Velocity in microchannel with temporal temperature variation

[big dream]

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$$\left(\frac{\partial f}{\partial y}\right)_{t,x}$$

$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$

 

[big dream]

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$$\left(\frac{\partial f}{\partial y}\right)_{t,x}$$

$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$

 

[Chestermiller]

$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
$$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
$$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$

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?

 

[big dream]

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?

$$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$

 

[Chestermiller]

$$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$

Wrong. You wrote down the original continuity equation incorrectly. And $\rho$ is not a function of T.

 

[big dream]

Wrong. You wrote down the original continuity equation incorrectly. And $\rho$ is not a function of T.

$$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$
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.

 

[Chestermiller]

$$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$
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.

The correct statement of the continuity equation is:
$$\frac{\partial \rho}{\partial t}+\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$

 

[big dream]

The correct statement of the continuity equation is:
$$\frac{\partial \rho}{\partial t}+\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$

$$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?

 

[Chestermiller]

$$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?

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?

 

[big dream]

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?

Yes, sir that would be very useful.

 

[Chestermiller]

Yes, sir that would be very useful.

The short answer is "intuition, based on many years of experience."

The long answer is: I recognized that, for an observer moving to the left with a constant velocity c, the spatial coordinate in his frame of reference, X=x+ct (the X axis is also moving to the left with the observer) has constant density at every value of X (i.e., at each value of X, the density is not changing with time), since $\rho=\rho(x+ct)=\rho(X)$. Basically, in the x frame of reference the density profile is traveling to the left like a wave moving at speed c, but in the X frame of reference, the density profile is stationary. This automatically led mathematically to the transformation of independent variables that we used.

 

[big dream]

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.

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 $$ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {\gamma}{1\over3}\nabla (\nabla \cdot {\bf u})$$
To $$\frac{\partial P}{\partial X}=\mu \frac{\partial ^2U}{\partial Y^2}\tag{2}$$
or directly take the first term $${\partial{\bf u}\over{\partial t}}=0$$ as the flow is steady. and $$\rho{\bf u} \cdot \nabla {\bf u}={\bf u} \cdot \nabla (\rho{\bf u})=0 $$
still could not eliminate this term, $${1\over3}\nabla (\nabla \cdot {\bf u})$$
or you have considered the momentum balance equation like this $$ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {g_{x}} $$ as $$\rho g_{x}=0$$
Or firstly use Galilean transformation then reduce the equation.I am little confused which way to show this momentum balance equation.

 

[Chestermiller]

I think I've helped enough on this thread already. I'll leave it up to you to work out the rest of these details.

 

[Chestermiller]

I changed my mind because of your post on that other forum. So, here goes.

If we apply our coordinate transformation to the momentum equation, we obtain for the X component the following:
$$U\frac{\partial U}{\partial X}+V\frac{\partial U}{\partial Y}=-\frac{1}{\rho}\frac{\partial p}{\partial X}+\gamma\left[\frac{\partial^2U}{\partial X^2}+\frac{\partial^2U}{\partial Y^2}\right]+\frac{\gamma}{3}\left[\frac{\partial ^2U}{\partial X^2}+\frac{\partial^2V}{\partial X\partial Y}\right]$$
OK so far??