# Velocity in microchannel with temporal temperature variation

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1. Nov 25, 2017

### big dream

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

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2. Nov 25, 2017

### Staff: Mentor

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, $$V=\left(-\frac{Y}{h}+3\left(\frac{Y}{h}\right)^2-2\left(\frac{Y}{h}\right)^3\right)\frac{ch}{\rho(X)}\frac{d\rho(X)}{dX}$$

3. Nov 25, 2017

### 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

4. Nov 26, 2017

### Staff: Mentor

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

5. Dec 15, 2017 at 3:43 PM

### big dream

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?

6. Dec 15, 2017 at 3:53 PM

### Staff: Mentor

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.

7. Dec 16, 2017 at 12:45 PM

### big dream

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

8. Dec 16, 2017 at 10:04 PM

### Staff: Mentor

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?

9. Dec 17, 2017 at 7:16 AM

### big dream

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?

10. Dec 17, 2017 at 7:56 AM

### Staff: Mentor

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

11. Dec 17, 2017 at 9:03 AM

### big dream

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

12. Dec 17, 2017 at 9:05 AM

### big dream

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

13. Dec 17, 2017 at 11:04 AM

### Staff: Mentor

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?

14. Dec 17, 2017 at 11:35 AM

### big dream

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

15. Dec 17, 2017 at 11:38 AM

### Staff: Mentor

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

16. Dec 17, 2017 at 12:20 PM

### big dream

$$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$
so $\rho$ is not the function of T so
$$\frac{\partial (\rho)}{\partial T}=0$$
$$c\frac{\partial (\rho)}{\partial X}$$ should be cancelled from both sides.

17. Dec 17, 2017 at 12:37 PM

### Staff: Mentor

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

18. Dec 17, 2017 at 1:17 PM

### big dream

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

19. Dec 17, 2017 at 1:35 PM

### Staff: Mentor

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?

Last edited: Dec 17, 2017 at 4:53 PM