7.2 Sublimation. The following assumptions are made in order to solve the problem: Sublimation Over a Flat Plate in a Parallel Flow

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1 7..1 Sublimation Over a Flat Plate in a Parallel Flow The following assumptions are made in order to solve the problem: The flat plate is very thin and so the thermal resistance along the flat plate can be neglected. The gas is incompressible, with no internal heat source in the gas. The sublimation problem is two-dimensional steady state. 1

2 v u, T, ω u Boundary layer y x qw constant Figure 7.3 Sublimation on a flat plate with constant heat flux.

3 The governing equations of the problem are u v + 0 x y (7.) u u u u +v ν x y y (7.3) T T T u +v α x y y (7.4) ω ω ω u + v D x y y (7.5) 3

4 The boundary conditions at the surface of the flat plate are u 0, m& y 0 (7.6) ρ D ω, 1 ω y y 0 (7.7) ω, y y 0 (7.8) m& ρ D ω v vw D, y 0 y T ω k hsv D qw, y 0 y y (7.9) (7.10) 4

5 Another reasonable, practical, representable boundary condition at the surface of the flat plate emerges by setting the mass fraction at the wall as the saturation mass fraction at the wall temperature. The mass fraction and the temperature at the surface of the flat plate have the following relationship: (7.11) ω AT + B, y 0 As y the boundary conditions are u u, T T, ω ω (7.1) Introducing the stream function ψ ψ (7.13) u v y x The momentum equation in terms of the stream function ψ ψ ψ ψ 3ψ ν y x y x y y3 (7.14) 5

6 Similarity variables u x ξ, η y, L ν Lξ θ k (T T ) qw ν Lξ / u Eq. (7.14) becomes f, ϕ f + ff ξ ( ψ ν u Lξ ρ hsg D(ω ω ) (7.15) qw ν Lξ / u f F ) ff (7.16) θ + Pr( f θ f θ ) Pr ξ ( f Θ θ F ) (7.17) ϕ + Sc( f ϕ f ϕ ) Sc ξ ( f Φ ϕ F ) (7.18) where prime ' represents partial derivative with respect to η, and all upper case variables represent partial derivative of primary similarity variable with respect to ξ. 6

7 F f θ ϕ, Θ, Φ ξ ξ ξ (7.19) It can be seen from eqs. (7.16) (7.18) that the similarity solution exists only if all variables defined in eq. (7.17) equal zero. In order to use eqs. (7.16) (7.18) to obtain solution of the sublimation problem, the supplemental equations about F, Θ, and Φ must be obtained. Taking partial derivatives of eqs. (7.16) (7.18) and neglecting the higher order term, one obtains f F ) F + Ff + F f ( f F (7.0) Θ + Pr( Fθ + f Θ F θ f Θ ) Pr ( f Θ θ F ) (7.1) Φ + Sc( Fϕ + f Φ F ϕ f Φ ) Sc ( f Φ ϕ F ) (7.) 7

8 Boundary conditions of equations (7.16)-(7.18) and eqs. (7.0)-(7.) f (ξ,0) 0, η 0 f (ξ,0) B ξ 1/ ϕ (ξ,0) ξ 3 / Φ (ξ,0), η 0 3 f (ξ, ) 1, η F (ξ,0) 0, η 0 F (ξ,0) 1 1 B ξ 3 ϕ (ξ,0) ξ 1/ Φ (ξ,0), η 0 1/ F (ξ, ) 0, η θ (ξ,0) + ϕ (ξ,0) 1, η 0 θ (ξ, ) 0, η Θ (ξ,0) + Φ (ξ,0) 0, η 0 Θ (ξ, ) 0, η (7.3) (7.4) (7.5) (7.6) (7.7) (7.8) (7.9) (7.30) (7.31) (7.3) 8

9 Ahsv 1 ϕ (ξ,0) θ (ξ,0) + ϕ sξ c p Le 1/, η 0 ϕ (ξ, ) 0, η Ahsv 1 ϕs Φ (ξ,0) Θ (ξ,0), η 0 3/ c p Le ξ Φ (ξ, ) 0, η qw B ρ hsvν ν L u (7.33) (7.34) (7.35) (7.36) (7.37) reflect the effect of injection velocity at the surface due to sublimation, and ϕs ρ hsv D (ω s ω ) qw ν L / u (7.38) 9

10 Figure 7.4 temperature and mass fraction distributions. 10

11 Figure 7.4 temperature and mass fraction distributions. Figure 7.5 Nusselt number based on convection and Sherwood number. 11

12 The local Nusselt number based on the total heat flux at the bottom of the 1/ flat plate is qw x Nu x k (Tw T ) (7.40) x T θ (ξ,0) 1/ Re x Tw T y y 0 θ (ξ,0) (7.41) ϕ (ξ, 0) Re1/x ϕ θ (ξ, 0) (7.4) Sherwood number Shx θ (ξ,0) Nusselt number based on the convective heat transfer coefficient Nu *x Re x x ω ω w ω y y 0 Figure 7.5 shows the effect of blowing velocity on the Nusselt number based on convective heat transfer and Sherwood number for ϕs0, i.e., the mass fraction of sublimable substance is equal to the saturation mass fraction corresponding to the incoming temperature. It can be seen that the effect of blowing velocity on mass transfer is stronger than that on heat transfer. 1

13 7.. Sublimation Inside an Adiabatic Tube Sublimation inside an adiabatic and externally heated tube will be analyzed in this and the next subsections. The physical model of the problem under consideration is shown in Fig The inner surface of a circular tube with radius R is coated with a layer of sublimable material which will sublime when gas flows through the tube. The fully-developed gas enters the tube with a uniform inlet mass fraction of the sublimable substance, ω0, and a uniform inlet temperature, T0. Since the outer wall surface is adiabatic, the latent heat of sublimation is supplied by the gas flow inside the tube; this in turn causes the change in the gas temperature inside the tube. 13

14 Figure 7.6 Sublimation in an adiabatic tube. 14

15 It is assumed that the flow inside the tube is incompressible laminar flow with constant properties. In order to solve the problem analytically, the following assumptions are made: The entrance mass fraction ω0 is assumed to be equal to the saturation mass fraction at the entry temperature T0. The saturation mass fraction can be expressed as a linear function of the corresponding temperature. The mass transfer rate is small enough that the transverse velocity components can be neglected. The fully-developed velocity profile in the tube is r u um 1 R (7.43) where um is the mean velocity of the gas flow inside the tube. 15

16 Neglecting axial conduction and diffusion, the energy and mass transfer equations are T T ur α r x r r (7.44) ω ω ur D r x r r (7.45) which are subjected to the following boundary conditions: T T0, x 0 ω ω 0, x 0 (7.46) (7.47) 16

17 T ω 0, r 0 r r T ω k ρ Dhsv, r R r r (7.48) (7.49) Equation (7.49) implies that the latent heat of sublimation is supplied as the gas flows inside the tube. Another boundary condition at the tube wall is obtained by setting the mass fraction at the wall as the saturation mass fraction at the wall temperature. ω AT + B, r R (7.50) 17

18 The following nondimensional variables are then introduced: um R r x α η ξ Le Re R RPe D ν T Tf ω ωf um R Pe θ ϕ α T0 T f ω0 ω f (7.51) Equations (7.44) (7.50) then become θ η (1 η ) ξ η θ η η ϕ 1 η (1 η ) ξ Le η ϕ η η (7.5) (7.53) 18

19 θ ϕ 1, ξ 0 (7.54) θ ϕ 0, η 0 η η (7.55) θ 1 ϕ, η 1 η Le η Ahsv ϕ θ, η 1 c p (7.56) (7.57) The heat and mass transfer equations (7.5) and (7.53) are independent but their boundary conditions are coupled by eqs. (7.56) and (7.57). The solution of eqs. (7.5) and (7.53) can be obtained via separation of variables. It is assumed that the solution of θ can be expressed as a product of the function of η and a function of ξ, i.e., θ Θ (η )Γ (ξ ) (7.58) 19

20 Substituting eq. (7.58) into eq. (7.5), the energy equation becomes d dθ Γ dη dη β Γ η (1 η )Θ (7.59) where β is the eigenvalue. Equation (7.59) can be rewritten as the following two ordinary differential equations: Γ + β Γ 0 (7.60) d dθ + β η (1 η )Θ 0 dη dη (7.61) The solution of eq. (7.60) is Γ C1e β ξ (7.6) 0

21 The boundary condition of eq. (7.61) at η0 is Θ (0) 0 The dimensionless temperature is then θ C1Θ (η )e β ξ (7.64) Similarly, the dimensionless mass fraction is ϕ C Φ (η )e (7.63) γ ξ (7.65) where Φ(η) satisfies d dφ + Le γ η (1 η )Φ 0 dη dη (7.66) 1

22 and the boundary condition of eq. (7.66) at η0 is Φ (0) 0 (7.67) Substituting eqs. (7.64) (7.65) into eqs. (7.56) (7.57), one obtains β γ (7.68) Ahsv Θ (1) Θ (1) Le (7.69) c Φ (1) Φ (1) p If we use any one of the eigenvalue βn and corresponding eigen functions Θn and Φn in eqs. (7.64) and (7.65), the solutions of eq. (7.5) and (7.53) become β n ξ θ C1Θ n (η )e ϕ C Φ n (η )e (7.70) β nξ (7.71) which satisfy all boundary conditions except those at ξ0.

23 In order to satisfy boundary conditions at ξ 0, one can assume that the final solutions of eqs. (7.5) and (7.53) are θ n 1 ϕ Gn Θ n (η )e n 1 β n ξ H n Φ n (η )e β nξ (7.7) (7.73) where and can be obtained by substituting eqs. (7.7) and (7.73) into eq. (7.54), i.e., 1 n 1 1 n 1 Gn Θ n (η ) (7.74) H n Φ n (η ) (7.75) 3

24 Due to the orthogonal nature of the eigeinfunctions Θn and Φn, expressions of Gn and Hn can be obtained by Θ n (1) 1 η (1 η ) Θ ( η ) d η + η (1 η )Φ n (η ) dη n 0 0 Φ n (1) Gn 1 Θ n (1) 0 η (1 η ) Θ n (η ) + ( Ahsv / c p ) Φ n (1) Φ n (η ) dη 1 Ahsv Θ n (1) Hn Gn c p Φ n (1) (7.76) (7.77) 4

25 The Nusselt number and Sherwood numbers are T k r r R R β nξ Nu Gn e Θ n (1) Tm Tw k θ m θ w n 1 ω D r r R R β nξ Sh H n e Φ n (1) ωm ωw D ϕ m ϕ w n 1 (7.78) (7.79) 5

26 7..3 Sublimation Inside a Tube Subjected to External Heating When the outer wall of a tube with a sublimable-material-coated inner wall is heated by a uniform heat flux qn (see Fig. 7.8), the latent heat will be supplied by part of the heat flux at the wall. The remaining part of the heat flux will be used to heat the gas flowing through the tube. The problem can be described by eqs. (7.43) (7.50) except that the boundary condition at the inner wall of the tube is replaced by ω T ρ hsv D + k q r r (7.80) where the thermal resistance of the tube wall is neglected because the tube wall and the coated layer are very thin. 6

27 Figure 7.8 Sublimation in a tube heated by a uniform heat flux. 7

28 The governing equations for sublimation inside a tube heated by a uniform heat flux can be nondimensionalized by using the dimensionless variables defined in eq. (7.51), except the following: hsg (ω ω i ) k (T T0 ) θ ϕ (7.81) q R c p q R where ωi is the saturation mass fraction corresponding to the inlet temperature T0, and the resultant dimensionless governing equations are θ η η ϕ 1 ϕ η (1 η ) η ξ Le η η θ 0, ξ 0 θ η (1 η ) ξ η (7.8) (7.83) (7.84) 8

29 ϕ ϕ 0, ξ 0 (7.85) θ ϕ 0, η 0 η η (7.86) θ 1 ϕ + 1, η 1 η Le η (7.87) Ahsv ϕ θ, η 1 c p (7.88) 9

30 The sublimation problem under consideration is not homogeneous because eq. (7.87) is a non-homogeneous boundary condition. The solution of the problem is consistent with its particular (fully-developed) solution and the solution of the corresponding homogeneous problem, i.e., θ (ξ,η ) θ 1 (ξ,η ) + θ (ξ,η ) (7.89) ϕ (ξ,η ) ϕ 1 (ξ,η ) + ϕ (ξ,η ) (7.90) While the fully-developed solutions of temperature and mass fraction, θ 1 (ξ,η ) and ϕ 1 (ξ,η ), respectively, must satisfy eqs. (7.8) (7.83) and (7.86) (7.88), the corresponding homogeneous solutions of the temperature and mass fraction θ (ξ,η ) and ϕ (ξ,η ) must satisfy eqs. (7.8), (7.83), (7.86), and (7.88) as well as the following conditions: θ θ 1 (0,η ), ξ 0 (7.91) ϕ ϕ 0 ϕ 1 (ξ,η ), ξ 0 (7.9) θ 1 ϕ + 0, η 1 η Le η (7.93) 30

31 The fully-developed profiles of the temperature and mass fraction are 1 θ1 1 + ahsv / c p 11Le ahsv / c p 18ahsv / c p 7 1 (7.94) 4ξ + η 1 4 η + ϕ 0 + 4(1 + ahsv / c p ) ahsv / c p 7Le ahsv / c p + 18 Le 11 1 ϕ1 4ξ + Leη 1 η + ϕ 0 (7.95) 1 + ahsv / c p 4 4(1 + ah / c ) sv p The solution of the corresponding homogenous problem can be obtained by separation of variables θ ϕ n 1 n 1 Gn Θ n (η )e β n ξ H n Φ n (η )e β nξ (7.96) (7.97) 31

32 Θ n (1) 1 η (1 η ) θ (0, η ) Θ ( η ) d η + η (1 η )ϕ (0,η )Φ n (η ) dη n 0 0 Φ (1) n Gn 1 Θ (1) n 0 η (1 η ) Θ n (η ) + ( ahsv / c p ) Φ n (1) Φ n (η ) dη 1 ahsv Θ n (1) Gn c p Φ n (1) Hn (7.98) (7.99) The Nusselt number based on the total heat flux at the external wall is q R Nu k (Tw T ) θw θ (1 + Ahsv / c p ) ahsv 11 β n ξ + 1+ Gn e 4 c p n 1 4 Θ (1) + Θ (1) n n β n (7.100) 3

33 The Nusselt number based on the convective heat transfer coefficient is R T θ Nu Tw Tm r r R θ w θ m η η 1 * + (1 + Ahsv / c p ) Gn e β n ξ Θ n (1) (7.101) n Ahsv 1 + cp n 1 Gn e β n ξ 4 Θ n (1) + Θ n (1) βn 33

34 The Sherwood number is R ω Sh ω w ω m r r R ϕ ϕ w ϕ m η η 1 Ahsv Ahsv β n ξ Le + (1 + ) H n e Φ n (1) cp c p n 1 Ahsv Ahsv 11 β n ξ Le + 1+ Gn e 4 cp c p n 1 (7.10) 4 Φ n (1) + β Le Φ n (1) n 34

35 When the heat and mass transfer are fully developed, eqs. (7.100)-(7.10) reduce to Ahsv 48 Nu c p 48 Nu 11 * 48 Sh 11 (7.103) (7.104) (7.105) 35

36 The variations of the local Nusselt number based on total heat flux along the dimensionless location ξ are shown in Fig It is evident from Fig. 7.9(a) that Nu increases significantly with increasing ( Ahsv / c p ). The Lewis number has very little effect on Nux when ( Ahsv / c p ) 0.1, but its effects become obvious in the region near the entrance when ( Ahsv / c p ) 1.0 and gradually diminishes in the region near the exit. The effect of ϕ 0 on Nu, as is seen from Fig. 7.9(b), has no apparent influence in almost the entire region when ( Ahsv / c p ) 1.0. When ( Ahsv / c p ) 0.1, Nux increases slightly when ξ is small. 36

37 Ahsv cp (a) ϕ 0 0 Ahsv cp (b) Le3.5 Figure 7.9 Nusselt number based on total heat flux 37

38 The variation of the local Nusselt number based on convective heat flux, Nu*, is shown in Fig. 7.10(a). Only a single curve is obtained, which implies that Nu* remains unchanged when the mass transfer parameters are varied. The value of Nu* is exactly the same as for the process without sublimation. Figure 7.10(b) shows the Sherwood number for various parameters. It is evident that (Ahsv/cp) and ϕ0 have no effect on Shx, but Le has an insignificant effect on Shx in the entry region. 38

39 Figure 7.10 Nusselt number based on convective heat flux and Sherwood number. 39

40 Example 7.1 Air flows through a circular tube that has a radius of R and is heated by external convection. The external convection heat transfer coefficient and fluid temperature are he and Te, respectively. The inner surface of the tube is coated by a layer of sublimable material. The fluid with a mass fraction of sublimable substance ω0 and a temperature T0 enters the tube with a velocity U. For the sake of simplicity, the flow inside the tube is assumed to be slug flow (uniform velocity). The heat and mass transfer inside the tube are assumed to be developing. Find the Nusselt number based on total heat transfer and convective heat transfer, as well as the Sherwood number. The thermal diffusivity and mass diffusivity is assumed to be the same, i.e., Le 1. 40

41 Solution The physical model of the problem is shown in Fig. 7.11, and the conservations of energy and species equations are T T α r x r r (7.106) ω ω D r x r r (7.107) Ur Ur with the following boundary conditions: T T0, x 0 ω ω 0, x 0 (7.108) (7.109) T ω 0, r 0 r r (7.110) T ω + ρ Dhsv he (Te T ), r R r r (7.111) 41 k

42 Introducing the following nondimensional variables, hr r x UR ξ Pe Bi e R RPe α k T T ω ω θ e ϕ e ω e ate + b Te T0 ωe ω0 η (7.11) where ωe is the saturation mass fraction corresponding to Te, the governing equations become η θ ξ η θ η η η ϕ ξ η ϕ η η θ ϕ 1, ξ 0 θ ϕ 0, η 0 η η (7.113) (7.114) (7.115) (7.116) 4

43 Ahsv θ ϕ + Biθ w, η 1 cp η η ϕ w θ w, η 1 (7.117) (7.118) Equations (7.113) and (7.114) can be solved using separation of variables, and the resulting temperature and mass fraction distributions are J1 ( β n ) J 0 ( β nη ) β nξ θ ϕ e n 1 β n J 0 ( β n ) + J1 ( β n ) (7.119) where J0 and J1 are the zeroth and first order Bessel functions. The Nusselt number based on the total heat supplied by the external fluid is Nu R he (Te Tw ) Biθ w k (Tw T ) θ θw (7.10) 43

44 The Nusselt number based on the heat transferred to the fluid inside the tube is R he (Te Tw ) Biθ w (7.10) Nu k (Tw T ) θ θw The Nusselt number based on the heat transferred to the fluid inside the tube is Nu * R T θ (Tw T ) r r R θ θ w η η 1 (7.11) The Sherwood number is Sh R ω ϕ ω w ω r r R ϕ ϕ w η η 1 (7.1) 44

45 Figure 7.11 Sublimation in a tube heated by external convection. 45

46 Figure 7.1 Effect of Biot number on Nu( Ahsv / c p 1) 46

47 Figure 7.13 Effect of Biot number on Nu* or Sh ( Ahsv / c p 1) 47

48 7..4 Sublimation with Chemical Reaction During combustion involving a solid fuel, the solid fuel may burn directly or it may be sublimated before combustion. In the latter case which will be discussed in this subsection gaseous fuel diffuses away from the solidvapor surface. Meanwhile, the gaseous oxidant diffuses toward the solid-vapor interface. Under the right conditions, the mass flux of vapor fuel and the gaseous oxidant meet and the chemical reaction occurs at a certain zone known as the flame. The flame is usually a very thin region with a color dictated by the temperature of combustion. 48

49 Figure 7.14 shows the physical model of the problem under consideration. The concentration of the fuel is highest at the solid fuel surface, and decreases as the location of the flame is approached. The gaseous fuel diffuses away from the solid fuel surface and meets the oxidant as it flows parallel to the solid fuel surface. Combustion occurs in a thin reaction zone where the temperature is the highest, and the latent heat of sublimation is supplied by combustion. The combustion of solid fuel through sublimation can be modeled as a steady-state boundary layer type flow with sublimation and chemical reaction. 49

50 Figure 7.14 Sublimation with chemical reaction. 50

51 To model the problem, the following assumptions are made: The fuel is supplied by sublimation at a steady rate. The Lewis number is unity, so the thermal and concentration boundary layers have the same thickness. The buoyancy force is negligible. 51

52 The governing equations in the boundary layer are ( ρ u ) ( ρ v) + 0 x y u u u u + v ν x y y y T ( ρ c p ut ) + ( ρ c p vt ) k x y y y (7.13) (7.14) (7.15) + m& o hc, o 5

53 ω o ( ρ uω o ) + ( ρ vω o ) ρd x y y y m& o (7.16) where h is enthalpy, and m& o is rate of oxidant consumption (kg/ m3-s). hc,o is the heat released by combustion per unit mass consumption of the oxidant (J/kg), which is different from the combustion heat defined in Chapter 3. ω o is mass fraction of the oxidant in the gaseous mixture. The corresponding boundary conditions of eqs. (7.13) (7.16) are u u, T T, ωo ωo, at y (7.17) m& ω f o u 0, v, 0 at y 0 ρ y (7.18) 53

54 u τw µ, y y 0 (7.19) T, y y 0 (7.130) qw k The exact solution of the heat and mass problem described by eqs. (7.13) (7.16) can be obtained using the boundary layer theory, which is very complex. It is useful here to introduce the results obtained by analogy between momentum and heat transfer. Multiplying eq. (7.16) by and adding the result to eq. (7.15), one obtains ρ u (c pt + ω o hc,o ) + ρ v(c pt + ω o hc,o ) (7.131) x y ω o T k + ρ Dh c,o y y y 54

55 Considering the assumption that Lewis number is unity, i.e., Le α / D 1, eq. (7.131) can be rewritten as ρ u (c pt + ω o hc,o ) + ρ v (c pt + ω o hc,o ) x y (7.13) ρ α ( c T + ω h ) p o c,o y y qw ρ α qw (c pt + ω o hc,o ), y y 0 τw (c pt + ω o hc,o ) w (c pt + ω o hc,o ) u τw c p (Tw T ) + hc,o (ω o, w ω o, ) u (7.133) (7.134) 55

56 qw m& f hsv + ql m& f B τw u (7.135) (7.136) c p (T Tw ) + hc,o (ω o, ω o, w ) B hsg + ql / m f (7.137) τw Cf ρ u / (7.138) m& f Cf ρ u B (7.139) 56

57 vw m& f ρ Cf u B (7.140) ( ρ v) w 1/ Re x ( ρ u ) (7.141) B 1/ B Re x C f (7.14) ln(1 + B ) B.6 B 0.15 (7.143) B 57

58 7.3 Chemical Vapor Deposition (CVD) Example 7. Air with a temperature of 7 C flows at 1 m/s over a 1-m long solid fuel surface with a temperature of 77 C. The concentration of the oxidant at the solid fuel surface is 0.1, and the heat released per unit mass of the oxidant consumed is 1000 kj/kg. The latent heat of sublimation for the solid fuel is 1500 kj/kg. Neglect the sensible heat required to raise the surface temperature of the solid fuel to sublimation temperature, and heat loss to the solid fuel. Estimate the average blowing velocity due to sublimation on the fuel surface. 58

59 7.3 Chemical Vapor Deposition (CVD) Solution: The mass fractions of the oxygen at the solid fuel surface and in the incoming air are, respectively, ωo,w 0.1and ωo, 0.1. The specific heat of gas, approximately taken Tave (Tw + T ) / 377 C, as specific heat of air at is cp1.063 kj/kg-k. The combustion heat per unit oxidant consumed is hc,o 1000 kj/kg. The latent heat of sublimation is hsv 1500 kj/kg. The density at the wall and the incoming temperatures are respectively ρw kg/m3 and ρ kg/m3. The viscosity at Tave is v 60.1x10-6 m/s. 59

60 7.3 Chemical Vapor Deposition (CVD) The transfer driving force can be obtained from eq. (7.137), i.e., Z c p (T Tw ) + hc,o (ω o, ω o, w ) hsv (7 77) ( ) The blowing parameter obtained from eq. (7.143) is B ln(1 + Z ) ln( ) Z The blowing velocity at the surface is obtained from eq. (7.141): ρ ρ 1/ vw Bu Re x 1/ B ( u ν ) x 1/ ρw ρw which can be integrated to yield the average blowing velocity: vw ρ 1/ B ( u ν L ) ρw 1/ ( ) m/s

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