Velocity in microchannel with temporal temperature variation

[Chestermiller]

View attachment 215530

Excellent. Now the only thing left to do is to determine the vertical velocity V. Any ideas?

 

[big dream]

we could use the continuity equation. But I don't know what to do with this

 

[Chestermiller]

we could use the continuity equation. But I don't know what to do with this View attachment 215538

If you're asking how to differentiate this with respect to X, everything in the expression is constant except $\rho (X)$

 

[big dream]

After getting the expression of V what to do with $\rho (x,t)$

 

[Chestermiller]

After getting the expression of V what to do with $\rho (x,t)$

What do you get for V(X,Y)?

 

[big dream]

$${\it mambiguous} \left( V =~\left(-\frac{3\cdot c \cdot Y ^{2}}{h }+
\frac{2\cdot c \cdot Y ^{3}}{h ^{2}}\right)\frac{\partial \rho \left(X
\right)}{\partial X }~\left[\left\{\frac{1-\frac{\rho \left(X \right)
}{\lambda }\int _{0}^{\lambda }{\it mambiguous} \left( \frac{\partial
\xi }{\rho \left(\xi \right)}, \right) }{\frac{\rho \left(X \right)}{
\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi
\right)}~}+c \right\}-\frac{1}{\lambda }\int _{0}^{\lambda }\frac{
\partial \xi }{\rho \left(\xi \right)}\left\{\frac{1}{\frac{\rho
\left(X \right)}{\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{
\rho \left(\xi \right)}~}+\frac{1-\frac{\rho \left(X \right)}{\lambda
}\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi \right)}~}{
\frac{\rho \left(X \right)\escirc 2}{\lambda }\int _{0}^{\lambda }
\frac{\partial \xi }{\rho \left(\xi \right)}~}~\right\}\right]
\right)$$

 

[Chestermiller]

$${\it mambiguous} \left( V =~\left(-\frac{3\cdot c \cdot Y ^{2}}{h }+
\frac{2\cdot c \cdot Y ^{3}}{h ^{2}}\right)\frac{\partial \rho \left(X
\right)}{\partial X }~\left[\left\{\frac{1-\frac{\rho \left(X \right)
}{\lambda }\int _{0}^{\lambda }{\it mambiguous} \left( \frac{\partial
\xi }{\rho \left(\xi \right)}, \right) }{\frac{\rho \left(X \right)}{
\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi
\right)}~}+c \right\}-\frac{1}{\lambda }\int _{0}^{\lambda }\frac{
\partial \xi }{\rho \left(\xi \right)}\left\{\frac{1}{\frac{\rho
\left(X \right)}{\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{
\rho \left(\xi \right)}~}+\frac{1-\frac{\rho \left(X \right)}{\lambda
}\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi \right)}~}{
\frac{\rho \left(X \right)\escirc 2}{\lambda }\int _{0}^{\lambda }
\frac{\partial \xi }{\rho \left(\xi \right)}~}~\right\}\right]
\right)$$

Yikes! That's nothing like what I get. Here is my development:

$$U(X,Y)\rho(X)=6c\left[\frac{1-\frac{\rho (X)}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}}{\frac{1}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}}\right]\left(\frac{Y}{h}-\left(\frac{Y}{h}\right)^2\right)+c\rho(X)$$So,
$$\frac{\partial (\rho U)}{\partial X}=\left(1-6\frac{Y}{h}+6\left(\frac{Y}{h}\right)^2\right)c\frac{d\rho(X)}{dX}$$
So, $$\frac{\partial (\rho V)}{\partial Y}=-\left(1-6\frac{Y}{h}+6\left(\frac{Y}{h}\right)^2\right)c\frac{d\rho(X)}{dX}$$
So, $$V=\left(-\frac{Y}{h}+3\left(\frac{Y}{h}\right)^2-2\left(\frac{Y}{h}\right)^3\right)\frac{ch}{\rho(X)}\frac{d\rho(X)}{dX}$$

 

[big dream]

sir, Thank you very much for your help and patience. last doubt how to simplify the expression of u(x,t)? i.e. thread #35

 

[Chestermiller]

In post # 35, $\rho(t, x)=\rho_0\left(1-\beta \Delta T \cos{}\frac{2\pi(x+ct)}{\lambda}\right)$ and
$$\frac{1}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}=\frac{1}{\rho_0}\frac{1}{2\pi}\int_0^{2\pi}{\frac{d\theta}{(1-\beta \Delta T\cos{\theta})}}$$

 

[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.

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]

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?

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]

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.

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]

Actually how you get continuity equation from this equation $$\frac{\partial \rho( U+c)}{\partial (X-ct)}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}$$

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]

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?

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?

 

[Chestermiller]

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?

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

 

[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??