- #1
TriKri
- 72
- 0
Hi,
I'm looking for a little bit of help in this matter, I'm trying to put up a general formula of the internal force/volume in a fluid in a spot, as a function of how the velocity varies locally around that spot.
What I started with was a formula I found in a book called "Physics Handbook - for Science and Engineering":
Now, do anyone know how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t?
Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
Multiplying with the density gives us
[tex]\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
where m is the mass, V is the volume, and ay is the acceleration in y-direction. Since m*a=f, we have:
[tex]\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
Now, if you have [tex]v_y=v_y(x,z,t)[/tex] instead, I guess the equation would look like
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
And to what I suppose, the force created (only by the inner friction of the liquid, i.e. the viscosity) would be the same, even if [tex]v_y[/tex] is a function of both x, y, z and t.
What I realized was this:
[tex]\frac{df_x}{dV}=\eta\left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right)[/tex]
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
[tex]\frac{df_z}{dV}=\eta\left(\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right)[/tex]
or
[tex]\frac{d\vec{f}}{dV}=\eta\left(\begin{array}{ccccc}
0&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&0&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&0
\end{array}\right)[/tex]
Now, is there any way to simplify this expression? Anyway, I worked a bit on it on my own and I found that it equals to
[tex]\eta\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
-
\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
I also found that
[tex]\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
=
(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/tex]
[tex]\text{(And also }=\vec{\nabla}(\vec{\nabla}\cdot\vec{v})\ -\ \vec{\nabla}\times(\vec{\nabla}\times\vec{v})\ )[/tex]
so that basically, one can write
[tex]\frac{d\vec{f}}{dV}
=
\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}
-
\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
Now I have showed you my work, how long I have come so far. I would really appreciate if someone could help me with the last part. The thing I have done here is that I have removed almost all x, y and z and written it as a vector field instead of as three different scalar fields. Hence trying to make it undependent of which coordinate system that is used. I just can't find a way to write that last vector without using x, y and z, and without making it depend on which coordinate system you use. I want to minimize it into one row as I have done with the bigger vector.
If someone knows a way to do this, I would be really thankful if that one could tell me. Or, if someone knows how the thing I wanted to know from the beginning (how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t) usually is written, that would be fine too.
I'm looking for a little bit of help in this matter, I'm trying to put up a general formula of the internal force/volume in a fluid in a spot, as a function of how the velocity varies locally around that spot.
What I started with was a formula I found in a book called "Physics Handbook - for Science and Engineering":
Differential equation for a Newtonian liquid flowing in the y-direction with speed vy = vy(x,t)
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}+\frac{f}{\rho}[/tex]
[tex]f[/tex]=external force in y-direction per volume
[tex]\eta[/tex]=coefficient of viscosity
Now, do anyone know how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t?
Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
Multiplying with the density gives us
[tex]\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
where m is the mass, V is the volume, and ay is the acceleration in y-direction. Since m*a=f, we have:
[tex]\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
Now, if you have [tex]v_y=v_y(x,z,t)[/tex] instead, I guess the equation would look like
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
And to what I suppose, the force created (only by the inner friction of the liquid, i.e. the viscosity) would be the same, even if [tex]v_y[/tex] is a function of both x, y, z and t.
What I realized was this:
[tex]\frac{df_x}{dV}=\eta\left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right)[/tex]
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
[tex]\frac{df_z}{dV}=\eta\left(\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right)[/tex]
or
[tex]\frac{d\vec{f}}{dV}=\eta\left(\begin{array}{ccccc}
0&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&0&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&0
\end{array}\right)[/tex]
Now, is there any way to simplify this expression? Anyway, I worked a bit on it on my own and I found that it equals to
[tex]\eta\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
-
\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
I also found that
[tex]\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
=
(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/tex]
[tex]\text{(And also }=\vec{\nabla}(\vec{\nabla}\cdot\vec{v})\ -\ \vec{\nabla}\times(\vec{\nabla}\times\vec{v})\ )[/tex]
so that basically, one can write
[tex]\frac{d\vec{f}}{dV}
=
\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}
-
\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
Now I have showed you my work, how long I have come so far. I would really appreciate if someone could help me with the last part. The thing I have done here is that I have removed almost all x, y and z and written it as a vector field instead of as three different scalar fields. Hence trying to make it undependent of which coordinate system that is used. I just can't find a way to write that last vector without using x, y and z, and without making it depend on which coordinate system you use. I want to minimize it into one row as I have done with the bigger vector.
If someone knows a way to do this, I would be really thankful if that one could tell me. Or, if someone knows how the thing I wanted to know from the beginning (how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t) usually is written, that would be fine too.