The Equation of Continuity — Question 10

Section 14-8 · Physics pages 2210–2211

Question 10 A moving control surface counts relative crossing speed A uniform incompressible stream has density ρ, area A and constant laboratory speed v to the right. A control region extends from a fixed left plane to a right plane moving right at speed u, with 0 < u < v. There is no flux through its side boundary. v u fixed moving Control region between a fixed inlet plane and a moving outlet plane. (a) Find the mass inflow through the fixed plane, outflow through the moving plane, and mass accumulation inside the region. (b) Evaluate for ρ = 1000 kg/m3, A = 0.0020 m2, v = 3.0 m/s and u = 1.0 m/s. What error results from using laboratory speed at both planes?
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Question 10 - Solution A moving control surface counts relative crossing speed v u fixed moving (a) Derivation and calculation Flux is counted relative to each control surface: ˙min = ρAv , ˙mout = ρA(v −u) . dMCV dt = ˙min −˙mout = ρAu . dVCV dt = Au ⇒d(ρVCV) dt = ρAu. (b) Result and interpretation ˙min = 1000(.00200)(3.0) = 6.00 kg/s . ˙mout = 1000(.00200)(3.0 −1.0) = 4.00 kg/s . dMCV/dt = 2.00 kg/s . Using laboratory speed at both planes falsely predicts zero accumulation despite the growing control-region volume. Physical check When u →0, the region becomes fixed and the two mass fluxes become equal.

Question 10

A moving control surface counts relative crossing speed

A uniform incompressible stream has density ρ\rho, area AA and constant laboratory speed vv to the right. A control region extends from a fixed left plane to a right plane moving right at speed uu, with 0<u<v0<u<v. There is no flux through its side boundary.

Diagram shown in the original SVG worksheet.

  1. Find the mass inflow through the fixed plane, outflow through the moving plane, and mass accumulation inside the region.

  2. Evaluate for ρ=1000kg/m3\rho=1000\,\mathrm{kg/m^3}, A=0.0020m2A=0.0020\,\mathrm{m^2}, v=3.0m/sv=3.0\,\mathrm{m/s} and u=1.0m/su=1.0\,\mathrm{m/s}. What error results from using laboratory speed at both planes?

Question 10 - Solution

A moving control surface counts relative crossing speed

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Flux is counted relative to each control surface:ṁin=ρAv,ṁout=ρA(v−u).\boxed{\dot m_{\mathrm{in}}=\rho Av},\qquad\boxed{\dot m_{\mathrm{out}}=\rho A(v-u)}.dMCVdt=ṁin−ṁout=ρAu.\boxed{\frac{dM_{\mathrm{CV}}}{dt}=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}=\rho Au}.dVCVdt=Au⇒d(ρVCV)dt=ρAu.\frac{dV_{\mathrm{CV}}}{dt}=Au\Rightarrow\frac{d(\rho V_{\mathrm{CV}})}{dt}=\rho Au.

(b) Result and interpretation

ṁin=1000(.00200)(3.0)=6.00kg/s.\boxed{\dot m_{\mathrm{in}}=1000(.00200)(3.0)=6.00\,\mathrm{kg/s}}.ṁout=1000(.00200)(3.0−1.0)=4.00kg/s.\boxed{\dot m_{\mathrm{out}}=1000(.00200)(3.0-1.0)=4.00\,\mathrm{kg/s}}.dMCV/dt=2.00kg/s.\boxed{dM_{\mathrm{CV}}/dt=2.00\,\mathrm{kg/s}}.Using laboratory speed at both planes falsely predicts zero accumulation despite the growing control-region volume.

Physical check

When u→0u\to0, the region becomes fixed and the two mass fluxes become equal.