Supplementary Material
This document presents the mathematical proofs for the equations for Capillary Flow, and Stage 1, 2,
and 3 Pressures.
Article Title:
Theoretical Development and Critical Analysis of Burst Frequency Equations for Passive Valves on
Centrifugal Microfluidic Platforms
Where presented in the paper, the difference between the proofs shown and the equations from the
literature is indicated in boxed sections.
The sequence of the equations presented here may not be the same as that presented in the paper:
Journal:
Medical & Biological Engineering & Computing
[eq#] refers to the sequence of equation in this supplementary document
(#) refers to the sequence of equations in the paper
Authors:
Tzer Hwai Gilbert Thio, Salar Soroori, Fatimah Ibrahim, Wisam Alfaqheri, Norhayati Soin,
Lawrence Kulinsky, Marc Madou
Corresponding Author:
Fatimah Ibrahim
Medical Informatics & Biological Micro-electro-mechanical Systems (MIMEMS) Specialized
Research Laboratory, Department of Biomedical Engineering, Faculty of Engineering, University of
Malaya, 50603 Kuala Lumpur, Malaysia
Tel. No. (Office): + 603-7967-6818
Fax No.: +603 7967 4579
E-mail address: [email protected]
Page 1 of 33
Page 2 of 33
Calculation of the Burst Frequency for Capillary Valves on Centrifugal Microfluidic Platforms
Fluid Flow within an Infinitely Long Channel (Capillary Flow):
On a centrifugal microfluidic CD with capillary channels, there are two pressures present: The
centrifugal pressure, and the capillary pressure
The centrifugal pressure is due to the rotation of the CD:
Pcentrifugal = ρω 2 ∆r r
where ρ
ω
∆r
r
[eq1] (1)
is the density of the liquid
is the rotational speed of the CD in rounds per minute (rpm)
is the difference between the top and bottom of the liquid levels at rest with respect to
the centre of the CD
is the average distance of the liquid from the centre of the CD
Figure 1: Model of Rectangular Channel
The pressure that is due to the liquid-air surface tension, contact angle of the liquid, and the hydraulic
diameter of the capillary channel is known as:
Pcap =
where γ al
4γ al cosθ c
Dh
[eq2]
is the liquid-air surface tension
θc
is the contact angle of the liquid with respect to the solid wall of the channel
Dh
is the hydraulic diameter of the channel given as
4 × Area
4wh
2wh
Dh =
=
=
Perimeter w + w + h + h w + h
where w is the width, and h is the height of the channel
(13)
For the liquid to move towards the edge of a CD, the centrifugal pressure must be greater than the
capillary pressure. This allows us to calculate the burst frequency:
Figure 2: Model of Circular Channel (Solid & Sliced)
The total interfacial energy within a solid-liquid-air system is expressed as
From [eq1] and [eq2]
U T = Aslγ sl + Asaγ sa + Alaγ la
Pcentrifugal = ρω 2 ∆r r = Pcap
Rearranging yields:
where Asl , Asa , Ala
ω=
Pcap
γ sl , γ sa , γ la
[eq4] (3)
are the solid-liquid, solid-air, and liquid-air interface areas,
are the corresponding surface energies per unit area. A
ρ ∆r r
Alternatively:
rpm = ω ×
The surface energies per unit area are related by Young’s equation
30
π
=
Pcap  30 
 
ρ∆r r  π 
[eq3] (7)
Page 3 of 33
γ sl = γ sa − γ la cos θ c
[eq5]
Page 4 of 33
Substituting Young’s equation into [eq4] gives us
We can now consider Asl for various capillary shapes
U T = Asl (γ sa − γ la cos θ c ) + Asaγ sa + Ala γ la
UT
For a circular capillary
= ( Asl + Asa )γ sa + ( Ala − Asl cosθ c )γ la
The surface boundary between the solid and liquid can be found by
Asl = π D x
As all the terms in ( Asl + Asa )γ sa are constants, we can denote it as U 0
U T = U 0 + ( Ala − Asl cos θ c )γ la
[eq6] (4)
where D
x
(8)
is the diameter of a circular capillary
is the length of expansion along the liquid’s axis of movement
The volume of the liquid can be found as
The pressure in the capillary can be derived from the change of total interfacial energy of the solidliquid-air system with respect to the injected liquid volume:
Pcap = −
dU T
dV
 d
d
 d
 
Pcap = −  U 0 + 
Ala −
Asl cosθ c γ la 
dV
 
 dV
 dV
d
Since U0 is a constant,
U 0 = 0 . This simplifies [eq8] to
dV
d
 d

Pcap = γ la 
Asl cosθ c −
Ala 
dV
dV


[eq7] (2)
[eq8] (5)
d
Ala = 0 (as Ala which is the surface boundary between liquid and air in a
dV
capillary opening does not change just before reaching the end of the channel we can simplify [eq9]
to
Pcap =
Pcap
πD 2 x
(9)
4
Hence we can determine for a circular capillary
dAsl π D dx 4
=
=
dV
D
πD 2
dx
4
[eq11]
For a rectangular capillary
[eq9]
If we further apply
d
Asl cos θ c γ la
dV
dA
= sl cos θ c γ la
dV
V =
[eq10] (6)
The surface boundary can be determined as
Asl = 2( w + h) x
(10)
where w , h are the width and height of the capillary
x is the length of expansion along the liquid’s axis of movement
The volume of the liquid can be found as
V =wh x
Hence we can determine for a rectangular capillary
dAsl 2( w + h) dx
4
=
=
dVl
w h dx
Dh
(11)
[eq12] (12)
Alternatively, manipulating [eq12] gives
dAsl
( w + h)
=2
dVl
wh
dAsl
1 1 
= 2 + 
dVl
h w
Page 5 of 33
[eq13]
Page 6 of 33
Hence, for both Circular and Rectangular Capillary
Stage 1:
Substituting [eq11] into [eq10] yields
4 cos θ c γ la
for circular capillary
Pcap =
D
Substituting [eq12] into [eq10] yields
4 cosθ c γ la
Pcap =
for rectangular capillary
Dh
The liquid is in a capillary channel, stopped at an opening with a concave meniscus (the meniscus
changes from concave to flat)
[eq14]
According to [eq6] and [eq8], we need to determine Ala and Asl for U T = U 0 + ( Ala − Asl cos θ c )γ la ,
[eq15] (14)
and we also need to determine the volume of the liquid, V.
Different from Zeng et al (2000) eq (2)
Rh =
h2
This is different
αh
αh
CLh
Different from Chen et al (2008) eq (2)
Substituting [eq13] into [eq10] yields
1 1 
Pcap = 2 cosθ c γ la  + 
 h w
h/2
sin α h
h2
Rw =
[eq16] (15)
w/ 2
sin α w
w2
Substituting either [eq14] or [eq15] into [eq3] yields
rpm =
4γ al cos θ c  30 
 
ρ ∆r r Dh  π 
αw
αw
CLw
w2
[eq17]
Figure 3: Mathematical Model for Stage 1 Meniscus
Ala can be found by multiplying the curve length of height, CLh , and the curved length of width, CLw :
Ala = CLh × CLw
Page 7 of 33
[eq18]
Page 8 of 33
Let the curved lengths have angles of 2α w and 2α h , and curve radii of Rw and Rh respectively.
We can derive the following from the figure:
h/2
- see Figure 3
[eq19]
Rh =
sin α h
w/2
- see Figure 3
[eq20]
Rw =
sin α w
hα h
CLh = Rh × 2α h =
- see Figure 3
[eq21]
sin α h
wα w
CLw = Rw × 2α w =
- see Figure 3
[eq22]
sin α w
Hence substituting [eq21] and [eq22] into [eq18] yields,
hwα hα w
Ala =
sin α h sin α w
[eq23]
ATriangle
Amh
ASector
Figure 5: Mathematical Model for Stage 1 Liquid Surface
Asl can be found by taking the surface of solid-liquid within the capillary if there was no meniscus ( Ah
and Aw ) and subtracting the surface of liquid-air due to the meniscus ( Amh and Amw ). We assume the
liquid to progress up to length L from the point of reference (right up to the opening).
3D View
Asl = 2( Ah + Aw − Amh − Amw )
- see Figure 4
[eq24]
Where Ah = hL
- see Figure 4
[eq25]
Aw = wL
- see Figure 4
[eq26]
Amh = ASector − ATriangle
- see Figure 5
[eq27]
Where ASector = angle of curvature × radius of curvature
Top View
ASector
- see Figure 3
2α
2
2
= h (π ) × Rh = α h Rh
2π
2
w
Aw
Amw
And
L
Side View
 h2 
h 2α h
 =
ASector = α h 
4 sin α h
 sin α h 
1
- see Figure 3
ATriangle = × height × base
2
h
h2
1
× Rh cos α h × h = ×
cos α h
Triangle =
2
2 sin α h
ATriangle =
h
Ah
Amh
Amh =
h 2 cosα h
4 sin α h
[eq28]
[eq29]
h 2α h
h 2 cos α h
−
2
4 sin α h
4 sin α h
[eq30]
w 2α w
w 2 cos α w
−
2
4 sin α w
4 sin α w
[eq31]
Similarly
L
Figure 4: Model for Stage 1 Fluid Flow
Amw =
Page 9 of 33
Page 10 of 33
The volume V can be found by taking the volume of liquid within the capillary if there was no
meniscus ( Vhw ) and subtracting the volume of displaced liquid due to the meniscus ( Vmw and Vmh ). We
assume the liquid to progress up to length L from the point of reference:
Hence substituting [eq25], [eq26], [eq30], and [eq31] into [eq24] yields,

h 2α h
h 2 cos α h
w 2α w
w 2 cos α w 

Asl = 2 hL + wL −
+
−
+
2
2
4 sin α h
4 sin α w 
4 sin α h
4 sin α w




h2  α h
w2  α w


− cos α h  −
− cos α w 
Asl = 2 (h + w)L −
4 sin α h  sin α h
 4 sin α w  sin α w


V = Vhw − Vmw − Vmh
- see Figure 6
where Vwh = hwL
[eq36]
 w αw
w cos α w 
h
Vmw = Amw × h = 
−
2
4 sin α w 
α
4
sin
w

 h 2α h
h 2 cos α h  wα w

Vmh = Amh × CLw = 
−
2
4 sin α h  sin α w
 4 sin α h
2
Now, substituting [eq23] and [eq32] into [eq6] yields,
U T = U 0 + ( Ala − Asl cos θ c )γ la
[eq33]




h2  α h



− cos α h 
(h + w)L −
 hwα α

4 sin α h  sin α h

h w
UT = U 0 + 
− 2 
cos
θ
c γ la

2
α
α
sin
sin


α
h
w


− w

 w − cos α w 


 4 sin α w  sin α w




2


 αw

w

− cos α w 
(h + w)L −
4 sin α w  sin α w
 hwα hα w 



U T = U 0 − 2 cosθ c γ la 
+ γ la 

2


αh
 sin α h sin α w 
− h


− cosα h 

 4 sin α h  sin α h

[eq35]
[eq32]
2
- see Figure 3, 4, 5, 6 [eq37]
- see Figure 3, 4, 5, 6 [eq38]
Hence substituting [eq36], [eq37], and [eq38] into [eq35] yields
[eq34] (16)
 w 2α w
w 2 cos α w   h 2α h
h 2 cos α h  wα w
h − 

V = hwL − 
−
−
2
2


4 sin α w   4 sin α h
4 sin α h  sin α w
 4 sin α w

 αh

h 2 wα w
hw2  α w


− cos α w  −
− cos α h 
V = hwL −
4 sin α w  sin α w
 4 sin α h sin α w  sin α h

[eq39] (17)
Finally, we can apply [eq34] and [eq39] into:
dU T
[eq40]
PStage1 = −
dV
To simplify the equation, as Ala , Asl and V are functions of α w and α h , we can obtain the burst
pressure by letting f to be a function of α w and α h . Hence
PStage1 = −
Point of
Reference
for L
Vhw
Vmw
Vmh
PStage1
PStage1
Vhw – Vmw
Vhw – Vmw – Vmh
dU T
dU T
df
=−
dV
dV
df
[eq41]
 ∂U T
  ∂U T

 ∂U T


∂α w  
∂α h 
∂α h  ∂f
 ∂U T
 + 

+

 ∂α 


∂α w   ∂f
w

 ∂f
  ∂f


∂α w  
∂α h 
∂α h 

= −
=−
 ∂V

 ∂V
  ∂V


∂α h  ∂f
∂α w  
∂α h 
 ∂V
 
 ∂α 

+

 ∂α  + 
w
w

 ∂f

 ∂f
  ∂f

 ∂α h 
 ∂α w   ∂α h 
−1
 ∂U T
 +  ∂U T
  ∂f
 ∂f
 

 
  ∂α  ∂α  
α
α
∂
∂
w
h 
w 
h




[eq42]
=−
−1
 ∂V
 +  ∂V
  ∂f
 ∂f
 
 ∂α   ∂α   ∂α  ∂α  
w
h 
w 
h




Figure 6: Mathematical Model for Stage 1 Liquid Volume
Page 11 of 33
Page 12 of 33
Now from [eq34], we can derive
∂U T
∂U T
and
:
∂α w
∂α h
Now from [eq39], we can derive


w2  α w
∂ 
γ la − 2 cos θ c γ la

− cos α w 
−
∂α w  4 sin α w  sin α w


∂U T hwα h ∂  α w 
cos θ c γ la w 2 ∂  α w
cos α w 



γ la +
=
−
∂α w sin α h ∂α w  sin α w 
2
∂α w  sin 2 α w sin α w 

∂U T
hwα h ∂  α w 
w 2 cos θ c ∂  α w


 + γ la
= γ la
− cot α w 
∂α w
sin α h ∂α w  sin α w 
2
∂α w  sin 2 α w


∂U T
hwα h  sin α w − α w cos α w 
w 2 cos θ c  sin 2 α w − α w 2 sin α w cos α w

 + γ la

= γ la
+ csc 2 α w 


∂α w
sin α h 
2
sin 2 α w
sin 4 α w



∂U T
∂
=
∂α w ∂α w
 hwα hα w

 sin α h sin α w
 1
α w cos α w 
α w 2 cos α w
w 2 cos θ c  1
1


 2
+
 sin α − sin 2 α  + γ la
 sin α − sin 3 α
2
sin 2 α w
w
w 
w
w


∂U T
hwα h  1
w 2 cos θ c  1
α cos α 
α cos α 


= γ la
− w 2 w  + γ la
− w 2 w 

∂α w
sin α h  sin α w
sin α w  sin α w
sin α w 
sin α w 
hwα h
∂U T
= γ la
sin α h
∂α w
 w 2 cos θ c hwα h
∂U T
= γ la 
+
∂α w
sin α h
 sin α w
∂U T
∂α h
∂U T
∂α h
 1
α h cos α h

 sin α − sin 2 α
h
h

α h cos α h
∂U T hwα w γ la  1

=
−
∂α h
sin α w  sin α h
sin 2 α h
 h 2 cos θ c γ la  1
α h 2 cos α h
+
 2

 sin α − sin 3 α
2
h
h


 h 2 cos θ c γ la  2
α h 2 cos α h
+
 2

 sin α − sin 3 α
2
h
h


α h cos α h  h 2 cos θ c γ la  1
α h cos α h 
hwα w γ la  1

+


=
−
−
sin α w  sin α h
sin 2 α h 
sin 2 α h  sin α h
sin 2 α h 
 h 2 cos θ c hwα w  sin α h − α h cos α h 


= γ la 
+

sin α w 
sin 2 α h
 sin α h

∂U T
∂α h
 αw

 αh

h 2 wα w
∂


− cos α w  −
− cos α h 
α
α
α
α
α
sin
∂
4
sin
sin
sin
w
w
h
w 
h




 ∂  αw 
∂V
hw2 ∂  α w
h2w  α h




=−
− cot α w  −
− cos α h 
∂α w
4 ∂α w  sin 2 α w
 ∂α w  sin α w 
 4 sin α h  sin α h
hw2
∂V
∂
=−
∂α w
∂α w 4 sin α w
 sin 2 α w − α w 2 sin α w cos α w

 sin α w − α w cos α w 
h2 w  α h



+ ccs 2α w  −
− cos α h 
4
sin α w
sin 2 α w

 4 sin α h  sin α h


 1
α w 2 cos α w
α w cos α w 
∂V
hw 2  1
h2w  α h
1 

−


=−
−
+
− cos α h 
−
∂α w
4  sin 2 α w
sin 3 α w
sin 2 α w  4 sin α h  sin α h
sin 2 α w 
 sin α w
∂V
hw2
=−
∂α w
4
α 2 cos α w 
∂V
hw 2  2
h2w

−
=−
− w 3
∂α w
4  sin 2 α w
sin α w  4 sin α h
 αh
 1
α w cos α w 
 sin α − cos α h  sin α − sin 2 α 
h
w


w 
 1
α w cos α w 
α w cos α w 
hw 2  1
h2w  αh
∂V

−


=−
−
− cos α h 
−
∂α w
2 sin α w  sin α w
sin 2 α w  4 sin α h  sin α h
sin 2 α w 
 sin α w
 hw 2
∂V
h2w
= −
+
∂α w
 2 sin α w 4 sin α h
[eq43]
 αh
 sin α w − α w cos α w

− cos α h 
α
sin
sin 2 α w
h






[eq45]

h 2 wα w ∂  1  α h
∂V

=−
− cos α h 

∂α h
4 sin α w ∂α h  sin α h  sin α h

2


αh
h wα w ∂
cos α h
∂V


=−
−
∂α h
4 sin α w ∂α h  sin 2 α h sin α h 
 hwα hα w

 sin α h sin α w
∂U T hwα w γ la
=
sin α w
∂α h
∂U T
∂α h






∂ 
h2  αh
γ la − 2 cos θ c γ la

− cos α h 
−
α
α
α
∂
4
sin
sin
h 
h 
h


cos α h 
hwα w ∂  α h 
h 2 cos θ c γ la ∂   α h


γ la −
=
−
− 
sin α w ∂α h  sin α h 
2
∂α h   sin 2 α h sin α h 

hwα w ∂  α h 
h 2 cos θ c γ la ∂  α h


γ la +
=
− cot α h 
sin α w ∂α h  sin α h 
2
∂α h  sin 2 α h


hwα wγ la  sin α h − α h cos α h  h 2 cos θ cγ la  sin 2 α h − α h 2 sin α h cos α h

 +

=
+ ccs 2α h 
sin α w 
sin 2 α h
2
sin 4 α h



∂U T
∂
=
∂α h ∂α h
∂U T
∂α h
 sin α w − α w cos α w


sin 2 α w





∂V
∂V
and
:
∂α w
∂α h

h 2 wα w ∂  α h
∂V

=−
− cot α h 
∂α h
4 sin α w ∂α h  sin 2 α h


h 2 wα w  sin 2 α h − α h 2 sin α h cos α h
∂V

=−
+ csc2 α h 
∂α h
4 sin α w 
sin 4 α h

α 2 sin α h cosα h
h 2 wα w  1
1
∂V

=−
− h
+
4 sin α w  sin 2 α h
sin 4 α h
sin 2 α h
∂α h


+
sin 2 α h 




1



α 2 cosα h 
h 2 wα w  2
∂V


=−
− h 3
∂α h
4 sin α w  sin 2 α h
sin α h 
h 2 wα w  sin α h α h cos α h 
∂V


=−
−
∂α h
2 sin α w  sin 3 α h
sin 4 α h 
 sin α h α h cos α h 
 2

 sin α − sin 2 α 
h
h 

 sin α h − α h cos α h 
h 2 wα w
∂V


=−

∂α h
2 sin α h sin α w 
sin 2 α h

h 2 wα w
∂V
=−
2 sin α w sin α h
∂α h
[eq44]
Page 13 of 33
[eq46]
Page 14 of 33
Now if we assume that α w is changing in such a way that the meniscus now starts to changes from
concave to flat (on its way to becoming convex); while α h remains constant, we can simplify [eq42]
to:
 ∂U T


∂α w 

PStage1
=−
[eq47]
∆α h = 0
 ∂V

 ∂α 
w

Now, substituting [eq43] and [eq45] into [eq47] yields
PStage1
 w cosθ c hwα h
+
sin α h
 sin α w
 sin α w − α w cosα w 


sin 2 α w


=−
 hw2
 sin α w − α w cos α w 
h2w  α h


−
+
− cos α h 
sin 2 α w


 2 sin α w 4 sin α h  sin α h
 w 2 cosθ c hwα h
+
sin α h
 sin α w
γ la 
PStage1 =
hw 2
h2w
+
2 sin α w 4 sin α h



Asl = 2( Ah + Aw + Amh + Amw )
Where Ah = hL
Aw = wL
Now, if we assume that α h approaches 0, then we will have
 w 2 cosθ c hwα h
+
sin α h
 sin α w
γ la 
PStage1
α h →0
PStage1 α
h →0
PStage1 α
h →0
=
2
2
hw
h w
+
2 sin α w 4 sin α h



 αh


− cos α h 
sin
α
h

 αh→0
 w 2 cosθ c
 2 sin α w 
= γ la 
+ hw 

2
α
sin
w

 hw 
2γ  w

= la  cosθ c + sin α w 
w h

Note:
αh
sin α h
The derivative of the equations for this is similar to that of Stage 1. The differences are as such:
For Stage 1: we subtract away the meniscus Area and Volume
For Stage 2: we add on the meniscus Area and Volume
 w 2 cosθ c

+ hw 
 sin α w

 hw 


 2 sin α w 
2
- see Figure 4
[eq51]
- see Figure 4
- see Figure 4
[eq52]
[eq53]
Amh =
h 2α h
h 2 cos α h
−
2
4 sin α h
4 sin α h
- see Figure 4
[eq54]
Amw =
w 2α w
w 2 cos α w
−
2
4 sin α w
4 sin α w
- see Figure 4
[eq55]
γ la 
=
[eq50]
Asl can be found by taking the surface of solid-liquid within the capillary if there was no meniscus ( Ah
and Aw ) and adding the surface of liquid-air due to the meniscus ( Amh and Amw ). We assume the liquid
to progress up to length L from the point of reference (right up to the opening).
[eq48]
 αh


− cos α h 
 sin α h

The liquid is in the capillary channel, stopped at the opening. The meniscus becomes convex from
the flat transitory stage and starts to expand beyond the opening with the convex profile
Ala is the same as from Stage 1:
hwα hα w
Ala =
sin α h sin α w
2
γ la 
Stage 2:
Hence substituting [eq52], [eq53], [eq54], and [eq55] into [eq51] yields,
[eq49] (20)

h 2α h
h 2 cos α h
w 2α w
w 2 cos α w 

Asl = 2 hL + wL +
−
+
−
2
2
4 sin α h
4 sin α h
4 sin α w
4 sin α w 




h2  αh
w2  α w


Asl = 2(h + w)L +
− cos α h  +
− cos α w  [eq56]
4
sin
sin
4
sin
sin
α
α
α
α
h 
h
w 
w



=1
α h →0
Page 15 of 33
Page 16 of 33
Now, substituting [eq50] and [eq56] into [eq6] yields
Hence




h2  αh



− cosα h 
(h + w)L +

 hwα hα w
4 sin α h  sin α h


UT = U0 + 
− 2
cosθ c γ la

2
α
α
sin
sin
 αw



h
w

+ w

− cos α w 




α
α
4
sin
sin
w
w








w2  α w

− cos α w 
(h + w)L +
4 sin α w  sin α w
 hwα hα w 



U T = U 0 − 2 cosθ c γ la 
+ γ la 

2
 αh

 sin α h sin α w 
+ h


− cos α h 
 4 sin α h  sin α h


PStage 2 = −
PStage 2 = −
where Vwh = hwL
 w 2α w
w 2 cos α w 
h
Vmw = Amw × h = 
−
2
4 sin α w 
 4 sin α w
 h 2α h
h 2 cos α h  wα w

Vmh = Amh × CLw = 
−
2
4 sin α h  sin α w
 4 sin α h
−1
 ∂V
 
  ∂f
 ∂f
 
 ∂α  +  ∂V ∂α   ∂α  ∂α  
w
h 
w 
h




Now from [eq57], we can derive
[eq58]
[eq65]
∂U T
∂U T
and
:
∂α w
∂α h

∂U T
∂ 
∂  hwα hα w 
w2  α w

γ la − 2 cosθ c γ la

− cos α w 
=
+
∂α w  4 sin α w  sin α w
∂α w ∂α w  sin α h sin α w 

∂U T hwα h ∂  α w 
cosθ c γ la w 2 ∂  α w
cos α w 



γ la −
=
−
∂α w sin α h ∂α w  sin α w 
2
∂α w  sin 2 α w sin α w 

∂U T
hwα h ∂  α w 
w 2 cosθ c ∂  α w
 2

 − γ la
= γ la
− cot α w 
∂α w
sin α h ∂α w  sin α w 
2
∂α w  sin α w


∂U T
hwα h  sin α w − α w cos α w 
w 2 cosθ c  sin 2 α w − α w 2 sin α w cos α w

 − γ la

= γ la
+ csc 2 α w 
4
α
∂α w
sin α h 
sin 2 α w
2
sin
w



[eq59]
[eq60]
[eq61]
α cos α 
α 2 cos α w
∂U T
hwα h  1
w 2 cosθ c  1
1 

 2

= γ la
− w 2 w  − γ la
− w 3
+
2
∂α w
sin α h  sin α w
sin α w 
2
sin
α
sin
α
sin
α w 
w
w

α cos α 
α cos α 
∂U T
hwα h  1
w 2 cos θ c  1


= γ la
− w 2 w  − γ la
− w 2 w 
∂α w
sin α h  sin α w
sin α w  sin α w
sin α w 
sin α w 
Hence substituting [eq59], [eq60], and [eq61] into [eq58] yields
 w 2α w
w 2 cosα w   h 2α h
h 2 cos α h  wα w
h + 

V = hwL + 
−
−
2
2
4 sin α w   4 sin α h
4 sin α h  sin α w
 4 sin α w

 αh

h 2 wα w
hw2  α w


V = hwL +
− cos α w  +
− cos α h 
α
α
α
4 sin α w  sin α w
4
sin
sin
sin
h
w 
h


−1
 ∂U T
 +  ∂U T
  ∂f
 ∂f
 









∂α w  
∂α h   ∂α w  ∂α h  



[eq64]
[eq57]
The volume V can be found by taking the volume of liquid within the capillary if there was no
meniscus ( Vhw ) and adding the volume of displaced liquid due to the meniscus ( Vmw and Vmh ). We
assume the liquid to progress up to length L from the point of reference:
V = Vhw + Vmw + Vmh
dU T
dU T
df
=−
dV
dV
df
 w 2 cosθ c hwα h
∂U T
= γ la  −
+
sin α h
∂α w
 sin α w
 sin α w − α w cosα w 


sin 2 α w


[eq66]
[eq62]
Finally, we can apply [eq57] and [62] into
PStage 2 = −
dU T
dV
[eq63]
To simplify the equation, as Ala , Asl and V are functions of α w and α h , we can obtain the burst
pressure by letting f to be a function of α w and α h .
Page 17 of 33
Page 18 of 33
∂U T
∂
=
∂α h ∂α h
 hwα hα w 

∂  h2  α h

γ la − 2 cosθ c γ la

− cos α h 

∂α h  4 sin α h  sin α h
 sin α h sin α w 

2
h cosθ c γ la ∂  α h
cos α h 
∂U T hwα w ∂  α h 

γ la −

−
=

∂α h  sin 2 α h sin α h 
∂α h sin α w ∂α h  sin α h 
2

h 2 wα w ∂  1  α h
∂V

=
− cos α h 

∂α h 4 sin α w ∂α h  sin α h  sin α h

2
h wα w ∂  α h
cos α h 
∂V


=
−
∂α h 4 sin α w ∂α h  sin 2 α h sin α h 

∂U T hwα w ∂  α h 
h 2 cosθ c γ la ∂  α h


γ la −
=
− cot α h 
∂α h sin α w ∂α h  sin α h 
2
∂α h  sin 2 α h

2
2




∂U T hwα wγ la sin α h − α h cos α h
h cosθ c γ la sin α h − α h 2 sin α h cosα h

 −

=
+ ccs 2α h 
2
4
∂α h
sin α w 
sin α h
2
sin α h




h 2 wα w ∂  α h
∂V
 2
=
− cot α h 
∂α h 4 sin α w ∂α h  sin α h


h 2 wα w  sin 2 α h − α h 2 sin α h cos α h
∂V

=
+ csc 2 α h 
∂α h 4 sin α w 
sin 4 α h

 1
α cos α

− h 2 h
sin
α
sin α h
h

∂U T hwα wγ la  1
α cos α

=
− h 2 h
∂α h
sin α w  sin α h
sin α h
α 2 sin α h cos α h
h 2 wα w  1
∂V
1

=
− h
+
∂α h 4 sin α w  sin 2 α h
sin 4 α h
sin 2 α h
∂U T hwα wγ la
=
∂α h
sin α w
 h 2 cos θ c γ la
 −
2

2
 h cosθ c γ la
 −
2

 1
α 2 cos α h
1
 2
− h 3
+
2
sin
α
sin
α
sin
αh
h
h

 2
α 2 cos α h 
 2

− h 3
sin α h 
 sin α h
 1
α cos α
α cos α  h 2 cos θ c γ la  1


− h 2 h  −
− h 2 h
2
sin
sin
α
α
sin
sin
sin α h
α
α
h
h
h 
h


2
 h cosθ c hwα w  sin α h − α h cos α h 
∂U T


= γ la  −
+
∂α h
sin α w 
sin 2 α h
 sin α h

∂U T hwα wγ la
=
∂α h
sin α w



α 2 cos α h
h 2 wα w  2
∂V

=
− h 3
∂α h 4 sin α w  sin 2 α h
sin α h






 sin α h α h cos α h
h 2 wα w
∂V

=
−
∂α h 2 sin α w sin α h  sin 2 α h
sin 2 α h
[eq67]
 sin α h − α h cos α h
h 2 wα w
∂V

=
∂α h 2 sin α h sin α w 
sin 2 α h

 ∂  αw 
∂V
hw 2 ∂  α w
h2w  α h
 2



=
− cot α w  +
− cos α h 
∂α w
4 ∂α w  sin α w
4
sin
α
sin
α
h 
h
 ∂α w  sin α w 


 sin α w − α w cosα w 
hw 2  sin 2 α w − α w 2 sin α w cosα w
h2w  α h
∂V



=
+ ccs 2α w  +
− cos α h 
4
4 
sin α w
sin 2 α w
∂α w

 4 sin α h  sin α h

 1
α cos α 
α 2 cos α w
∂V
hw 2  1
h2w  αh
1 
 2
+

=
− w 3
+
− cos α h 
− w 2 w 
2


∂α w
4  sin α w
sin
α
sin α w
sin α w  4 sin α h  sin α h
sin α w 
w

2
2






αh
α cos α
α 2 cos α w
∂V
hw
2
h w
1

+

=
− w 3
− cos α h 
− w 2 w 
∂α w
4  sin 2 α w
sin α w  4 sin α h  sin α h
sin α w 
 sin α w
∂V  hw 2
h2w
=
+
∂α w  2 sin α w 4 sin α h



[eq69]
PStage 2
∆α h = 0
 ∂U T


∂α w 
= −
 ∂V

 ∂α 

w
[eq70]
Now, substituting [eq66] and [eq68] into [eq70] yields
 w 2 cosθ c hwα h  sin α w − α w cos α w 


γ la  −
+
sin α h 
sin 2 α w
 sin α w

PStage 2 = −
 hw2
 sin α w − α w cos α w 
h2w  αh


+
− cos α h 

sin 2 α w
 

 2 sin α w 4 sin α h  sin α h
 αh
 1
α cos α 

− cos α h 
− w 2 w 
sin α w 
 sin α h
 sin α w
 αh
 sin α w − α w cos α w 


− cos α h 
α
sin
sin 2 α w
h


 



Now if we assume that α w is changing in such a way that the meniscus changes from almost flat to
convex, while α h remains constant, we can simplify [eq65] to:

 αh

h 2 wα w
∂V
∂
hw 2  α w
∂


=
− cos α w  +
− cos α h 
α
α
α
α
∂α w ∂α w 4 sin α w  sin α w
∂
4
sin
sin
sin
w
h
w 
h


 1
α w cos α w 
h2w


 sin α − sin 2 α  + 4 sin α
w
h
w 




h 2 wα w  sin α h α h cos α h
∂V

=
−
∂α h 2 sin α w  sin 3 α h
sin 4 α h
∂V
∂V
and
:
Now from [eq62], we can derive
∂α w
∂α h
∂V
hw 2
=
∂α w 2 sin α w



 w 2 cosθ c hwα h
+
sin α h
 sin α w
γ la  −
[eq68]
PStage 2 = −
Page 19 of 33
hw2
h2w
+
2 sin α w 4 sin α h



 αh


− cos α h 
sin
α
h


[eq71]
Page 20 of 33
Now, if we assume that α h approaches 0, then we will have
 w 2 cosθ c hwα h
+
sin α h
 sin α w
γ la  −
PStage 2
α h →0
PStage 2 α
h →0
PStage 2 α
h →0
PStage 2
α h →0
=−
hw 2
h2w
+
2 sin α w 4 sin α h



 αh


− cos α h 
sin
α
h

 αh→0
 w 2 cosθ c
 2 sin α w 
= −γ la  −
+ hw 

2
α
sin
w

 hw 
2γ  w

= − la  − cosθ c + sin α w 
w  h

2γ la  w

=
 cos θ c − sin α w 
w h

Stage 3:
The liquid has expanded beyond the border of the opening (opening with an angle β), with a
convex meniscus
 w 2 cosθ c

+ hw 
 sin α w

γ la  −
=−
 hw 2 


 2 sin α w 
3D View
Vhw
( Vmw + Vmh )
Vow
[eq72] (22)
Mistake in Chen et al (2008) eq (15)
Ala
Top View
XC
Note:
αh
sin α h
=1
α h →0
L
X CL
w
CLw
Aw
Aow
L
X CL
cos β
Amw
Side View
h
Ah
Aoh
CLh
Amh
Figure 7: Model for Stage 2
Page 21 of 33
Page 22 of 33
Asl can be found by taking the surface of solid-liquid within the capillary if there was no meniscus ( Ah
and Aw ) and adding the surface of solid-liquid due to the opening ( Aoh and Aow ), and due to the
X CL
cos β
meniscus ( Amh and Amw ).
β
Asl = 2( Ah + Aw + Aoh + Aow + Amh + Amw )
X CL
CLw
w
- see Figure 7
[eq73]
Where Ah = hL
- see Figure 7
[eq74]
Aw = wL
- see Figure 7
[eq75]
- see Figure 7, 8
[eq76]
- see Figure 7, 8
[eq77]
 X 
Aoh =  CL  × h
 cos β 
Aow = ( X CL ) × w + X CL × ( X CL tan β )
Amh = ASector − ATriangle
Aow
[eq78]
Amw
Where ASector = angle of curvature × radius of curvature
Figure 8: Model for Stage 2 Meniscus
ASector =
X CL
cos β
αw
αw
2
X CL tan β
β
And
X CL
w/ 2
CLw
w/ 2
β
Rw
2α h
(π ) × Rh 2 = α h Rh 2
2π
 h2 
h 2α h
 =
ASector = α h 
4 sin α h
 sin α h 
1
ATriangle = × height × base
2
1
h h2
ATriangle = × Rh cos α h × h = ×
cos α h
2
2 sin α h
ATriangle =
X CL tan β
Amh =
Rw cos α w
h 2 cosα h
4 sin α h
[eq79]
[eq31g]
h 2α h
h 2 cos α h
h2
−
=
2
4 sin α h
4 sin α h
4 sin α h
 αh


− cos α h 
 sin α h

[eq80]
Amw = ASector − ATriangle
[eq81]
Where ASector = angle of curvature × radius of curvature
Figure 9: Mathematical Model for Stage 2 Meniscus
According to [eq6] and [eq8], we need to determine Ala and Asl for U T = U 0 + ( Ala − Asl cos θ c )γ la ,
and we also need to determine the volume of the liquid, V.
ASector
ASector
Page 23 of 33
2α w
(π )× Rw 2 = α w Rw 2
- see Figure 9
2π
2
 w + X CL tan β 
 = α  w + 2 X CL tan β
= α w  2
w

sin α w
2 sin α w



2
(w + 2 X CL tan β ) α w
=
2 sin α w
ASector =
( )



2
[eq82]
Page 24 of 33
And
ATriangle
ATriangle
ATriangle
ATriangle =
Amw =
Amw =
hα h
sin α h
(w + 2 X CL tan β )α w
CLw = Rw × 2α w =
sin α w
1
× height × base
[eq83]
2
1
- see Figure 7
= × (Rw cos α w ) × 2 w + X CL tan β
2
2
 w + X CL tan β 
 cos α × w + X tan β
=  2
w
CL

2
sin α w


 w + 2 X CL tan β 
 w + 2 X CL tan β 
 cos α w × 
= 

2
sin
2
α


w


CLh = Rh × 2α h =
ATriangle =
(( )
( )
)
(( )
(w + 2 X CL tan β )2 cos α w
)
Ala =
−
4 sin α w

αw
 sin α − cos α w 
w


[eq85]
Hence substituting [eq74], [eq75], [eq76], [eq77], [eq80], and [eq85] into [eq73] yields:
2

 αh

X
 hL + wL + CL h + X CL w + X CL 2 tan β + h

− cos α h  
cos β
4 sin α h  sin α h


Asl = 2

2


 (w + 2 X CL tan β )  α w

+
−
cos
α
w
 sin α


α
4
sin
w
w




2


(
w + 2 X CL tan β )  α w


h
 L(h + w) + X CL 

− cos α w  
 cos β + w + X CL tan β  +
4 sin α w



 sin α w

Asl = 2
 [eq86]
2

 αh
h




+
−
cos
α
h
 4 sin α  sin α

h
h




Ala can be found by multiplying the curve length of height, CLh , and the curved length of width, CLw :
Ala = CLh × CLw
Note: this is the same method as in [eq18]
- see Figure 7
h(w + 2 X CL tan β )α hα w
sin α h sin α w
[eq92]
Now, substituting [eq86] and [eq92] into [eq6] yields,
4 sin α w
(w + 2 X CL tan β )2 
[eq91]
Note: The derivation of [eq89] and [eq91] are similar to that of [eq20] and [eq22], except with different
representation of “w”.
(w + 2 X CL tan β )2 α w (w + 2 X CL tan β )2 cos α w
4 sin 2 α w
- see Figure 9
Substituting [eq90] and [eq91] into [eq87] yields:
[eq84]
4 sin α w
[eq90]




 L(h + w) + X CL  h + w + X CL tan β  


 cos β










2


β
α
h
(
w
+
2
X
tan
)
(
w
+
2
X
tan
)
β
α
α


CL
w
CL
h w

− 2 +
− cos α w   cos θ c γ la
UT = U 0 + 


sin α h sin α w
4 sin α w
 sin α w
 



2


 αh



h


+
−
cos
α
h


 4 sin α  sin α

h 
h







 
h
 L(h + w) + X CL 
 cos β + w + X CL tan β  





2

 
hα h α w 
 (w + 2 X CL tan β )  α w

U T = U 0 + γ la (w + 2 X CL tan β )
 − 2γ la cos θ c  +
 sin α − cos α w  
sin α h sin α w 
4 sin α w
w


 




h2  αh


 + 4 sin α  sin α − cos α h 

h 
h



[eq93] (24)
Mistake in Chen et al (2008) eq (16)
[eq87]
Let the curved lengths have angles of 2α w and 2α h , and curve radii of Rw and Rh respectively.
We can derive the following:
h/2
[eq88]
Rh =
sin α h
w + X tan β
(w + 2 X CL tan β )
CL
=
- see Figure 9
[eq89]
Rw = 2
sin α w
2 sin α w
( )
Page 25 of 33
Page 26 of 33
The volume V can be found by taking the volume of liquid within the capillary ( Vhw ) and adding the
volume of liquid at the opening (Vow ) and the volume of the liquid in the meniscus ( Vmw and Vmh ). We
assume the liquid to progress up to length Xc from the point of reference (and XCL = XC – L):
V = Vhw + Vow + (Vmw + Vmh )
Where Vwh = hwL
(
)
2
Vow = Aow h = X CL w + X CL tan β h
[eq94]
- see Figure 5, 6, 7
[eq95]
- see Figure 5, 6, 7
[eq96]
pressure as follows:
Pburst = −
4 sin α w

 
(w + 2 X CL tan β )2  α w
 hwL + X CL w + X CL 2 tan β h +

 sin α − cos α w h 
4 sin α w

w


V =

 (w + 2 X CL tan β )α w
h2  α h




cos
+
−
α
h
 4 sin α  sin α

sin
α
h
h
w




2
 




(
)
w
2
X
tan
+
β
α
2
CL
w
 h  wL + X CL w + X CL tan β +

− cos α w  
 
4 sin α w
 sin α w
 
V =
 [eq99] (25)
2

 αh
 (w + 2 X CL tan β )α w

h


+

 sin α − cos α h 
4
sin
sin
α
α
h 
h
w



(
∂U T
∂X CL
∂U T
∂X CL
Hence substituting [eq95], [eq96], [eq97], and [eq98] into [eq94] yields:
)
∂U T
∂X CL
dU T
=−
∂V
dV
∂X CL
Now from [eq93], we can derive
(w + 2 X CL tan β )2 

αw
- see Figure 5, 6, 7
[eq97]
 sin α − cos α w h
w


 (w + 2 X CL tan β )α w
h2  α h

- see Figure 5, 6, 7 [eq98]
= Amh × CLw =
− cos α h 
4 sin α h  sin α h
sin α w

Vmw = Amw × h =
Vmh
- see Figure 5
To simplify [eq7] for our purpose, as Ala , Asl and V are functions of XCL, we can obtain the burst
[eq100]
∂U
:
∂X CL



 h


 X CL 

+ w + X CL tan β 
β

cos



hα h α w 


∂
γ
θ
=
−
2
cos
 γ la (2 X CL tan β )



la
c
2
∂X CL  
sin α h sin α w 
 + (w + 2 X CL tan β )  α w − cos α   
w 
 sin α


4 sin α w
w




 h

+ w + 2 X CL tan β


hα h α w
 cos β

= γ la (2 tan β )
− 2γ la cos θ c 
 αw

2(w + 2 X CL tan β )
sin α h sin α w
+
(2 tan β )
− cos α w  

4 sin α w
 sin α w



w 2 X CL
 1

+ +
tan β



h
h
cos
β
2
tan
2
γ
α
α
β
γ
∂U T


= hw la h w
− la cos θ c 
 w sin α h sin α w
 αw
 
∂X CL
w
X
2
w
tan
β


CL
+



 sin α  h + h tan β  sin α − cos α w  

w
w

 

Now from [eq99], we can derive
[eq101]
∂V
:
∂X CL
2
 

 h  wL + X CL w + X CL 2 tan β + (w + 2 X CL tan β )  α w − cos α w  
 sin α
 

α
4
sin
∂V
∂
w
w



=


∂X CL ∂X CL 
 (2 X CL tan β )α w

h2  α h
 + 4 sin α  sin α − cos α h 

sin
α
h 
h
w



Mistake in Chen et al (2008) eq (17)
 

 h w + 2 X CL tan β + 2(w + 2 X CL tan β ) (2 tan β ) α w − cos α w  



4 sin α w
∂V
 sin α w
 
= 

∂X CL 
 (2 tan β )α w

h2  αh


cos
+
−
α
h



 sin α
w

 4 sin α h  sin α h

  2 X CL
 
tan β  2 X CL
 α w
 1 +
tan β +
tan β 
− cos α w  
1 +

sin α w 
w
w
 sin α w
∂V
 
= hw 

∂X CL

 h α w tan β  α h



+
−
cos
α
h




 2w sin α w sin α h  sin α h

Page 27 of 33
[eq102]
Page 28 of 33
Substituting [eq101] and [eq102] into [eq100] yields:
Pburst = −
Pburst
dU T
=−
dV

w 2 X CL
 1

tan β
+ +



h
2γ α α tan β
2γ
 cos β h

hw la h w
− la cos θ c 

 w sin α h sin α w


w
α
2
X
β
tan
w


CL
w
+

 
β
α
+
tan
−
cos



w
 sin α  h

h
 sin α w
w
 


  2 X CL
 
tan β  2 X CL
 α w
 1 +
− cos α w  
tan β +
tan β 
1 +

α
w
sin α w 
w
sin

w
 
hw


 h α w tan β  α h


− cos α h 
+

α
α
α
2
w
sin
sin
sin
w
h 
h



 2γ la α hα w tan β

 w sin α sin α

h
w

= −
 2γ
 1
 
w 2 X CL
tan β  w 2 X CL
 α w
+ +
tan β +
tan β 
− cos α w  
− la cos θc 
 +

h
sin α w  h
h
 sin α w
 w
 
 cos β h
  2 X CL
tan β
× 1 +
tan β +

w
sin
αw

 h α w tan β
 2 X CL
 α w
tan β 
− cos α w  +
1 +
w

 sin α w
 2 w sin α w sin α h
 αh


− cos α h  
 sin α h
 
If we consider the pressure when XCL and α h both approach 0, [eq103] then reduces to:
(Note: Also See Alternative)
Pburst

w
 1

+ +



cos
h
β
2γ α tan β 2γ la

  
tan β
cos θ c 
= −  la w
−
 ×  1 + sin α
 w sin α w


α
w
β
tan
w
 

w
w






+
− cos α w  

 sin α  h  sin α

w
w

 


Pburst = −
2γ la
w
Pburst = −
2γ la
w
Pburst = −
2γ la
w
Pburst = −
2γ la
w
−1
[eq103] (26)
Similar to Chen et al (2008) eq (18)
Pburst
 αw
 

− cos α w  
α
sin
w

 
 α sin β
 1

w
sin β
 w  α w
w

− cos θ c 
+ +
− cos α w   
 
 sin α w cos β

 cos β h sin α w cos β  h  sin α w


  cos β




α
sin
β


w


+
− cos α w  


  cos β sin α w cos β  sin α w







 α sin β





α
w
sin β  w 
w
 w
− cos θ c 1 + cos β +
− cos α w   
 


sin α w  h  sin α w
h
 sin α w
   multiplied with cos β



cos β






 cos β + sin β  α w − cos α w  






sin α w  sin α w
 



 α w sin β

 αw

w
sin
β

− cos θ c − cos θ c  cos β +
− cos α w   



h
sin α w  sin α w
 sin α w





sin β  α w



cos β +
− cos α w 


sin α w  sin α w





α w sin β
− cos θ c


sin α w
w

− cos θ c 


 h
sin β  α w
 cos β +


− cos α w 
sin
sin
α
α
w 
w





α sin β


cos θ c − w
2γ la  w
sin α w

=−
− cos θ c +


w
h

 
sin β  α w


− cos β +
− cos α w 

sin α w  sin α w

 

[eq104] (27)
Mistake in Chen et al (2008) eq (20)
Page 29 of 33
Page 30 of 33
−1
Should the aspect ratio of w/h approaches 0 (h >> w), the equation then reduces to
Pburst


α sin β


cos θ c − w
2γ 
sin α w

= la 
w 
 αw
 
sin
β
 cos β +

− cos α w  
 
sin α w  sin α w
 
[eq105] (28)
Alternative
If we consider the case as α h approaches 0, [eq103] then reduces to:
Pburst
 2γ
 1
w 2 X CL
tan β
= − − la cos θ c 
+ +
tan β +
αw
β
cos
sin
w
h
h


Restriction:
Special Note (see Figure 9):
As [eq104] (and subsequently [eq105]) is derived based on the hidden assumption that β = α w , applying
β = α w to [eq104] yields:
Pburst


β sin β


cos θ c −
w
2γ
sin β

= − la − cosθ c +
w  h

 
sin β  β


− cos β +
− cos β  
sin β  sin β
 


Pburst




2γ la  w
cos θ c − β

=−
− cos θ c +

w
h

 β
 

− cos β + 
− cos β  
 sin β
 


 
 w 2 X CL
 α w
tan β 
− cos α w  
 +
h
h
 sin α w
 
  2 X CL
 
tan β  2 X CL
 α w
tan β +
tan β 
× 1 +
− cos α w  
1 +

α
w
sin
w
sin
α

w 
w
 

−1
Pburst
[eq106]
Pburst
Pburst
Pburst




2γ la  w
cos θ c − β 
=−
− cos θ c +
w  h
 β 
−


 sin β  


2γ  w
 cos θ c

= − la − cos θ c − 
− 1 sin β 
w  h
 β



2γ la  w
 cos θ c

=
− 1 sin β 
 cos θ c + 
w h
 β



2γ la  w sin β 
=
 cos θ c − sin β 
 +
β 
w  h

[eq107]
[eq108] (29)
[eq107] and [eq108] differs from to Chen et al (2008) eq (20)
Should the aspect ratio of w/h approaches 0 (h >> w), [eq107] equation then reduces to
 cos θ c


− 1 sin β
 β


2γ la  sin β 
=
 cos θ c − sin β 

w  β 

Pburst =
Pburst
2γ la
w
[eq109]
[eq110] (30)
[eq109] and [eq110] differs from Man et al (1998) eq (13)
Page 31 of 33
Page 32 of 33
Special Consideration of Meniscus Angle Conditions
(a) Stage 1
αw
θc
θc
αw
α w = 90° − θ c
(b) Stage 2
θc
αw
θc
α w = 90° − θ c
αw
αw
β
θc
(c) Stage 3
θc
αw
α w + (θ C − β ) =
αw =
π
2
π
2
− (θ C − β )
Figure 10 Geometry of meniscus angle for various stages
Applying the Meniscus angle of Stage 1 from Figure 10 into [eq19] and [eq20] yields
h
[eq111] (18)
Rh =
2 cosθ C
w
Rw =
[eq112] (19)
2 cosθ C
Applying the Meniscus angle of Stage 1 from Figure 10 into [eq49] yields
2γ  w 
PStage1 α →0 = la  + 1 cosθ c
h
w h 
[eq113] (21)
Applying the Meniscus angle of Stage 2 from Figure 10 into [eq72] yields
PStage 2 α
h →0
=
2γ la  w 
 − 1 cosθ c
w h 
[eq114] (23)
Page 33 of 33