Flux — Question 1

Question 1

An inclined opening measures the normal component of a flow

A uniform fluid velocity is $\mathbf v=(3,4,0)$ m/s. An oriented planar opening has area $.200$ m$^2$ and unit normal $\hat{\mathbf n}=(4,-3,0)/5$. Define signed volume flux as $Q=\int\mathbf v\cdot\hat{\mathbf n}\,dA$.

1. Find Q and explain how it can vanish despite nonzero speed.

2. Rotate the normal within the xy plane to maximize positive Q. Give the new normal and maximum flux.

Question 1 - Solution

An inclined opening measures the normal component of a flow

(a) Method and calculation

$$\mathbf v\cdot\hat{\mathbf n}=(12-12)/5=0,\quad Q=A\mathbf v\cdot\hat{\mathbf n}=0.$$$$|\mathbf v|=\sqrt{3^2+4^2}=5.00\,\mathrm{m/s}.$$The flow is tangent to this opening, so no fluid crosses it.

(b) Interpretation and further calculation

$$Q=Av\cos\theta\le Av,\quad\hat{\mathbf n}_{\max}=\mathbf v/v=(3,4,0)/5.$$$$Q_{\max}=(.200)(5.00)=1.00\,\mathrm{m^3/s}.$$

Physical check

Flux uses the angle with the surface normal; using the angle with the plane would interchange sine and cosine.