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With reference to a standard Cartesian (x, y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are separated by distance, $\mathrm{h}$, is given by the expression $u=-\frac{h^{2}}{8 \mu} \frac{d p}{d x}\left[1-4\left(\frac{y}{h}\right)^{2}\right]$ In this equation, the $\mathrm{y}=0$ axis lies equidistant between the plates at a distance $\mathrm{h} / 2$ from the two plates, $\mathrm{p}$ is the pressure variable and $\mu$ is the dynamic viscosity term. The maximum and average velocities are, respectively

$\begin{array}{l}
\text { (A) } u_{\max }=-\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{2}{3} u_{\max } \\
\text { (B) } u_{\max }=\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{2}{3} u_{\max } \\
\text { (C) } u_{\max }=-\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{3}{8} u_{\max } \\
\text { (D) } u_{\max }=\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{3}{8} u_{\max }
\end{array}$



Question ID - 156438 | SaraNextGen Top Answer

With reference to a standard Cartesian (x, y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are separated by distance, $\mathrm{h}$, is given by the expression $u=-\frac{h^{2}}{8 \mu} \frac{d p}{d x}\left[1-4\left(\frac{y}{h}\right)^{2}\right]$ In this equation, the $\mathrm{y}=0$ axis lies equidistant between the plates at a distance $\mathrm{h} / 2$ from the two plates, $\mathrm{p}$ is the pressure variable and $\mu$ is the dynamic viscosity term. The maximum and average velocities are, respectively

$\begin{array}{l}
\text { (A) } u_{\max }=-\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{2}{3} u_{\max } \\
\text { (B) } u_{\max }=\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{2}{3} u_{\max } \\
\text { (C) } u_{\max }=-\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{3}{8} u_{\max } \\
\text { (D) } u_{\max }=\frac{h^{2}}{8 \mu} \frac{d p}{d x} \text { and } u_{\text {average }}=\frac{3}{8} u_{\max }
\end{array}$

1 Answer
127 votes
Answer Key / Explanation : (A) -

$u_{\max }=-\frac{h^{2}}{8 \mu} \frac{d p}{d x}$ and $u_{\text {average }}=\frac{2}{3} u_{\max }$

127 votes


127