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Excellent. Now the only thing left to do is to determine the vertical velocity V. Any ideas?big dream said:
Excellent. Now the only thing left to do is to determine the vertical velocity V. Any ideas?big dream said:
If you're asking how to differentiate this with respect to X, everything in the expression is constant except ##\rho (X)##big dream said:we could use the continuity equation. But I don't know what to do with this View attachment 215538
What do you get for V(X,Y)?big dream said:After getting the expression of V what to do with ##\rho (x,t)##
Yikes! That's nothing like what I get. Here is my development:big dream said:$${\it mambiguous} \left( V =~\left(-\frac{3\cdot c \cdot Y ^{2}}{h }+
\frac{2\cdot c \cdot Y ^{3}}{h ^{2}}\right)\frac{\partial \rho \left(X
\right)}{\partial X }~\left[\left\{\frac{1-\frac{\rho \left(X \right)
}{\lambda }\int _{0}^{\lambda }{\it mambiguous} \left( \frac{\partial
\xi }{\rho \left(\xi \right)}, \right) }{\frac{\rho \left(X \right)}{
\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi
\right)}~}+c \right\}-\frac{1}{\lambda }\int _{0}^{\lambda }\frac{
\partial \xi }{\rho \left(\xi \right)}\left\{\frac{1}{\frac{\rho
\left(X \right)}{\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{
\rho \left(\xi \right)}~}+\frac{1-\frac{\rho \left(X \right)}{\lambda
}\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi \right)}~}{
\frac{\rho \left(X \right)\escirc 2}{\lambda }\int _{0}^{\lambda }
\frac{\partial \xi }{\rho \left(\xi \right)}~}~\right\}\right]
\right)$$
How could the system be in steady state when X=x+ct which is also related to time? If you have some time would you explain it?Chestermiller said:Before doing anything else, I need you to prove to yourself that, if we make the following transformation of variables,
t = T
y = Y
x = X - ct
v = V(X,Y)
u = U(X,Y) - c
p = P(X)
our equations reduce to:
$$\rho=\rho(X)$$
$$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=0\tag{1}$$
$$\frac{\partial P}{\partial X}=\mu \frac{\partial ^2U}{\partial Y^2}\tag{2}$$
subject to the boundary conditions ##U=c## and ##V=0## at ##Y=0,h##. These are equivalent to the equations that an observer who is traveling at a constant speed c in the negative x direction would write. This observer would conclude that, as reckoned from his frame of reference, the system is at steady state, with all parameters functions of X and Y only.
It's at steady state as reckoned from the frame of reference of an observer who is moving to the left with a speed equal to c.big dream said:How could the system be in steady state when X=x+ct which is also related to time? If you have some time would you explain it?
Actually how you get continuity equation from this equation $$\frac{\partial \rho( U+c)}{\partial (X-ct)}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}$$Chestermiller said:Before doing anything else, I need you to prove to yourself that, if we make the following transformation of variables,
t = T
y = Y
x = X - ct
v = V(X,Y)
u = U(X,Y) - c
p = P(X)
our equations reduce to:
$$\rho=\rho(X)$$
$$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=0\tag{1}$$
$$\frac{\partial P}{\partial X}=\mu \frac{\partial ^2U}{\partial Y^2}\tag{2}$$
subject to the boundary conditions ##U=c## and ##V=0## at ##Y=0,h##. These are equivalent to the equations that an observer who is traveling at a constant speed c in the negative x direction would write. This observer would conclude that, as reckoned from his frame of reference, the system is at steady state, with all parameters functions of X and Y only.
In post #11, I asked you to prove to yourself that, with the transformation of variables I indicated, the equations transform the way I say they do. In post #12, you indicated that you had successfully done this. Now you tell me that you had not done this (and don't know how to). How can I trust you?big dream said:Actually how you get continuity equation from this equation $$\frac{\partial \rho( U+c)}{\partial (X-ct)}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}$$
after using Galilean transformation I got,Chestermiller said:In post #11, I asked you to prove to yourself that, with the transformation of variables I indicated, the equations transform the way I say they do. In post #12, you indicated that you had successfully done this. Now you tell me that you had not done this (and don't know how to). How can I trust you?
You have not carried out the transformation correctly. The starting equations are $$T=t$$$$X=x+ct$$and $$Y=y$$big dream said:after using Galilean transformation I got,
$$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}$$
now as## \rho## becomes the function of X. so $$\frac{\partial \rho}{\partial T}$$ must be zero. but when I show it to the teacher he told that X =x+ct so it is also a function of t, differentiation w.r.t. T may not be zero.
He asked me to show mathematical evidence. How it becomes a steady state?
Sir, I thought maybe my perception of the transformation of the equation is wrong, or my explanation. Is there something wrong with my explanation?
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$Chestermiller said:You have not carried out the transformation correctly. The starting equations are $$T=t$$$$X=x+ct$$and $$Y=y$$
For any function f of position and time, we start out by writing:$$df=\left(\frac{\partial f}{\partial T}\right)_{X,Y}dT+\left(\frac{\partial f}{\partial X}\right)_{T,Y}dX+\left(\frac{\partial f}{\partial Y}\right)_{T,X}dY$$From this, it follows that:
$$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial t}\right)_{x,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial t}\right)_{x,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial t}\right)_{x,y}$$
But, in our problem, $$\left(\frac{\partial T}{\partial t}\right)_{x,y}=1$$$$\left(\frac{\partial X}{\partial t}\right)_{x,y}=c$$and $$\left(\frac{\partial Y}{\partial t}\right)_{x,y}=0$$
Therefore, $$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}+c\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
Now it's your turn. What do you get for $$\left(\frac{\partial f}{\partial x}\right)_{t,y}$$and$$\left(\frac{\partial f}{\partial y}\right)_{t,x}$$
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$Chestermiller said:You have not carried out the transformation correctly. The starting equations are $$T=t$$$$X=x+ct$$and $$Y=y$$
For any function f of position and time, we start out by writing:$$df=\left(\frac{\partial f}{\partial T}\right)_{X,Y}dT+\left(\frac{\partial f}{\partial X}\right)_{T,Y}dX+\left(\frac{\partial f}{\partial Y}\right)_{T,X}dY$$From this, it follows that:
$$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial t}\right)_{x,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial t}\right)_{x,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial t}\right)_{x,y}$$
But, in our problem, $$\left(\frac{\partial T}{\partial t}\right)_{x,y}=1$$$$\left(\frac{\partial X}{\partial t}\right)_{x,y}=c$$and $$\left(\frac{\partial Y}{\partial t}\right)_{x,y}=0$$
Therefore, $$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}+c\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
Now it's your turn. What do you get for $$\left(\frac{\partial f}{\partial x}\right)_{t,y}$$and$$\left(\frac{\partial f}{\partial y}\right)_{t,x}$$
OK. Now of we substitute ##\rho## for f in the equation $$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}+c\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$big dream said:$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
$$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$Chestermiller said:OK. Now of we substitute ##\rho## for f in the equation $$\left(\frac{\partial f}{\partial t}\right)_{x,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}+c\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
what do we get?
And, if we substitute ##\rho V## for f in the equation $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
what do we get?
And, if we substitute ##\rho (U-c)## for f in the equation $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
what do we get?
Wrong. You wrote down the original continuity equation incorrectly. And ##\rho## is not a function of T.big dream said:$$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$
$$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$Chestermiller said:Wrong. You wrote down the original continuity equation incorrectly. And ##\rho## is not a function of T.
The correct statement of the continuity equation is:big dream said:$$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$
so ##\rho## is not the function of T so
$$\frac{\partial (\rho)}{\partial T}=0$$
$$c\frac{\partial (\rho)}{\partial X}$$ should be canceled from both sides.
$$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?Chestermiller said:The correct statement of the continuity equation is:
$$\frac{\partial \rho}{\partial t}+\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$
That is the coordinate transformation we defined.big dream said:$$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?
Yes, sir that would be very useful.Chestermiller said:That is the coordinate transformation we defined.
Or are you asking how I was able to recognize that this particular transformation of independent variables would bring about such great simplification in the analysis of the problem?
The short answer is "intuition, based on many years of experience."big dream said:Yes, sir that would be very useful.
Have you considered any scale like U=u/c , V=v/c, X= x/L, Y=y/L to reduce the momentum balance equation from $$ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {\gamma}{1\over3}\nabla (\nabla \cdot {\bf u})$$Chestermiller said:Before doing anything else, I need you to prove to yourself that, if we make the following transformation of variables,
t = T
y = Y
x = X - ct
v = V(X,Y)
u = U(X,Y) - c
p = P(X)
our equations reduce to:
$$\rho=\rho(X)$$
$$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=0\tag{1}$$
$$\frac{\partial P}{\partial X}=\mu \frac{\partial ^2U}{\partial Y^2}\tag{2}$$
subject to the boundary conditions ##U=c## and ##V=0## at ##Y=0,h##. These are equivalent to the equations that an observer who is traveling at a constant speed c in the negative x direction would write. This observer would conclude that, as reckoned from his frame of reference, the system is at steady state, with all parameters functions of X and Y only.