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Flow Through a Circular Tube
5. Flow Through a Circular Tube
Cylindrical coordinate
Control system ΔV = 2πr·Δr·L
Postulates: vr = vθ = 0, vz = f(vr)
Momentum balance:
When we divide above equation by 2πLΔr and
take the limit as Δr → 0, we get”
5. Flow Throught a Circular Tube
In the next step, we get:
Therefore the momentum
flux distribution is equal
After simplifications:
Above equation may be
integrated to give:
Consequently C1 must be zero,
for otherwise the momentum
flux would be infinite at the axis
of the tube.
5. Flow Throught a Circular Tube
Newton's law of viscosity for this situation:
This first-order separable differential equation may be integrated to give:
The constant C2 is evaluated from the boundary condition:
at r = R,
The velocity distribution is:
vz = 0
Flow through an annulus
Use Cylindrical Coordinates, which are
the natural coordinates for the
descripion of position in a tube.
Consider a steady state laminar flow of
a fluid of constant density p in a long
tube of lenght L and radius R (assume L
>>R and ignore end effects). Shell of
thickness ∆ r and lenght L, considering z
as the flow direction. We now consider
the force balance.
Force balance
Equations
 Momentum balance over a think cylindrical shell
 Integrating
 Substituting the Newton’s Law and integrating
 Constant are determined using BC:s
Other results
•
Average velocity
•
Comparison with flow in a slit – Problem 2B.3 and 2B.5
•
Valid for laminar flow
6. Flow throught an Annulus
When we make a momentum balance over a thin cylindrical shell of liquid, we
arrive at the following differential equation:
The C1 cannot be determined immediately, since we have no information about the
momentum flux at the fixed surface r = κR and r = R. All we know is that there
will be a maximum in the velocity curve at some (as yet unknown) plane r = λR at
which the momentum flux will be zero. That is,
The momentum –flux distribution
and velocity distribution for the
upward flow in a cylindrical
annulus. Note that the momentum
flux changes sign at the same
value of r for which the velocity
has a maximum.
When the solve this equation for C1 and substitute it we get:
6. Flow throught an Annulus
When we substitute Newton’s law of viscosity, τrz = -µ(d vz /dr) and we have
Integration of this first-order separable differential equation then gives
Evaoulation of two constans of integration, λ and C2 by using the no-slip condition on each solid boundary:
Substitution of these boundary conditions to previous equation we have:
The momentum-flux distribution and the velocity distribution is receiving from:
6. Flow throught an Annulus
Now we can get other reults of interest:
(i)
The maximum velocity is
(ii)
The average velocity is given by
(iii)
The mass rate of flow is ω = πR2 (1 – κ2) ρ<v2>, or
(iv)
The force exerted by the fluid on the solid surface is obtained by summing the forces acting on the inner and
outer cylinders, as follows
The reader should explain the chouice of signs in front of the shear stresses above and also give an interp retation of the final
result. The equations derived above are valid only for laminar flow. The laminar-turbulent transition occurs in the
neighborhood of Re = 2000, with the Reynolds number defined as Re = 2R (1 – κ)<v2>ρ/µ
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7. Flow of Two Adjacent Immiscible
Fluids
Two immiscible, incompressible liquids are flowing in the z direction in a horizontal thin slit of length L and width W under
the influence of a horizontal pressure gradient (po – pL)/L/ The fluid flow rates are adjusted so that the slit is half filled with
fluid I (the more dense phase) and half filled with fluid II (the less dense phase). The fluids are flowing sufficiently slowly
that no instabilities occur–that is, that the interface remains exactly planar. It is desired to find the momentum- flux and
velocity distributions.
A differential momentum balance leads to the following differential equation for the momentum flux:
The equation is obtained for both phase I and phase II. Integration of previous equation for the two regions gives
7. Flow of Two Adjacent Immiscible
Fluids
Flow of two immiscible fluids between a pair of
horizontal plates under the influence of a
pressure gradient.
We may immediately make use of one of the
boundary
conditions–namely,
that
the
momentum flux is continuous through the fluid–
fluid interface:
This tells us that C1I = C1II, hence we drop the
superscript and call both integration contans C1
When Newton’s law of viscosity is substituted into previous equation we get
Those two equations can be integrated to give
7. Flow of Two Adjacent Immiscible
Fluids
The three integration constants
can be determined from the
following no-slip boundary
conditions:
When these three boundary
conditions are applied, we get
three simultaneous equations
for the integration constants :
From these three equations
we get
The resulting momentumflux and velocity profiles are
In both viscosities are the same,
then the velocity distribution is
parabolic, as one would expect
for a pure fluid flowing between
parallel plates
The average velocity in each layer can be obtained and the results are
From the velocity and momentum–flux
distributions given above, one can also
calculate the maximum velocity, the velocity
at the interface, the plane of zero shear
stress, and the drag on the walls of the slit.
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