File: enmf_529_HW_01.5_sol'n_complete_4print.nb
Date: 11/7/12 17:34:45
ENMF 529 - 2011 Fall term, Homework 1
Problem 5 - Solution
0. Preparing Mathematica for executing commands in this document (loading additional packages, if
needed, symbolizing, ...)
1. Computing the spere's volume and the applicable "scaling law"
Volume of sphere
V=
4
3
p R3 ;
V ê. 8R -> L<
4 L3 p
3
Answer: The buoyancy is proportional to L3
2. A slightly different and more interesting problem:
How does the ratio of immersion depth, h, to the radius, R, scale with size
Volume of the submerged sphere cap, as a function of the immersion depth h ; R radius of the sphere
Vc @h_D =
1
3
p h2 H3 R - hL;
Following the Archimedes law we are solving, for h , the equality” weight_of_displaced_liquid ==
weight_of_sphere”
Vc@hD rliqud g == V rs g ,
i.e.,
1 h2 p H3 R - hL r
4
3
liqud == 3 p R rs
3
Note that the following solution returns 3 roots, 2 of which are complex (i.e., not applicable).
Only the first solution appears to be real. For the sake of brevity we assume
rs = h rliqud ; h
<1
res = Simplify@
Solve@HVc @hD rliqud == V rs L ê. rs -> h rliqud , hD, Assumptions -> 8R > 0<D
R
::h Ø R I-1 + 2 h + 2
:h Ø
:h Ø
1
2
1
2
R 2+
H-1 + hL h M
1+Â
I-1 + 2 h + 2
R 2+
1-Â
I-1 + 2 h + 2
1ê3
- R K-1 + 2 h + 2
3
H-1 + hL h M
1ê3
3
H-1 + hL h M
1ê3
H-1 + hL h O
1ê3
>,
+ J1 - Â
3 N K-1 + 2 h + 2
H-1 + hL h O
+ J1 + Â
3 N K-1 + 2 h + 2
H-1 + hL h O
Microsensors Dynamics Lab. Contact: S. Spiewak, [email protected] , ph. (403) 220 8976
1ê3
1ê3
>,
>>
Page 1
File: enmf_529_HW_01.5_sol'n_complete_4print.nb
Date: 11/7/12 17:34:45
“Extracting” the real (1st) solution from the above result
hsoln = h ê.resP1T
R
RI-1 + 2 h + 2
H-1 + hL h M
1ê3
- R K-1 + 2 h + 2
H-1 + hL h O
1ê3
Sought ratio
r = SimplifyB
hsoln
R
F
1
1I-1 + 2 h + 2
H-1 + hL h M
1ê3
- K-1 + 2 h + 2
H-1 + hL h O
1ê3
Answer: The ratio does not depend upon “R”, i.e., the size parameter “L”
NOTE: In this solution we have neglected the surface tension of the liquid. It plays significant role as the size
decreases.
3. The above problem, however with the surface tension taken into account
Finding potential solutions We modify the equation based on the Archimedes law by subtracting the buoyancy due to the surface tension,
FB @hD, from the weight of the sphere: weight_of_displaced_liquid == weight_of_sphere - FB @hD
Vc@hD rliqud g == V rs g - FB@hD,
4 p R3r g - F @hD
s
B
3
i.e.,
1 h2 p H3 R - hL r
liqud g ==
3
FB @hD= Dp p IR2 - HR - hL2 M
Entering into Mathematica the expression for the Surface Tension Force
FB @h_D = SimplifyADp p IR2 - HR - hL2 ME
-h p Hh - 2 RL Dp
where
y is a constant depending upon the liquid and surface properties of the sphere involved
Dp = y
2
R
;
Showing the expression for the “reduced weight due to surface tension”
V rs g - FB @hD
4
3
g p R 3 rs +
2 h p Hh - 2 RL y
R
Solving for h
NoteDynamics
that theLab.
following
returns
3 roots - we ,will
to investigate
them
Microsensors
Contact:solution
S. Spiewak,
[email protected]
ph.have
(403) 220
8976
Page 2
File: enmf_529_HW_01.5_sol'n_complete_4print.nb
Date: 11/7/12 17:34:45
Note that the following solution returns 3 roots - we will have to investigate them
res = Simplify@
Solve@HVc @hD rliqud == V rs - FB @hDL, hD, Assumptions -> 8R > 0<D;
Dimensions@resD
83, 1<
Inspecting all solutions and eliminating meaningless ones
Quick inspection of one of the solutions (#1)
resP1T
:h Ø
1
18 R rliqud
I18 IR2 rliqud - 2 yM +
I18 IR4 r2liqud + 4 y2 MM ë IR6 r2liqud Hrliqud - 2 rs L + 2 I-4 y3 + , I-R4 r2liqud IR8 r2liqud
Hrliqud - rs L rs + 3 R4 r2liqud y2 + 4 R2 Hrliqud - 2 rs L y3 + 12 y4 MMMM
18 IR6 r2liqud Hrliqud - 2 rs L + 2 I-4 y3 + , I-R4 r2liqud IR8 r2liqud Hrliqud - rs L rs +
3 R4 r2liqud y2 + 4 R2 Hrliqud - 2 rs L y3 + 12 y4 MMMM
1ê3
1ê3
+
M>
Simplifying all three solutions and representing each of them as a single Mathematica expression - we do not
know which of them is/are meaningful, so we will need to test them by, e.g., plotting or using “Manipulate”
hsoln1 = Simplify@h ê.resP1TD;
hsoln2 = Simplify@h ê.resP2TD;
hsoln3 = Simplify@h ê.resP3TD;
Defining and inspecting the ratio h / R
... first, defining some representative numerical values
numSubst1 = 8y -> 0.073 , rs -> 7800 µ 9.81, rliqud -> 1000 µ 9.81<;
In the line above: specific density of steel is rs -> 7800 kg ë m 3 ; specific density of water is
rliqud -> 1000 kg ë m 3
Clear@r, RD
... defining the ratio h / R for each of the solutions
r1 @R_D = SimplifyB
1
2
18 R rliqud
hsoln1
R
F
I18 IR2 rliqud - 2 yM +
I18 IR4 r2liqud + 4 y2 MM ë IR6 r2liqud Hrliqud - 2 rs L + 2 I-4 y3 + , I-R4 r2liqud IR8 r2liqud
Hrliqud - rs L rs + 3 R4 r2liqud y2 + 4 R2 Hrliqud - 2 rs L y3 + 12 y4 MMMM
18 IR6 r2liqud Hrliqud - 2 rs L + 2 I-4 y3 + , I-R4 r2liqud IR8 r2liqud Hrliqud - rs L
rs + 3 R4 r2liqud y2 + 4 R2 Hrliqud - 2 rs L y3 + 12 y4 MMMM
r2 @R_D = SimplifyB
hsoln2
R
1ê3
1ê3
+
M
F;
Microsensors Dynamics Lab. Contact: S. Spiewak, [email protected] , ph. (403) 220 8976
Page 3
File: enmf_529_HW_01.5_sol'n_complete_4print.nb
r3 @R_D = SimplifyB
Date: 11/7/12 17:34:45
hsoln3
R
F;
Note: The ratio h / R appears to depend upon R, i.e., upon the size parameter “L”. Below we inspect this
ratio for the 3 obtained solutions. If the ratio is >1 or the solution has a significant imaginary component, the
sphere sinks
ManipulateA
ChopAGridA99"R = ", 103 N@RD, " mm"=,
8"r1 HRL = ", r1 @RD<,
8"r2 HRL = ", r2 @RD<,
8"r3 HRL = ", r3 @RD<
=E ê. numSubstE,
99R, 1.2 µ 10-3 =, 10-4 , 2 µ 10-3 =
E
R
0.0012
R =
r1 HRL =
r2 HRL =
r3 HRL =
1.2
1.29286
-30.1004
0.801737
mm
The above inspection suggets that only the 3rd solution is meaningful
Final plot of the immersion‐to‐radius ratio as a function of R (size of the sphere)
PlotAChopAr3 A10-3 RE ê. numSubst1 ê. numSubst1E,
8R, 0.01, 1.5<, AxesLabel -> 8"R, @mmD", "hêR"<E
hêR
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
1.2
1.4
R, @mmD
Microsensors Dynamics Lab. Contact: S. Spiewak, [email protected] , ph. (403) 220 8976
Page 4