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Velocity in microchannel with temporal temperature variation

  1. Nov 25, 2017 #41
    $${\it mambiguous} \left( V =~\left(-\frac{3\cdot c \cdot Y ^{2}}{h }+
    \frac{2\cdot c \cdot Y ^{3}}{h ^{2}}\right)\frac{\partial \rho \left(X
    \right)}{\partial X }~\left[\left\{\frac{1-\frac{\rho \left(X \right)
    }{\lambda }\int _{0}^{\lambda }{\it mambiguous} \left( \frac{\partial
    \xi }{\rho \left(\xi \right)}, \right) }{\frac{\rho \left(X \right)}{
    \lambda }\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi
    \right)}~}+c \right\}-\frac{1}{\lambda }\int _{0}^{\lambda }\frac{
    \partial \xi }{\rho \left(\xi \right)}\left\{\frac{1}{\frac{\rho
    \left(X \right)}{\lambda }\int _{0}^{\lambda }\frac{\partial \xi }{
    \rho \left(\xi \right)}~}+\frac{1-\frac{\rho \left(X \right)}{\lambda
    }\int _{0}^{\lambda }\frac{\partial \xi }{\rho \left(\xi \right)}~}{
    \frac{\rho \left(X \right)\escirc 2}{\lambda }\int _{0}^{\lambda }
    \frac{\partial \xi }{\rho \left(\xi \right)}~}~\right\}\right]
    \right)$$
     

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  2. Nov 25, 2017 #42
    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 #43
    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 #44
    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 #45
    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 #46
    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 #47
    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 #48
    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 #49
    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 #50
    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 #51
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
     
  12. Dec 17, 2017 at 9:05 AM #52
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
     
  13. Dec 17, 2017 at 11:04 AM #53
    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 #54
    $$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$
     
  15. Dec 17, 2017 at 11:38 AM #55
    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 #56
    $$\frac{\partial (\rho U)}{\partial X}-c\frac{\partial (\rho)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=\frac{\partial \rho}{\partial T}+c\frac{\partial \rho}{\partial X}$$
    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 #57
    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 #58
    $$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?
     
  19. Dec 17, 2017 at 1:35 PM #59
    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
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