Forced Convection (I) Lecture 28
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Forced Convection (I) Lecture 28 Notes by MIT Student (and MZB) 4/18/2014 To motivate looking into forced convection, we will quickly review concentration polarization. Consider the system shown in Figure 1. Figure 1: Steady state concentration profile of diffusion through a porous stagnant media The current per area is given by: πΌπΌ ππππ = −ππππππ (π₯π₯ = 0) π΄π΄ ππππ At steady state the concentration profile is linear (as shown in Figure 1). The current is therefore given by: πΌπΌ ππππππ(ππΜ − πππ π ) = π΄π΄ πΏπΏ The limiting current occurs when πππ π = 0. πΌπΌππππππ ππππππππΜ = π΄π΄ πΏπΏ 1 The I-V curve, Figure 2, shows the effects of polarization. Figure 2: I-V curve showing the effect of polarization. In order to reduce the losses due to concentration polarization, the limiting current, Ilim, should be increased. Two ways to increase the limiting current are to increase ππΜ or to use forced convection. Forced convection introduces a flow over the reacting surface. This flow leads to a boundary layer which is critical to the concentration profile and therefore the reaction. Figure 3 is a schematic of a boundary layer caused by flowing over a reacting surface. Figure 3: Boundary layer caused by flowing over a reacting surface The flux in the y-direction is given by the diffusion in the y-direction πΉπΉπ¦π¦ = −π·π· ππππ ππππ 2 The length scale in the y-direction is δ(x) so the flux can be approximated as: πΉπΉπ¦π¦ ~ − π·π· ππΜ − πππ π πΏπΏ(π₯π₯) Because πΏπΏ(π₯π₯) βͺ πΏπΏ, transport is typically faster in forced convection. In order to model the concentration we will examine the 2-D steady state species conservation equation. ππππ ππππ ππ 2 ππ ππ 2 ππ π’π’π₯π₯ + π’π’π¦π¦ = π·π· οΏ½ 2 + 2 οΏ½ ππππ ππππ πππ₯π₯ πππ¦π¦ Applying that there is only flow in the x-direction, π’π’π¦π¦ = 0, and performing scaling analysis on the right hand side of the conservation equation: Applying ππ 2 ππ ππ ~ , πππ₯π₯ 2 πΏπΏ2 ππ 2 ππ ππ ~ 2 πππ¦π¦ πΏπΏ(π₯π₯)2 πΏπΏ(π₯π₯) βͺ πΏπΏ Therefore: ππ 2 ππ ππ ππ ππ 2 ππ ~ βͺ ~ πππ₯π₯ 2 πΏπΏ2 πΏπΏ(π₯π₯)2 πππ¦π¦ 2 ππ 2 ππ ππ 2 ππ βͺ πππ₯π₯ 2 πππ¦π¦ 2 This simplifies the conservation equation to: π’π’π₯π₯ ππππ ππ 2 ππ = π·π· 2 ππππ πππ¦π¦ We will now examine 2 flow profiles, the first is plug flow and the second is Poiseuille flow with the Leveque approximation. In plug flow the velocity profile is just given by a constant, π’π’π₯π₯ = π’π’0 . In Poiseuille flow, the velocity profile can be solved by solving the Navier-Stokes equation with a channel of height, H (shown in the appendix). π¦π¦ π¦π¦ 2 π’π’π₯π₯ = 6π’π’οΏ½ οΏ½ − οΏ½ οΏ½ οΏ½ π»π» π»π» 3 Applying the Leveque approximation using π¦π¦~πΏπΏ(π₯π₯) βͺ π»π», the quadratic term is negligible and therefore the velocity profile simplifies to: π’π’π₯π₯ = 6π’π’οΏ½ π¦π¦ π»π» First we will scale the conservation equation with each of these velocity profiles to learn how the boundary layers scale with x. Plug Flow π’π’0 ππππ ππ 2 ππ = π·π· 2 πππ¦π¦ ππππ Poiseuille Flow with Leveque Approx. 6π’π’οΏ½ ππππ ππ 2 ππ οΏ½ π¦π¦οΏ½ = π·π· 2 πππ¦π¦ π»π» ππππ ππΜ ππΜ π’π’0 ~π·π· π₯π₯ πΏπΏ(π₯π₯)2 ππΜ ππΜ π’π’οΏ½ οΏ½ πΏπΏ(π₯π₯)οΏ½ ~π·π· π₯π₯ πΏπΏ(π₯π₯)2 π»π» π·π·π·π· πΏπΏ(π₯π₯)~οΏ½ π’π’0 3 π»π»π»π»π»π» πΏπΏ(π₯π₯)~ οΏ½ π’π’οΏ½ πΏπΏ(π₯π₯)2 ~ π·π·π·π· π’π’0 πΏπΏ(π₯π₯)3 ~ π»π»π»π»π»π» π’π’οΏ½ For the plug flow case, the boundary layer scales with the square root of x but using Poiseuille flow with the Leveque approximation the boundary layer scales with the cube root of x. The analytical solution can also be found for each of these cases. There is not an intrinsic length scale for this PDE so a similarity variable should be used to solve each. Plug Flow π’π’0 ππππ ππ 2 ππ = π·π· 2 ππππ πππ¦π¦ ππ(π₯π₯, 0) = πππ π = 0 ππ(π₯π₯, π¦π¦ → ∞) = ππΜ ππ(0, π¦π¦) = ππΜ Poiseuille Flow with Leveque Approx. 6π’π’οΏ½ ππππ ππ 2 ππ οΏ½ π¦π¦οΏ½ = π·π· 2 π»π» ππππ πππ¦π¦ ππ(π₯π₯, 0) = πππ π = 0 ππ(π₯π₯, π¦π¦ → ∞) = ππΜ ππ(0, π¦π¦) = ππΜ 4 The similarity variable that should be chosen is: ππ = π¦π¦ π’π’0 οΏ½ 2 π·π· π₯π₯ Plugging this into the governing equation gives: ππππ ππ 2 ππ 0 = 2 + 2ππ ππππ ππππ The similarity variable that should be chosen is: 3 2π’π’οΏ½ ππ = π¦π¦ οΏ½ 3π»π»π»π»π»π» Plugging this into the governing equation gives: 0= ππ(ππ = 0) = 0 ππ(ππ = 0) = 0 ππ(ππ → ∞) = ππΜ See appendix for derivation The solution to this differential equation is an error function: Where erf(π₯π₯) = π¦π¦ π’π’0 ππ = ππΜ erf οΏ½ οΏ½ οΏ½ 2 π·π· π₯π₯ 2 π₯π₯ −π π 2 ∫ ππ ππππ √ππ 0 ππ 2 ππ ππππ + 3 ππ2 2 ππππ ππππ ππ(ππ → ∞) = ππΜ See appendix for derivation The general solution to this differential equation is a regularized gamma function: Where 1 2π’π’οΏ½π¦π¦ 3 ππ = ππΜ P οΏ½ , οΏ½ 3 3π»π»π»π»π»π» P(ππ, π₯π₯) = π₯π₯ ∫0 π π ππ−1 ππ −π π ππππ ∞ ∫0 π π ππ−1 ππ −π π ππππ Now that the concentration profiles have been found, an estimate of the boundary layer thickness can be found. A good approximation for when the boundary layer ππ ends is ππΜ = 0.99. Applying the concentration ratio for the plug flow, the boundary layer comes out to: π·π·π·π· πΏπΏ = 3.643οΏ½ π’π’0 For the Poiseuille Flow with Leveque approximation, the boundary layer comes out to: 3 π»π»π»π»π»π» πΏπΏ = 1.607 οΏ½ π’π’οΏ½ 5 Any time there is flow over a reactive surface there is a balance between fuel utilization and power density. As the velocity increases, the power density increases because there is less of an effect of concentration polarization but more fuel is unused. Conversely, when the velocity is small, the boundary layer is larger so more fuel is used but there is not as large a power density because concentration polarization diminishes the returns. Figure 4 shows this effect. Figure 4: The figure on the right shows the boundary layer for two different velocities π’π’2 > π’π’1 and the figure on the left shows the corresponding I-V curve showing the effect of the velocity on the concentration polarization. 6 Appendix Derivation of Poiseuille flow ππ 2 π’π’π₯π₯ 1 ππππ = πππ¦π¦ 2 ππ ππππ π’π’π₯π₯ (π¦π¦ = 0) = 0 π’π’π₯π₯ (π¦π¦ = π»π») = 0 Where P is the pressure and μ is the viscosity and the boundary conditions come from no slip on the surface. ππππ ππππ will be constant so the solution to the differential equation is: Applying π’π’π₯π₯ (π¦π¦ = 0) = 0, B = 0. π’π’π₯π₯ = Applying π’π’π₯π₯ (π¦π¦ = π»π») = 0 π’π’π₯π₯ = The average velocity is given by: 1 ππππ 2 π¦π¦ + π΄π΄π΄π΄ + π΅π΅ 2ππ ππππ π΄π΄ = − 1 ππππ 2 π»π» ππππ π¦π¦ − π¦π¦ 2ππ ππππ 2ππ ππππ π’π’οΏ½ = π’π’οΏ½ = 1 π»π» οΏ½ π’π’ ππππ π»π» 0 π₯π₯ 1 π»π» 1 ππππ 2 π»π» ππππ οΏ½ οΏ½ π¦π¦ − π¦π¦οΏ½ ππππ π»π» 0 2ππ ππππ 2ππ ππππ π’π’οΏ½ = The velocity is therefore π»π» ππππ 2ππ ππππ 1 1 ππππ 3 π»π» ππππ 2 οΏ½ π»π» − π»π» οΏ½ π»π» 6ππ ππππ 4ππ ππππ π’π’οΏ½ = − 1 ππππ 2 π»π» 12 ππ ππππ ππππ 12 ππ = − 2 π’π’οΏ½ ππππ π»π» 7 π¦π¦ π¦π¦ 2 π’π’π₯π₯ = 6π’π’οΏ½ οΏ½ − οΏ½ οΏ½ οΏ½ π»π» π»π» Derivation of ππ and the governing concentration profile for plug flow: ππππ ππ 2 ππ π’π’0 = π·π· 2 ππππ πππ¦π¦ Assume a similarity variable exists in the form: ππ = π¦π¦ ππ(π₯π₯) Plugging this into the differential equation: π’π’0 ππππ ππππ ππ 2 ππ ππππ 2 = π·π· 2 οΏ½ οΏ½ π’π’0 ππππ πππ₯π₯ ππππ πππ¦π¦ ππππ π¦π¦ ππ′ (π₯π₯) ππ 2 ππ 1 2 οΏ½− οΏ½ = π·π· 2 οΏ½ οΏ½ ππ(π₯π₯) ππ(π₯π₯) ππππ ππ(π₯π₯) ππππ ππ 2 ππ π’π’0 ππππ 0 = 2 + οΏ½ οΏ½ ππ(π₯π₯)ππ′ (π₯π₯)ππ ππππ ππππ π·π· For the similarity variable to exist neither x nor y can appear in the governing π’π’ equation so οΏ½ π·π·0 οΏ½ ππ(π₯π₯)ππ′ (π₯π₯) must be equal to a constant. To simplify the math later this value can be set to 2. οΏ½ π’π’0 οΏ½ ππ(π₯π₯)ππ′ (π₯π₯) = 2 π·π· ππ(π₯π₯)ππ′ (π₯π₯) = 2 π·π· π’π’0 ππ(π₯π₯)2 π·π· = 2 π₯π₯ 2 π’π’0 Therefore: π·π· ππ(π₯π₯) = 2οΏ½ π₯π₯ π’π’0 ππ = π¦π¦ π’π’0 οΏ½ 2 π·π· π₯π₯ 8 ππ 2 ππ ππππ 0 = 2 + 2 ππ ππππ ππππ Derivation of ππ and the governing concentration profile for Poiseuille flow with the Leveque approximation: 6π’π’οΏ½ ππππ ππ 2 ππ π¦π¦ = π·π· 2 π»π» ππππ πππ¦π¦ Assume a similarity variable exists in the form: ππ = π¦π¦ ππ(π₯π₯) Plugging this into the differential equation: ππ 2 ππ ππππ 2 6π’π’οΏ½ ππππ ππππ π¦π¦ = π·π· 2 οΏ½ οΏ½ π»π» ππππ πππ₯π₯ ππππ πππ¦π¦ 6π’π’οΏ½ ππππ π¦π¦ 2 ′ ππ 2 ππ 1 2 οΏ½− οΏ½ οΏ½ ππ (π₯π₯)οΏ½ = π·π· 2 οΏ½ οΏ½ π»π» ππππ ππ(π₯π₯) ππππ ππ(π₯π₯) 0= ππππ ππ 2 ππ 6π’π’οΏ½ + οΏ½ οΏ½ ππ(π₯π₯)2 ππ′ (π₯π₯)ππ2 2 ππππ ππππ π»π»π»π» For the similarity variable to exist neither x nor y can appear in the governing οΏ½ 6π’π’ equation so οΏ½π»π»π»π»οΏ½ ππ(π₯π₯)2 ππ′ (π₯π₯) must be equal to a constant. To simplify the math later this value can be set to 3. 6π’π’οΏ½ οΏ½ οΏ½ ππ(π₯π₯)2 ππ′ (π₯π₯) = 3 π»π»π»π» ππ(π₯π₯)2 ππ′ (π₯π₯) = π»π»π»π» 2π’π’οΏ½ ππ(π₯π₯)3 π»π»π»π» = π₯π₯ 3 2π’π’οΏ½ 3 Therefore: ππ(π₯π₯) = οΏ½ 3π»π»π»π» π₯π₯ 2π’π’οΏ½ 3 2π’π’οΏ½ ππ = π¦π¦ οΏ½ 3π»π»π»π»π»π» 9 ππ 2 ππ ππππ 0 = 2 + 3 ππ2 ππππ ππππ 10 MIT OpenCourseWare http://ocw.mit.edu 10.626 Electrochemical Energy Systems Spring 2014 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
