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Homework Help: Velocity in microchannel with temporal temperature variation

  1. Dec 17, 2017 #51
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
     
  2. Dec 17, 2017 #52
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial x}\right)_{t,y}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial x}\right)_{t,y}$$
    $$\left(\frac{\partial f}{\partial x}\right)_{t,y}=\left(\frac{\partial f}{\partial X}\right)_{T,Y}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial T}\right)_{X,Y}\left(\frac{\partial T}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial X}\right)_{T,Y}\left(\frac{\partial X}{\partial y}\right)_{t,x}+\left(\frac{\partial f}{\partial Y}\right)_{T,X}\left(\frac{\partial Y}{\partial y}\right)_{t,x}$$
    $$\left(\frac{\partial f}{\partial y}\right)_{t,x}=\left(\frac{\partial f}{\partial Y}\right)_{T,X}$$
     
  3. Dec 17, 2017 #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?
     
  4. Dec 17, 2017 #54
    $$\left(\frac{\partial (\rho U)}{\partial X}\right)_{T,Y}+\left(\frac{\partial (\rho V)}{\partial Y}\right)_{T,X}=\left(\frac{\partial \rho}{\partial T}\right)_{X,Y}+2c\left(\frac{\partial \rho}{\partial X}\right)_{T,Y}$$
     
  5. Dec 17, 2017 #55
    Wrong. You wrote down the original continuity equation incorrectly. And ##\rho## is not a function of T.
     
  6. Dec 17, 2017 #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.
     
  7. Dec 17, 2017 #57
    The correct statement of the continuity equation is:
    $$\frac{\partial \rho}{\partial t}+\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$
     
  8. Dec 17, 2017 #58
    $$\frac{\partial \rho(X)}{\partial T}=0$$ There t=T and X=x+ct. why so?
     
  9. Dec 17, 2017 #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
  10. Dec 18, 2017 #60
    Yes, sir that would be very useful.
     
  11. Dec 18, 2017 #61
    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.
     
  12. Jan 2, 2018 #62
    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.
     
  13. Jan 2, 2018 #63
    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.
     
  14. Jan 3, 2018 #64
    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??
     
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