Fluid Mechanics Questions & Answers  












In this problem, it is stated that the center velocity is measured using a pitotstatic tube. So, in this case what is measured is not $q_\infty$ but $u$ in the middle of the pipe. Recall from the class notes that we derived an expression for $u$ as a function of $u_{\rm b}$ for fullydeveloped laminar flow in a pipe:$$ \frac{u}{u_{\rm b}}=2\left(1\frac{r^2}{R^2} \right) $$ If the flow is laminar, then use the latter to find $u_{\rm b}$ from the center velocity (velocity at ${r=0}$). For fullydeveloped turbulent flow, recall that the velocity profile in the pipe is almost uniform and varies only near the walls. Therefore, if the flow is turbulent, you can assume that the velocity in the center of the pipe is equal to the bulk velocity. Also, keep in mind that the flow in the pipe is air, not H$_2$O. Therefore, you cannot use directly the expression $u_{r=0}=\sqrt{2g\Delta {H}}$ to find the center velocity. You need to draw the Pitot tube, analyze the problem with Pascal's law, and then find out a new expression for the center velocity as a function of the height difference. Hint: the height difference in mm H$_2$O is proportional to the pressure difference between the pitot and static tubes, which itself is related to the difference between the stagnation and static pressure of the air.




Any liquid will “lose memory” of its initial shape when forces are applied to it. It doesn't matter what the initial shape was, when a liquid is deformed by forces, it will not go back unassisted (by itself) to its initial shape. But, a solid doesn't behave this way. A solid has a memory of its initial shape. After forces stop being applied, the solid will go back unassisted to its initial shape. It's a good question but there are several typo errors and it could have been formulated a bit better. I'll give you 1 point bonus boost for it.



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