4. Introduction to Heat & Mass Transfer

4. Introduction to Heat & Mass Transfer This section will cover the following concepts: A rudimentary introduction to mass transfer. Mass transfer from a molecular point of view. Fundamental similarity of heat and mass transfer. Application of mass transfer concepts: - Evaporation of a liquid layer - Evaporation of a liquid droplet 4. Heat & Mass Transfer 1 AER 1304 ÖLG

Mass-Transfer Rate Laws: Mass Transfer: Fick s Law of Diffusion: Describes, in its basic form, the rate at which two gas species diffuse through each other. For one-dimensional binary diffusion: ṁ A mass flow of A per unit area = Y A (ṁ A +ṁ B) mass flow of A associated with bulk flow per unit area dy A ρd AB dx mass flow of A associated with molecular diffusion (4.1) 4. Heat & Mass Transfer 2 AER 1304 ÖLG

A is transported by two means: (a) bulk motion of the fluid, and (b) molecular diffusion. Mass flux is defined as the mass flowrate of species A per unit area perpendicular to the flow: ṁ A = ṁ A /A (4.2) ṁ A has the units kg/(s m2 ). The binary diffusivity, orthemolecular diffusion coefficient, D AB is a property of the mixture and has units of m 2 /s. 4. Heat & Mass Transfer 3 AER 1304 ÖLG

In the absence of diffusion: ṁ A = Y A (ṁ A +ṁ B)=Y A ṁ Bulk flux of species A (4.3a) where ṁ is the mixture mass flux. The diffusional flux adds an additional component to the flux of A: ρd AB dy A dx Diffusional flux of A, ṁ A,diff (4.3b) 4. Heat & Mass Transfer 4 AER 1304 ÖLG

Note that the negative sign causes the flux to be postive in the x-direction when the concentration gradient is negative. Analogy between the diffusion of heat (conduction) and molecular diffusion. - Fourier s law of heat conduction: Q x = k dt (4.4) dx Both expressions indicate a flux (ṁ A,diff or Q x) being proportional to the gradient of a scalar quantity [(dy A /dx) or(dt/dx)]. 4. Heat & Mass Transfer 5 AER 1304 ÖLG

A more general form of Eqn 4.1: ṁ A = Y A ( ṁ A + ṁ B) ρd AB Y A (4.5) symbols with over arrows represent vector quantities. Molar form of Eqn 4.5 Ṅ A = χ A ( Ṅ A + Ṅ B) cd AB χ A (4.6) where Ṅ A,(kmol/(sm2 ), is the molar flux of species A; χ A is mole fraction, and c is the molar concentration, kmol/m 3. 4. Heat & Mass Transfer 6 AER 1304 ÖLG

Meanings of bulk flow and diffusional flux can be better explained if we consider that: ṁ mixture mass flux = ṁ A species A mass flux + ṁ B species B mass flux (4.7) If we substitute for individual species fluxes from Eqn 4.1 into 4.7: ṁ = Y A ṁ dy A ρd AB dx + Y Bṁ dy B ρd BA dx (4.8a) 4. Heat & Mass Transfer 7 AER 1304 ÖLG

Or: ṁ =(Y A + Y B )ṁ dy A ρd AB dx ρd dy B BA dx (4.8b) For a binary mixture, Y A + Y B =1; then: ρd AB dy A dx diffusional flux of species A ρd BA dy B dx diffusional flux of species B =0 (4.9) In general ṁ i =0 4. Heat & Mass Transfer 8 AER 1304 ÖLG

Some cautionary remarks: - We are assuming a binary gas and the diffusion is a result of concentration gradients only (ordinary diffusion). - Gradients of temperature and pressure can produce species diffusion. - Soret effect: species diffusion as a result of temperature gradient. - In most combustion systems, these effects are small and can be neglected. 4. Heat & Mass Transfer 9 AER 1304 ÖLG

Molecular Basis of Diffusion: We apply some concepts from the kinetic theory of gases. - Consider a stationary (no bulk flow) plane layer of a binary gas mixture consisting of rigid, non-attracting molecules. - Molecular masses of A and B are identical. - A concentration (mass-fraction) gradient exists in x-direction, and is sufficiently small that over smaller distances the gradient can be assumed to be linear. 4. Heat & Mass Transfer 10 AER 1304 ÖLG

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Average molecular properties from kinetic theory of gases: v Z A Mean speed of species A = molecules Wall collision frequency of A 8kB T 1/2 πm A molecules per unit area = 1 4 λ Mean free path = na v V 1 2π(ntot /V )σ 2 (4.10a) (4.10b) (4.10c) a Average perpendicular distance from plane of last collision to plane where next collision occurs = 2 3 λ (4.10d) 4. Heat & Mass Transfer 12 AER 1304 ÖLG

where - k B : Boltzmann s constant. - m A : mass of a single A molecule. - (n A /V ): number of A molecules per unit volume. - (n tot /V ): total number of molecules per unit volume. - σ: diameter of both A and B molecules. Net flux of A molecules at the x-plane: ṁ A = ṁ A,(+)x dir ṁ A,( )x dir (4.11) 4. Heat & Mass Transfer 13 AER 1304 ÖLG

In terms of collision frequency, Eqn 4.11 becomes ṁ A Net mass flux of species A = (ZA) x a Number of A crossing plane x originating from plane at (x a) m A (ZA) x+a Number of A crossing plane x originating from plane at (x+a) m A (4.12) Since ρ m tot /V tot, then we can relate Z A to mass fraction, Y A (from Eqn 4.10b) Z Am A = 1 4 n A m A m tot ρ v = 1 4 Y Aρ v (4.13) 4. Heat & Mass Transfer 14 AER 1304 ÖLG

Substituting Eqn 4.13 into 4.12 ṁ A = 1 4 ρ v(y A,x a Y A,x+a ) (4.14) With linear concentration assumption dy dx = Y A,x a Y A,x+a 2a = Y A,x a Y A,x+a 4λ/3 (4.15) 4. Heat & Mass Transfer 15 AER 1304 ÖLG

From the last two equations, we get ṁ A = ρ vλ 3 dy A dx Comparing Eqn. 3.16 with Eqn. 3.3b, D AB is (4.16) D AB = vλ/3 (4.17) Substituting for v and λ, along with ideal-gas equation of state, PV = nk B T D AB = 2 k 3 B T 1/2 T 3 π 3 m A σ 2 (4.18a) P 4. Heat & Mass Transfer 16 AER 1304 ÖLG

or D AB T 3/2 P 1 (4.18b) Diffusivity strongly depends on temperature and is inversely proportional to pressure. Mass flux of species A, however, depends on ρd AB, which then gives: ρd AB T 1/2 P 0 = T 1/2 (4.18c) In some practical/simple combustion calculations, the weak temperature dependence is neglected and ρd is treated as a constant. 4. Heat & Mass Transfer 17 AER 1304 ÖLG

Comparison with Heat Conduction: We apply the same kinetic theory concepts to the transport of energy. Same assumptions as in the molecular diffusion case. v and λ have the same definitions. Molecular collision frequency is now based on the total number density of molecules, n tot /V, Z = Average wall collision frequency per unit area = 1 4 ntot V v (4.19) 4. Heat & Mass Transfer 18 AER 1304 ÖLG

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In the no-interaction-at-a-distance hard-sphere model of the gas, the only energy storage mode is molecular translational (kinetic) energy. Energy balance at the x-plane; Net energy flow in x direction = kinetic energy flux with molecules from x a to x kinetic energy flux with molecules from x+a to x ke is given by Q x = Z (ke) x a Z (ke) x+a (4.20) ke = 1 2 m v2 = 3 2 k BT (4.21) 4. Heat & Mass Transfer 20 AER 1304 ÖLG

heat flux can be related to temperature as Q x = 3 2 k BZ (T x a T x+a ) (4.22) The temperature gradient dt dx = T x+a T x a 2a (4.23) Eqn. 4.23 into 3.22, and definitions of Z and a Q x = 1 n 2 k B vλ dt (4.24) V dx 4. Heat & Mass Transfer 21 AER 1304 ÖLG

Comparing to Eqn. 4.4, k is k = 1 n 2 k B V vλ (4.25) In terms of T and molecular size, k = k 3 B π 3 mσ 4 1/2 T 1/2 Dependence of k on T (similar to ρd) (4.26) k T 1/2 (4.27) - Note: For real gases T dependency is larger. 4. Heat & Mass Transfer 22 AER 1304 ÖLG

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Species Conservation: One-dimensional control volume Species A flows into and out of the control volume as a result of the combined action of bulk flow and diffusion. Within the control volume, species A may be created or destroyed as a result of chemical reaction. The net rate of increase in the mass of A within the control volume relates to the mass fluxes and reaction rate as follows: 4. Heat & Mass Transfer 24 AER 1304 ÖLG

dm A,cv dt Rate of increase of mass A within CV -where =[ṁ AA] x Mass flow of A into CV [ṁ AA] x+ x Mass flow of A out of the CV + ṁ A V Mass prod. rate of A by reaction (4.28) - ṁ A is the mass production rate of species A per unit volume. - ṁ A is defined by Eqn. 4.1. 4. Heat & Mass Transfer 25 AER 1304 ÖLG

Within the control volume m A,cv = Y A m cv = Y A ρv cv, and the volume V cv = A x; Eqn. 4.28 A x (ρy A) = A Y A ṁ Y A ρd AB t x x A Y A ṁ Y A ρd AB x x+ x +ṁ A A x (4.29) Dividing by A x and taking limit as x 0, (ρy A ) = Y A ṁ Y A ρd AB +ṁ A t x x (4.30) 4. Heat & Mass Transfer 26 AER 1304 ÖLG

For the steady-flow, ṁ A d dx Y A ṁ ρd AB dy A dx =0 (4.31) Eqn. 4.31 is the steady-flow, one-dimensional form of species conservation for a binary gas mixture. For a multidimensional case, Eqn. 4.31 can be generalized as ṁ A Net rate of species A production by chemical reaction ṁ A Net flow of species A out of control volume =0 (4.32) 4. Heat & Mass Transfer 27 AER 1304 ÖLG

Some Applications of Mass Transfer: The Stefan Problem: 4. Heat & Mass Transfer 28 AER 1304 ÖLG

Assumptions: - Liquid A in the cylinder maintained at a fixed height. - Steady-state - [A]intheflowinggasislessthan[A]atthe liquid-vapour interface. - B is insoluble in liquid A Overall conservation of mass: ṁ (x) =constant=ṁ A +ṁ B (4.33) 4. Heat & Mass Transfer 29 AER 1304 ÖLG

Since ṁ B =0,then ṁ A = ṁ (x) =constant (4.34) Then, Eqn.4.1 now becomes ṁ A = Y A ṁ A ρd AB dy A dx (4.35) Rearranging and separating variables ṁ A ρd AB dx = dy A 1 Y A (4.36) 4. Heat & Mass Transfer 30 AER 1304 ÖLG

Assuming ρd AB to be constant, integrate Eqn. 4.36 ṁ A ρd AB x = ln[1 Y A ]+C (4.37) With the boundary condition Y A (x =0)=Y A,i (4.38) We can eliminate C; then Y A (x) =1 (1 Y A,i )exp ṁ A x ρd AB (4.39) 4. Heat & Mass Transfer 31 AER 1304 ÖLG

The mass flux of A, ṁ A, can be found by letting Y A (x = L) =Y A, Then, Eqn. 4.39 reads ṁ A = ρd AB L 1 ln YA, 1 Y A,i (4.40) Mass flux is proportional to ρd, andinversely proportional to L. 4. Heat & Mass Transfer 32 AER 1304 ÖLG

Liquid-Vapour Interface: Saturation pressure P A,i = P sat (T liq,i ) (4.41) Partial pressure can be related to mole fraction and mass fraction Y A,i = P sat(t liq,i ) P MW A MW mix,i (4.42) -TofindY A,i we need to know the interface temperature. 4. Heat & Mass Transfer 33 AER 1304 ÖLG

In crossing the liquid-vapour boundary, we maintain continuity of temperature T liq,i (x =0 )=T vap,i (x =0 + )=T (0) (4.43) and energy is conserved at the interface. Heat is transferred from gas to liquid, Q g i.someofthis heats the liquid, Q i l, while the remainder causes phase change. Q g i Q i l = ṁ(h vap h liq )=ṁh fg (4.44) or Q net = ṁh fg (4.45) 4. Heat & Mass Transfer 34 AER 1304 ÖLG

Droplet Evaporation: 4. Heat & Mass Transfer 35 AER 1304 ÖLG

Assumptions: - The evaporation process is quasi-steady. - The droplet temperature is uniform, and the temperature is assumed to be some fixed value below the boiling point of the liquid. - The mass fraction of vapour at the droplet surface is determined by liquid-vapour equilibrium at the droplet temperature. - We assume constant physical properties, e.g., ρd. 4. Heat & Mass Transfer 36 AER 1304 ÖLG

Evaporation Rate: Same approach as the Stefan problem; except change in coordinate sysytem. Overall mass conservation: ṁ(r) =constant=4πr 2 ṁ (4.46) Species conservation for the droplet vapour: ṁ A = Y A ṁ A ρd AB dy A dr (4.47) 4. Heat & Mass Transfer 37 AER 1304 ÖLG

Substitute Eqn. 4.46 into 4.47 and solve for ṁ, ṁ = 4πr 2 ρd AB dy A 1 Y A dr (4.48) Integrating and applying the boundary condition -yields Y A (r = r s )=Y A,s (4.49) Y A (r) =1 (1 Y A,s)exp[ ṁ/(4πρd AB r) exp [ ṁ/(4πρd AB r s )] (4.50) 4. Heat & Mass Transfer 38 AER 1304 ÖLG

Evaporation rate can be determined from Eqn. 4.50 by letting Y A = Y A, for r : ṁ =4πr s ρd AB ln (1 YA, ) (1 Y A,s ) (4.51) In Eqn. 4.51, we can define the dimensionless transfer number, B Y, 1+B Y 1 Y A, 1 Y A,s (4.52a) 4. Heat & Mass Transfer 39 AER 1304 ÖLG

or Then the evaporation rate is B Y = Y A,s Y A, 1 Y A,s (4.52b) ṁ =4πr s ρd AB ln (1 + B Y ) (4.53) Droplet Mass Conservation: dm d = ṁ (4.54) dt where m d is given by m d = ρ l V = ρ l πd 3 /6 (4.55) 4. Heat & Mass Transfer 40 AER 1304 ÖLG

where D = 2r s,andv is the volume of the droplet. Substituting Eqns 4.55 and 4.53 into 4.54 and differentiating dd dt = 4ρD AB ρ l D ln (1 + B Y ) (4.56) Eqn 4.56 is more commonly expressed in term of D 2 rather than D, dd 2 dt = 8ρD AB ρ l ln (1 + B Y ) (4.57) 4. Heat & Mass Transfer 41 AER 1304 ÖLG

Equation 4.57 tells us that time derivative of the square of the droplet diameter is constant. D 2 varies with t with a slope equal to RHS of Eqn.4.57. This slope is defined as evaporation constant K: K = 8ρD AB ρ l ln (1 + B Y ) (4.58) Droplet evaporation time can be calculated from: 0 D 2 o dd 2 = t d 0 Kdt (4.59) 4. Heat & Mass Transfer 42 AER 1304 ÖLG

which yields t d = D 2 o/k (4.60) We can change the limits to get a more general relationship to provide a general expression for the variation of D with time: D 2 (t) =D 2 o Kt (4.61) Eq. 4.61 is referred to as the D 2 law for droplet evaporation. 4. Heat & Mass Transfer 43 AER 1304 ÖLG

D 2 law for droplet evaporation: 4. Heat & Mass Transfer 44 AER 1304 ÖLG