# Multiphase Flow - Energy Coupling Parameter

I have a general question regarding energy coupling in multi phase flow. My question comes is based on the text from this book:

(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?

Thanks.

• Upvoted because the question provides appropriate context and shows your thought process so far. – Brian Drummond Mar 17 '17 at 11:46
• The only thing I can see is that equation 2.66 in the book uses the "approximately equal" symbol ($\sim$) rather than the "exactly equals" symbol ($=$). Maybe the author just takes a big liberty with that and assumes the specific enthalpy is roughly equal across both phases (basically what your last equation says). – Carlton Mar 17 '17 at 17:38
• @Carlton I think you may be right, can't see any other way – MathsIsHard Mar 18 '17 at 11:01
• I'd agree with @Carlton that direct solutions to two phase heat transfer take a lot of liberties. My experience has been that once we developed decent CFD, researchers largely gave up trying to analytically solve two phase heat transfer problems because you had to make too many assumptions and the results frequently failed to be predictive. For example, semi-empircal flow boiling models frequently fail to agree on the order of magnitude of the heat transfer. – ericksonla Mar 19 '17 at 13:56

## 1 Answer

No answers, I agree with Carlton in the comments, some very large liberties have been taken.