- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
In my notes there is the following example:
One dimensional incompressible flow in a channel of constant section.
View attachment 4414
$$\overline{u}=(u(x, t), 0) \ \ \ \ \ p(x, y, t)=p(x)$$
incompressible flow: $\partial_x u=0$
Euler equations: $\rho_0 \partial_t u=-\partial_x p$
$$\Rightarrow \partial_x^2 p=0 \Rightarrow p(x)=p_1-\left (\frac{p_1-p_2}{L}\right )x$$
$$\partial_t u=-\frac{1}{\rho_0}\partial_x p \Rightarrow u=u(t)=\frac{p_1-p_2}{\rho_0 L}t+\text{ constant }$$
That the flow is incompressible is equivalent to the following:
Is this the same as $\partial_x u=0$ ?? (Wondering)
The Euler equations for imcompressible fluids are the following:
Are these equations the same as $\rho_0 \partial_t u=-\partial_x p$ ?? (Wondering)
In my notes there is the following example:
One dimensional incompressible flow in a channel of constant section.
View attachment 4414
$$\overline{u}=(u(x, t), 0) \ \ \ \ \ p(x, y, t)=p(x)$$
incompressible flow: $\partial_x u=0$
Euler equations: $\rho_0 \partial_t u=-\partial_x p$
$$\Rightarrow \partial_x^2 p=0 \Rightarrow p(x)=p_1-\left (\frac{p_1-p_2}{L}\right )x$$
$$\partial_t u=-\frac{1}{\rho_0}\partial_x p \Rightarrow u=u(t)=\frac{p_1-p_2}{\rho_0 L}t+\text{ constant }$$
That the flow is incompressible is equivalent to the following:
- $div \overrightarrow{u}=0$ everywhere
- $\frac{D\rho }{Dt}=0$
Is this the same as $\partial_x u=0$ ?? (Wondering)
The Euler equations for imcompressible fluids are the following:
- $\rho \frac{\partial}{\partial{t}}\overrightarrow{u}+\rho \overrightarrow{u} \cdot \nabla \overrightarrow{u}+\nabla p=0$
- $\frac{\partial}{\partial{t}}\rho+\overrightarrow{u}\cdot \nabla \rho=0$
- $div \overrightarrow{u}=0$
Are these equations the same as $\rho_0 \partial_t u=-\partial_x p$ ?? (Wondering)