The Equation of Continuity — Question 9

Section 14-8 · Physics pages 2208–2209

Question 9 A local velocity field must preserve parcel volume A candidate incompressible two-dimensional flow is vx = ax and vy = by, where a = 3.0 s−1 and b is unknown. There is no motion or variation in the third direction. Consider a small rectangle of fluid with side lengths δx and δy. δx δy Initial parcel; boundary velocity differences determine its area change. (a) Find b by requiring that the rectangle's area remain constant to first order in a short time dt. (b) If instead b = −1.0 s−1, find the instantaneous fractional rate of volume change and the density rate for a material parcel of density 1000 kg/m3.
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Question 9 - Solution A local velocity field must preserve parcel volume δx δy (a) Derivation and calculation Over a short interval, δx′ = δx(1 + a dt), δy′ = δy(1 + b dt). δA′ δA = (1 + a dt)(1 + b dt) = 1 + (a + b)dt + O(dt2). DV/Dt = 0 ⇒a + b = 0 ⇒b = −3.00 s−1 . (b) Result and interpretation For b = −1.00 s−1, 1 V DV Dt = a + b = 2.00 s−1 . m = ρV = constant ⇒V Dρ Dt + ρDV Dt = 0. Dρ Dt = −ρ(a + b) = −(1000)(2) = −2000 kg m−3 s−1 . Physical check The alternate field describes expansion with decreasing density and cannot represent an incompressible parcel.

Question 9

A local velocity field must preserve parcel volume

A candidate incompressible two-dimensional flow is vx=axv_x=ax and vy=byv_y=by, where a=3.0s−1a=3.0\,\mathrm{s^{-1}} and bb is unknown. There is no motion or variation in the third direction. Consider a small rectangle of fluid with side lengths δx\delta x and δy\delta y.

Diagram shown in the original SVG worksheet.

  1. Find bb by requiring that the rectangle's area remain constant to first order in a short time dtdt.

  2. If instead b=−1.0s−1b=-1.0\,\mathrm{s^{-1}}, find the instantaneous fractional rate of volume change and the density rate for a material parcel of density 1000kg/m31000\,\mathrm{kg/m^3}.

Question 9 - Solution

A local velocity field must preserve parcel volume

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Over a short interval,δx′=δx(1+adt),δy′=δy(1+bdt).\delta x^{\prime}=\delta x(1+a\,dt),\quad\delta y^{\prime}=\delta y(1+b\,dt).δA′δA=(1+adt)(1+bdt)=1+(a+b)dt+O(dt2).\frac{\delta A^{\prime}}{\delta A}=(1+a\,dt)(1+b\,dt)=1+(a+b)dt+O(dt^2).DV/Dt=0⇒a+b=0⇒b=−3.00s−1.D V/Dt=0\Rightarrow a+b=0\Rightarrow\boxed{b=-3.00\,\mathrm{s^{-1}}}.

(b) Result and interpretation

For b=−1.00s−1b=-1.00\,\mathrm{s^{-1}},1VDVDt=a+b=2.00s−1.\boxed{\frac1V\frac{DV}{Dt}=a+b=2.00\,\mathrm{s^{-1}}}.m=ρV=constant⇒VDρDt+ρDVDt=0.m=\rho V=\text{constant}\Rightarrow V\frac{D\rho}{Dt}+\rho\frac{DV}{Dt}=0.DρDt=−ρ(a+b)=−(1000)(2)=−2000kgm−3s−1.\boxed{\frac{D\rho}{Dt}=-\rho(a+b)=-(1000)(2)=-2000\,\mathrm{kg\,m^{-3}\,s^{-1}}}.

Physical check

The alternate field describes expansion with decreasing density and cannot represent an incompressible parcel.