Flux — Question 5

Question 5

A tent-shaped surface preserves projected flux despite extra area

A roof-shaped surface extends a distance L in y and spans $-a\le x\le a$. Its height is $z=h(1-|x|/a)$, with $a,h,L>0$. Orient both panels upward. A uniform flow $v_0\hat{\mathbf z}$ crosses the surface.

1. Find each panel area and its normal projection factor, then the total flux.

2. Calculate the area-weighted mean value of $\cos\theta$ over the entire surface and discuss the limit $h/a\to\infty$.

Question 5 - Solution

A tent-shaped surface preserves projected flux despite extra area

(a) Method and calculation

$$A_{\mathrm{panel}}=L\sqrt{a^2+h^2},\quad\cos\theta=\frac a{\sqrt{a^2+h^2}}.$$$$Q=2v_0A_{\mathrm{panel}}\cos\theta=2aLv_0.$$

(b) Interpretation and further calculation

$$A_{\mathrm{total}}=2L\sqrt{a^2+h^2},\quad\langle\cos\theta\rangle_A=\frac Q{v_0A_{\mathrm{total}}}=\frac a{\sqrt{a^2+h^2}}.$$$$h/a\to\infty:\quad\langle\cos\theta\rangle_A\to0,\quad Q=2aLv_0.$$Added tilted area compensates exactly for a smaller normal component.

Physical check

Net flux is set by oriented projection, not by unweighted surface area.