Hello, I have a general question regarding energy coupling in multi phase flow. My question comes is based on the text from this book:(adsbygoogle = window.adsbygoogle || []).push({});

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:

Thus, the coupling parameter will be:

- 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)

$$

\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?

Thanks.

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# Multiphase Flow in Fluid Dynamics - Energy Coupling

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