Ideal Fluids in Motion — Question 10

Section 14-7 · Physics pages 2190–2191

Question 10 A tracer must follow the fluid closely enough to measure it A tracer particle is used to photograph a nominally horizontal flow over an observation time T = 0.50 s. Relative to the liquid it settles at ws = 0.0020 m/s. Its velocity response to a sudden change of fluid velocity has a supplied mismatch model ∆v(t) = ∆v(0)e−t/τ with τ = 0.10 s. t (s) ∆v/∆v(0) .50 1 The velocity mismatch decays rapidly, but settling displacement continues accumulating. (a) Find its settling displacement during the exposure and the fraction of its initial velocity mismatch remaining after T. (b) The desired vertical-path accuracy is 0.10 mm. Does this tracer meet it? State what these calculations say about identifying photographed particle tracks with fluid pathlines.
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Question 10 - Solution A tracer must follow the fluid closely enough to measure it t (s) ∆v/∆v(0) .50 1 (a) Derivation and calculation ∆y = wsT = (.0020)(.50) = .00100 m = 1.00 mm . ∆v(T) ∆v(0) = e−T/τ = e−5 = .00673795 = 0.673795% . (b) Result and interpretation ∆y ϵy = 1.00 mm .100 mm = 10 . The tracer fails the vertical-path criterion even though its final velocity mismatch is small. small response lag̸ ⇒negligible accumulated slip. A photographed track approximates a fluid pathline only within the permitted buoyancy and inertia errors. Physical check At the given settling speed, the maximum exposure for a .100 mm settling error is only .0500 s.

Question 10

A tracer must follow the fluid closely enough to measure it

A tracer particle is used to photograph a nominally horizontal flow over an observation time T=0.50sT=0.50\,\mathrm{s}. Relative to the liquid it settles at ws=0.0020m/sw_s=0.0020\,\mathrm{m/s}. Its velocity response to a sudden change of fluid velocity has a supplied mismatch model Δv(t)=Δv(0)e−t/τ\Delta v(t)=\Delta v(0)e^{-t/\tau} with τ=0.10s\tau=0.10\,\mathrm{s}.

Diagram shown in the original SVG worksheet.

  1. Find its settling displacement during the exposure and the fraction of its initial velocity mismatch remaining after TT.

  2. The desired vertical-path accuracy is 0.10mm0.10\,\mathrm{mm}. Does this tracer meet it? State what these calculations say about identifying photographed particle tracks with fluid pathlines.

Question 10 - Solution

A tracer must follow the fluid closely enough to measure it

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Δy=wsT=(.0020)(.50)=.00100m=1.00mm.\boxed{\Delta y=w_sT=(.0020)(.50)=.00100\,\mathrm m=1.00\,\mathrm{mm}}.Δv(T)Δv(0)=e−T/τ=e−5=.00673795=0.673795%.\boxed{\frac{\Delta v(T)}{\Delta v(0)}=e^{-T/\tau}=e^{-5}=.00673795=0.673795\%}.

(b) Result and interpretation

Δyϵy=1.00mm.100mm=10.\frac{\Delta y}{\epsilon_y}=\frac{1.00\,\mathrm{mm}}{.100\,\mathrm{mm}}=\boxed{10}.The tracer fails the vertical-path criterion even though its final velocity mismatch is small.small response lag⇏negligible accumulated slip.\text{small response lag}\ \not\Rightarrow\ \text{negligible accumulated slip}.A photographed track approximates a fluid pathline only within the permitted buoyancy and inertia errors.

Physical check

At the given settling speed, the maximum exposure for a .100 mm settling error is only .0500 s.