Chapter 9, Problem 40P

Assume the flow conditions given in Example 9.3. Plot δ, δ*, and τ_w versus x/L for the plate.

Example 9.3 TURBULENT BOUNDARY LAYER ON A FLAT PLATE: APPROXIMATE SOLUTION USING $\frac{1}{7}$-POWER VELOCITY PROFILE

Water flows at U = 1 m/s past a flat plate with L = 1 min the flow direction. The boundary layer is tripped so it becomes turbulent at the leading edge. Evaluate the disturbance thickness, δ, displacement thickness, δ*, and wall shear stress, τ_w, at x = L. Compare with laminar flow maintained to the same position. Assume a $\frac{1}{7}$-power turbulent velocity profile.

Given: Flat-plate boundary-layer flow; turbulent flow from the leading edge. Assume $\frac{1}{7}$-power velocity profile.

Find: (a) Disturbance thickness, δ_L.

(b) Displacement thickness, ${\delta }_L^{*}$.

(c) Wall shear stress, τ_w(L).

(d) Comparison with results for laminar flow from the leading edge.

Solution: Apply results from the momentum integral equation.

Governing equations:

$\begin{array}{l}\frac{\delta }{x}=\frac{0.382}{R{e}_x^{1/5}}(9.26)\\ \delta *={\displaystyle {\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy}(9.1)\\ C_f=\frac{{\tau }_w}{\frac{1}{2}\rho U^2}=\frac{0.0594}{R{e}_x^{1/5}}(9.27)\end{array}$

At x = L, with ν = 1.00 × 10⁻⁶ m²/s for water (T = 20°C),

$R{e}_L=\frac{UL}{v}=1\frac{\text{m}}{\text{s}}\times 1\text{m}\times \frac{\text{s}}{{10}^{-6}{\text{m}}^{\text{2}}}={10}^6$

From Eq. 9.26,

${\delta }_L=\frac{0.382}{R{e}_L^{1/5}}L=\frac{0.382}{{({10}^6)}^{1\text{/}5}}\times 1\text{m}=0.0241\text{m}\text{or}{\delta }_L=24.1\text{mm}\underset{\leftarrow }{{\delta }_L}$

Using Eq. 9.1, with u/U = (y/δ)^1/7 = η^1/7, we obtain

$\begin{array}{l}{\delta }_L^{*}={\displaystyle {\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy={\delta }_L{\displaystyle {\int }_0^1\left(\frac{u}{U}\right)d\left(\frac{y}{\delta }\right)={\delta }_L}{\displaystyle {\int }_0^1(1-{\eta }^{1/7})d\eta =}}{\delta }_L{\left[\eta -\frac{7}{8}{\eta }^{8\text{/}7}\right]}_0^1\\ {\delta }_L^{*}=\frac{{\delta }_L}{8}=\frac{24.1\text{mm}}{8}=3.01\text{mm}\underset{\leftarrow }{{\delta }_L^{*}}\end{array}$

From Eq. 9.27,

$\begin{array}{l}{C}_f=\frac{0.0594}{{({10}^6)}^{1\text{/}5}}=0.00375\\ {\tau }_w=C_f\frac{1}{2}\rho U^2=0.00375\times \frac{1}{2}\times 999\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times {(}^1\frac{{\text{m}}^2}{{\text{s}}^2}\times \frac{\text{N}\cdot {\text{s}}^{\text{2}}}{\text{kg}\cdot \text{m}}\\ {\tau }_w=1.87{\text{N/m}}^{\text{2}}\underset{\leftarrow }{{\tau }_w(L)}\end{array}$

For laminar flow, use Blasius solution values. From Eq. 9.13 (on the web),

${\delta }_L=\frac{5.0}{\sqrt{R{e}_L}}L=\frac{5.0}{{({10}^6)}^{1\text{/}2}}\times 1\text{m}=0.005\text{m}\text{or 5}\text{.00}\text{mm}$

From Example W9.1, δ*/δ = 0.344, so

${\delta }^{*}=0.344\delta =0.344\times 5.0\text{mm}=1.72\text{mm}$