Question by Student #201227102
 I am a student of your Fluid Mechanics class 2. I have two question for you. For 3 Q of Assignment 4, Considering the force applied to the blade is positive in $y$-direction, is it right that the force applied fluid is negative in $y$-direction? As a result, $F$ is finally negative. is that correct? I'm so sorry to bother you with late questions. I am looking forward to your answer.
Yes you are right, $F$ is finally negative. The force applied to the blade by the fluid is in the negative $y$-direction. Thus, because $F$ is drawn as pointing in the positive $y$-direction in the diagram, the answer will be $F=$  - ???? N.
 11.22.13
 Question by Student #201027110
 I am currently in morning class of Fluid Mechanics. I have a basic question between Reynolds and Drag coefficient. In class you said $\Pi_1 = f(\Pi_2)$ and then says $C_{\rm D} = f({\rm Re}_{D})$. Is it correct? why is not $C_{\rm D} = g({\rm Re}_{D})$ First $f$ and second $f$ is not same? Thank you very much :)
You are correct: the two $f()$ functions are not the same, so this is a bit confusing... I should have said $C_{\rm D}=f({\rm Re}_D)$ and $\Pi_1=g(\Pi_2)$.. This would have been clearer. I'll do this next year ;) Thanks for pointing this out.
 Question by Student #201027110
 I am currently in morning class of Fluid Mechanics. Mid term is coming soon. We have had many examples in class but I want to learn more various examples. So if you don't mind could you recommend some books having various examples to me?
I can recommend "Fluid Mechanics" by Fox and Macdonald and Pritchard. But any introductory textbook in fluid mechanics should be fine: what I cover in this course is standard..
 Question by Student #201027110
Reviewing example, I find something doesn't make sense. So I e-mail to you with attached files:
First, There is negative sign in equation because direction of acceleration in manometer is opposite to positive direction of $x$-axis(-> +). Can I understand this like relation $g$ (constant of gravity) and $y$-axis? Second, Force due to Helium is zero. It's just appear because density of He is too much low? I'm looking forward to hearing from you. :)
 1. The sign of the acceleration is negative because the acceleration felt by the fluid is in the negative $x$-direction. 2. Yes, the force due to helium can be taken as negligible compared to the force due to water because the density of helium is much less than the density of water.
Have a nice weekend!
 Question by Student #201027128
Could you check my think in next sentence?
in assignment problem $F$ is defined as the force action on the blade in the positive $y$- direction. This means Force acting on fluid in negative $y$ direction, so Body Force is $$\int_V B dV= F \vec{j}$$ is that right?
Yes, that is right!
 Question by Student #201027128
 Dr. Parent I have question about assignment 6 - 3. After integration, ${\cal P}_{\rm L}$ and ${\cal P}_{\rm R}$ is same because the shear stress is independent of $z$. But question is $\frac{1}{2}{\cal P}_{\rm L}={\cal P}_{\rm R}$ — it is very strange.
Yes, you are correct that the shear stress is not a function of $z$. However, the power corresponds to the integral of the shear stress times the area where the shear stress acts times the velocity where the shear stress acts:$${\cal P}_{\rm L}=\int_{r=0}^{r=R} \left(\tau_{zr}\right)_{z=0} \, 2 \pi r \, \omega_{\rm L} r \, dr$$$${\cal P}_{\rm R}=\int_{r=0}^{r=R} \left(\tau_{zr}\right)_{z=H} \,2 \pi r \, \omega_{\rm R} r \, dr$$In the latter note that ${\cal P}_{\rm L} \ne {\cal P}_{\rm R}$ because $\omega_{\rm L} \ne \omega_{\rm R}$.
 Question by Student 201027128 assignment 9 question 6 you teached that. but I don't understand well, use pascal's law and bernouil equation final equation is $q_\infty$ =$\sqrt{2g\triangle{h}}$. but I don't know how can start that $q_\infty$=$u_b$ ?? pleast teach me
 12.13.13
In this problem, it is stated that the center velocity is measured using a pitot-static 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 fully-developed 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 fully-developed 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.
 Question by Student 201227128 Professor I have a question about first calss. The question is about the characteristics of metarial's phases. In the calss, when you compared Solids and Liquids, you said that the Liquids are deformed permanently when forces are acted on them. But i thought that in general case the liquids also can get back to the undeformed state. For example the water or mercury drops can maintain their shape after that the forces are acted. ( Of course these meterials have surface force, but all of the meterials has intermolecular forces.) For this reason I just think the state that the Liquids are deformed forever when the forces are acted is not correct. But why did professor said like that?
 09.03.14
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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