Question 4 of 5: Turbulent inlet region of a two-dimensional channel

Question 4: Turbulent inlet region of a two-dimensional channel (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. U_∞(δ); the inlet length as L/h and the centre-line speed there; the streamwise pressure gradient at x = L; and the wall shear stress at x = L.

Approach. Conservation of mass written with the displacement thickness gives the core acceleration in one line; substituting the given growth law at δ = h/2 fixes the inlet length; Bernoulli in the irrotational core then converts that acceleration into a pressure gradient, and the aid-sheet skin-friction law gives the wall stress.

1. Part (a) — write mass conservation with the displacement thickness. The volume flux per unit width entering at x = 0 is $ U_o h $. At any station x the two boundary layers each remove a flux $ U_\infty\delta^{*} $ from what a uniform core of speed $ U_\infty $ would carry — that is the definition of $ \delta^{*} $ — so $$U_o h=\int_0^h u\,dy=U_\infty\left(h-2\delta^{*}\right)$$ With the given shape factor $ \delta^{*}=\delta/8 $ this becomes $$\boxed{\;U_\infty(\delta)=\frac{U_o h}{h-\delta/4}=\frac{U_o}{1-\dfrac{\delta}{4h}}\;}$$ The core must accelerate as the layers thicken; it does so hyperbolically, slowly at first and then sharply as δ approaches its limit.
2. Part (b), first half — evaluate the centre-line velocity at the end of the inlet. At x = L the layers meet, $ \delta = h/2 $, and the displaced height is $ h-\delta/4=h-h/8=7h/8 $, so $$\boxed{\;U_\infty(L)=\frac{U_o}{1-1/8}=\frac{8}{7}U_o=\frac{8}{7}(5)=5.714\ \text{m/s}\;}$$ The core has gained 14.3 % of its speed — an increase that is fixed by the shape factor alone and is independent of the fluid or the flow rate. (Downstream of L the profile relaxes to the fully developed turbulent shape, whose centre-line speed is only a little higher.)
3. Part (b), second half — invert the growth law for L. Substituting $ \mathrm{Re}_x=U_\infty x/\nu $ into the given law and solving for x when $ \delta = h/2 $, $$\begin{aligned} \delta&=\frac{0.2\,x}{\left(U_\infty x/\nu\right)^{1/7}} =0.2\,x^{6/7}\left(\frac{\nu}{U_\infty}\right)^{1/7}\ \Longrightarrow\quad L&=\left[\frac{h/2}{0.2\left(\nu/U_\infty\right)^{1/7}}\right]^{7/6} \end{aligned}$$ Using $ U_\infty = 5.714 $ m/s at the end of the inlet, $ (\nu/U_\infty)^{1/7}=(1.75\times10^{-7})^{1/7}=0.1083 $ and $$L=\left[\frac{1.0\times10^{-3}}{0.2(0.1083)}\right]^{7/6} =\left(0.04616\right)^{7/6}=0.02765\ \text{m}=27.6\ \text{mm}$$ $$\boxed{\;\frac{L}{h}=\frac{27.65}{2.0}=13.8\;}$$ As a self-consistency check, $ \mathrm{Re}_L=U_\infty L/\nu=1.58\times10^{5} $, giving $ 0.2L/\mathrm{Re}_L^{1/7}=1.00\times10^{-3} $ m — the assumed δ recovered exactly. An entrance length of about 14 channel heights is typical of turbulent duct flow, and an order of magnitude shorter than the laminar estimate $ L/h \approx 0.04\,\mathrm{Re}_h = 400 $ would be.
4. Part (c), first half — differentiate the core velocity. The pressure gradient follows from how fast the core is accelerating, so first $$\begin{aligned} \frac{dU_\infty}{d\delta}&=\frac{U_o h/4}{\left(h-\delta/4\right)^{2}} =\frac{U_o h/4}{\left(7h/8\right)^{2}}=\frac{16\,U_o}{49\,h}\ &=\frac{16(5)}{49(0.002)}=816.3\ \text{s}^{-1} \end{aligned}$$ and, differentiating $ \delta \propto x^{6/7} $, $$\frac{d\delta}{dx}=\frac{6}{7}\frac{\delta}{x}\bigg|_{x=L} =\frac{6}{7}\cdot\frac{1.0\times10^{-3}}{0.02765}=0.03101$$ so by the chain rule $ dU_\infty/dx=816.3\times0.03101=25.31\ \text{s}^{-1} $.
5. Part (c), second half — convert acceleration into a pressure gradient. Assumption (iv) makes the core irrotational, so Bernoulli holds along the centre-line and $ p+\tfrac12\rho U_\infty^{2} $ is constant; differentiating, $$\frac{dp}{dx}=-\rho\,U_\infty\frac{dU_\infty}{dx}=-(1000)(5.714)(25.31)$$ $$\boxed{\;\frac{dp}{dx}\bigg|_{x=L}=-1.45\times10^{5}\ \text{Pa/m}=-145\ \text{kPa/m}\;}$$ The gradient is favourable (falling pressure), as it must be while the core is speeding up. Over the whole 27.6 mm inlet the total pressure drop is of order 4 kPa, most of it concentrated near x = L where the acceleration is steepest.
6. Part (d) — apply the aid-sheet skin-friction law at x = L. With $ \mathrm{Re}_L=1.579\times10^{5} $, so $ \mathrm{Re}_L^{1/7}=5.529 $, $$C_{fx}=\frac{0.0266}{\mathrm{Re}_x^{1/7}}=\frac{0.0266}{5.529}=4.811\times10^{-3}$$ $$\tau_w=C_{fx}\cdot\tfrac12\rho U_\infty^{2} =4.811\times10^{-3}\times\tfrac12(1000)(5.714)^{2}$$ $$\boxed{\;\tau_w=78.5\ \text{Pa}\;}$$ The friction velocity is $ u_\tau=\sqrt{\tau_w/\rho}=0.280 $ m/s, about 4.9 % of the core speed — the ratio one expects for a turbulent layer at this Reynolds number, and a useful sanity check on the arithmetic.
7. Note the two length scales the answers imply. With $ \delta^{*}=\delta/8=0.125 $ mm and $ \theta=(7/72)\delta=0.0972 $ mm at x = L, the shape factor is $ H=\delta^{*}/\theta=9/7=1.286 $ — the classic value for a turbulent 1/7-power profile, and far below the 2.59 of a laminar Blasius layer. That is the quantitative reason the turbulent inlet is so much shorter: the fuller profile displaces less flow, so the core has to accelerate less, and the layer reaches h/2 in a few tens of channel heights.