ME 150 – Heat and Mass Transfer
Chap. 16: Convection with Phase Change
Boiling and Condensation Boiling = Evaporation at a solid-liquid interface qʹ′ʹ′ = h ⋅ (Ts − Tsat ) = h ⋅ ΔTe
ΔTe = Excess Temperature
Modes of Boiling: defined by the kind of flow: - Pool boiling (natural/free convection) - Boiling with forced convection (e.g. pipe flow) defined by the temperature range: - Saturated boiling: Tfluid = Tsat - Subcooled boiling: Tfluid < Tsat
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Pool Boiling – Boiling Curve Experiment of Nukiyama: q = I ⋅V
h=
qʹ′ʹ′ = q
AW
TW = f ( RW )
RW = V
I
qʹ′ʹ′ (TW − Tsat )
h is determined by electric properties I, V and the wire surface AW Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Regions of the Boiling Curve Nucleate boiling
Transition
Film boiling
qʹ′ʹ′ [W / m 2 ]
C
6
10
A: Start nucleate boiling
P
q ʹ′ʹ′
B: Start nucleate jet boiling
W m2 ⋅ K
105
B
P: maximum h
D
C: maximum heat flux, end of nucleate boiling
104
D: Leidenfrost point, start of film boiling
103 1
A
5
10
30
120
1000
ΔTe = Ts − Tsat (°C)
Prof. Nico Hotz
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Example: Boiling of methanol in a horizontal tube Photographien von Prof. J.W. Westwater, University of Illinois at Champaign-Urbana
1. Nucleate boiling (jets and columns)
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Chap. 16.1: Pool Boiling
2. Transition boiling
3. Film boiling
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Free convection boiling (ΔTe < 5 K) Before the onset of nucleation, heat transfer only due to (natural) convection Temperature distribtion inside of liquid depending on height z Top surface is superheated (T0 - Tsat) laminar : turbulent :
5 4
qʹ′ʹ′ ∝ (ΔT )
4 3
qʹ′ʹ′ ∝ (ΔT )
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Formation of Bubbles Surface of the vapor bubble has a surface tension: equilibrium of forces: Pressure inside
=
Pressure outside
+
Surface tension
π ⋅R 2 ⋅ Pb = π ⋅R 2 ⋅ Pl + 2π ⋅ R ⋅ γ
To overcome the surface tension, there has to be an excess pressure inside the bubble:
ΔP = Pb − Pl =
2 ⋅γ R
Excess pressure requires an excess temperature of the liquid
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Nucleate Pool Boiling Empirical: Interaction between surface properties and bubble formation in the liquid Correlation by Rohsenov: 12
⎡ g ⋅ ( ρl − ρv ) ⎤ ʹ′ ʹ′ qs = µl ⋅ hlv ⋅ ⎢ ⎥ γ ⎣ ⎦ µl hlv ρ γ l, v C, n
⎛ c p ,l ⋅ ΔT ⎞ ⎜⎜ ⎟ n ⎟ ⎝ C ⋅ hlv ⋅ Prl ⎠
3
Viscosity of liquid Evaporation enthalpy Density Surface tension Indices for liquid and vapor experimental constants
Prof. Nico Hotz
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Constants for the correlation of Rohsenov C = Interation between fluid and surface n = Fluid property
Chap. 16.1: Pool Boiling
Combination Fluid/ Surface
C
n
Water/Copper scored polished
0.0068 0.0130
1.0 1.0
Water/Steel polished
0.0130
1.0
Water/Nickel
0.0060
1.0
Water/Platinum
0.0130
1.0
n-Pentane/Copper polished
0.0154
1.7
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Critical / Maximum Heat Flux If heat flux is higher: transition to film boiling (ΔT > 1000 K) ⎡ γ ⋅g ⋅ ( ρl − ρv ) ⎤ ʹ′ʹ′ = C ⋅ hlv ⋅ ρv ⎢ qmax ⎥ 2 ρ v ⎣ ⎦
14
C = 0.149 for large horizontal plates
Minimal Heat Flux (at Leidenfrost point) If heat flux is smaller: collapse of film boiling ⎡ γ ⋅g ⋅ ( ρl − ρv ) ⎤ ʹ′ʹ′ = C ⋅ hlv ⋅ ρv ⋅ ⎢ qmin ⎥ 2 ⎣ ( ρl + ρv ) ⎦
Prof. Nico Hotz
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C = 0.09 for large horizontal plates
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ME 150 – Heat and Mass Transfer
Chap. 16.1: Pool Boiling
Film Boiling Nusselt correlation for cylinder or sphere with diameter D 3
⎡ g ⋅ ( ρ l − ρ v ) ⋅ hlvʹ′ ⋅ D ⎤ h ⋅D N u D = conv = C ⋅ ⎢ ⎥ kv ν ⋅ k ⋅ ( T − T ) ⎣ v v s sat ⎦
1
4
Cylinder: C = 0.62 Sphere: C = 0.67
Correction for latent heat: hlvʹ′ = hlv + 0.80 ⋅ c p ,v ⋅ (Ts − Tsat )
For Ts > 300°C: thermal radiation has to be considered as well
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Chap. 16.2: Forced Convection Boiling
Boiling in a vertical pipe: Two-Phase Flow
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Chap. 16.3: Condensation
Heat Transfer with Condensation Occurs when Tw < Tsat Modes of condensation
Homogeneous condensation:
Direct condensation: Spray of vapor entering liquid
Mixture of hot humid gas with cold gas, formation of fog
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ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Laminar Film Condensation on a Vertical Plate Assumptions (Analysis by Nusselt): • laminar film flow, constant properties of fluid • Gas phase is pure vapor at Tsat, no heat conduction in the ∂u vapor =0 • no friction between vapor and liquid, i.e.
∂y
y =δ
• no thermal boundary layer inside the vapor • no convective transport (heat and momentum) inside the liquid boundary layer
Prof. Nico Hotz
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Chap. 16.3: Condensation
System to be considered: - Velocity profile without gradient at outer surface - Temperature profile inside boundary layer is linear
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Mathematical model for liquid film:
Mx
⎛ ∂ u ∂ u ⎞ dp ∂ 2u ⎟⎟ = − + µ ⋅ ρl ⎜⎜ u +v − ρl ⋅ g 2 ∂x ∂ y dx ∂ y ⎝ ⎠ ρv ⋅g =0 ( convection negligible )
E
⎛ ∂ u ∂ u ⎞ ∂ 2T ⎟⎟ = α f ⋅ ρ⋅ c p ⋅ ⎜⎜ u +v ∂x ∂ y ⎠ ∂ y2 ⎝ =0 ( convection negligible )
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Momentum equation: ∂ 2u g = − ⋅ ( ρl − ρ v ) 2 ∂ y µl
With boundary conditions: y = 0: u =0
y =δ :
∂u =0 ∂y
Solution: 2 g ⋅ ( ρ l − ρ v ) ⋅ δ 2 ⎡ y 1 ⎛ y ⎞ ⎤ u( y) = ⋅ ⎢ − ⋅ ⎜ ⎟ ⎥ µl ⎢⎣ δ 2 ⎝ δ ⎠ ⎥⎦
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ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Calculation of mass flux in the film (per unit width):
1
m ( x) = b
δ ( x)
ρ ⋅ gl ⋅ (ρl − ρv )⋅ δ 3 ∫0 ρ f ⋅ u( y) ⋅ dy = Γ( x) = 3 ⋅ µl
δ(x) is unknown, can be determined using the energy balance: ∂T qʹ′sʹ′ ⋅ b⋅ dx = − k ⋅ ⋅ b ⋅ dx = hlv ⋅ dm ∂y Conduction
2
qʹ′sʹ′ =
Condensati on
1 dm dΓ ⋅ ⋅ hlv = hlv ⋅ b dx dx
Calculate temperature profile:
E
∂ 2T =0 ∂y 2
Boundary conditions:
Prof. Nico Hotz
y = 0:
T (0) = Ts
y =δ:
T (δ ) = Tsat 18
ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Solution: linear profile
⎛ T − T ⎞ T = ⎜ sat s ⎟ ⋅ y + Ts ⎝ δ ⎠
Heat flux:
⎛ dT qʹ′sʹ′ = kl ⋅ ⎜⎜ ⎝ dy
Substituting in
d Γ kl ⋅ (Tsat − Ts ) = dx δ ⋅ hlv
2
Substituting with
⎞ T −T ⎟⎟ = kl ⋅ sat s δ ⎠ y =0
1
δ 3 ⋅ dδ =
kl ⋅ µl ⋅ (Tsat − Ts ) ⋅ dx g ⋅ ρ l ⋅ (ρ l − ρ v )⋅ hlv
Prof. Nico Hotz
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ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Integration leads to solution for δ(x): 14
⎡ 4 ⋅ kl ⋅ µl ⋅ (Tsat − Ts ) ⎤ δ ( x) = ⎢ ⋅ x⎥ ⎣ g ⋅ ρl ⋅ (ρl − ρv )⋅ hlv ⎦
With convective heat transfer inside the film is included, we can use an effective latent heat hlv‘ hlvʹ′ = hlv ⋅ (1 + 0.68 ⋅ Ja )
Jakob Number :
Ja =
c p ,l ⋅ (Tsat − Ts ) hlv
Calculation of h value using Fourier‘s Law: ⎛ dT ⎞ ⎟⎟ − kl ⋅ ⎜⎜ k ⎝ dy ⎠ y =0 h( x ) = = l Ts − Tsat δ ( x)
⎡ g ⋅ ρ l ⋅ (ρ l − ρ v )⋅ kl 3 ⋅ hlvʹ′ ⎤ h( x) = ⎢ ⎥ ( ) 4 ⋅ µ ⋅ T − T ⋅ x l sat s ⎣ ⎦
Prof. Nico Hotz
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20
ME 150 – Heat and Mass Transfer
Chap. 16.3: Condensation
Averaged h value: H ⎡ g ⋅ ρ l ⋅ (ρ l − ρ v )⋅ kl 3 ⋅ hlvʹ′ ⎤ 1 h = ∫ h( x) ⋅ dx = 0.943 ⋅ ⎢ ⎥ ( ) H 0 µ ⋅ T − T ⋅ H l sat s ⎣ ⎦
14
=
4 ⋅ h x=H 3
Nusselt correlation for laminar film condensation on a vertical plate with height H 3
Nu H =
14
⎡ g ⋅ ρl ⋅ (ρl − ρv )⋅ hlvʹ′ ⋅ H ⎤ h ⋅H = 0.943 ⋅ ⎢ ⎥ ( ) kl µ ⋅ k ⋅ T − T l l sat s ⎣ ⎦
Calculation of heat transfer rate and mass transfer rate q = h ⋅ A ⋅ (Tsat − Ts )
m =
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q h ⋅ A ⋅ (Tsat − Ts ) = hlvʹ′ hlvʹ′ 21
ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Prof. Nico Hotz
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