Analysis of droplet flash evaporation in vacuum
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Analysis of droplet flash evaporation in vacuum 1. Introduction The ability to predict to predict droplet evaporation rate and diameter of nozzle is important to in many application such as spray combustion, spray cooling and many more. Assumptions: 1. One-dimensional conduction in the radial direction. Figure 1: Water droplet Water droplet of initial radius R o and temperature T o is spay in the vac- uum of saturation temperature of T s . R is the transient radius of the droplet. In spherical coordinates the radial form of the heat flux vector and Fouri- ers law is 1 r 2 ∂ ∂r κr 2 ∂T ∂r = ρc p ∂T ∂t lets assume 1
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Transcript of Analysis of droplet flash evaporation in vacuum
Analysis of droplet flash evaporation in vacuum
1. Introduction
The ability to predict to predict droplet evaporation rate and diameterof nozzle is important to in many application such as spray combustion,spray cooling and many more.
Assumptions: 1. One-dimensional conduction in the radial direction.
Figure 1: Water droplet
Water droplet of initial radius Ro and temperature To is spay in the vac-uum of saturation temperature of Ts. R is the transient radius of the droplet.
In spherical coordinates the radial form of the heat flux vector and Fouri-ers law is
1
r2∂
∂r
(κr2
∂T
∂r
)= ρcp
∂T
∂t
lets assume
1
θ =T − TsTo − Ts
∂θ
∂t=
1
To − Ts∂T
∂t
∂θ
∂r=
1
To − Ts∂T
∂r
1
r2∂
∂r
(r2∂T
∂r
)=
1
α
∂T
∂t
1
r2∂
∂r
(r2(To − Ts)
∂θ
∂r
)=
1
α(To − Ts)
∂θ
∂t
1
r2∂
∂r
(r2∂θ
∂r
)=
1
α
∂θ
∂t
Initial conditionT (r, 0) = To
orθ(r, 0) = 1
AndR(0) = Ro
Boundary condition 1.∂T
∂r
∣∣∣∣r=0
= 0
∂θ
∂r
∣∣∣∣r=0
= 0
2.T |r=R = Ts
orθ|r=R = 0
3.
−4πR2κ∂T
∂r
∣∣∣∣r=R
= −4πR2ρdR
dtL
κ∂T
∂r
∣∣∣∣r=R
= ρdR
dtL
2
κ∂T
∂r
∣∣∣∣r=R
= ρdR
dtL
∂T
∂r
∣∣∣∣r=R
=ρL
κ
dR
dt
(To − Ts)∂θ
∂r
∣∣∣∣r=R
=ρL
κ
dR
dt
∂θ
∂r
∣∣∣∣r=R
=ρL
κ(To − Ts)dR
dt
∂θ
∂r
∣∣∣∣r=R
= CodR
dt
where
Co =ρl
κ(To − Ts)
1
r2∂
∂r
(r2∂θ
∂r
)=
1
α
∂θ
∂t
According to the separation of variables method we assume a solution ofthe form
θ(r, t) = X(r)Y (t)
1
r2∂
∂r
(r2∂(XY )
∂r
)=
1
α
∂(XY )
∂t
1
r2∂
∂r
(r2Y
∂X
∂r
)=X
α
∂Y
∂t
Y
r2∂
∂r
(r2dX
dr
)=X
α
dY
dt
Y
r2d
dr
(r2dX
dr
)=X
α
dY
dt
1
Xr2d
dr
(r2dX
dr
)=
1
αY
dY
dt
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let λ is constant
1
Xr2d
dr
(r2dX
dr
)=
1
αY
dY
dt= −λ2
1
Xr2d
dr
(r2dX
dr
)= −λ2
and1
αY
dY
dt= −λ2
dY
Y= −λ2αdt
Y = C1e−λαt ....(1)
1
Xr2d
dr
(r2dX
dr
)= −λ2
d
dr
(r2dX
dr
)= −λ2Xr2
2rdX
dr+ r2
d2X
dr2+ λ2Xr2 = 0
d2X
dr2+
2
r
dX
dr+ λ2X = 0
let X(r) = u(r)v(r)
u(r) = e−12
∫2rdr
u(r) =1
r
d2v
dr2+ v
(λ2 − 4
r2− 1
2
(−2)
r2
)= 0
d2v
dr2+ vλ2 = 0
v(r) = C2 cosλr + C3 sinλ ....(2)
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X(r) = u(r)v(r)
X(r) = C2cosλr
r+ C3
sinλr
r
θ(r, t) = X(r)Y (t)
θ(r, t) = C1e−λαt
(C2
cosλr
r+ C3
sinλr
r
)From BC 1.
∂θ
∂r
∣∣∣∣r=0
= 0
C2 = 0
θ(r, t) = C1C3e−λαt sinλr
r
θ(r, t) = Ae−λαtsinλr
rwhere A = C1C3
From BC 2.
θ|r=R = 0
θ(R, t) = Ae−λαtsinλR
R= 0
λR = nπ
for n = 1,2,3,....
λ =nπ
R
θn(r, t) = Ane−λαtsinλnr
r
θ(r, t) =∞∑n=1
θn(r, t)
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θ(r, t) =∞∑n=1
Ane−λαtsinλnr
r
To determine the An’s we use the given initial condition, i.e.
θ(r, 0) = 1
∞∑n=1
Ansinλnr
r= 1
∞∑n=1
An1
rsin
nπr
R= 1
∞∑n=1
Angn = 1 where gn =1
rsin
nπr
R
R∫0
gm
∞∑n=1
Angndr =
R∫0
gmdr
Am
R∫0
g2mdr =
R∫0
gmdr
Am =
R∫0gmdr
R∫0g2mdr
An =
R∫0gndr
R∫0g2ndr
An =
R∫0
1
rsin
nπr
Rdr
R∫0
(1
rsin
nπr
R
)2
dr
θ(r, t) =∞∑n=1
Ane−λαt(
1
rsin
nπr
R
)
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