Fundamentals of Fluid Mechanics Questions and Answers  
Question by AME536A Student
I am getting stuck on problem #4 on the homework (the canoe question). I am able to do part A and it gives me the correct answer for part c, but I am having trouble with the integral on part B. The task is to integrate the forces instead of using Archimedes principle. The basic premise is pressure multiplied by area, so starting off, we can say the area integral for the submerged part of the boat is $$ \int_{\theta_0}^{\theta_1} \int_{0}^x rd \theta dx $$ (I couldn't get Latex to work with subscripts in the integral bounds. It should be $\theta^{}_{0}$ and $\theta^{}_{1}$). Next we can use Pascal's law to find the pressure: $$ P = P^{}_{atm} + \rho^{}_{w}g(H-y) $$ Here is where I am getting stuck. Originally I had the integral set up using Y and integrating from 0 to H multiplied by the area to get force: $$ \int_{\theta}^\theta \int_{0}^x \int_{0}^H r(P^{}_{atm} + \rho^{}_{w}g(H-y)) \sin(\theta) d \theta dx dy $$ This doesn't seem like the right way to integrate to find the force. There isn't really a y direction, it's just backward along the radius. So, my thought was to find Y as a function of $\theta$: $$ y = R(1-\sin(\theta)) $$ So the integral becomes $$ \int_{\theta}^\theta \int_{0}^x r(P^{}_{atm} + \rho^{}_{w}g(H-R(1-\sin(\theta))) \sin(\theta) d \theta dx $$ Neither of the ways I've done the problem thus far are resulting in the correct answer. I'm unsure of what is going wrong. Are either of the two ways I mentioned the correct way to find the force applied on the bottom of the canoe?
It's difficult to say where the problem is because you don't define $y$ and $\theta_0$ and $\theta_1$. Is $y$ pointing up or down? What is $\theta_0$? This should be made clear through text or figure. You seem to be on the right track thus. There's probably a small mistake in the setup of the integral. Have you tried integrating only half of the canoe and then multiplying by 2? This will get rid of one potential error related to the angles.. PS. I fixed your problem in your post. To see how I did it, right click on the equation and choose “Show Math As” -> “TEX Commands”.
Question by AME536A Student
I have a quick question about Problem 6, based on the figure provided, do we need to take into account of mercury surface tension on the side walls of the tank ?
If there is enough information to take this into account, then yes you have to. If not you have to mention this in the assumptions.
Question by AME536A Student
In question #2 for Assignment #3, can we assume that the duct that the fluid is traveling through is square?
No, you shouldn't make this assumption. Your answer should work for any type of duct as long as the intake is aligned with the $y$ axis.
Question by AME536A Student
In Assignment #6, Question #3 :
(1) Is it okay to use the conservation of mass equation twice, one for the water and one for the air? therefore we have $$\frac{dm_A}{dt}=...$$ $$\frac{dm_L}{dt}=...$$
(2) Can we assume that $e_L$ is constant with time (or the water temperature is constant) so that: $$\frac{dm_L e_L}{dt}= e_L\frac{dm_L}{dt}$$
Yes this is correct. However, if you do assume that the water temperature is constant in time, you need to justify this assumption. Why would this be the case? Explain. Further, only one question per post in the future.
Question by AME536A Student
On Assignment #6, Question #3 I am stuck on part B of the tank problem. We're tasked to find the temp. once all of the water has left the tank. Using the energy transport equation and mass conservation equation I've derived the following: $$ (e^{}_{a}-h^{}_r)dm^{}_{a} + (e^{}_{L}-h^{}_{L})dm^{}_{L} + m^{}_{a}de^{}_{a} = 0 $$ which can be re-written as follows in preparation for integration: $$ \frac{1}{m^{}_{a}}dm^{}_{a} + \frac{(e^{}_{L}-h^{}_{L})} {m^{}_{a}(e^{}_{a}-h^{}_r)}dm^{}_{L} + \frac{1} {e^{}_{a}-h^{}_r}de^{}_{a} = 0 $$ I'm confused on how to proceed. We can argue that water temperature is constant, so $e^{}_{l}$ and $h^{}_{l}$ won't be a problem when integrating, but all of the other variables in the second term can't be integrated with respect to $m^{}_{L}$. What is the next step?
At constant volume, there is a relationship between $dm_{\rm a}$ and $dm_{\rm l}$ (if a bit of air mass is added to the volume, you can find exactly what the bit of water mass must be removed). Use this relationship to get rid of $dm_{\rm l}$.
Question by AME536A Student
There is something controversial concerning the assignment number 5 question 4. In class, you told us that we have to prove with Bernoulli that the velocity at the exit is equal to the velocity at the inlet meaning that from the mass conservation the area in both inlet and exit are also equal.
If we have to follow the figure as mentioned, the area at the exit is bigger and more specifically $A_{exit}=d \frac{B}{cos(30)}=dL$ meaning that $q_{exit}=12.0284 $ m/s leading to different results.
The cross-sectional area of the reflected snow can (in fact, must!) be found through Bernoulli. It's indicated in the question statement that there is no friction, so Bernoulli's equation certainly applies. Also note that measuring cross-sectional area of the reflected snow can't be done with certainty given the drawing dimensions (in the figure, it wasn't clear what $d$ of the reflected snow stood for exactly). Also, $B/\cos(30)$ is not the width of the reflected snow — build a 3D model of the snow with a sheet of paper to convince yourself of this. Albeit coming a bit later than I would like, thank you for the question: I can see how the dimension $d$ of the reflected snow was not clear in this respect and made a change in the figure in consequence.
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