Velocity in microchannel with temporal temperature variation

[big dream]

a microchannel of length 2L and width h in the thermal cycling region. the temperature profile

...(1)
the cyclic temperature profile leads to a time dependent density

...(2)
using the mass conservation equation i.e.

...(3)
and momentum balance equation i.e.

...(4)
we have to find the exact solution of u, v & p

The attempt at a solution
equ(1) →

equ(2) →

length scales are normalised as

so equ(2) becomes

no slip & no penetration boundary condition at the walls (y=0,h) ; and constant pressure at channel entrance(x=-L) and exit(x=L). for small reynolds number eq(3) becomes

this equation normalised as

as

and from

we can get

&

is it the right process?

 

[Chestermiller]

a microchannel of length 2L and width h in the thermal cycling region. the temperature profile View attachment 214021...(1)
the cyclic temperature profile leads to a time dependent density View attachment 214023 ...(2)
using the mass conservation equation i.e. View attachment 214024...(3)
and momentum balance equation i.e. View attachment 214025 ...(4)
we have to find the exact solution of u, v & p

The attempt at a solution
equ(1) → View attachment 214026
equ(2) → View attachment 214027
length scales are normalised as View attachment 214028
so equ(2) becomes View attachment 214029
View attachment 214036
no slip & no penetration boundary condition at the walls (y=0,h) ; and constant pressure at channel entrance(x=-L) and exit(x=L). for small reynolds number eq(3) becomes View attachment 214030
this equation normalised as View attachment 214031
as View attachment 214032
and from View attachment 214033
we can get View attachment 214034
& View attachment 214035 is it the right process?

What is the exact statement of the problem?

 

[big dream]

 

[Chestermiller]

I'm a bit confused over the problem statement. It says, "There are no spatial temperature gradients present along the length of the micro channel," and yet the temperature equation they provide indicates otherwise. I'm having trouble articulating what is happening physically in this problem. Are you supposed to assume that the length scale of the periodic temperature variations (lambda) is small compared to the distance 2L?

Please articulate what you interpret is happening in the problem. Is it that the density variations are causing fluid flow pumping in the x direction?

 

[big dream]

sorry this is the real physical problem

 

[Chestermiller]

The two fluid flow problems you have presented here are extremely different, and I would approach them entirely differently. I am assuming that you really want to solve the problem involving the moving temperature wave, indicated in the first post. I would assume that the spatial wavelength $\lambda$ is small compared to the length of the tube 2L, so that, for all intents and purposes, the tube could be assumed infinitely long. I would then change the frame of reference of the observer such that he is moving along with the temperature wave at the speed c. In that case, he would observe a density variation that is a function only of x. This then would reduce the problem to a steady state flow problem.

I'm prepared to reveal more hints on how to implement the approach I described above involving a moving observer if what I said is not sufficiently clear.

 

[big dream]

Sir, for the second problem is my process right? Or I have to solve in another way. More preciously if this process is wrong, how I can get the exact solution of velocities?

 

[Chestermiller]

Sir, for the second problem is my process right? Or I have to solve in another way. More preciously if this process is wrong, how I can get the exact solution of velocities?

As I said, in my judgment, because of the combined x and t variation, you are not going to get it solved the way you are doing it. But, the method I was explaining will work.

From the frame of reference of an observer moving to the left at velocity c (i.e., moving coordinate system), you have $$T=T_0+\Delta T \cos{2\pi \frac{X}{\lambda}}$$ where X now replaces $x+ct$. So now the density is a function only of X. As reckoned from this moving coordinate system, the continuity equation becomes: $$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial y}=0$$where U and V are the velocity components reckoned from the frame of reference of the moving observer.

Boundary conditions on the flow are now: $U=+c$ and V= 0 at y=h and y=0.

Questions??

 

[Chestermiller]

It is too bad that @big dream has not returned. I find this a very interesting problem. Is there anyone else out there who is interested in pursuing the solution to this problem?

 

[big dream]

sir,
I was sick. so now should I put stream function? or I have to normalise continuity equation first.

 

[Chestermiller]

sir,
I was sick. so now should I put stream function? or I have to normalise continuity equation first.

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.

 

[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:
$$\rho=\rho(X)$$
$$\frac{\partial (\rho U)}{\partial X}+\frac{\partial (\rho V)}{\partial Y}=0$$
$$\frac{\partial P}{\partial X}=\mu \frac{\partial ^2U}{\partial Y^2}$$
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.

Yes, sir I got it. Now ?

 

[Chestermiller]

Yes, sir I got it. Now ?

OK. Starting with Eqn. 2, in post #11, solve for U in terms of dP/dX using the boundary conditions at y = 0, h on U. What do you obtain?

 

[big dream]

boundary condition: U= c at Y= 0,h

 

[Chestermiller]

boundary condition: U= c at Y= 0,h
View attachment 215375

Close, but this doesn't satisfy the boundary condition at Y=h. Please try once more.

 

[big dream]

 

[Chestermiller]

View attachment 215378

Much better. Now please take the integrated average of this between 0 and h to get the average axial velocity $\bar{U}(X)$

 

[big dream]

 

[Chestermiller]

View attachment 215382

OK. Now divide by h to get Ubar.

After that, please integrate the continuity equation (Eqn. 1 in post #11) between Y=0 and Y = h. What do you get?

Unfortunately, I have to leave for a few hours now, but I'll be back later.

 

[big dream]

Should I put the value of U and then evaluate?

 

[Chestermiller]

View attachment 215404
Should I put the value of U and then evaluate?

No. This is not done correctly. You forgot to put in the limits of integration on V, and you did not recognize that the integral sign and the partial derivative could be interchanged. Please try again.

 

[big dream]

is this correct

this kind of solution coming

 

[Chestermiller]

is this correct
View attachment 215427
this kind of solution coming View attachment 215428

Still not correct. The correct result is $$\frac{d}{d X}\left(\rho \int_0^h{UdY}\right)=[\rho V]_0^h=0$$
So, we have: $$\frac{d[\rho(X)\bar{U}(X)]}{dX}=0$$where $$\bar{U}(X)=\frac{1}{h}\int_0^h{UdY}$$
Now, integration with respect to X yields: $$\rho(X)\bar{U}(X)=k$$where k is a constant. Or equivalently, $$\bar{U}(X)=\frac{k}{\rho(X)}$$
Also, in post #18, you showed that $$\bar{U}=-\frac{1}{12\mu}\frac{dP}{dX}h^2+c$$If we combine these two equations, we obtain:
$$\frac{dP}{dX}=-\frac{12\mu}{h^2}\left[\frac{k}{\rho(X)}-c\right]$$

OK so far?

 

[big dream]

Still not correct. The correct result is $$\frac{d}{d X}\left(\rho \int_0^h{UdY}\right)=[\rho V]_0^h=0$$
So, we have: $$\frac{d[\rho(X)\bar{U}(X)]}{dX}=0$$where $$\bar{U}(X)=\frac{1}{h}\int_0^h{UdY}$$
Now, integration with respect to X yields: $$\rho(X)\bar{U}(X)=k$$where k is a constant. Or equivalently, $$\bar{U}(X)=\frac{k}{\rho(X)}$$
Also, in post #18, you showed that $$\bar{U}=-\frac{1}{12\mu}\frac{dP}{dX}h^2+c$$If we combine these two equations, we obtain:
$$\frac{dP}{dX}=-\frac{12\mu}{h^2}\left[\frac{k}{\rho(X)}-c\right]$$

OK so far?

It's great

 

[Chestermiller]

It's great

The next step is to determine the value of the constant k. This is related to how P has to be varying with X in order to satisfy the end constraint at the two ends of the channel. Thoughts?

 

[big dream]

The next step is to determine the value of the constant k. This is related to how P has to be varying with X in order to satisfy the end constraint at the two ends of the channel. Thoughts?

at the center of the channel, P is zero. and P is constant at the entrance & exit of the channel.
maybe

I am not sure

 

[Chestermiller]

The reservoir pressures are presumably the same at the two far ends of the channel at +L and -L. So there are no long range pressure variations along the length of the channel in the X direction. Therefore, the pressure P must be periodic in the X direction, with a period of $\lambda$. Thus, $P(X+\lambda)=P(X)$; and, in particular, $P(\lambda)=P(0)$. How can this be used to in conjunction with the last equation in post #23 to determine the value of the constant k?

 

[Chestermiller]

View attachment 215445

Integrate both sides of the last equation in post #23 from $X = 0$ to $X = \lambda$. What do you get?

 

[big dream]

Integrate both sides of the last equation in post #23 from $X = 0$ to $X = \lambda$. What do you get?

 

[Chestermiller]

View attachment 215452

Not even close. $$k=\frac{c}{\left[\frac{1}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}\right]}$$
Next, please substitute this into the last equation of post #23 to get dP/dX. What do you obtain?

 

[big dream]

what is /xi here?

 

[Chestermiller]

xi is a constant of integration to help us remember (in our subsequent analysis) that the value of the integral is a constant, independent of X. Please factor the c out of the brackets in this equation (it hurts my eyes).

The next step is substitute this expression for dP/dX back into our equation for U(X,T) in post #16 to get our final result for U. What do you get?

 

[big dream]

xi is a constant of integration to help us remember (in our subsequent analysis) that the value of the integral is a constant, independent of X. Please factor the c out of the brackets in this equation (it hurts my eyes).

The next step is substitute this expression for dP/dX back into our equation for U(X,T) in post #16 to get our final result for U. What do you get?

 

[Chestermiller]

View attachment 215497

I get the following: $$U(X,Y)=6c\left[\frac{1-\frac{\rho (X)}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}}{\frac{\rho (X)}{\lambda}\int_0^{\lambda}{\frac{d\xi}{\rho(\xi)}}}\right]\left(\frac{Y}{h}-\left(\frac{Y}{h}\right)^2\right)+c$$

As reckoned from the laboratory frame of reference, what does this give for u(t,x,y)?

 

[big dream]