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Multiphase Flow in Fluid Dynamics - Energy Coupling

  1. Mar 17, 2017 #1
    Hello, I have a general question regarding energy coupling in multi phase flow. My question comes is based on the text from this book:

    https://books.google.co.uk/books?id=CioXotlGMiYC&pg=PA33&lpg=PA33&dq=thermal+coupling+parameter+continuous+dispersed&source=bl&ots=s9uWpatRmM&sig=-aRSlKoQXHc7gTlZTsl30ROncfo&hl=en&sa=X&ved=0ahUKEwjh6-DfhdzSAhVELMAKHfKOBHAQ6AEIGjAA#v=onepage&q=thermal coupling parameter continuous dispersed&f=false

    (page 33)

    This is kind of obscure so I'll provide a quick background:

    Basically, the premise of the question is that there is a flow of fluid, the fluid has a continuous phase, and within the continuous there exists a dispersed phase, so something like water flowing with little droplets of oil inside the water. The idea is that, suppose the water is flowing through a pipe and into a control volume. The coupling parameter will compare the effect of the dispersed phase on the control volume to the continuous phases effect. So, a flow of fluid with a continuous phase containing n drops per volume flows into a cubic control volume of length L (a volume of $L^3$).

    Energy coupling is related to temperature. So, in energy coupling the two things that are to be compared are:
    • The energy (heat) released by the droplets (the convective heat transfer from droplets to water)
    • The energy provided by the continuous phase (the heat energy of the water)
    Thus, the coupling parameter will be:
    \Pi_{\text{Energy}} = \frac{\text{Heat Released by Droplets}}{\text{Enthalpy Flux from Water}}
    The heat released by the droplets comes from Newtons law of cooling:
    mC_d\frac{dT_d}{dt} = Nu k_c \pi D (T_c - T_d)
    D is the diameter of the droplet, Tc and Td are the temperatures of the continuous/dispersed phases. Nu is the Nusselt number. Again, there are n droplets per unit volume therefore the total heat released from the droplets is:
    nL^3Nu k_c \pi D (T_c - T_d)
    The heat flux due to the continuous will just be:
    \rho_c u C_c T_c L^2
    Which is the density of the continuous phase multiplied by the volume flowing per unit time multiplied by specific heat capacity and it's temperature. The coupling parameter is therefore:
    \Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d)}{\rho_c u C_c T_c L^2}

    Ok, so that's all fine and dandy. The issue is when they simplify the above formula. So there exists a thermal response time constant (I do not believe it matters where it comes from, this is basically an algebraic exercise at this point):
    \tau_T = \frac{\rho_dC_dD^2}{12k_c}
    This can be included in the above equation such that :
    \Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d) \rho_d C_d D^2 \tau_T}{\rho_c u C_c T_c L^2 \tau_T }
    This can be cleaned up, Nu/2 can be approximated to 1 at low Reynolds numbers. piD^2/6 multiplied by density of a droplet (spherical) gives the mass of that droplet, multiply that by n and you end up with the total droplet density per unit volume. The above equation simplifies to:
    \Pi_{\text{Energy}}= \frac{T_dCLC_d}{uC_c T_C \tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)
    Where C is the ratio of total dispersed phase density and continuous phase density.

    Here is the problem, the book quotes the final result as:
    \Pi_{\text{Energy}}= \frac{CL}{u\tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)
    Which implies that:
    \frac{C_dT_d}{C_c T_c} = 1

    I don't see any justification for that though...does anyone know where the extra terms went?

  2. jcsd
  3. Mar 18, 2017 #2
    I worked on a Physics Forums thread very similar to this about a month ago. It was a mass transfer problem rather than a heat transfer problem, but the basic setup was the same. Here is the reference: https://www.physicsforums.com/threads/cocurrent-diffusion-cant-be-this-hard.904956/
    Here is the corresponding analysis for your problem:
    If the total volumetric flow rate of continuous and discontinuous phases is F and the volume fraction of disperse phase is f, then Eqns. 1 and 2 can be rewritten as:
    $$f\rho_D C_D\frac{dT_D}{dt}=\alpha \phi\tag{1}$$
    $$(1-f)\rho_C C_C\frac{dT_C}{dt}=-\alpha \phi\tag{2}$$where t is the cumulative residence time (t = V/F), ##\alpha## is the heat transfer surface per unit volume and ##\phi## is the heat flux through the interface. The heat flux is given by:
    $$\phi=\frac{k_CNu}{2R}(T_C-T_D)\tag{3}$$where R is the discontinuous phase drop radius.
    The discontinuous phase drop radius R in Eqn. 3 is related to the volume fraction of drops f and the surface area per unit volume ##\alpha## by the equations:
    $$4\pi R^2 n=\alpha\tag{4}$$
    $$\frac{4}{3}\pi R^3 n=f\tag{5}$$where n is the number of drops per unit volume. If we divide Eqn. 5 by Eqn. 4, we obtain:
    If we add equation 1 and 2, we find that $$f\rho_D C_DT_D+(1-f)\rho_C C_CT_C=f\rho_D C_DT_{D0}+(1-f)\rho_C C_CT_{C0}\tag{7}$$
    Combining Eqns. 1-3 and 6 yields:
    $$\frac{dT_D}{dt}=\frac{k_C\ Nu\ \alpha^2}{6f^2\rho_D C_D}(T_C-T_D)\tag{8}$$
    $$\frac{dT_C}{dt}=-\frac{k_C\ Nu\ \alpha^2}{6f(1-f)\rho_C C_C}(T_C-T_D)\tag{9}$$
    If we subtract Eqns. 8 and 9, we obtain: $$\frac{d\ln(T_D-T_C)}{dt}=-\left[\frac{1}{f\rho_D C_D}+\frac{1}{(1-f)\rho_C C_C}\right]\frac{k_C\ Nu\ \alpha^2}{6f}\tag{10}$$
    Hope this helps.
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