Question #121456
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.
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Expert's answer
2020-06-11T16:38:44-0400

a) First we consider the mass of a fluid which leaves region QQ by flowing across an element in δS\delta S of SS in time δt\delta t . This quantity is exactly that which is contained in a small cylinder of cross section δS\delta S of length (vnˇ).δt(v*ň).\delta t hence the rate at which fluid leaves qq by flowing across element SδS\delta is ρ(vnˇ)δS\rho(v*ň)\delta S .

Summing over all such elementsδS\delta S , we obtain the rate of flow of fluid coming out of qq across the entire surface SS . Hence the rate at which mass flow out of region qq is ρ(vnˇ)δS=(ρv).nˇδS\int \rho(v*ň)\delta S=\int(\rho v).ň\delta S .

b) By Gauss divergence theorem div(ρv) deltaV\int div(\rho v)\ delta V , the mass MM of the fluid possesed by the volume VV of the fluid is M=ρδVM=\int\rho\delta V where ρ=ρ(x,y,z,t)\rho = \rho(x,y,z,t) with (x,y,z)(x,y,z) the Cartesian co coordinates of a general point of qq , a fixed region of space. Since the space coordinates are independent of timett ,therefore the rate of increase of mass within VV is δMδt\delta M\over \delta t== δδt\delta \over\delta t == (( ρdV\int \rho dV)) == \int δρδt\delta \rho \over \delta tdV*dV .

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

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

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


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