Diyas on the Bagmati river
"The water of the stream, as revealed by the leaves, flowed
faster in the middle of the stream than near the edges (banks)" [1]. This is due
to the fact that the friction with the stream-bed -- and with the stream-banks -- slows
down the flow of the water, and the effect of the friction dissipates the
further away you get from the bed or bank. This observation helps to
understand the picture above.
Caption: Oil lamps offered by devotees illuminate the Bagmati River
flowing through the premises of the Pashupatinath Temple during the Bala
Chaturdashi festival in Kathmandu, on Monday, November 28, 2016.
The image above was captioned 'river of light' in the Hindu, but
the original description is above.
What I see in the image is the obvious streamlines in the middle
of the river. On second thoughts, the ‘streamlines’ near the centre of the
river could just be the result of the time lapse image with a single diya. Or
multiple diyas all tracing the same path, although the latter appears less
likely.
But I also would argue that the concentration of light on both
sides of the river near both banks is because the speed of the river goes to
zero or close to zero on either side (being a maximum close to the middle of
the river). Since it is a time lapse picture the slow-moving candles near the
bank should give most light.
More prosaically, the oil lamps are being floated into the river
by hand and not by boat so that would be a simpler explanation of why most of
the oil lamps are near the banks.
Can we argue that the diyas will move some distance transversely
towards the middle of the river – but not very far? If so, what is the width of
this transverse region? Looking at the image there does seem to be such a
region with an almost constant width.
The first reason (slow speed near the river edge) is easy to
prove; the second argument – that the diyas are confined within a given width –
is harder (for me). I try to give some
plausible arguments and then try to make them numerical, and it’s
in the writing of equations, solving them and comparing them with the image
that I run into trouble. Still, I’m giving it a go. However, a cautionary note:
that the diyas are confined in a given width of the river near the banks seems
plausible – but it cannot be a definite conclusion. There may be circumstances
in which that confinement does not occur. It is also possible that the
transverse velocity of the diyas is much less than the stream velocity (and
that seems to happen according to simulations, as detailed below), so that in a
given time, the diya would travel a much greater distance downstream than
transversely towards midstream – and this would give rise to a similar
appearance of a ‘width’ to the region containing diyas, and a largely unlit
midstream region.
|
|
Suppose the diya is close to the
bank, where the velocity gradient is the greatest.
Since the velocity of the stream
at point C is higher than at point B (assuming the diya is close to the Left
Bank), by Bernoulli’s law there is a ‘lift’ which should push the diya towards the middle
of the stream.
The difference in velocity across
the diya is 2Dv
which yields a corresponding transverse pressure gradient:
Dp
= (r/2)( 2Dv)2
Where 2Dv = (dv/dz) dd,
Where dd is the
diameter of the diya.
The pressure gradient will push
the diya towards the centre of the river.
But there should also be an
opposing drag - which could be expressed by Stoke’s law or by the
square-velocity law.
Anyway, the region of high
velocity gradient is close to the bank, and the diyas should mostly stay there
in that region.
The flow of the river could be
laminar or turbulent, and the (transverse) velocity profile across the river is
different in the two cases.
So far, so (qualitatively) good.
To go beyond, one must answer
several questions:
1 1) How
fast does the river flow?
2 2) Is
the flow of the river laminar or turbulent?
3 3) How
does the velocity profile and the velocity gradient change the transverse
pressure?
4 4) What
are the relevant parameters of the diya itself?
5 5) Is
the drag described by Stoke’s law or by the square-velocity law? Plug in to
write the equation of motion.
6 6) What
boundary conditions are needed to solve the equation of motion?
1)
How fast does the river flow?
What is the speed of a
river or stream? The following three sources give an indication:
(i)
As mentioned earlier, current speeds can
get up to about 3 m/s, but most rivers have currents less than 1 m/s; many
"fast" streams are actually flowing about 0.3 m/s, and large rivers
have deceptively fast currents [2].
(ii)
The speed of a river varies from close to
0 m/s to 3.1 m/s [3].
(iii)
In mountain streams, water may reach
speeds of 5-10 m/s [4].
In general, the speed that a stream
reaches is terminal velocity obtained by equating the inertial force acting on
a unit volume of water with the shear stress acting on the water due to the
stream bed and its banks [5], where the shear stress t is proportional to the
square of the stream velocity v:
V
= C/(ds)1/2
Where d is the depth of the stream, for a
stream whose breadth b >20 d, and s is the slope of the stream bed ( s =
tan(µ)),
and C is Chezy’s constant. Manning found that Chezy’s constant is given by:
C
= d1/6/n,
where n is Manning’s roughness factor that
characterizes the ‘roughness’ and the ‘waviness’ of the stream, and lies in the
range 0.01 - 0.1 for most regular open
channels.
Substituting:
V
= d2/3s1/2/n
In general, Manning’s formula
gives the speed of the stream that results from a particular value of slope S
in terms of the hydraulic radius Rh:
V
=
where
Rh is the hydraulic radius (ratio of area to perimeter, A/P) which
reduces to the depth of the river, as above. The value of n varies depending
upon the roughness and the ‘waviness’ of the river: n = 0.02-0.025 for a straight river [6],
n = 0.03 - 0.05 for a meandering river
[7], n = 0.075 – 0.150 for a weedy stream [8].
For n = 0.02 and Rh
= 1 m, and S = 0.001, v = 1.6 m/s; while if the slope is 10X (S = 0.01), then V
= 5.0 m/s.
Most river stream-beds have
angles less than 5°
( slope s = 0.087) [5].
2)
Is the stream flow laminar or turbulent?
First, one must address
the question of whether the flow in a stream or river can be laminar at all,
and if so, when.
We would need to know the depth d
of the river because that is the number that decides the hydraulic radius Rh
if the breadth b of the river is sufficiently large compared to the depth: b >> d.
Assuming it has a rectangular
shape (breadth b, depth d):
Rh
= bd/(b+d)
And the Reynolds number
is defined in terms of the hydraulic radius:
Re
= vmRh/m
Consider 2 cases: (i) b
>> d and (ii) b = d.
In the 1st
case: the hydraulic radius reduces to: Rh = d, the depth of the river.
If we take vm
= 1 m/s and d = 1 m, m
= 1x10-6 m2/s,
then Re = 106 and the flow is turbulent.
It will remain turbulent unless the depth d is
reduced to about 2 mm!
Alternatively, if the
depth is 1 m, then the flow will be laminar only if the speed of the water is extremely
slow ( ~mm/sec).
If w = d, then Dh =
d/3. And the trend is just about the same.
The transition Reynolds
number - for a wide river - below which flow is laminar is Ret =
1500 [8].
That is, if b >> d, for flow to be laminar there is a limit on the product of vd: vd < 1.5x10-3.
The conclusion is that
unless a stream/river is extremely shallow (depth in mm’s) or slow-moving
(speed in mm/sec), the flow is bound to be turbulent.
3)
How does the velocity profile and the
velocity gradient change the transverse pressure?
a a)
Laminar flow b) Turbulent flow.
Laminar flow [9]:
Turbulent flow [9]:
a a) So
if we assume laminar flow, the derivative becomes:
dv/dr
= -{2rvm)/R2
Note that the velocity
gradient for laminar flow at r = R (the river’s edge) is nonzero, so the
pressure moving the diya towards the centre of the stream starts out as
nonzero.
So the pressure across
a diya of diameter d is:
P
= (r/2)[(2ddrvm)/R2]2
Where vd is
the transverse velocity of the diya.
Plugging in values a
bit arbitrarily, assume the diya dia is d = 0.06 m (6 cms), the radius of the
river R =30 m, while the density of water r
is 1000, and its kinematic viscosity m
[10] is 1.0x10-6 m2/s at 20 °C.
b b) Suppose
the river is turbulent and the velocity profile is given by:
v = vm [1 – (r/R)]1/7
Differentiating:
dv/dr
= [vm/(7R)] [1 – r/R]-6/7
In the case of
turbulent flow, at r = R (the river’s edge), the velocity gradient and thus the
tranverse pressure gradient are both nonzero.
In fact, the velocity
gradient is about 3X larger for the turbulent flow than for the laminar flow
(at the same speed vm = 1 m/s), as is seen in the Figure below. Also,
the velocity
gradient decreases as
one approaches the middle of the stream (r = 0).
So, the pressure at
that point is about 10X more for turbulent than for laminar flow.
Realistically, this factor would be even higher because the speed of laminar
flow would be lower
than for turbulent
flow.
So the pressure is:
P
= (r/2)[dvm/(7R)]2
[1 – r/R]-12/7
It turns out that the
velocity gradient is higher for turbulent flow for values of r/R < 0.074 and
for r/R > 0.951; and laminar flow has higher gradient for 0.074 < r/R < 0.951.<0 .951.="" o:p="">
(Near the centre of the
river (r = 0) the velocity gradient is zero for laminar flow, but nonzero for
turbulent flow).
Near the banks, for r/R > 0.951, the velocity gradient is greater for
turbulent flow than it is for laminar flow.
One might think that,
even with turbulent flow, right near the river bank, where the velocity drops
to zero, the flow would be laminar. Apparently, such an argument is made –
but for v® 0 near the river bed. Specifically, the
thickness of this viscous sublayer is given by:
d = 11.6 m/v*,
Where v* is:
v*
= (t/r)1/2 = (gdS)1/2
and the numerical value
is ~1
mm. Possibly this is too small compared to the irregularity of the river banks
– or maybe it is not applicable at all. However that may be, the laminar sublayer [11] does not
seem to apply to river banks.
According to Hickin [11],
even in a turbulent stream, near the banks (or the bed) where the velocity is
low enough, this is due to the significant effects of viscous drag, and the flow in this sub-layer may be laminar and
described by Stoke’s law. Hickin does not specify how wide this viscous
sub-layer is, but we can argue that near the banks both the depth and the velocity may be
low enough that the Reynolds number drops below 1500. However, the numerical
magnitude estimated above is 1 mm thickness, near the bed, and if the same applies to the banks, it may be neglected.
1)
What are the relevant parameters of the diya
itself?
Floating earthen diya: size 5.5” x
5.5” weight = 260 -350 gms [12].
Density of dry clay: 1.6 gm/cc
[13].
Assume that the diya is an open-topped
cylinder with radius r = 7 cms, height h = 2 cms, and uniform thickness t = 0.9
cms, with a density of rd
= 1.6 gm/cc:
Its mass is m = prd
(r2 + 2hr) t = 1.6 (154 + 88) (0.9) = 348 gms.
Note: the total height of the diya is: 2.9 cms.
The volume V displaced by the diya
is 446 cc.
The weight of the diya is 350 gms,
so the force acting on it is: Fg = mg = (0.35)(9.80) = 3.43 N
The buoyant force acting on the
diya is: Fb = r
gV = (4.46x10-4)(1000) (9.80) = 4.37 N.
The submerged fraction of the diya
is given by the ratio 3.43/4.37 = 0.785, and so the submerged depth is:
(0.785)(2.9) = 2.3 cms.
The area subject to drag of the diya is half
of the front cross-section: (prh/2)
= (1.57)(2.3)(7.0) = 25.3 cm2.
Note that we can define
a Reynolds number for the diya:
Red
= 2vdrd/m
However, this cannot be
used to define whether the flow of the stream is laminar or turbulent – that gets
decided by the Reynolds number defined above.
It can be used to
decide whether the drag faced by the diya is closer to Stoke’s law or square law
(Stoke’s law is derived and is only valid for a sphere).
The plot of the drag
coefficient Cd vs Red , for a circular cylinder, is given
below [14]:
The trend is that the
drag coefficient is close to 2 for Re>10, and increases as Re decreases (Cd
µ
1/Re ) to as much as 365 for Re = 0.01, the latter being close to the Stoke’s
law limit.
1)
Drag and Equation of motion:
Stoke’s law applies in the case of a
sphere: it does not apply for the case of a circular cylinder.
As discussed above, the
stream flow is most likely to be turbulent: it can only be laminar if the
stream is very slow-moving or very shallow. However, just for completeness, the laminar flow case
is also discussed. At fairly low speeds, the drag experienced by the diya is
closer to the Stoke’s law limit. At higher speeds, it is given by the square
law form of drag.
a)
The equation of motion, for laminar flow
and square law drag, is:
(m/A)(dvd
/dt) = (r/2)[(2ddvm)r/R2]2
– (rdCD
(vd)2/2)
Since vd is
the transverse velocity of the diya, it is given by: vd = dr/dt. The
initial conditions are: r (0) = R & vd (0) = 0.
Numerically, m/A =
0.35/(25.3x10-4) = 138. First term on r.h.s.: (500)[14/900]2
= 2.42x10-4 r2 (assuming vm = 1 m/s). 2nd
term on r.h.s.: (1600)(2)/2 = 1600 (vd)2 for Cd
= 2.
This gives:
138(d2r/dt2) = 8.89 x10-6 r2 – 1600
(dr/dt)2
This is a second-order
differential equation, and it is probably nonlinear because of the r2 and
the (vd)2 term.
At t = 0, vd
= dr/dt = 0 and r = R = 30, so the 2nd term on the r.h.s. is zero,
and the initial acceleration is: a = dvd/dt = 5.8 x10-5
m/s2 (assuming vm = 1 m/s). If we take 10 m/s, a = 0.058
m/s2 .
So, this differential
equation is easy to solve numerically – trying to get an analytic solution, I
was unsuccessful. I tried some forms in which the diya velocity either asymptotically reaches a fixed
velocity and another in which it increases from zero, peaks and again decreases to zero - but I was
not lucky.
b)
Let us assume the flow is laminar and the
diya is ‘close to spherical’ and apply Stoke’s law instead (this means the low
Red limit):
(m/A)(dvd
/dt) = (r/2)[(2ddvm/R]2[1
– (2r/R)] – (6hvd/rd)
c) For the turbulent
flow case, and square law drag, the equation of motion is:
(m/A)(dvd
/dt) = (r/2)[dvm/(7R)]2
[1 – r/R]-12/7 – (rCD (vd)2/2)
where Cd is
the drag coefficient (with a value of ~2).
d) For turbulent flow
and Stoke’s law:
(m/A)(dvd
/dt) = (r/2)[dvm/(7R)]2
[1 – r/R]-12/7 – (6hvd/rd)
So, all these cases
have been solved numerically by writing a BASIC program. The results are
plotted below:
a a) Laminar
flow, square law:
a b) Laminar
flow, Stoke’s law limit:
a c) Turbulent
flow, square law:
a d) Turbulent
flow, Stoke’s law limit:
Note: in the turbulent flow, I set the initial position at r = 0.999Rs, because at exactly r = Rs, there is a problem...
Interestingly,
only in the cases (b) (of laminar flow and Stoke’s law limit) and d) (of turbulent
flow and Stoke’s law limit) is the drag sufficient to ensure that the velocity
of the diya decreases significantly. This would be require if the diyas are to
be confined to a narrow region close to the river banks. In the square law
limit cases the drag coefficient is too low, and the diyas rapidly
cover
the entire width of the stream – the only difference being that in the laminar
flow case (a) they slow down a bit midstream.
OK: Bottomline: The image of the diyas on the Bagmati
gives a value of r/R = 1.5/8 = 0.19.
So, about one-fifth of the stream
is covered by diyas in the image.
Can the explanation of transverse
pressure being opposed by viscous drag explain this observation?
Possibly, for the turbulent flow
case (which is most likely), if the viscous drag is sufficiently high as to approach the Stoke's law limit (case d).
To really verify this explanation
one would need much more extensive observation to check what value of Red
and Cd are appropriate to explain the observed width. And that rests upon the presumption that the image taken is a steady-state situation, and not one evolving to a different image where diyas cover the whole river!
I suspect that the speed with which the diyas move transversely is less than the speed of the stream, so that most of the river will be unlit... (assuming that there is no shortage of diyas!)...
However, I don't know this to be true...
I suspect that the speed with which the diyas move transversely is less than the speed of the stream, so that most of the river will be unlit... (assuming that there is no shortage of diyas!)...
However, I don't know this to be true...
Armchair analysis is, as usual, insufficient.
Links & References:
11)
“It’s a drag to have a boundary
layer” Peter Garrison Flying Magazine (Oct.2005) pp. 97-99
4 4)
Air and Water: The Biology and Physics of Life's Media Mark W.
Denny (Princeton Univ.Press, 1993) p.7
5 5)
“River hydraulics and the channel” Edward Hickins pp.4-10
6 6)
Leeds Univ.: CIVE2400 Fluid Mechanics: Section 2:
Open Channel Hydraulics p.14
8 8)
Mechanics of Fluids” Merle C. Potter,
David C. Wiggert, Bassem H. Ramadan (A1 books, 2016) p.332
111) Ch.5:
“The nature of turbulence and velocity distributions in rivers” p.3 Edward Hickin
114) “The steady flow of a viscous fluid past a
circular cylinder”, S.C.R.Dennis and M.Shimshoni (1965 report) p.28
No comments:
Post a Comment