The velocity distribution in the boundary layer is given as \(\fr
The velocity distribution in the boundary layer is given as \(\frac{u}{U}=\frac{y}{\delta }\). The dimensionless-profile shape factor is
A. 1/3
B. 3
C. 2
D. 1/2
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Right Answer is: B
SOLUTION
Concept:
Dimensionless profile shape factor (H) = Displacement thickness/Momentum thickness = δ*/θ
\({{\delta }^{*}}=\mathop{\int }_{0}^{\delta }\left( 1-\frac{u}{U} \right)dy,\theta =\mathop{\int }_{0}^{\delta }\frac{u}{U}\left( 1-\frac{u}{U} \right)dy\)
\(\frac{u}{U}=\frac{y}{\delta }\)
\({{\delta }^{*}}=\mathop{\int }_{0}^{\delta }\left( 1-\frac{y}{\delta } \right)dy=\left\{ y-\frac{{{y}^{2}}}{2\delta } \right\}_{0}^{\delta }=\delta -\frac{{{\delta }^{2}}}{2\delta }=\frac{\delta }{2}\)
\(\theta =\mathop{\int }_{0}^{\delta }\frac{u}{U}\left( 1-\frac{u}{U} \right)dy=\mathop{\int }_{0}^{\delta }\frac{y}{\delta }\left( 1-\frac{y}{\delta } \right)dy\)
\(\theta =\mathop{\int }_{0}^{\delta }\left( \frac{y}{\delta }-\frac{{{y}^{2}}}{{{\delta }^{2}}} \right)dy\)
\(=\left\{ \frac{{{y}^{2}}}{2\delta }-\frac{{{y}^{3}}}{3{{\delta }^{2}}} \right\}_{0}^{\delta }\)
\(\Rightarrow \frac{\delta }{2}-\frac{\delta }{3}=\frac{\delta }{6}\)
\(H=\frac{{{\delta }^{*}}}{\theta }=\frac{\frac{\delta }{2}}{\frac{\delta }{6}}=3\)
So, shape factor (H) = 3