# How to infer from the volume fraction function whether a flow is transient or steady?

Hello dear engineering community.

I have a specific question about multiphase flows. I was wondering if anybody could help me with it, please.

Assume, I have a one dimensional water-air flow. Air is the dispersed phase. Let's say, it's a vertical flow. Water phase flow is incompressible.

QUESTION: if volume fraction (either of liquid or gas phase) is time dependent, then does it mean that that the flow is necessarily transient (either liquid or gas, or the entire mixture flow)? Or it can be steady?

• Dear Fred. I appreciate your carefulness about the way questions are written on this website. Nonetheless, I would like my question to be written as I wrote it and any suggestions for corrections should me discussed with me at first. I understand that rules of this website may prohibit the form of writing I opted for my question. In this case, I encourage you to raise a question of banning me from this website. – Ivan Nepomnyashchikh Aug 7 '19 at 15:39

I think I've a partial answer to my question.

1) Proof that liquid phase flow is steady if liquid volume fraction is time independent.

Continuity equation for incompressible liquid phase: $$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$ Let's say that volume fraction is not time dependent and only spatially dependent; and assume that velocity is still spatially and time dependent. Then we'll have:$$\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$ $$\frac{d \alpha_L}{dx}·v_L+ \frac{\partial v_L}{\partial x}· \alpha_L=0$$ $$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{\partial v_L}{\partial x}·\frac{1}{v_L}$$ Now we have that the entire left hand side is only spatially dependent, whereas the entire right hand side is assumed to be both spatially and time dependent. But it can't be. If we know for sure that the left hand side is not time dependent and is equal to the right hand side, then the right hand side must also not be time dependent: $$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{d v_L}{d x}·\frac{1}{v_L}$$ Therefore, it follows that if liquid volume fraction is not time dependent then the liquid flow is steady.

The same discussion should work for the gaseous phase (but I am not sure about that).

2) Proof that liquid phase flow is transient if liquid volume fraction is time independent.

Continuity equation for incompressible liquid phase: $$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$ Let's say that volume fraction is both time and space dependent. Then assume that velocity can be only spatially dependent. Prove that it can't be. Let's do further math manipulations with liquid continuity: $$\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L+\frac{d v_L}{d x}·\alpha_L=0$$ $$\frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)=-\frac{d v_L}{d x}$$ Now, even if both of $$\alpha_L$$ partial derivatives are constants and with the assumption that $$v_L=f(x)$$, the left hand side is still time dependent because $$\alpha_L=f(x,t)$$.

Let's write an integral formulation: $$v_L=-\int \frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)dx$$ From the integral formulation it is explicitly seen that since the integrand is time dependent, the result will also be time dependent. It means that if $$\alpha_L=f(x,t)$$, $$v_L$$ cannot be a function of space only.

Therefore, it follows that if liquid volume fraction is time dependent then the liquid flow is transient.

The same discussion should work for the gaseous phase (but I am not sure about that).