/Jos ef_Stefan
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion The molar-average velocity of the mixture can be calculated as follows: 𝑢 ∗ = σ ?=1 𝑁 ? ? 𝑢 ? σ ?=1 𝑁 ? ? = ෍ ?=1 𝑁 ? ? 𝑢 ? where ○ N is the number of components in the mixture ○ u* is the molar-average velocity of the mixture [m/s] ○ C i is the concentration of component i [mol i /m 3 ] ○ u i is the absolute velocity of component i with respect to a stationary coordinate system [m/s] ○ C i u i is the moles of component i that pass through a unit cross section perpendicular to u i [mol i m -2 s -1 ] ○ σ ?=1 𝑁 ? ? = ? is the total concentration of the mixture including all components [mol/m 3 ] ○ y i is the mole fraction of component i [mol i /mol] 51
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CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion If using a stationary coordinate system , the molar flux of A is given by the following equation: ? ? mol A s ⋅ m 2 = ? ? mol A m 3 ⋅ 𝑢 ? m s where: ○ u A is the absolute velocity of component A with respect to the stationary coordinate system [m/s] Fick’s law applied directly without correction for bulk motion will not accurately estimate the flux even for binary systems unless the system is very dilute or equimolar counter-diffusion occurs. Transport of A molecules through stationary B molecules using a stationary coordinate system.
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CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion If using a coordinate system that moves relative to the molar-average velocity , the molar flux is given by the following equation: 𝐽 ? mol A s ⋅ m 2 = ? ? mol A m 3 ⋅ 𝑢 ? − 𝑢 ∗ m s where ○ u A is the absolute velocity of component A with respect to the stationary coordinate system [m/s] ○ u* is the average molar velocity [m/s] This flux only includes diffusion and is what is given by Fick’s law: 𝐽 ? = −?? ?? ?? ? ?? Transport of A molecules through stationary B molecules using a coordinate system that moves relative to the molar-average velocity.
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion We are typically interested in determining the molar flux using stationary coordinates : ? ? = 𝐽 ? + ? ? 𝑢 ∗ = ? ? 𝑢 ? For a binary system, the molar-average velocity is given by the following equation: 𝑢 ∗ = ? ? 𝑢 ? + ? ? 𝑢 ? = 1 ? ( ถ ? ? 𝑢 ? 𝑁 ? + ? ? 𝑢 ? 𝑁 ? ) Fick’s law can be used for J A : 𝐽 ? = −?? ?? ?? ? ?? Combining these equations gives the Fick’s rate equation (or Fick’s law with bulk motion ): ? ? = −?? ?? ?? ? ?? + ? ? ? ? + ? ? 54
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion Information about the flux of component B, or at least its relationship to A, is needed to solve this equation. In the previous example, the air had no source or sink, so its flux can be assumed to be zero, which gives the following relationship: ? ? = −?? ?? ?? ? ?? + ? ? ? ? ? ? = − ?? ?? 1 − ? ? ?? ? ?? This equation can be integrated to solve for N A . Note that the concentration gradient will not be linear due to the 1 – y A term. The equation reduces to Fick’s law when the concentration of A is very low ( y A ≈ 0 ). 55 L Air C A = C A * C A = 0 A [m 2 ] Saturated with A
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Fick’s First Law of Diffusion Depending on the selected units of concentration and the reference frame, Fick’s law can be written in many different ways. Some of the most common forms are shown in the following table. 56 Flux Gradient Fick’s law (binary system) ? ? ∇? ? ? ? = −?? ?? ∇? ? + ? ? ? ? + ? ? 𝑛 ? ∇? ? 𝑛 ? = −𝜌? ?? ∇? ? + ? ? 𝑛 ? + 𝑛 ? 𝐽 ? ∇? ? 𝐽 ? = −?? ?? ∇? ? ? ? ∇? ? ? ? = −?? ?? ∇? ?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 57
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Analogous Form Heat Momentum Mass ሶ? ? ′′ = −? ?𝑇 ?? Recast in terms of the gradient of the concentration of energy , by multiplying both sides by ρ c p : ሶ? ? ′′ = − ? 𝜌? 𝑝 ? 𝜌? 𝑝 𝑇 ?? 𝜏 ?? = −? ?𝑢 ? ?? Recast in terms of the gradient of the concentration of momentum by multiplying both sides by ρ : 𝜏 ?? = − ? 𝜌 ? 𝜌𝑢 ? ?? 𝐽 ? ? = −? ?? ?? ? ?? The flux equation is already written in terms of the concentration of species . The units of the gradient are kg m 3 ⋅ J kg ⋅ K ⋅ K ⋅ 1 m = J m 3 ⋅ m The units of the gradient are kg m 3 ⋅ m s ⋅ 1 m = kg ⋅ m s ⋅ m 3 ⋅ m = N ⋅ s m 3 ⋅ m The units of the gradient are mol A m 3 ⋅ m The thermal diffusivity is defined as follows: 𝛼 = ? 𝜌? 𝑝 J m ⋅ K ⋅ s ⋅ m 3 kg ⋅ kg ⋅ K J = m 2 s The momentum diffusivity , also known as the kinematic viscosity , is defined as follows: ? = ? 𝜌 Pa ⋅ s kg/m 3 = N ⋅ s m 2 ⋅ m 4 N ⋅ s 2 = m 2 s The mass diffusivity , also known as the diffusion coefficient , is defined as follows: ? ?? m 2 s 58
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Analogous Form Flux Quantity Proportionality Constant Gradient Heat ሶ? ? ′′ = heat flux −? thermal conductivity × ?𝑇 ?? temperature Momentum 𝜏 ?? = stress −? viscosity × ?𝑢 ? ?? velocity Mass 𝐽 ? 𝑦 = mass flux −? ?? diffusion coefficient × ?? ? ?? concentration 59 Flux Quantity Diffusivity [m 2 /s] Gradient Heat ሶ? ? ′′ = heat flux −𝛼 thermal diffusivity × ? 𝜌? 𝑝 𝑇 ?? energy concentration Momentum 𝜏 ?? = stress −? kinematic viscosity × ? 𝜌𝑢 ? ?? momentum concentration Mass 𝐽 ? 𝑦 = mass flux −? ?? diffusion coefficient × ?? ? ?? mass concentration
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Analogous Form • From a mathematical standpoint the equations are the same: ○ A solution to one equation is a solution to all three equations. • This mathematical analogy does not mean that the three physical mechanisms are the same. ○ The three mechanisms of heat transfer by conduction, momentum transfer in viscous fluids, and mass transfer by diffusion are entirely different. • We can define three dimensionless groups based on the analogous forms of the equations that allow us to compare scales of the transport processes: 60 Prandtl Schmidt Lewis Pr = ? 𝛼 Sc = ? ? ?? Le = 𝛼 ? ?? Ludwig Prandtl (1875 – 1953) www.aps.org/units/dfd/resources/upload/pr andtl_vol58no12p42_48.pdf Warren K. Lewis (1882 – 1975) www.nasonline.org/publications/biographic al-memoirs/memoir-pdfs/lewis-warren- k.pdf Ernst Schmidt (1892 – 1975) www.mhtlab.uwaterloo.ca/courses/ece309/ schmidt.html
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier ’s Law of Heat Conduction • Newton ’s Law of Viscosity • Fick ’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 61
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms Complete form of Newton’s Law of Viscosity: The coefficient of “bulk “ viscosity 𝜅 is only significant in supersonic/hypersonic compressible flows. Therefore, we more commonly see the following form: If the flow is incompressible ∇ ∙ 𝑢 = 0 by the continuity equation. Newton’s Law of Viscosity 62 Ƹ𝜏 = −? ∇ 𝑢 + ∇ 𝑢 ? + 2 3 ? − 𝜅 ∇ ∙ 𝑢 𝛿 Ƹ𝜏 = −? ∇ 𝑢 + ∇ 𝑢 ? + 2 3 ? ∇ ∙ 𝑢 𝛿
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms Cartesian coordinates: Newton’s Law of Viscosity 63 𝜏 ?? = −2? 𝜕𝑢 ? 𝜕? + 2 3 ?∇ ∙ 𝑢 𝜏 ?? = −? 𝜕𝑢 ? 𝜕? + 𝜕𝑢 ? 𝜕? = 𝜏 ?? 𝜏 ?? = −2? 𝜕𝑢 ? 𝜕? + 2 3 ?∇ ∙ 𝑢 𝜏 ?? = −? 𝜕𝑢 ? 𝜕? + 𝜕𝑢 ? 𝜕? = 𝜏 ?? 𝜏 ?? = −2? 𝜕𝑢 ? 𝜕? + 2 3 ?∇ ∙ 𝑢 𝜏 ?? = −? 𝜕𝑢 ? 𝜕? + 𝜕𝑢 ? 𝜕? = 𝜏 ??
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms Cylindrical coordinates: Newton’s Law of Viscosity 64 𝜏 𝑟𝑟 = −? 2 𝜕𝑢 𝑟 𝜕? + 2 3 ?∇ ⋅ 𝑢 𝜏 𝑟𝜃 = −? ? 𝜕 𝜕? 𝑢 𝜃 ? + 1 ? 𝜕𝑢 𝑟 𝜕𝜃 = 𝜏 𝜃𝑟 𝜏 𝜃𝜃 = −2? 1 ? 𝜕𝑢 𝜃 𝜕𝜃 + 𝑢 𝑟 ? + 2 3 ?∇ ⋅ 𝑢 𝜏 𝜃? = −? 𝜕𝑢 𝜃 𝜕? + 1 ? 𝜕𝑢 ? 𝜕𝜃 = 𝜏 ?𝜃 𝜏 ?? = −? 2 𝜕𝑢 ? 𝜕? + 2 3 ?∇ ⋅ 𝑢 𝜏 ?𝑟 = −? 𝜕𝑢 𝑟 𝜕? + 𝜕𝑢 ? 𝜕? = 𝜏 𝑟?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms Spherical coordinates: Newton’s Law of Viscosity 65 𝜏 𝑟𝑟 = −? 2 𝜕𝑢 𝑟 𝜕? + 2 3 ?∇ ⋅ 𝑢 𝜏 𝑟𝜃 = −? ? 𝜕 𝜕? 𝑢 𝜃 ? + 1 ? 𝜕𝑢 𝑟 𝜕𝜃 = 𝜏 𝜃𝑟 𝜏 𝜃𝜃 = −2? 1 ? 𝜕𝑢 𝜃 𝜕𝜃 + 𝑢 𝑟 ? + 2 3 ?∇ ⋅ 𝑢 𝜏 𝜃𝜙 = −? sin 𝜃 ? 𝜕 𝜕𝜃 𝑢 𝜙 sin 𝜃 + 1 ? sin 𝜃 𝜕𝑢 𝜃 𝜕𝜙 = 𝜏 𝜙𝜃 𝜏 𝜙𝜙 = −2? 1 ? sin 𝜃 𝜕𝑢 𝜙 𝜕𝜙 + 𝑢 𝑟 ? + 𝑢 𝜃 ? cot 𝜃 + 2 3 ?∇ ⋅ 𝑢 𝜏 𝑟𝜙 = −? 1 ? sin 𝜃 𝜕𝑢 𝑟 𝜕𝜙 + ? 𝜕 𝜕? 𝑢 𝜃 ? = 𝜏 𝜙𝑟
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms General form: Cartesian coordinates: Cylindrical coordinates: Spherical coordinates: Fourier’s Law of Heat Conduction 66 Ԧ ሶ? ′′ = −?∇𝑇 ሶ? ? ′′ = −? 𝜕𝑇 𝜕? ሶ? ? ′′ = −? 𝜕𝑇 𝜕? ሶ? ? ′′ = −? 𝜕𝑇 𝜕? ሶ? 𝑟 ′′ = −? 𝜕𝑇 𝜕? ሶ? 𝜃 ′′ = − ? ? 𝜕𝑇 𝜕𝜃 ሶ? ? ′′ = −? 𝜕𝑇 𝜕? ሶ? 𝑟 ′′ = −? 𝜕𝑇 𝜕? ሶ? 𝜃 ′′ = − ? ? 𝜕𝑇 𝜕𝜃 ሶ? 𝜙 ′′ = − ? ? sin 𝜃 𝜕𝑇 𝜕𝜙
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Multidimensional Forms General form: Cartesian coordinates: Cylindrical coordinates: Spherical coordinates: Fick’s First Law of Diffusion 67 𝐽 ? = −?? ?? ∇? ? 𝐽 ?? = −?? ?? 𝜕? ? 𝜕? 𝐽 ?? = −?? ?? 𝜕? ? 𝜕? 𝐽 ?? = −?? ?? 𝜕? ? 𝜕? 𝐽 ?𝑟 = −?? ?? 𝜕? ? 𝜕? 𝐽 ?𝜃 = − ?? ?? ? 𝜕? ? 𝜕𝜃 𝐽 ?? = −?? ?? 𝜕? ? 𝜕? 𝐽 ?𝑟 = −?? ?? 𝜕? ? 𝜕? 𝐽 ?𝜃 = − ?? ?? ? 𝜕? ? 𝜕𝜃 𝐽 ?𝜙 = − ?? ?? ? sin 𝜃 𝜕? ? 𝜕𝜙
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 68
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes • The previously presented equations only describe the relationship between flux and the primary variable (i.e., the variable that is directly affected by the flux). • In reality, secondary variables also contribute to the flux. For example, the gradients of other concentrations contribute to the species flux in a multicomponent system. • Furthermore, a temperature gradient can cause a species mass flux (Soret effect) and a concentration gradient can cause a heat flux (Dufour effect). • The derivation of the fundamental equations for these secondary fluxes and their relationships are studied in irreversible/nonequilibrium thermodynamics . • In fact, the Onsager reciprocal relations show that the transport coefficients for reciprocal secondary fluxes must be equal, and Onsager won the Nobel prize for these developments. 69 https://en.wikipedia.org/wiki/Lars_Onsager Lars Onsager (1903 – 1976)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes Some of the most important secondary fluxes are summarized in the following table. 70 (Plawsky, 2020)
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes Suppose that we decompose the energy flux into the following components (other terms could also be included): The first term ሶ? ′′ ? is the familiar conductive heat flux (Fourier’s Law of Heat Conduction), which is the primary flux – gradient relationship. The second term ሶ? ′′ 𝐻 is the energy flux caused by interdiffusion , which occurs because species carry enthalpy/internal energy as they diffuse. where ഥ 𝐻 ? is the partial molar enthalpy of species i , and 𝐽 ? is the molar diffusion flux. 71 ሶ? ′′ = ሶ? ′′ ? + ሶ? ′′ 𝐻 + ሶ? ′′ ? + ሶ? ′′ ? + ሶ? ′′ 𝑃 ሶ? ′′ ? = −?∇𝑇 ሶ? ′′ 𝐻 = ෍ ?=1 𝑁 ഥ 𝐻 ? 𝐽 ?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes The third term ሶ? ′′ ? is the transfer of energy due to concentration gradients, which is normally called the Dufour effect . The fourth term ሶ? ′′ E is the energy flux caused by electrical conduction ( Peltier effect ) and electrical conduction within a temperature gradient ( Thomson effect ). The reverse of the Peltier effect is called the Seebeck effect and is the primary mechanism allowing temperature measurement in thermocouples. where Ԧ ? ? is the current density. The final term ሶ? ′′ 𝑃 is the energy flux caused by the pressure gradient . 72 ሶ? ′′ ? = −? ? ? ∇? ? ሶ? ′′ ? = −Π ? Ԧ ? ? + 𝜏 ? Ԧ ? ? ∇𝑇 ሶ? ′′ 𝑃 = −? 𝑃 ∇?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes The total species flux can also include various secondary components. Furthermore, the forms of the expressions depend on the selected units for the flux and concentration. For a pure multicomponent diffusion problem (only concentration gradients contribute to the species fluxes), the Maxwell-Stefan equation provides the general expression: This equation can be recast into the following form: 73 ∇? ? = ഥ ? 𝜌 ෍ ?=1 (?≠?) 𝑁 ? ? ? ? − ? ? ? ? ? ?? ? ? where ? = 1,2, … , ? − 1 ? ? = ෍ ?=1 𝑁−1 𝜌? ???,?? ∇? ? where ? = 1,2, … , ? − 1 ? ???,?? = 𝐹 −1 𝐹 ?? = ? ? ഥ ? ? ?𝑁 ? 𝑁 + ෍ ?=1 (?≠?) 𝑁 ? ? ഥ ? ? ?? ? ? 𝐹 ?? = ? ? ഥ ? 1 ? ?? ? ? − 1 ? ?𝑁 ? 𝑁 where ? ≠ ?
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Secondary Fluxes If other gradients contribute to the species flux, the expression becomes even more complex. For example, the species flux expression for a binary system in which pressure and temperature gradients also exist can be written as follows: Given the increased complexity of the equations and coupling between equations when secondary fluxes are included, it is very important only to include the fluxes that are expected to be significant. This is where you must apply your own judgement . 74 ? ? = −𝜌? ?? ∇? ? + ? ? ? 2 ? ?? ? ? ? ∇? − ? ? ? ∇ ln 𝑇
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Outline • Learning Outcomes • Introduction • Rate, Flow, Flux, and Gradient • Fourier’s Law of Heat Conduction • Newton’s Law of Viscosity • Fick’s First Law of Diffusion • Analogous Form • Multidimensional Forms • Secondary Fluxes • Self-Test • References 75
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/77 CHG 8116 – Chapter 2 | Jan Haelssig Self-Test 1. Describe the difference between a heat/momentum/mass transfer rate and a heat/momentum/mass flux. What are the units used for each? 2. What is the driving force for heat transfer? momentum transfer? mass transfer? 3. What is the proportionality constant for heat transfer? What kinds of materials tend to have high values for it? 4. What is the proportionality constant for momentum transfer? 5. What is the proportionality constant for mass transfer? What kinds of systems tend to have high values for it? 6. What are the different categories of fluids? Which ones tend to follow Newton’s law? 7. Why do we identify two directions for the momentum flux? 8. Why does the original formulation of Fick’s law not work for all systems? Why do Fourier’s law and Newton’s law not require the same correction? 76
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/77 CHG 8116 – Chapter 2 | Jan Haelssig References Bergman, T.L., Lavine, A.S., Incropera, F.P., & Dewitt, D.P. (2011). Fundamentals of Heat and Mass Transfer (7 th ed.). Wiley. Çengel, Y. (2003). Heat Transfer: A Practical Approach (2 nd ed.). McGraw-Hill. Plawsky, J. (2020). Transport Phenomena Fundamentals (4 th ed.). CRC Press. Welty, J., Rorrer, G.L., & Foster, D.G. (2015). Fundamentals of Momentum, Heat, and Mass Transfer (6 th ed.). Wiley. 77
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