# What is the Ratio of Thicknesses for a Beam Splitting on a Stationary Plate?

• mathmari
In summary, the conversation discusses the solution to a fluid mechanics problem involving a two-dimensional beam of thickness S and velocity c falling onto a stationary plate and separating into two sections with different velocities. The solution involves the application of the continuity equation, the conservation of mechanical energy (Bernoulli) equation, and the macroscopic momentum balance equation. However, there may be a sign or labeling error in the solution as a small angle ϕ should result in a larger S2 and smaller S1, rather than the opposite.
mathmari
Gold Member
MHB
Hello!

A two-dimensional beam of thickness S and velocity c (evenly distributed at the thickness) falls in stationary plate and get separated. Calculate the ratio of thicknesses S1/S2 as a function of the angle φ.

(see attachment)In my notes there is this solution:

Continuity equation:

Let c1, c2 the velocities of the sections 1, 2

$$0: \overrightarrow{u}=c\hat{e} \ \ , \ \ \overrightarrow{n}_0=-\hat{e}=-(-\sin \phi \hat{i}+\cos \phi \hat{j}) \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_0=-c \\ 1: \overrightarrow{u}=c_1 \overrightarrow{n}_1=c_1 \hat{i} \ \ , \ \ \overrightarrow{n}_1=\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_1=c_1 \\ 2: \overrightarrow{u}=c_2 \overrightarrow{n}_2=-c_2\hat{i} \ \ , \ \ \overrightarrow{n}_2=-\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_2=c_2$$

$$\int_{\partial{W_1}}\overrightarrow{u} \cdot \overrightarrow{n}dA=0 \Rightarrow -c \cdot (S \cdot 1)+c_1 \cdot (S_1 \cdot 1)+c_2 \cdot (S_2 \cdot 1)=0 \Rightarrow cS=c_1S+c_2S_2 \ \ \ \ \ (1)$$

Conservation of energ $\Rightarrow$ Bernoulli equation

$$\frac{1}{2}|\overrightarrow{u}|^2+\frac{p}{\rho_0}=\text{ constant along a stremline }$$

$$0 \rightarrow 1 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_1^2+\frac{p_a}{\rho_0} \Rightarrow c_1=c \\ 0 \rightarrow 2 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_2^2+\frac{p_a}{\rho_0} \Rightarrow c_2=c \\ \Rightarrow c_1=c_2=c \ \ \ \ \ (2)$$

$$(1) \wedge (2) \Rightarrow S=S_1+S_2 \ \ \ \ \ (3)$$

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

$$\int_{\partial_{W_1}}p \overrightarrow{n} dA=p_a \int_{\partial{W_1}}\overrightarrow{n}dA=0$$

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)=(\rho_0c^2S\cos \phi +\rho_0 c_1^2S_1-\rho_0c_2^2S_2)\hat{i}-\rho c^2S\sin \phi \hat{j}=-\overrightarrow{F}_{W_0}$$

$$F_{W_0, x}=-\rho_0c^2(S\cos \phi +S_1-S_2) \ \ \ \ \ (4a) \\ F_{W_0, y}=\rho c^2 S \sin \phi \ \ \ \ \ (4b)$$

Since the fluid is ideal we have that $$F_{W_0, x}=0 \Rightarrow S_2-S_1=S \cos \phi \ \ \ \ \ (5)$$

$$(3) \wedge (5) \Rightarrow \\S_1=\frac{S}{2}(1+\cos \phi) \\ S_2=\frac{S}{2}(1-\cos \phi) \\ \Rightarrow \frac{S_2}{S_1}=\frac{1-\cos \phi}{1+\cos \phi}$$
Could you explain to me the general idea of the solution? I haven't really understood that...Every time when we have such an exercise do we have to use the continuity equation and the Bernoulli equation?

Also could you explain to me why the following equality stand?

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)$$

#### Attachments

• ratio.png
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You are learning how to apply macroscopic balances to solve certain types of fluid mechanics problems (that are amenable to this type of approach). Which part of the approach don't you understand: (a) continuity equation, (b) conservation of mechanical energy (Bernoulli) equation, or (c) macroscopic momentum balance equation? You seem to be indicating that you are not comfortable with the application of the macroscopic momentum balance equation, correct?

Chet

There seems to be a sign or labeling error. A small ϕ (and therefore large cos ϕ) should make S2 large and S1 small, not the opposite.

## What is fluid mechanics and why is it important?

Fluid mechanics is the branch of physics that studies the properties and behavior of fluids, including liquids, gases, and plasmas. It is important because it helps us understand the forces and movements of fluids, which are essential in many areas such as engineering, meteorology, and biology.

## What is the ratio in fluid mechanics and how is it calculated?

The ratio in fluid mechanics is the relationship between two quantities, such as velocity and pressure, that affect the behavior of a fluid. It is calculated by dividing the value of one quantity by the value of the other quantity.

## How does the ratio impact the flow of fluids?

The ratio has a significant impact on the flow of fluids. In general, a high ratio indicates a more turbulent flow, while a low ratio indicates a more laminar flow. The ratio also affects the pressure and velocity distribution within the fluid, which can have an impact on the efficiency of systems that use fluids.

## What are some common applications of fluid mechanics?

Fluid mechanics has many practical applications in our daily lives. Some common examples include the design of aircraft and automobiles, the study of weather patterns, the analysis of blood flow in the human body, and the development of sustainable energy sources such as wind turbines and hydroelectric power plants.

## How is the study of fluid mechanics related to other branches of science?

Fluid mechanics is closely related to other branches of science, such as thermodynamics, electromagnetism, and materials science. It also has connections to fields such as biology, geology, and astrophysics. The principles of fluid mechanics are often used in combination with other scientific principles to solve complex problems and develop new technologies.

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