MHB Find Forces Given Air Pressure and Velocity

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The discussion revolves around calculating the forces acting on a support due to air pressure and velocity in a fluid dynamics scenario. The forces are derived from pressure and momentum integrals over the boundaries of the system, taking into account uniform pressure and velocity distributions at the inlet and atmospheric pressure at the exit. The equilibrium conditions indicate that the reaction forces, \( R_x \) and \( R_y \), must balance the forces generated by the fluid, ensuring static equilibrium. Participants seek clarification on the calculations of the integrals and the signs in the equilibrium equations, emphasizing the importance of understanding the directionality of forces. The conversation highlights the relationship between fluid dynamics principles and static equilibrium in mechanical systems.
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Hey! :o

Air of pressure $p_0$ and velocity $|\overrightarrow{u}_{A}|=|\overrightarrow{u}_{B}|=c$ enters the space $D$ from the sections $A, B$ of surface $S$. If at the orifices the distribution of the pressure and the velocity is uniform and at the exit $\Gamma$ the pressure is equal to the atmospheric one $P_a$, find the forces $R_x, R_y$ acting on the support $\Delta$ (air density $\rho_a$).

View attachment 4420

The solution that I found in my notes is the following:

$$\overrightarrow{F}_{W_0}=-\int_{\partial{W_1}}p\overrightarrow{n}dA-\int_{\partial{W_1}}\rho \overrightarrow{u}(\overrightarrow{u} \cdot \overrightarrow{n})dA$$

$\overrightarrow{F}_{W_0}$ is the force over the solid boundary $\partial_{W_0}$ (square ΑΒΓΔ)

Continuity equation: $$u_{\Gamma}S_{\Gamma}=u_AS_A+u_BS_B \Rightarrow u_{\Gamma}2S=cS+cS \Rightarrow u_{\Gamma}=u_A=u_B=c$$

$$\text{ Section } A : \overrightarrow{n}_A=\hat{i} \ \ , \ \ \overrightarrow{u}_{A}=-c\hat{i} \\ \text{ Section } B : \overrightarrow{n}_B=\frac{\sqrt{2}}{2}(\hat{i}-\hat{j}) \ \ , \ \ \overrightarrow{u}_{B}=c\left (-\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})\right ) \\ \text{ Section } \Gamma : \overrightarrow{n}_{\Gamma}=-\hat{j} \ \ , \ \ \overrightarrow{u}_{\Gamma}=-c\hat{j}$$

$$\overrightarrow{u}_{A} \cdot \overrightarrow{n}_{A}=-c \\ \overrightarrow{u}_{B} \cdot \overrightarrow{n}_{B}=-c \\ \overrightarrow{u}_{\Gamma} \cdot \overrightarrow{n}_{\Gamma}=c$$

$$\int_{\partial{W_1}}p\overrightarrow{n}dA=P_AS\hat{i}+P_BS\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})+P_{\Gamma}2S(-\hat{j})=P_0S\left (\hat{i}+\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})\right )-2P_aS\hat{j}=P_0S\left (1+\frac{\sqrt{2}}{2}\right )\hat{i}-\left (P_0S\frac{\sqrt{2}}{2}+2P_aS\right )\hat{j}$$

$$\int_{\partial{W_1}}\rho \overrightarrow{u}(\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_a [\overrightarrow{u}_A (\overrightarrow{u}_A \cdot \overrightarrow{n})S_A+\overrightarrow{u}_B (\overrightarrow{u}_B \cdot \overrightarrow{n})S_B+\overrightarrow{u}_{\Gamma} (\overrightarrow{u}_{\Gamma} \cdot \overrightarrow{n})S_{\Gamma}]=\rho_a [-c\hat{i}(-c)S-c\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})(-c)S-c\hat{j}c2S]=\rho_a c^2S\left (\hat{i}+\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})-2\hat{j}\right )=\rho_ac^2S\left (\left (1+\frac{\sqrt{2}}{2}\right )\hat{i}-\left ( \frac{\sqrt{2}}{2}+2\right )\hat{j}\right )$$

$$\overrightarrow{F}_{W_0}=-P_0S\left (1+\frac{\sqrt{2}}{2}\right )\hat{i}+\left (P_0S\frac{\sqrt{2}}{2}+2P_aS\right )\hat{j}-\rho_a c^2S\left (1+\frac{\sqrt{2}}{2}\right )\hat{i}+\rho_a c^2S\left (\frac{\sqrt{2}}{2}+2\right )\hat{j}$$

$$F_{W_0,x}=-\left (P_0+\rho_a c^2\right )S\left (1+\frac{\sqrt{2}}{2}\right )<0 \\ F_{W_0, y}=\left (\left ( P_2\frac{\sqrt{2}}{2}+2P_a\right ) +\rho_ac^2\left (\frac{\sqrt{2}}{2}+2\right )\right )S>0$$

Balance of the construction:

$$R_x+F_{W_0, x}=0 \Rightarrow R_x=-F_{W_0, x}>0 \\ -R_y+F_{W_0, y}=0 \Rightarrow R_y=F_{W_0, y}>0$$

How we have we found that $\overrightarrow{n}_B=\frac{\sqrt{2}}{2}(\hat{i}-\hat{j})$ ?? (Wondering)

Could also explain to me how we have calculated the integrals $\int_{\partial{W_1}}p\overrightarrow{n}dA$ and $\int_{\partial{W_1}}\rho \overrightarrow{u}(\overrightarrow{u}\cdot \overrightarrow{n})dA$ ?? (Wondering)
 

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Why does it stand that $R_x+F_{W_0, x}=0$ and $-R_y+F_{W_0, y}=0$ ?? (Wondering)
 
mathmari said:
Why does it stand that $R_x+F_{W_0, x}=0$ and $-R_y+F_{W_0, y}=0$ ?? (Wondering)

Hey! (Wave)

The wall is fixed and the system presses itself into the top left corner.
That means that $R_x$ and $R_y$ will be exactly as large as needed to counter the forces generated by the fluid.

Put otherwise, the system as a whole is in static equilibrium (what they termed "Balance of the system").
That means:
$$\sum F_x = 0\\
\sum F_y = 0 \\
\sum M = 0$$
(Thinking)
 
I like Serena said:
Hey! (Wave)

The wall is fixed and the system presses itself into the top left corner.
That means that $R_x$ and $R_y$ will be exactly as large as needed to counter the forces generated by the fluid.

Put otherwise, the system as a whole is in static equilibrium (what they termed "Balance of the system").
That means:
$$\sum F_x = 0\\
\sum F_y = 0 \\
\sum M = 0$$
(Thinking)

What is $M$ ?? (Wondering)

Could you explain to me the signs at the equation $-R_y+F_{W_0, y}=0$ ?? (Wondering) Why is at $R_y$ a minus and at $F_{W_0, y}$ a plus?? (Wondering)
 
mathmari said:
What is $M$ ?? (Wondering)

$M$ is the symbol for moments or torques, which make the system rotate. (Nerd)
It's not included in your problem though.

Could you explain to me the signs at the equation $-R_y+F_{W_0, y}=0$ ?? (Wondering) Why is at $R_y$ a minus and at $F_{W_0, y}$ a plus?? (Wondering)

More specifically, static equilibrium is often written as something like:
\begin{aligned}
\overset{+}{\rightarrow} &\sum_i F_{i,x} &=0 \\
+\!\!\uparrow &\sum_i F_{i,y} &=0 \\
\underset+\curvearrowleft &\sum_i M_{i,\text{arbitrary point}}&= 0
\end{aligned}

The arrows indicate which direction we count as plus.
Since the direction of $R_y$ (as it is drawn) is in the opposite direction of the arrow for vertical forces, we give it a minus sign.
Apparently $F_{W_0, y}$ is going in the direction of the arrow (up), although I don't see it in the drawing. (Wasntme)
 
Which force is $F_{W_0, y}$ ?? (Wondering)
 
mathmari said:
Which force is $F_{W_0, y}$ ?? (Wondering)

From the problem statement:
[box=yellow]$\overrightarrow{F}_{W_0}$ is the force over the solid boundary $\partial_{W_0}$ (square ΑΒΓΔ) [/box]

That is, the force $F_{W_0, y}$ is the vertical component of the force that applies to the box due to fluid movements caused by differences in pressure.
 
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