Fundamentals of Fluid Mechanics Questions and Answers  
Question by AME536A Student
Control volumn.PNG
The attachment Control volumn.PNG is no longer availableI have a question about prob #3 in the assignment 3. You said that the both density of water and air are everywhere the same and does not change in time. The equation for Tank 1 and Tank 2 is as follows.
Tank 1: $\frac{d}{dt}\int_{Vtank1,w} \rho_w dVtank1,w$ + $\frac{d}{dt}\int_{Vtank1,a} \rho_a dVtank1,a$ - $\dot{m}_{air1}$ + $\dot{m}_{12}$ = 0
Tank 2: $\frac{d}{dt}\int_{Vtank2,w} \rho_w dVtank2,w$ + $\frac{d}{dt}\int_{Vtank2,a} \rho_a dVtank2,a$ - $\dot{m}_{air2}$ - $\dot{m}_{12}$ - $\dot{m}_{Win}$ + $\dot{m}_{Wout}$= 0
I tried to isolate $\dot{m}_{12}$ in the equation of tank 1, and then I substitute tank 1 equation into tank 2. And then, I got as follows.
$\rho_w \frac{d}{dt}\int_{Vtank1,w}dVtank1,w$ + $\rho_a \frac{d}{dt}\int_{Vtank1,a} dVtank1,a$ + $\rho_w \frac{d}{dt}\int_{Vtank2,w} dVtank2,w$ + $\rho_a\frac{d}{dt}\int_{Vtank2,a} dVtank2,a$ -$\dot{m}_{air2}$ -$\dot{m}_{air1}$- $\dot{m}_{Win}$ + $\dot{m}_{Wout}$= 0
$\frac{d}{dt}\int_{Vtank2,w} dVtank2,w$= 0.005m3/s = - $\frac{d}{dt}\int_{Vtank2,a} dVtank2,a$
I think that in order for me to get $\dot{m}_{Wout}$, I need the relationship between $\frac{d}{dt}\int_{Vtank1,w}dVtank1,w$ and $\frac{d}{dt}\int_{Vtank1,a} dVtank1,a$
How can I obtain the relationship? I will attach my control volumn.
12.14.20
This is too hard to read and is giving me a headache. Please put long equations on their own line using the \$\$ environment instead of the \$ environment. Also, I can't read $dVtank1$ and the like. This is not properly typeset. Use subscripts and superscripts as needed.
Question by AME536A Student
Control volumn.PNG
I apologize for this. I have a question about prob #3 in the assignment 3. You said that the both density of water and air are everywhere the same and does not change in time. The equation for Tank 1 and Tank 2 is as follows.
Tank 1: $$ \rho_{w}\frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}+\rho_{a}\frac{d}{dt}\int_{V_{tank1,a}}dV_{tank1,a}-\dot{m}_{air1}+\dot{m}_{12}=0$$
Tank 2: $$ \rho_{w}\frac{d}{dt}\int_{V_{tank2,w}}dV_{tank2,w}+\rho_{a}\frac{d}{dt}\int_{V_{tank2,a}}dV_{tank2,a}-\dot{m}_{air2}-\dot{m}_{12}-\dot{m}_{Win}+\dot{m}_{Wout}=0$$
I tried to isolate $\dot{m}_{12}$ in the equation of tank 1, and then I substitute $\dot{m}_{12}$ into tank 2. And then, I got as follows.
$$\rho_{w}\frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}+\rho_{a}\frac{d}{dt}\int_{V_{tank1,a}}dV_{tank1,a}+\rho_{w}\frac{d}{dt}\int_{V_{tank2,w}}dV_{tank2,w}+\rho_{a}\frac{d}{dt}\int_{V_{tank2,a}}dV_{tank2,a}-\dot{m}_{air1}-\dot{m}_{air2}-\dot{m}_{Win}+\dot{m}_{Wout}=0$$
$$ \frac{d}{dt}\int_{V_{tank2,w}}dV_{tank2,w}=0.005m3/s=-\frac{d}{dt}\int_{V_{tank2,a}}dV_{tank2,a}$$
I think that in order for me to get $\dot{m}_{Wout}$, I need the relationship between $\frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}$ and $\frac{d}{dt}\int_{V_{tank1,a}}dV_{tank1,a}$
How can I obtain the relationship? I will attach my control volumn.
You need another equation that fixes the volume of each tank (sum of air volume and water volume) to a constant in time.
Question by AME536A Student
I got another equation through the relationship of $$\frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}=-\frac{d}{dt}\int_{V_{tank1,a}}dV_{tank1,a}$$
But, when the above relationship is applied to the equation combining the tank 1 equation with tank 2 equation, I have two unknown parameters
(either $$ \frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}$$ or $$\frac{d}{dt}\int_{V_{tank1,a}}dV_{tank1,a}$$, $\dot{m}_{W_{out}}$).
How can I find $ \frac{d}{dt}\int_{V_{tank1,w}}dV_{tank1,w}$?
You need another control volume around the air ducts linking tank 1 and tank 2.
Question by AME536A Student
I have a question about prob #3 in the assignment 6. Are the both density of water and air everywhere the same and does not change in time, too?
In A6Q3, you can take the density of water as constant, but the density of the air is not.
Question by AME536A Student
Dr. Parent, any idea when will you be adding the final grades to UAccess?
12.19.20
I added them just the day after the exam. But I may not have clicked the “Post” button. Check if you can see them now. The grade distribution is as follows:
A:  4 Students
B:  3 Students
E:  1 Student
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