# 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