Question

#1 The Equation of Continuity: Consider Figure 1 which illustrates a small mathernat ical box in a fluid. The basic idea behind the equation of continuity is that the rate of mass flow into the box must equal the time rate of change of the mass in the box (pur)l Figure 1: A small mathematical box in a fluid A) Consider just the 2-faces of the box for now. Figure 1 shows ρυ.Ε entering at and leaving at + ρ is the fhnd density which could depend on position and u is ther component of the speed of the fhì which also could depend on position). What are the units of the quantity pu.,? What are the units of this term if it is multiplied by an area? (B) The rate of mass flow into the box at z is given by+pmus dzdy and the rate of mass flow out of the box at - dr is given by -(pu) dzdy. Use what you know about derivatives to show that the net mass flow into the box is given by dV where dV -dudydz is the volume of the box (C) The net mass flow into the box, Equation 1, should equal the time rate of change of mass in the volume, dV. Use the definition of density to show that, ot ot (D) The one-dimensional equation of continuity is now dV Show that similar analysis including mass flow through the top and bottom as well as the front and back yields the following (E) The term in square brackets looks like a divergence! Recall a divergence is given by + ал a Use this to write the equation of continuity in its standard form sume that the density fluctuations and velocity fluctuations are small and derive a ized equation of continuity. Recall the definition of the condensation, ρ-Po(1 + s). ρο is the ambient (F) Now we want to as

Answer 1

(A) u x is the velocity of flow along x-axis p is density of fluid then momentum = m v = u beta v = then rho/v = rho v_x rightdoublearrow momentum along per unit volume when it is multiple by area it represents momentum along x-axis (b) the mass flow out of per x = dx = -(rho v_x) x dy dz we by using the stokes theorem integral_S A d s = integral_v nabla A dv ds - surface element dV - volume element \r\n from -ol_S (rho v_x) dy dz = -integral_v nabla (rho u x) dx dy dz along x-axis sine u x is then only nable rho u x = partial differential/partial differential x (rho u x) then -rho u x dx dy = -partial differential/partial differential x (rho u x) dV (c) we can prove that rho m/rho t = rho p/rho t dv the net mass flow outside in all direction as given by
$from alom x-axs Sinc ux.@stnn 0%\g 3 9 d 2 dヒ d L -3운 Hence p-oved$