- #1

Guillem_dlc

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- 15

- Homework Statement
- A two-dimensional jet of ##10\, \textrm{cm}## height sprays water at ##10\, \textrm{m}/\textrm{s}## which impacts, without friction, on an inclined plate...

DATA: ##\rho =1000\, \textrm{kg}/\textrm{m}^3##, ##c_1=10\, \textrm{m}/\textrm{s}##, ##h_1=10\, \textrm{cm}=0,10\, \textrm{m}##

Working hypothesis

- Stationary regime

- Uniform properties

- Incompressible flow

- Relevant Equations
- Continuity equation, Bernoulli equation

I was looking at an example of fluid mechanics and I don't understand this.

Statement figures:

CONTINUITY EQUATION

$$\left. \dfrac{dm}{dt}\right]_{MC}=(\dot{m}_2+\dot{m}_3)-\dot{m}_1=0$$

$$\dot{m}_1=\dot{m}_2+\dot{m}_3$$

$$\rho c_1A_1=\rho c_2A_2+\rho c_3A_3$$

$$\rho c_1 h1=\rho c_2 a1+\rho c_3 b1 \rightarrow \textrm{unit width}$$

$$\boxed{c_1h=c_2a+c_3b}$$

GENERALISED BERNOULLI EQUATION WITHOUT MACHINE

1-2

$$\dfrac{p_2-p_1}{\rho}+\dfrac{c_2^2-c_1^2}{2}+g(z_2-z_1)+w_{\textrm{roz}}=0 \,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l}

p_1=p_2=p_{\textrm{atm}}=0 \\

z_1=z_2=0 \\

w_{\textrm{roz}}=0

\end{array}\right.$$

1-3

$$\dfrac{c_2^2-c_1^2}{2}=0\rightarrow c_1=c_2$$

$$\dfrac{p_3-p_1}{\rho}+\dfrac{c_3^2-c_1^2}{2}+g(z_3-z_1)+w_{\textrm{roz}}=0\,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{

\begin{array}{l}

p_1=p_3=p_{\textrm{atm}}=0 \\

z_1=z_3=0 \\

w_{\textrm{roz}}=0

\end{array}\right.$$

$$\dfrac{c_3^2-c_1^2}{2}=0\rightarrow c_1=c_3 \quad \boxed{c_1=c_2=c_3=c}$$

Then,

$$c_1h=c_2a+c_3b$$

$$c_1=c_2=c_3=c$$

$$\boxed{h=a+b}$$

Equation Quantity Movement 1-2

$$\overrightarrow{F}_{\textrm{paret}\rightarrow \textrm{fluid}}=(\overrightarrow{c}_2\dot{m}_2+\overrightarrow{c}_3\dot{m}_3)-\overrightarrow{c}_1\dot{m}_1$$

x-axis

$$F_{p\rightarrow f(x)}=c_{2x}\rho c_{2x}A_2+(-)c_{3x}\rho c_{3x}A_3-c_{1x}\rho c_{1x}A_1$$

$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$

$$\cos (\alpha)h=a-b$$

y-axis

$$F_{p\rightarrow f(y)}=c_{2y}\rho c_{2y}A_2+c_{3y}\rho c_{3y}A_3-(-)c_{1y}\rho c_{1y}A_1$$

$$F_{p\rightarrow f(y)}=c^2\rho h\sin (\alpha)$$

Then,

$$\left\{ \begin{array}{l}

h = a+b\\

\cos (\alpha)h=a-b

\end{array}

\right.$$

$$\alpha :=\dfrac{45\pi}{180}=0.785$$

$$h:=0.1$$

We solve the system for ##a## and ##b##:

$$(0.08536,0.01464)$$

$$\rho := 1000$$

$$c1:=10$$

$$Fpfz:=\rho \cdot c1^2\cdot h\cdot \sin (\alpha)=7071\, \textrm{N}$$

What I don't understand is the following (which has appeared before):

x-axis:

$$\tau A=0\rightarrow \textrm{Frictionless}$$

$$\cos (\alpha)h=a-b$$

$$\alpha :=\dfrac{45\pi}{180}=0.785$$

So the ##h## would be about the size of the 1, wouldn't it?

Statement figures:

CONTINUITY EQUATION

$$\left. \dfrac{dm}{dt}\right]_{MC}=(\dot{m}_2+\dot{m}_3)-\dot{m}_1=0$$

$$\dot{m}_1=\dot{m}_2+\dot{m}_3$$

$$\rho c_1A_1=\rho c_2A_2+\rho c_3A_3$$

$$\rho c_1 h1=\rho c_2 a1+\rho c_3 b1 \rightarrow \textrm{unit width}$$

$$\boxed{c_1h=c_2a+c_3b}$$

GENERALISED BERNOULLI EQUATION WITHOUT MACHINE

1-2

$$\dfrac{p_2-p_1}{\rho}+\dfrac{c_2^2-c_1^2}{2}+g(z_2-z_1)+w_{\textrm{roz}}=0 \,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l}

p_1=p_2=p_{\textrm{atm}}=0 \\

z_1=z_2=0 \\

w_{\textrm{roz}}=0

\end{array}\right.$$

1-3

$$\dfrac{c_2^2-c_1^2}{2}=0\rightarrow c_1=c_2$$

$$\dfrac{p_3-p_1}{\rho}+\dfrac{c_3^2-c_1^2}{2}+g(z_3-z_1)+w_{\textrm{roz}}=0\,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{

\begin{array}{l}

p_1=p_3=p_{\textrm{atm}}=0 \\

z_1=z_3=0 \\

w_{\textrm{roz}}=0

\end{array}\right.$$

$$\dfrac{c_3^2-c_1^2}{2}=0\rightarrow c_1=c_3 \quad \boxed{c_1=c_2=c_3=c}$$

Then,

$$c_1h=c_2a+c_3b$$

$$c_1=c_2=c_3=c$$

$$\boxed{h=a+b}$$

Equation Quantity Movement 1-2

$$\overrightarrow{F}_{\textrm{paret}\rightarrow \textrm{fluid}}=(\overrightarrow{c}_2\dot{m}_2+\overrightarrow{c}_3\dot{m}_3)-\overrightarrow{c}_1\dot{m}_1$$

x-axis

$$F_{p\rightarrow f(x)}=c_{2x}\rho c_{2x}A_2+(-)c_{3x}\rho c_{3x}A_3-c_{1x}\rho c_{1x}A_1$$

$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$

$$\cos (\alpha)h=a-b$$

y-axis

$$F_{p\rightarrow f(y)}=c_{2y}\rho c_{2y}A_2+c_{3y}\rho c_{3y}A_3-(-)c_{1y}\rho c_{1y}A_1$$

$$F_{p\rightarrow f(y)}=c^2\rho h\sin (\alpha)$$

Then,

$$\left\{ \begin{array}{l}

h = a+b\\

\cos (\alpha)h=a-b

\end{array}

\right.$$

$$\alpha :=\dfrac{45\pi}{180}=0.785$$

$$h:=0.1$$

We solve the system for ##a## and ##b##:

$$(0.08536,0.01464)$$

$$\rho := 1000$$

$$c1:=10$$

$$Fpfz:=\rho \cdot c1^2\cdot h\cdot \sin (\alpha)=7071\, \textrm{N}$$

What I don't understand is the following (which has appeared before):

x-axis:

$$\tau A=0\rightarrow \textrm{Frictionless}$$

$$\cos (\alpha)h=a-b$$

$$\alpha :=\dfrac{45\pi}{180}=0.785$$

So the ##h## would be about the size of the 1, wouldn't it?