# Mass conservation in fluid mechanics

1. Jan 21, 2006

### r4nd0m

Part of the mechanics course I'm taking this semester are also fluids, but the material our teacher gave us to this topic is very poor (but unfortunately I haven't found a better source). The problem is that there are many "magic formulas", which come just out of nowhere - without any explanation.
I have a problem with the very first chapter and that is the mass conservation in fluids.
So if we have a fluid in a pipe with constant area S,
fluid has density $$\varrho(x,t)$$ (depending on time and location), and velocity v(x,t) then the mass of the fluid in volume $$V = S*\Delta x$$ is $$m = S*\Delta x * \varrho$$
And now the formula I don't understand occurs:
$$\frac{\partial (\varrho*S*\Delta x)}{\partial t} = \varrho(x,t)*v(x,t)*S - \varrho(x+ \Delta x,t)*v(x+ \Delta x,t)*S$$
Could someone plaese explain me how can we come to that equation?

Last edited: Jan 21, 2006
2. Jan 21, 2006

### arildno

1.Mass conservation implies that the only way in which the mass within the pipe region of length $\bigtriangleup{x}$ and cross-sectional area S is that mass either leaves or enters the region (there is no mass source or sink within the region).
2. The change of mass within the region from time t to $t+\bigtriangleup{t}$ is clearly given by:
$$\bigtriangleup{m}=(\rho(x,t+\bigtriangleup{t})-\rho(x,t))S\bigtriangleup{x}\approx\frac{\partial\rho}{\partial{t}}S\bigtriangleup{t}\bigtriangleup{x}$$
3. However, the amount of mass entering the region from the "x"-side during the same time interval is given by:
$\rho(x,t)v(x,t)S\bigtriangleup{t}$
whereas the amount of mass leaving at the $x+\bigtriangleup{x}$-side is given by:
$\rho(x+\bigtriangleup{x},t)v(x+\bigtriangleup{x},t)S\bigtriangleup{t}$
Thus, we have an alternative expression for $\bigtriangleup{m}$:
$$\bigtriangleup{m}=\rho(x,t)v(x,t)S\bigtriangleup{t}-\rho(x+\bigtriangleup{x},t)v(x+\bigtriangleup{x},t)S\bigtriangleup{t}\approx{-}\frac{\partial}{\partial{x}}(\rho{v})S\bigtriangleup{x}\bigtriangleup{t}$$
4. Setting these expressions for the change in mass equal to each other yields finally:
$$\frac{\partial\rho}{\partial{t}}=-\frac{\partial}{\partial{x}}(\rho{v})$$

Last edited: Jan 21, 2006
3. Jan 22, 2006

### r4nd0m

thanks a lot, this was really helpful.

But unfortunately I am stuck again - now with turbulent flow:
$$\frac {\partial \rho v_{x}}{\partial t} + \frac {\partial \rho v_{x}^2}{\partial x} = -\frac {\partial p}{\partial x} + \rho g_{x} - \frac {\lambda \rho V_{x}^2}{2D}$$

there are some symbols that are unknown to me - D is probably hydraulic diameter, but I don't know what $$V_{x}$$ and $$\lambda$$ stand for. And another thing - further in the text when solving some problems the left side of the equation is 0 - why?
Any ideas?