A1 — Boundary layer and drag on a flat plate: 23-Chem-B1 Transport Phenomena, May 2017, Q1

Question 1: A1 — Boundary layer and drag on a flat plate (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.

Quantity	Symbol	Value
Free-stream velocity	$U$	$50\ \text{ft/s}$
Kinematic viscosity of air	$\nu$	$1.8\times10^{-4}\ \text{ft}^2/\text{s}$
Distance / plate length	$x=L$	$5\ \text{ft}$
Transition Reynolds number	$Re_{c}$	$3.2\times10^{5}$
Air state (for density)	$P,\,T$	$14.7\ \text{psia},\ 70\ \text{°F}=529.7\ \text{°R}$

Find. (a) the boundary-layer thickness $\delta$ at $x=5$ ft treating the layer as turbulent; (b) $\delta$ when a laminar leading length precedes the turbulent layer; (c) the drag force on the plate for the part-(a) treatment.

Fig. A1: The boundary layer thickens from the leading edge. At $x=5$ ft the local Reynolds number is well past transition, so the layer is turbulent; part (b) accounts for the thin laminar run before $x_c$.

Approach. Evaluate the local Reynolds number (it lands firmly in the turbulent range), apply the turbulent flat-plate thickness correlation for part (a), for part (b) join a turbulent layer to the laminar leading length through a virtual origin that matches the momentum thickness at transition, and close with the average turbulent drag coefficient for part (c).

Reynolds number at the station. $$Re_x=\frac{U x}{\nu}=\frac{(50)(5)}{1.8\times10^{-4}}=1.39\times10^{6}.$$ Since $Re_x\gg Re_c=3.2\times10^{5}$, the layer at $x=5$ ft is turbulent, and part (a) treats it as turbulent from the leading edge.
(a) Turbulent boundary-layer thickness. The one-seventh-power correlation gives $\delta=0.376\,x\,Re_x^{-1/5}$, so $$\delta_a=\frac{0.376(5)}{(1.39\times10^{6})^{1/5}}=\frac{1.88}{16.93}\ \Longrightarrow\ \boxed{\delta_a\approx0.111\ \text{ft}=1.33\ \text{in}.}$$
(b) Locate transition and the laminar thickness there. Transition occurs at $x_c=\nu\,Re_c/U=(1.8\times10^{-4})(3.2\times10^{5})/50=1.152$ ft. Over $0
(b) Momentum thickness carried across transition. The momentum deficit (drag accumulated so far) is continuous through transition, so the turbulent layer starts from the laminar momentum thickness $\theta_c=0.664\,x_c/\sqrt{Re_c}=0.664(1.152)/565.7=1.35\times10^{-3}$ ft, not from zero.
(b) Virtual origin of the turbulent layer. For the 1/7-power profile $\theta=\tfrac{7}{72}\delta$, so a turbulent layer grown from a virtual origin a distance $\xi$ upstream has $\theta=\tfrac{7}{72}(0.376)\,\xi^{4/5}(\nu/U)^{1/5}=2.98\times10^{-3}\,\xi^{4/5}$ (ft). Setting this equal to $\theta_c$ gives $\xi_c=(1.35\times10^{-3}/2.98\times10^{-3})^{5/4}=0.372$ ft: the turbulent layer behaves as if it had begun $0.372$ ft before $x_c$. At $x=5$ ft its effective length is $\xi=5-1.152+0.372=4.22$ ft, so $$\delta_b=0.376\,\xi^{4/5}\Big(\frac{\nu}{U}\Big)^{1/5}=0.376(4.22)^{0.8}\Big(\frac{1.8\times10^{-4}}{50}\Big)^{0.2}\ \Longrightarrow\ \boxed{\delta_b\approx0.097\ \text{ft}=1.16\ \text{in}.}$$ (Matching the thickness $\delta$ itself at $x_c$ instead of $\theta$ gives $0.095$ ft, practically the same.) Recognising the laminar leading run makes the layer thinner (it grows more slowly than a turbulent layer would from the leading edge).
(c) Air density and dynamic pressure. From the ideal-gas law with $R_{\text{air}}=1716\ \text{ft}\!\cdot\!\text{lbf}/(\text{slug}\!\cdot\!\text{°R})$ and $P=14.7(144)=2117\ \text{lbf/ft}^2$, $$\rho=\frac{P}{R_{\text{air}}T}=\frac{2117}{(1716)(529.7)}=2.33\times10^{-3}\ \text{slug/ft}^3,\qquad \tfrac12\rho U^2=2.91\ \text{lbf/ft}^2.$$
(c) Drag from the average turbulent friction coefficient. For the turbulent plate, $C_D=0.072\,Re_L^{-1/5}=0.072/16.93=4.25\times10^{-3}$. The drag on one side of the plate, per unit width $b$, is $$F_D=C_D\left(\tfrac12\rho U^2\right)L\,b=(4.25\times10^{-3})(2.91)(5)\,b\ \Longrightarrow\ \boxed{F_D\approx0.062\ \text{lbf per ft of width (one side)}.}$$

Quantity	Result
(a) Turbulent BL thickness at 5 ft	$\delta_a\approx0.111\ \text{ft}\ (1.33\ \text{in})$
(b) Thickness with laminar leading length	$\delta_b\approx0.097\ \text{ft}\ (1.16\ \text{in})$
Transition location	$x_c=1.15\ \text{ft}$
(c) Drag (one side, per unit width)	$F_D\approx0.062\ \text{lbf/ft}$ ($C_D=4.25\times10^{-3}$)

Check — plate width and side count

The statement gives no plate width, so the “total” drag is reported per unit width; multiply by the actual span $b$ (ft) for the total, and double it if both faces are wetted. The value is for the turbulent (part-a) treatment; the laminar-plus-turbulent split of part (b) would lower $C_D$ slightly and hence the drag.

Concept & Theory

Topic. External laminar/turbulent boundary layers on a flat plate; friction drag.

Key relations. Laminar (Blasius) $\delta=5.0\,x\,Re_x^{-1/2}$, $C_{fx}=0.664\,Re_x^{-1/2}$; turbulent (1/7-power) $\delta=0.376\,x\,Re_x^{-1/5}$, average $C_D=0.072\,Re_L^{-1/5}$; drag $F_D=C_D\,(\tfrac12\rho U^2)\,A$.

Why this works. The Reynolds number sets the regime; here $Re_x=1.4\times10^{6}$ is four times the transition value, so the layer is turbulent at the station and the 1/7-power law applies. The laminar leading run of part (b) is short ($x_c=1.15$ ft of the 5 ft), yet it thins the layer because a laminar layer grows more slowly ($\delta\propto x^{1/2}$) than a turbulent one ($\delta\propto x^{4/5}$).

Common pitfall. Applying the Blasius laminar formula at $x=5$ ft (it is only valid up to transition), or forgetting that the given “total drag” needs a plate area — without a width, report drag per unit width and state it.

Source. Welty et al., Momentum, Heat and Mass Transfer, Ch. 12 (boundary-layer flow and flat-plate drag).

Question 1 of 6: A1 — Boundary layer and drag on a flat plate

Question 1: A1 — Boundary layer and drag on a flat plate (25 marks)