Find Forces Given Air Pressure and Velocity

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In summary: Nerd)In summary, the problem involves a fluid entering a space with uniform pressure and velocity, and exiting with atmospheric pressure. The forces acting on a fixed wall within this space are calculated using equations for force and continuity, taking into account the direction and magnitude of the pressure and velocity. The resulting forces are counteracted by the support, and the overall system is in static equilibrium.
  • #1
mathmari
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Hey! :eek:

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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  • #2
Why does it stand that $R_x+F_{W_0, x}=0$ and $-R_y+F_{W_0, y}=0$ ?? (Wondering)
 
  • #3
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)
 
  • #4
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)
 
  • #5
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)
 
  • #6
Which force is $F_{W_0, y}$ ?? (Wondering)
 
  • #7
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.
 

Related to Find Forces Given Air Pressure and Velocity

1. What is the relationship between air pressure and velocity?

The relationship between air pressure and velocity is known as Bernoulli's principle. This principle states that as the velocity of a fluid (such as air) increases, the pressure decreases. This is because the faster moving fluid molecules exert less force on the surface they come in contact with, resulting in a lower pressure.

2. How is air pressure measured?

Air pressure is typically measured using a device called a barometer. This instrument uses either mercury or aneroid (mechanical) methods to measure the atmospheric pressure. The unit of measurement for air pressure is typically expressed in kilopascals (kPa) or millibars (mb).

3. Can air pressure affect the velocity of an object?

Yes, air pressure can affect the velocity of an object. When an object is moving through a fluid (such as air), the force of air pressure can act upon the object and change its velocity. This is known as aerodynamic force and is a crucial factor in the design of objects such as airplanes and cars.

4. How does altitude affect air pressure and velocity?

As altitude increases, the air pressure decreases due to a decrease in the number of air molecules in the atmosphere. This decrease in air pressure can also affect the velocity of an object, as there is less air resistance at higher altitudes. In fact, some high-altitude areas are commonly used for activities such as skydiving and base jumping due to the lower air resistance.

5. What factors can influence air pressure and velocity?

There are several factors that can influence air pressure and velocity, including temperature, altitude, and the shape and size of objects moving through the air. Changes in these factors can result in changes in air pressure and velocity, which can have a significant impact on the behavior of fluids and the objects moving through them.

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