Answer to Question #121456 in Calculus for Roger

Question

2. Suppose that a fluid has density p (function of space and time) and velocity v with no sources or sinks. (A) Show that the rate of change of the mass m of the fluid contained in a region Q is dm d! du. -SITE (B) Suppose further that, if the fluid crosses the boundary, show that dt -- S6w) - neds. (©) Using your results in Q2(A) and 02(B), show that де V. (e) + at = 0. (D) Why the contimity equation for water is given by V. (pv) = 0.

Answer 1

a) First we consider the mass of a fluid which leaves region $Q$ by flowing across an element in $\delta S$ of $S$ in time $\delta t$ . This quantity is exactly that which is contained in a small cylinder of cross section $\delta S$ of length $(v*ň).\delta t$ hence the rate at which fluid leaves $q$ by flowing across element $S\delta$ is $\rho(v*ň)\delta S$ .

Summing over all such elements$\delta S$ , we obtain the rate of flow of fluid coming out of $q$ across the entire surface $S$ . Hence the rate at which mass flow out of region $q$ is $\int \rho(v*ň)\delta S=\int(\rho v).ň\delta S$ .

b) By Gauss divergence theorem $\int div(\rho v)\ delta V$ , the mass $M$ of the fluid possesed by the volume $V$ of the fluid is $M=\int\rho\delta V$ where $\rho = \rho(x,y,z,t)$ with $(x,y,z)$ the Cartesian co coordinates of a general point of $q$ , a fixed region of space. Since the space coordinates are independent of time$t$ ,therefore the rate of increase of mass within $V$ is $\delta M\over \delta t$$=$ $\delta \over\delta t$ $=$ $($ $\int \rho dV$$)$ $=$ $\int$ $\delta \rho \over \delta t$$*dV$ .

c) The total rate of change of mass is , and thus from (2a) and (2b) we get,

$\int$ $\delta \rho \over \delta t$ $+diq(\rho v)=0$ .

d) When the motion of fluid is steady, then $\delta \rho \over \delta t$$=0$ . Thus the equation of continuity becomes$div(\rho v)=0$ . Which is the same for homogenous and incompressible fluid.