Velocity in microchannel with temporal temperature variation

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big dream
a microchannel of length 2L and width h in the thermal cycling region. the temperature profile
Equation: T_Wall = T0 + ΔT cos(2πx/λ + 2πct/λ).
...(1)
the cyclic temperature profile leads to a time dependent density
Equations: rho(t) = rho_0 - beta(T - T_0), and rho_0 = rho(T_0), where rho is density, T is temperature and beta is a thermal expansion coefficient.
...(2)
using the mass conservation equation i.e.
Equation: partial of rho with respect to t equals negative rho times the divergence of v, the continuity equation in differential form.
...(3)
and momentum balance equation i.e.
Momentum balance equation: rho times (partial v / partial t plus v dot grad v) equals minus grad p plus mu(T) times (laplacian v plus grad(div v) / 3).
...(4)
we have to find the exact solution of u, v & p

The attempt at a solution
equ(1) →
Wall temperature equation: T_Wall = T_0 + ΔT · cos(kc + ωt), with T_Wall on the left and T_0, ΔT, k, c and t on the right, written as typeset mathematics.

equ(2) →
Equation: partial derivative of rho with respect to t equals minus rho times the divergence of v, which equals beta omega Delta T sin(kx + omega t).

length scales are normalised as
Scaling relations: x = eta L, y = xi h, u* = u/(omega L), v* = v/(omega h), defining dimensionless coordinates and velocities.

so equ(2) becomes
Equation: partial u* over partial eta plus partial v* over partial xi equals negative beta delta T sin(k L eta + c xi t), over rho_0 minus beta delta T cos(k L eta + c xi t), equals chi(s c eta).

Two-line set of equations: a partial derivative of u* with respect to eta plus a partial derivative of v* with respect to xi equals minus Lambda epsilon sin(k L eta + omega tau) divided by one minus Lambda epsilon cos(k L eta + omega tau), which equals chi(scaled).

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
Equation: nabla p = mu nabla squared v, a vector Laplacian relation between pressure gradient and velocity, as in Stokes flow.

this equation normalised as
Equation: partial derivative of p* with respect to eta equals the second partial derivative of u* with respect to xi squared.

as
Equation: u* equals the partial derivative of p* with respect to eta, multiplied by xi(1 - xi) divided by 2.

and from
Equation: integral from 0 to 1 of u* d-xi equals chi, where the integrand is the dimensionless velocity u* and the variable of integration is xi.

we can get
Equation for normalised streamwise velocity: u* = 6ξ(1 − ξ)·ln(1 − A·cos(kLη + ωt)), where ξ and η are dimensionless coordinates.

&
Equation: v* = 2 xi times (minus Lambda epsilon sin(kL eta + c tau)) divided by (1 minus Lambda epsilon cos(kL eta + c tau)).
is it the right process?

 
Attachments
  • Equation: T_Wall = T0 + ΔT cos(2πx/λ + 2πct/λ).
    upload_2017-10-30_14-21-20.png
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  • Equations: rho(t) = rho_0 - beta(T - T_0), and rho_0 = rho(T_0), where rho is density, T is temperature and beta is a thermal expansion coefficient.
    upload_2017-10-30_14-22-42.png
    915 bytes · Views: 1,281
  • Equation: partial of rho with respect to t equals negative rho times the divergence of v, the continuity equation in differential form.
    upload_2017-10-30_14-23-38.png
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  • Momentum balance equation: rho times (partial v / partial t plus v dot grad v) equals minus grad p plus mu(T) times (laplacian v plus grad(div v) / 3).
    upload_2017-10-30_14-25-21.png
    1.3 KB · Views: 1,250
  • Wall temperature equation: T_Wall = T_0 + ΔT · cos(kc + ωt), with T_Wall on the left and T_0, ΔT, k, c and t on the right, written as typeset mathematics.
    upload_2017-10-30_14-29-10.png
    978 bytes · Views: 1,211
  • Equation: partial derivative of rho with respect to t equals minus rho times the divergence of v, which equals beta omega Delta T sin(kx + omega t).
    upload_2017-10-30_14-29-35.png
    1.1 KB · Views: 1,221
  • Scaling relations: x = eta L, y = xi h, u* = u/(omega L), v* = v/(omega h), defining dimensionless coordinates and velocities.
    upload_2017-10-30_14-30-20.png
    916 bytes · Views: 1,220
  • Equation: partial u* over partial eta plus partial v* over partial xi equals negative beta delta T sin(k L eta + c xi t), over rho_0 minus beta delta T cos(k L eta + c xi t), equals chi(s c eta).
    upload_2017-10-30_14-30-55.png
    1.8 KB · Views: 1,231
  • Equation: nabla p = mu nabla squared v, a vector Laplacian relation between pressure gradient and velocity, as in Stokes flow.
    upload_2017-10-30_14-37-9.png
    443 bytes · Views: 1,241
  • Equation: partial derivative of p* with respect to eta equals the second partial derivative of u* with respect to xi squared.
    upload_2017-10-30_14-45-34.png
    675 bytes · Views: 1,196
  • Equation: u* equals the partial derivative of p* with respect to eta, multiplied by xi(1 - xi) divided by 2.
    upload_2017-10-30_14-49-1.png
    842 bytes · Views: 1,208
  • Equation: integral from 0 to 1 of u* d-xi equals chi, where the integrand is the dimensionless velocity u* and the variable of integration is xi.
    upload_2017-10-30_14-49-22.png
    646 bytes · Views: 1,219
  • Equation for normalised streamwise velocity: u* = 6ξ(1 − ξ)·ln(1 − A·cos(kLη + ωt)), where ξ and η are dimensionless coordinates.
    upload_2017-10-30_14-51-27.png
    1.1 KB · Views: 1,213
  • Equation: v* = 2 xi times (minus Lambda epsilon sin(kL eta + c tau)) divided by (1 minus Lambda epsilon cos(kL eta + c tau)).
    upload_2017-10-30_14-53-56.png
    1.1 KB · Views: 1,224
  • Two-line set of equations: a partial derivative of u* with respect to eta plus a partial derivative of v* with respect to xi equals minus Lambda epsilon sin(k L eta + omega tau) divided by one minus Lambda epsilon cos(k L eta + omega tau), which equals chi(scaled).
    upload_2017-10-30_14-55-45.png
    1.6 KB · Views: 1,263
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big dream said:
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?
 
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?
 
sorry this is the real physical problem
Two-panel figure: (a) schematic of a DNA plug in a horizontal microchannel of height h and length 2L with temperature T(t) = T0 + ΔT cos ωt; (b) temperature T in °C versus time t in s over one PCR cycle, showing ideal dashed and smoothed solid profiles stepping between about 95, 55, and 72 °C.
Velocity in microchannel with temporal temperature variation
Printed page of text describing a time-dependent density driving fluid flow and dispersion, with equation (1): partial rho over partial t equals minus rho times the divergence of v, equals beta delta T omega sin omega t.
Printed page of a physics paper showing equations (2) to (5) for dimensionless velocity fields u* and v* in a microchannel with temporal temperature variation, plus the momentum balance equation (3).
Handwritten derivation titled 'The attempt at a solution' with normalized momentum balance equations, a velocity profile u* = -(dp*/dz) ξ(1-ξ)/2, an integral of u* over ξ, and a final expression for u* involving ε, sin τ and cos τ, with the note 'ξ is missing'.
 
Attachments
  • Two-panel figure: (a) schematic of a DNA plug in a horizontal microchannel of height h and length 2L with temperature T(t) = T0 + ΔT cos ωt; (b) temperature T in °C versus time t in s over one PCR cycle, showing ideal dashed and smoothed solid profiles stepping between about 95, 55, and 72 °C.
    IMG_20171115_141125.jpg
    25.6 KB · Views: 601
  • Velocity in microchannel with temporal temperature variation
    IMG_20171115_141217.jpg
    49.4 KB · Views: 552
  • Printed page of text describing a time-dependent density driving fluid flow and dispersion, with equation (1): partial rho over partial t equals minus rho times the divergence of v, equals beta delta T omega sin omega t.
    IMG_20171115_145801.jpg
    30.6 KB · Views: 643
  • Printed page of a physics paper showing equations (2) to (5) for dimensionless velocity fields u* and v* in a microchannel with temporal temperature variation, plus the momentum balance equation (3).
    IMG_20171115_145838.jpg
    46.4 KB · Views: 575
  • Handwritten derivation titled 'The attempt at a solution' with normalized momentum balance equations, a velocity profile u* = -(dp*/dz) ξ(1-ξ)/2, an integral of u* over ξ, and a final expression for u* involving ε, sin τ and cos τ, with the note 'ξ is missing'.
    IMG_20171026_151233.jpg
    40.1 KB · Views: 618
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.
 
Last edited:
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?
 
big dream said:
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??
 
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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?
 
sir,
I was sick. so now should I put stream function? or I have to normalise continuity equation first.
 
big dream said:
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.
 
Last edited:
Chestermiller said:
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 ?
 
big dream said:
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?
 
boundary condition: U= c at Y= 0,h
Equation: U = (1 / (2 * mu)) * (dP/dX) * (Y^2 - hY) + c * (Y/h + 1), giving velocity U as a function of Y.
 
Attachments
  • Equation: U = (1 / (2 * mu)) * (dP/dX) * (Y^2 - hY) + c * (Y/h + 1), giving velocity U as a function of Y.
    upload_2017-11-21_18-23-31.png
    1.2 KB · Views: 1,069
big dream said:
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.
 
Equation V = -(h/ρ) times the integral from 0 to h of (∂(ρ·U)/∂X) dY, with V on the left side.

Should I put the value of U and then evaluate?
 
Attachments
  • Equation V = -(h/ρ) times the integral from 0 to h of (∂(ρ·U)/∂X) dY, with V on the left side.
    upload_2017-11-22_1-44-33.png
    1.1 KB · Views: 1,081
big dream said:
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.
 
Last edited:
is this correct
Equation: d/dY of the integral from 0 to h of V dY, minus V times dh/dY, equals minus 1/rho times the integral from 0 to h of the partial derivative of (rho times U) with respect to X, dY.

this kind of solution coming
Equation: 0 = -∂(ρ U) h(Y) / (ρ ∂(X)), with ρ, U, h(Y) and X as shown.
 
Attachments
  • Equation: d/dY of the integral from 0 to h of V dY, minus V times dh/dY, equals minus 1/rho times the integral from 0 to h of the partial derivative of (rho times U) with respect to X, dY.
    upload_2017-11-22_10-55-10.png
    1.4 KB · Views: 1,051
  • Equation: 0 = -∂(ρ U) h(Y) / (ρ ∂(X)), with ρ, U, h(Y) and X as shown.
    upload_2017-11-22_10-56-35.png
    963 bytes · Views: 1,034
big dream said:
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?
 
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Chestermiller said:
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
 
big dream said:
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?
 
Chestermiller said:
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
Mathematical expression: integral from 0 to h of the partial derivative of P with respect to X, with respect to Y, written as an integral sign with limits 0 and h and differential dY.

I am not sure
 
Attachments
  • Mathematical expression: integral from 0 to h of the partial derivative of P with respect to X, with respect to Y, written as an integral sign with limits 0 and h and differential dY.
    upload_2017-11-22_20-7-44.png
    689 bytes · Views: 495
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 said:
Integrate both sides of the last equation in post #23 from ##X = 0## to ##X = \lambda##. What do you get?
Equation k = -c·ρ(X), with k on the left and a minus sign, c and ρ(X) multiplied on the right, shown in italic typeset mathematics.
 
Attachments
  • Equation k = -c·ρ(X), with k on the left and a minus sign, c and ρ(X) multiplied on the right, shown in italic typeset mathematics.
    upload_2017-11-22_23-10-30.png
    524 bytes · Views: 941
big dream said:
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?
 

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