Convective mass transfer

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1 Errata

• Slide 10 gives 0.322 as the heat-column coefficient. Use 0.332.

  This agrees with the velocity and mass columns.

• Slide 12 gives 0.332 as the average laminar plate coefficient. Use 0.664 in the average Sherwood-number correlation

  ¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3$$\overline{\mathrm{Sh}}_L =0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3}$$

• Slide 13 uses 𝑑/𝑑𝑥$d/dx$ for the Reynolds-analogy gradients. Use 𝑑/𝑑𝑦$d/dy$.

  These gradients are normal to the wall.

• Slide 17 labels the subliming solid as carbon dioxide and supplies a vapour pressure of 4740 Pa without a surface temperature. Using the 293 K air temperature for solid CO2$_2$ is physically inconsistent: NIST phase data place its triple point at 216.58 K. Its solid-vapour correlation gives 4740 Pa at about 163.5 K. Example 1 below instead uses a hypothetical volatile solid with explicitly stipulated properties and surface temperature. Its numerical answer must not be used for dry ice.

• Slide 20 omits the humidity and surface temperature needed for an evaporation rate. Example 4 below specifies dry air and a surface maintained at 35 degrees Celsius. An earlier worked solution used 6221 Pa at that temperature; use 5624 Pa. This pressure error was in the worked solution, rather than a printed value in the PDF question.

2 Reading list

• Çengel and Ghajar (2015) and Geankoplis (2003) for heat and mass transfer correlations.
• Bird et al. (2006) and Welty et al. (2001) for transport balances and diffusivity definitions.

Examples use mol unless stated otherwise.

3 Choose a convective-transfer model

Given conditions	Model and checks
Transfer between gas and liquid bulks	Use the two-film model, Eq. 1 and Eq. 2; put both resistances on the same basis.
A known interface concentration and one flowing phase	Use 𝑁𝐴 =𝑘𝑐(𝑐𝐴,𝑠 −𝑐𝐴,𝑏)$N_A=k_c(c_{A,s}-c_{A,b})$ with the appropriate geometry and local or average Sherwood number.
A matching heat-transfer correlation is available	Use the heat/mass analogy only with matching geometry, flow and boundary conditions.
Bulk composition changes along a tube	Integrate a stream balance instead of using inlet concentration everywhere.
Buoyancy rather than imposed flow drives transfer	Choose a mass-buoyancy correlation and separate composition and temperature effects.

4 Deriving the two-film resistance

Convective mass transfer

Species transport involving molecular diffusion and bulk fluid motion. The flow changes the concentration field and the gradient driving interfacial transfer.

Consider steady absorption of A from a gas into a liquid. A thin gas film and a thin liquid film supply the resistance. Each bulk phase is well mixed. Take 𝑁𝐴 >0$N_A>0$ from gas to liquid, and neglect accumulation, reaction and interfacial resistance. The same molar flux crosses both films.

Figure 1: Two films meet at a gas-liquid interface. Positive species flux runs from the gas bulk through the gas film and liquid film into the liquid bulk.

The profile heights are schematic and use separate composition scales for the two phases. Equal chemical potential at the interface does not require equal gas and liquid mole fractions.

Let 𝑏$b$ mean bulk and 𝑖$i$ mean interface. Define the coefficients on a mole-fraction basis, in mol m−2$^{-2}$ s−1$^{-1}$ per unit mole-fraction difference:

𝑁𝐴=𝑘𝑦(𝑦𝐴,𝑏−𝑦𝐴,𝑖)𝑁𝐴=𝑘𝑥(𝑥𝐴,𝑖−𝑥𝐴,𝑏)𝑦𝐴,𝑖=𝑚𝑥𝐴,𝑖.$$\begin{aligned} N_A&=k_y(y_{A,b}-y_{A,i})\\ N_A&=k_x(x_{A,i}-x_{A,b})\\ y_{A,i}&=m x_{A,i}. \end{aligned}$$(1)

The slope 𝑚$m$ comes from the local equilibrium relation. It is not a ratio of diffusivities. Solve the liquid-film equation for its interface value, then substitute into the gas-film equation:

𝑥𝐴,𝑖=𝑥𝐴,𝑏+𝑁𝐴𝑘𝑥𝑁𝐴=𝑘𝑦[𝑦𝐴,𝑏−𝑚(𝑥𝐴,𝑏+𝑁𝐴𝑘𝑥)]𝑦𝐴,𝑏−𝑚𝑥𝐴,𝑏=𝑁𝐴(1𝑘𝑦+𝑚𝑘𝑥).$$\begin{aligned} x_{A,i}&=x_{A,b}+\frac{N_A}{k_x}\\ N_A&=k_y\left[y_{A,b}-m\left(x_{A,b}+\frac{N_A}{k_x}\right)\right]\\ y_{A,b}-m x_{A,b}&=N_A\left(\frac1{k_y}+\frac m{k_x}\right). \end{aligned}$$

Define the overall coefficient 𝐾𝑦$K_y$ using the gas-basis driving force:

𝑁𝐴=𝐾𝑦(𝑦𝐴,𝑏−𝑚𝑥𝐴,𝑏)$$N_A=K_y(y_{A,b}-m x_{A,b})$$

1𝐾𝑦=1𝑘𝑦+𝑚𝑘𝑥$$\frac1{K_y}=\frac1{k_y}+\frac m{k_x}$$(2)

The factor 𝑚$m$ converts the liquid-film resistance to the gas basis. Comparing 1/𝑘𝑦$1/k_y$ directly with 1/𝑘𝑥$1/k_x$ would omit that conversion. A concentration-basis coefficient in m s⁻¹ cannot be substituted without converting its driving force.

For example, take 𝑘𝑦 =0.020$k_y=0.020$, 𝑘𝑥 =0.010$k_x=0.010$ in the units above, 𝑚 =2$m=2$, 𝑦𝐴,𝑏 =0.100$y_{A,b}=0.100$ and 𝑥𝐴,𝑏 =0.020$x_{A,b}=0.020$. Then

1𝐾𝑦=50+200=250 m2 s mol−1𝐾𝑦=0.0040 mol m−2 s−1𝑁𝐴=0.0040[0.100−2(0.020)]=2.4×10−4 mol m−2 s−1.$$\begin{aligned} \frac1{K_y}&=50+200=250\ \mathrm{m^2\,s\,mol^{-1}}\\ K_y&=0.0040\ \mathrm{mol\,m^{-2}\,s^{-1}}\\ N_A&=0.0040[0.100-2(0.020)]\\ &=2.4\times10^{-4}\ \mathrm{mol\,m^{-2}\,s^{-1}}. \end{aligned}$$

The liquid film contributes 80 percent of the total gas-basis resistance. Recovering 𝑥𝐴,𝑖 =0.044$x_{A,i}=0.044$ and 𝑦𝐴,𝑖 =0.088$y_{A,i}=0.088$ gives the same flux through each film and satisfies 𝑦𝐴,𝑖 =2𝑥𝐴,𝑖$y_{A,i}=2x_{A,i}$. Reducing 𝑚$m$ to 0.2 while keeping the coefficients fixed changes the resistances to 50 and 20; the gas film then controls most of the resistance. Solubility and the chosen basis matter.

Two-film transfer obeys 𝑦∗𝐴 =𝑚𝑥𝐴$y_A^*=mx_A$. Which expression gives the overall gas-mole-fraction resistance?

• 1/𝐾𝑦 =1/𝑘𝑦 +1/𝑘𝑥$1/K_y=1/k_y+1/k_x$.
• 1/𝐾𝑦 =1/𝑘𝑦 +𝑚/𝑘𝑥$1/K_y=1/k_y+m/k_x$. ✓
• 𝐾𝑦 =𝑘𝑦 +𝑘𝑥$K_y=k_y+k_x$.
• 1/𝐾𝑦 =𝑚/𝑘𝑦 +1/𝑘𝑥$1/K_y=m/k_y+1/k_x$.

Explanation

Use the equilibrium slope to express both film resistances in terms of gas mole fraction:

1𝐾𝑦=1𝑘𝑦+𝑚𝑘𝑥$$\frac{1}{K_y}=\frac{1}{k_y}+\frac{m}{k_x}$$

The factor multiplying the liquid-film resistance converts a liquid mole-fraction difference to the equivalent gas mole-fraction difference. Both resistances must use the same basis before they are added.

5 Deriving external-flow mass transfer

Concentration boundary layer

The region near an interface where species concentration differs appreciably from its outer-fluid value. It has no sharp physical edge.

Schmidt number

The dimensionless ratio Sc =𝜈/𝐷𝐴𝐵$\mathrm{Sc}=\nu/D_{AB}$, where 𝜈 =𝜇/𝜌$\nu=\mu/\rho$ is the fluid’s kinematic viscosity, or momentum diffusivity, and 𝐷𝐴𝐵$D_{AB}$ is the species diffusivity. Here 𝜇$\mu$ is dynamic viscosity and 𝜌$\rho$ is fluid density.

For Sc ≪1$\mathrm{Sc}\ll1$, species diffuse faster than momentum and the concentration boundary layer is thicker than the velocity boundary layer. For Sc ≫1$\mathrm{Sc}\gg1$, species diffuse more slowly and the concentration layer is thinner. At Sc ≈1$\mathrm{Sc}\approx1$, the layers have comparable thicknesses under matching laminar-plate conditions. A liquid with 𝜈 =10−6 m2 s−1$\nu=10^{-6}\ \mathrm{m^2\,s^{-1}}$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$ has Sc =1000$\mathrm{Sc}=1000$ and a thin concentration layer. Replacing species diffusivity by thermal diffusivity would give Prandtl number, which answers a different transport question.

Lewis number

The dimensionless ratio

Le=𝛼𝐷𝐴𝐵=ScPr,$$\mathrm{Le}=\frac{\alpha}{D_{AB}}=\frac{\mathrm{Sc}}{\mathrm{Pr}},$$

where 𝛼$\alpha$ is the fluid’s thermal diffusivity and 𝐷𝐴𝐵$D_{AB}$ is the species diffusivity, both in m2 s−1$\mathrm{m^2\,s^{-1}}$.

For Le ≪1$\mathrm{Le}\ll1$, concentration disturbances diffuse faster than thermal disturbances. For Le ≫1$\mathrm{Le}\gg1$, heat diffuses faster and the thermal layer extends farther than the concentration layer under matching flow conditions. At Le ≈1$\mathrm{Le}\approx1$, matching heat and mass models can have similar scalar profiles, which supports a heat and mass transfer analogy when their boundary conditions also match. For 𝛼 =10−7 m2 s−1$\alpha=10^{-7}\ \mathrm{m^2\,s^{-1}}$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$, Le =100$\mathrm{Le}=100$; treating heat and species diffusion as equally fast would be a poor approximation. The air-water calculation below illustrates the near-unity case. Neither case permits setting ℎ =𝑘𝑐$h=k_c$, since those coefficients have different units.

These diffusivity ratios help explain the relative thicknesses of the boundary layers. To calculate the mass transfer rate, we now need the concentration gradient at the surface. Consider fluid flowing over a wet flat plate, as shown in Figure 2. Let 𝑥$x$ measure distance along the plate in the flow direction, and let 𝑧$z$ measure perpendicular distance from the plate into the fluid, with 𝑧 =0$z=0$ at the surface. At a fixed position 𝑥$x$, the concentration falls from 𝑐𝐴,𝑠$c_{A,s}$ at the wet surface to 𝑐𝐴,∞$c_{A,\infty}$ in the free stream. The slope of this concentration profile at 𝑧 =0$z=0$ gives the surface gradient in Fick’s law. Below, we use that gradient to obtain the mass transfer coefficient.

Figure 2: Concentration boundary layer over a wet plate and the local concentration profile at a marked downstream position. The layer edge is schematic; the dashed oblique tangent shows the negative wall gradient for evaporation.

5.1 The coefficient and Sherwood number

Mass transfer coefficient

A coefficient relating species flux to a specified interfacial-to-bulk driving difference. On a molar-concentration basis,

𝑁𝐴=𝑘𝑐(𝑐𝐴,𝑠−𝑐𝐴,𝑏)$$N_A=k_c(c_{A,s}-c_{A,b})$$

and 𝑘𝑐$k_c$ has units m s−1$\mathrm{m\,s^{-1}}$; other driving-force bases use different units.

Use Sherwood number, first introduced in the spherical-diffusion calculation, to express the coefficient on a dimensionless basis. It corresponds to Nusselt number with species diffusivity replacing thermal conductivity in the appropriate coefficient definition.

Write the outgoing molar flux 𝑁𝐴$N_A$ in mol m−2$^{-2}$ s−1$^{-1}$ as a coefficient times a concentration difference. With constant total molar concentration 𝑐$c$, its mole-fraction form is equivalent:

𝑁𝐴=𝑘𝑐(𝑐𝐴,𝑠−𝑐𝐴,∞)$$N_A=k_c(c_{A,s}-c_{A,\infty})$$

𝑁𝐴=𝑘𝑦(𝑦𝐴,𝑠−𝑦𝐴,∞),𝑘𝑦=𝑐𝑘𝑐$$N_A=k_y(y_{A,s}-y_{A,\infty}),\quad k_y=c k_c$$(3)

The concentration-basis coefficient 𝑘𝑐$k_c$ has units m s−1$^{-1}$. The source calls the mole-fraction coefficient 𝑘𝑥$k_x$, but this topic uses 𝑘𝑦$k_y$ for gas composition and reserves 𝑘𝑥$k_x$ for liquid composition. Do not mix a coefficient defined per mole fraction with one defined per concentration.

At low normal mass flux, Fick’s wall law is

𝑁𝐴=−𝐷𝐴𝐵[∂𝑐𝐴∂𝑧]𝑠.$$N_A=-D_{AB}\left[\frac{\partial c_A}{\partial z}\right]_s.$$

Equating it with Eq. 3 gives

𝑘𝑐=−𝐷𝐴𝐵[∂𝑐𝐴∂𝑧]𝑠𝑐𝐴,𝑠−𝑐𝐴,∞$$k_c=\frac{-D_{AB}\left[\frac{\partial c_A}{\partial z}\right]_s}{c_{A,s}-c_{A,\infty}}$$(4)

The coefficient packages the wall gradient. Multiplication by a characteristic length 𝐿$L$ gives the dimensionless Sherwood number,

Sh=𝑘𝑐𝐿𝐷𝐴𝐵$$\mathrm{Sh}=\frac{k_c L}{D_{AB}}$$

Sh=−[∂𝑐𝐴∂𝑧]𝑠𝑐𝐴,𝑠−𝑐𝐴,∞𝐿$$\mathrm{Sh}=\frac{-\left[\frac{\partial c_A}{\partial z}\right]_s}{\frac{c_{A,s}-c_{A,\infty}}{L}}$$(5)

It compares the actual gradient with a reference gradient over 𝐿$L$. Large Sherwood means a steep wall gradient. It is analogous to Nu =ℎ𝐿/𝑘$\mathrm{Nu}=hL/k$, where ℎ$h$ is the heat-transfer coefficient in W m−2$^{-2}$ K−1$^{-1}$. It is not a universal enhancement ratio with a stagnant value of one. A stagnant sphere has 𝑘𝑐 =𝐷𝐴𝐵/𝑅$k_c=D_{AB}/R$, hence Sh =2$\mathrm{Sh}=2$ when 𝐿 =2𝑅$L=2R$.

A correlation gives Sh =100$\mathrm{Sh}=100$ for 𝐿 =0.10$L=0.10$ m and 𝐷𝐴𝐵 =2 ×10−5 m2 s−1$D_{AB}=2\times10^{-5}\ \mathrm{m^2\,s^{-1}}$. What is 𝑘𝑐$k_c$?

• 0.20 m s−1$0.20\ \mathrm{m\,s^{-1}}$.
• 5 m s−1$5\ \mathrm{m\,s^{-1}}$.
• 2 ×10−6 m s−1$2\times10^{-6}\ \mathrm{m\,s^{-1}}$.
• 0.020 m s−1$0.020\ \mathrm{m\,s^{-1}}$. ✓

Explanation

Rearrange the Sherwood-number definition:

𝑘𝑐=Sh𝐷𝐴𝐵𝐿$$k_c=\frac{\mathrm{Sh}D_{AB}}{L}$$

Substituting the given values,

𝑘𝑐=100(2×10−5)0.10=0.020 m s−1$$k_c=\frac{100(2\times10^{-5})}{0.10}=0.020\ \mathrm{m\,s^{-1}}$$

Use the species diffusivity in this mass transfer calculation.

5.2 Boundary-layer equations and local results

For steady two-dimensional laminar flow over a flat plate, assume constant properties, negligible streamwise diffusion, zero pressure gradient, no species reaction, no heat generation and low species injection. Continuity gives

∂𝑢∂𝑥+∂𝑣∂𝑧=0.$$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial z}=0.$$

Here 𝑢$u$ and 𝑣$v$ are velocity components in m s−1$^{-1}$. The other balances are

𝑢𝑢𝑥+𝑣𝑢𝑧=𝜈𝑢𝑧𝑧𝑢𝑇𝑥+𝑣𝑇𝑧=𝛼𝑇𝑧𝑧𝑢(𝑐𝐴)𝑥+𝑣(𝑐𝐴)𝑧=𝐷𝐴𝐵(𝑐𝐴)𝑧𝑧$$\begin{aligned} u u_x+v u_z&=\nu u_{zz}\\[0.5em] u T_x+v T_z&=\alpha T_{zz}\\[0.5em] u (c_A)_x+v (c_A)_z&=D_{AB}(c_A)_{zz} \end{aligned}$$(6)

Subscripts on variables in Eq. 6 denote derivatives. At 𝑧 =0$z=0$, impose 𝑢 =𝑣 =0$u=v=0$, 𝑇 =𝑇𝑠$T=T_s$ and 𝑐𝐴 =𝑐𝐴,𝑠$c_A=c_{A,s}$. Far away, 𝑢 =𝑈$u=U$, 𝑇 =𝑇∞$T=T_\infty$ and 𝑐𝐴 =𝑐𝐴,∞$c_A=c_{A,\infty}$. The common advection operator and matching boundary conditions explain the analogy. If 𝜈 =𝛼 =𝐷𝐴𝐵$\nu=\alpha=D_{AB}$, the normalized velocity, temperature and concentration profiles coincide. Blasius gives the wall velocity gradient

∂𝑢∂𝑧∣𝑧=0  =0.33206𝑈√𝑈𝜈𝑥.$$\left.\frac{\partial u}{\partial z}\right|_{z=0} \;=0.33206 U\sqrt{\frac{U}{\nu x}}.$$

Thus, at Sc =1$\mathrm{Sc}=1$, Sh𝑥 =0.332Re1/2𝑥$\mathrm{Sh}_x=0.332\mathrm{Re}_x^{1/2}$ with Re𝑥 =𝑈𝑥/𝜈$\mathrm{Re}_x=Ux/\nu$. The usual Schmidt correction gives

Sh𝑥=0.332Re1/2𝑥Sc1/3𝑘𝑐,𝑥=0.332𝐷𝐴𝐵√𝑈𝜈𝑥Sc1/3$$\begin{gathered} \mathrm{Sh}_x=0.332\mathrm{Re}_x^{1/2}\mathrm{Sc}^{1/3}\\ k_{c,x}=0.332 D_{AB}\sqrt{\frac U{\nu x}}\mathrm{Sc}^{1/3} \end{gathered}$$(7)

For a constant surface concentration the coefficient varies as 𝑥−1/2$x^{-1/2}$. Its length average is

¯¯¯¯𝑘𝑐=𝐿−1∫𝐿0𝑘𝑐,𝑥 𝑑𝑥.$$\bar k_c=L^{-1}\int_0^L k_{c,x}\,dx.$$

The integral is 2√𝐿$2\sqrt L$, so the average is twice the trailing-edge value:

¯¯¯¯¯¯¯Sh𝐿=¯¯¯¯𝑘𝑐𝐿𝐷𝐴𝐵=0.664Re1/2𝐿Sc1/3$$\overline{\mathrm{Sh}}_L=\frac{\bar k_c L}{D_{AB}} =0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3}$$(8)

Use the laminar expression for Re𝐿 <5 ×105$\mathrm{Re}_L<5\times10^5$ and Sc >0.5$\mathrm{Sc}>0.5$.

5.3 Reynolds analogy

Mass Stanton number

The dimensionless species-transfer coefficient

St𝐷=𝑘𝑐𝑈=ShRe Sc,$$\mathrm{St}_D=\frac{k_c}{U}=\frac{\mathrm{Sh}}{\mathrm{Re}\,\mathrm{Sc}},$$

using a concentration-basis 𝑘𝑐$k_c$ in m s⁻¹ and the same speed and length throughout. It compares surface transfer velocity with stream velocity and corresponds to the thermal Stanton number.

Small St𝐷$\mathrm{St}_D$ means weak transfer relative to the flowing species capacity per area; larger values mean stronger transfer on that basis. For 𝑘𝑐 =0.01 m s−1$k_c=0.01\ \mathrm{m\,s^{-1}}$ and 𝑈 =2 m s−1$U=2\ \mathrm{m\,s^{-1}}$, St𝐷 =0.005$\mathrm{St}_D=0.005$. This is not directly the fraction absorbed by a whole pipe, which also depends on length and diameter through Number of transfer units. A mole-fraction coefficient such as 𝑘𝑦$k_y$ cannot be divided by 𝑈$U$ here without converting its basis.

At Pr =1$\mathrm{Pr}=1$, normalize the profiles as

𝑢𝑈=𝑇−𝑇𝑠𝑇∞−𝑇𝑠.$$\frac{u}{U}=\frac{T-T_s}{T_\infty-T_s}.$$

Differentiate normal to the wall, multiply by 𝐶𝑝𝜇 =𝑘$C_p\mu=k$, and recognize the wall shear 𝜏𝑤 =𝜇𝑢𝑧∣𝑠$\tau_w=\mu u_z|_s$ and Fourier wall heat flux. This gives ℎ =𝐶𝑝𝜏𝑤/𝑈$h=C_p\tau_w/U$. Defining the skin-friction coefficient

𝐶𝑓=𝜏𝑤12𝜌𝑈2$$C_f=\frac{\tau_w}{\frac{1}{2}\rho U^2}$$

gives

𝐶𝑓2=St𝐻=ℎ𝜌𝑈𝐶𝑝$$\frac{C_f}{2}=\mathrm{St}_H=\frac h{\rho U C_p}$$

𝐶𝑓2=St𝐷=𝑘𝑐𝑈$$\frac{C_f}{2}=\mathrm{St}_D=\frac{k_c}{U}$$(9)

The second equality requires Sc =1$\mathrm{Sc}=1$ and matching transport conditions. The wall-normal gradients determine surface shear and heat flux. The dimensionless forms are

St𝐻=NuRePr$$\mathrm{St}_H=\frac{\mathrm{Nu}}{\mathrm{Re}\mathrm{Pr}}$$

and

St𝐷=ShReSc.$$\mathrm{St}_D=\frac{\mathrm{Sh}}{\mathrm{Re}\mathrm{Sc}}.$$

For a pipe the Fanning friction factor is the same shear normalization, while the Darcy factor is four times it. This distinction prevents a factor-of-four error. Do not replace wall skin friction by the total drag coefficient of a cylinder or sphere, where pressure drag can dominate. The simple analogy extends approximately to appropriate turbulent flows when momentum and scalar transport have comparable effective diffusivities; unity molecular ratios alone do not prove every turbulent profile identical.

5.4 Chilton-Colburn analogy and Lewis relation

Colburn mass factor

The dimensionless factor 𝑗𝐷 =St𝐷Sc2/3$j_D=\mathrm{St}_D\mathrm{Sc}^{2/3}$, using concentration-basis mass Stanton number. In matching flows it can be compared with the Colburn heat factor, 𝑗𝐻 =St𝐻Pr2/3$j_H=\mathrm{St}_H\mathrm{Pr}^{2/3}$. Equality of the factors is an empirical analogy with geometry and transport restrictions.

The empirical transfer factors relax the unity-property condition:

𝑗𝐻=St𝐻Pr2/3$$j_H=\mathrm{St}_H\mathrm{Pr}^{2/3}$$

𝑗𝐷=St𝐷Sc2/3,𝑗𝐻≈𝑗𝐷$$j_D=\mathrm{St}_D\mathrm{Sc}^{2/3},\quad j_H\approx j_D$$(10)

For suitable plates and pipes these factors also approximate 𝐶𝑓/2$C_f/2$. The transfer-to-transfer relation can remain useful when total drag cannot be used, provided geometry, boundary conditions and transport assumptions match. It is not a guarantee for rough packed beds or high blowing. The source quotes approximate ranges 0.6 <Pr <60$0.6<\mathrm{Pr}<60$ and 0.6 <Sc <3000$0.6<\mathrm{Sc}<3000$; individual correlations may impose tighter ranges. Equating factors and cancelling 𝑈$U$ yields

ℎ𝑘𝑐=𝜌𝐶𝑝(ScPr)2/3$$\frac h{k_c}=\rho C_p\left(\frac{\mathrm{Sc}}{\mathrm{Pr}}\right)^{2/3}$$

ℎ=𝜌𝐶𝑝𝑘𝑐Le2/3$$h=\rho C_p k_c\mathrm{Le}^{2/3}$$(11)

With water vapour in air at 298 K, 𝐷𝐴𝐵 =2.5 ×10−5$D_{AB}=2.5\times10^{-5}$ and 𝛼 =2.18 ×10−5$\alpha=2.18\times10^{-5}$ m2$^2$ s−1$^{-1}$ give Le =0.872$\mathrm{Le}=0.872$ and Le2/3 =0.9127$\mathrm{Le}^{2/3}=0.9127$. Dropping this roughly nine-percent correction gives the Lewis relation ℎ ≈𝜌𝐶𝑝𝑘𝑐$h\approx\rho C_p k_c$. It supports wet-bulb and humidification models for air-water; other pairs need their own Lewis number.

The Chilton-Colburn heat and mass analogy applies, and Le =1$\mathrm{Le}=1$. Which relation follows for concentration-basis 𝑘𝑐$k_c$?

• ℎ =𝜌𝐶𝑝𝑘𝑐$h=\rho C_p k_c$. ✓
• ℎ =𝑘𝑐$h=k_c$.
• ℎ =𝜌𝐶𝑝/𝑘𝑐$h=\rho C_p/k_c$.
• ℎ =𝑘𝑐𝐷𝐴𝐵$h=k_cD_{AB}$.

Explanation

The heat and mass transfer analogy gives

ℎ=𝜌𝐶𝑝𝑘𝑐Le2/3$$h=\rho C_p k_c\mathrm{Le}^{2/3}$$

With Lewis number equal to one,

ℎ=𝜌𝐶𝑝𝑘𝑐$$h=\rho C_p k_c$$

The density and heat capacity are still needed. The heat and mass transfer coefficients have different units and cannot be set equal.

6 Classroom examples 1 to 5

All examples use mol rather than kmol. Multiply a concentration or rate in kmol by 1000 to obtain mol. The molar gas constant is 𝑅 =8.314$R=8.314$ J mol−1$^{-1}$ K−1$^{-1}$. The molar rate is ˙𝑛𝐴$\dot n_A$ in mol s−1$^{-1}$ and mass rate is ˙𝑚𝐴 =𝑀𝐴˙𝑛𝐴$\dot m_A=M_A\dot n_A$ in kg s−1$^{-1}$. An average coefficient carries a bar; 𝐴𝑠$A_s$ is transfer area in m2$^2$.

Volatile solid

Air flows at 2 m s⁻¹ over a hypothetical volatile solid A with an exposed area of 10−3 m2$10^{-3}\ \mathrm{m^2}$. The measured sublimation rate is 2.29 ×10−4 mol s−1$2.29\times10^{-4}\ \mathrm{mol\,s^{-1}}$. Both the air and the surface are maintained at 293 K; external heating supplies the latent heat. The pressure is 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$ and incoming air contains negligible A.

For this model solid, use the stipulated equilibrium vapour pressure 𝑝𝐴,𝑠 =4740 Pa$p_{A,s}=4740\ \mathrm{Pa}$ and diffusivity in air 𝐷𝐴𝐵 =1.5 ×10−5 m2 s−1$D_{AB}=1.5\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$. Use air properties 𝜈 =1.55 ×10−5 m2 s−1$\nu=1.55\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$, Pr =0.71$\mathrm{Pr}=0.71$, 𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$ and molar mass 𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$.

Determine the concentration-basis mass transfer coefficient and estimate the heat transfer coefficient using the Chilton-Colburn analogy. This adaptation replaces the inconsistent CO2$_2$ interface state in slide 17.

Solution

Data

Quantity	Value
Exposed area	𝐴𝑠 =10−3 m2$A_s=10^{-3}\ \mathrm{m^2}$
Measured molar rate	˙𝑛𝐴 =2.29 ×10−4 mol s−1$\dot n_A=2.29\times10^{-4}\ \mathrm{mol\,s^{-1}}$
Gas temperature	𝑇 =293 K$T=293\ \mathrm K$
Gas pressure	𝑃 =101300 Pa$P=101300\ \mathrm{Pa}$
Species diffusivity	𝐷𝐴𝐵 =1.5 ×10−5 m2 s−1$D_{AB}=1.5\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Kinematic viscosity	𝜈 =1.55 ×10−5 m2 s−1$\nu=1.55\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface equilibrium pressure	𝑝𝐴,𝑠 =4740 Pa$p_{A,s}=4740\ \mathrm{Pa}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Specific heat capacity	𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Air molar mass	𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$

These values are stipulated for the hypothetical solid and air.

Assumptions

• Steady sublimation over the exposed area.
• Negligible A in the bulk air.
• Ideal gas behaviour and constant supplied properties.
• Low normal mass flux and matching heat and mass transfer conditions.

Calculations

The gas-side surface concentration follows from ideal-gas equilibrium:

𝑐𝐴,𝑠=𝑝𝐴,𝑠𝑅𝑇=47408.314(293)=1.9458 mol m−3$$c_{A,s}=\frac{p_{A,s}}{RT} =\frac{4740}{8.314(293)}=1.9458\ \mathrm{mol\,m^{-3}}$$

The measured rate equals coefficient times area times concentration driving force:

˙𝑛𝐴=𝑘𝑐𝐴𝑠(𝑐𝐴,𝑠−𝑐𝐴,∞)$$\dot n_A=k_cA_s(c_{A,s}-c_{A,\infty})$$

Since the bulk concentration is negligible, solve for 𝑘𝑐$k_c$.

From Eq. 3,

𝑘𝑐=2.29×10−410−3×1.9458=0.1177 m s−1$$k_c=\frac{2.29\times10^{-4}}{10^{-3}\times1.9458} =0.1177\ \mathrm{m\,s^{-1}}$$

The Schmidt number uses momentum and molecular diffusivities:

Sc=𝜈𝐷𝐴𝐵=1.55×10−51.5×10−5=1.0333$$\mathrm{Sc}=\frac{\nu}{D_{AB}} =\frac{1.55\times10^{-5}}{1.5\times10^{-5}}=1.0333$$𝜌=101300×0.0298.314×293$$\rho=\frac{101300\times0.029}{8.314\times293}$$

𝜌=1.2059 kg m−3$$\rho=1.2059\ \mathrm{kg\,m^{-3}}$$(12)

The Chilton-Colburn relation converts the measured mass coefficient to a heat coefficient:

ℎ=𝜌𝐶𝑝𝑘𝑐(ScPr)2/3$$h=\rho C_pk_c\left(\frac{\mathrm{Sc}}{\mathrm{Pr}}\right)^{2/3}$$

Substituting the air properties gives

ℎ=1.2059(1007)(0.1177)(1.03330.71)2/3=183.55 W m−2 K−1$$h=1.2059(1007)(0.1177)\left(\frac{1.0333}{0.71}\right)^{2/3} =183.55\ \mathrm{W\,m^{-2}\,K^{-1}}$$

The surface fraction is about 0.0468, so the low-flux approximation is plausible but not exact.

External heating maintains the specified surface temperature. The estimated heat coefficient describes how this flow would respond to a small temperature difference; it does not supply latent heat when the air and surface are isothermal.

Answer

The concentration-basis mass transfer coefficient is 𝑘𝑐 ≃0.118 m s−1$k_c\simeq0.118\ \mathrm{m\,s^{-1}}$. The corresponding heat transfer coefficient is ℎ ≃184 W m−2 K−1$h\simeq184\ \mathrm{W\,m^{-2}\,K^{-1}}$.

Naphthalene sheet

A stream of air at 100 kPa pressure and 290 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 1 m with a velocity of 20 m/s and surface temperature of 310K. Calculate:

• The average mass transfer coefficient over the flat plate
• The rate of loss of naphthalene from the surface per unit width

Data @ 𝑇𝑓 =300 K$T_f=300\ \mathrm{K}$:

• 𝐷𝐴𝐵 =6 ×10−6 m2/s$D_{AB}=6\times10^{-6}\ \mathrm{m^2/s}$ (naphthalene vapour in air)
• 𝜈 =1.5 ×10−5 m2/s$\nu=1.5\times10^{-5}\ \mathrm{m^2/s}$ (kinematic viscosity of air)
• 𝐶𝐴𝑠 =1 ×10−5 kmol/m3$C_{As}=1\times10^{-5}\ \mathrm{kmol/m^3}$ (conc. of naphthalene at the air-solid interface)

Solution

Figure 3: Air flows along the naphthalene sheet. The velocity boundary layer is the same type as for a heated plate; the concentration boundary layer replaces the thermal boundary layer in the mass transfer analogy.

Use the concentration boundary layer in Figure 2, with the leading edge at the start of the sheet. The one-metre dimension is along the flow; width remains unspecified.

Data

• Sheet length 𝐿 =1 m$L=1\ \mathrm m$

• Air speed 𝑈 =20 m s−1$U=20\ \mathrm{m\,s^{-1}}$

• Film diffusivity 𝐷𝐴𝐵 =6 ×10−6 m2 s−1$D_{AB}=6\times10^{-6}\ \mathrm{m^2\,s^{-1}}$

• Film kinematic viscosity 𝜈 =1.5 ×10−5 m2 s−1$\nu=1.5\times10^{-5}\ \mathrm{m^2\,s^{-1}}$

• Surface concentration

  𝑐𝐴,𝑠=10−5 kmol m−3=0.0100 mol m−3$$c_{A,s}=10^{-5}\ \mathrm{kmol\,m^{-3}}=0.0100\ \mathrm{mol\,m^{-3}}$$

Assumptions

• Steady transfer from a uniform saturated surface.
• Negligible naphthalene in the incoming air.
• Constant properties and negligible normal injection.
• The source estimate treats the boundary layer as turbulent from the leading edge. A separate estimate allows natural transition.

Calculations

Use length-based flow and transfer groups:

Re𝐿=𝑈𝐿𝜈,Sc=𝜈𝐷𝐴𝐵$$\mathrm{Re}_L=\frac{UL}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}$$

The groups are

Re𝐿=201.5×10−5=1.3333×106$$\mathrm{Re}_L=\frac{20}{1.5\times10^{-5}}=1.3333\times10^6$$

and Sc =2.5$\mathrm{Sc}=2.5$.

For the source’s fully turbulent model, first determine the average Sherwood number, then recover the coefficient from

¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿.$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L}.$$

The transfer area per unit width is 𝐿$L$, so the loss per unit width is

˙𝑛𝐴𝑏=¯¯¯¯𝑘𝑐𝐿(𝑐𝐴,𝑠−𝑐𝐴,∞).$$\frac{\dot n_A}{b}=\bar k_cL(c_{A,s}-c_{A,\infty}).$$

Substitution gives

¯¯¯¯¯¯¯Sh𝐿=0.0366Re0.8𝐿Sc1/3$$\overline{\mathrm{Sh}}_L=0.0366\mathrm{Re}_L^{0.8}\mathrm{Sc}^{1/3}$$¯¯¯¯¯¯¯Sh𝐿=3945.3$$\overline{\mathrm{Sh}}_L=3945.3$$

Use the same one-metre length in the dimensional conversion:

¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿=3945.3(6×10−6)1.0=0.02367 m s−1$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L} =\frac{3945.3(6\times10^{-6})}{1.0}=0.02367\ \mathrm{m\,s^{-1}}$$˙𝑛𝐴width=¯¯¯¯𝑘𝑐𝐿𝑐𝐴,𝑠$$\frac{\dot n_A}{\text{width}}=\bar k_c Lc_{A,s}$$

˙𝑛𝐴𝑏=2.367×10−4 mol s−1 m−1$$\frac{\dot n_A}{b}=2.367\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$$(13)

Here 𝑏$b$ denotes sheet width in metres.

The table uses 0.037, which changes this estimate by about one percent.

For an untripped plate, the laminar starting length is

𝑥𝑡𝑟=5×105𝜈𝑈=0.375 m.$$x_{tr}=\frac{5\times10^5\nu}{U}=0.375\,\mathrm{m}.$$

The mixed average relation

¯¯¯¯¯¯¯Sh𝐿=(0.037Re0.8𝐿−871)Sc1/3$$\overline{\mathrm{Sh}}_L=(0.037\mathrm{Re}_L^{0.8}-871)\mathrm{Sc}^{1/3}$$

gives 2806.4, ¯¯¯¯𝑘𝑐 =0.01684$\bar k_c=0.01684$ m s⁻¹ and ˙𝑛𝐴/width =1.684 ×10−4$\dot n_A/\text{width}=1.684\times10^{-4}$ mol s⁻¹ m⁻¹.

This is the preferred estimate when transition occurs naturally at the stated Reynolds number.

Both answers are conditional on the specified flow model; they are not interchangeable rounding choices.

With 𝑀𝐴 =0.128$M_A=0.128$ kg mol⁻¹ the source fully turbulent estimate corresponds to 2.62 kg day⁻¹ per metre of width.

Answer

With the source’s fully turbulent model, ¯¯¯¯𝑘𝑐 =0.0237 m s−1$\bar k_c=0.0237\ \mathrm{m\,s^{-1}}$ and the loss per unit width is 2.37 ×10−4 mol s−1 m−1$2.37\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$. For an untripped plate with natural transition, the preferred estimates are 0.0168 m s−1$0.0168\ \mathrm{m\,s^{-1}}$ and 1.68 ×10−4 mol s−1 m−1$1.68\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$.

Wet-bulb thermometer

Dry air at 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$ blows across a thermometer whose bulb has been covered with a dampened wick. The classical “wet bulb” thermometer indicates a steady-state temperature is reached by a small amount of liquid water evaporating into a large reservoir of unsaturated vapour-gas mixture. The thermometer reads 290 K. At this temperature, the following properties were evaluated:

Property	Value
Vapour pressure of water	1.94 ×103 Pa$1.94\times10^3\ \mathrm{Pa}$
Density of air	1.219 kg/m3$^3$
Latent heat of vaporisation of water	2461 kJ/kg
Prandtl number	0.71
Schmidt number	0.61
Specific heat of air	1.006 kJ/kg.K

What is the temperature of the dry air?

Solution

Data

Quantity	Value
Wet-bulb temperature	𝑇𝑠 =290 K$T_s=290\ \mathrm K$
Surface vapour pressure	𝑝𝐴,𝑠 =1940 Pa$p_{A,s}=1940\ \mathrm{Pa}$
Air density	𝜌 =1.219 kg m−3$\rho=1.219\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1006 J kg−1 K−1$C_p=1006\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Schmidt number	Sc =0.61$\mathrm{Sc}=0.61$
Latent heat	𝜆 =2.461 ×106 J kg−1$\lambda=2.461\times10^6\ \mathrm{J\,kg^{-1}}$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

Assumptions

• Steady wick temperature and evaporation rate.
• Dry incoming air and saturated gas at the wet surface.
• Radiation and conduction through the thermometer stem are neglected.
• Sensible energy transport by injected vapour and blowing are neglected.
• Constant properties and matching heat and mass transfer conditions.

Calculations

At steady state convection supplies evaporation’s latent heat.

With zero incoming vapour,

ℎ(𝑇∞−𝑇𝑠)=𝜆𝑀𝐴𝑘𝑐𝑐𝐴,𝑠$$h(T_\infty-T_s)=\lambda M_A k_c c_{A,s}$$

𝑘𝑐ℎ=1𝜌𝐶𝑝(PrSc)2/3$$\frac{k_c}{h}=\frac1{\rho C_p} \left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3}$$(14)

The ratio is 9.023 ×10−4$9.023\times10^{-4}$ m3$^3$ K/J, and

𝑐𝐴,𝑠=19408.314×290=0.80465 mol m−3.$$c_{A,s}=\frac{1940}{8.314\times290}=0.80465\,\mathrm{mol\,m^{-3}}.$$

The energy balance gives the temperature rise above the wet bulb:

𝑇∞−𝑇𝑠=𝜆𝑀𝐴𝑐𝐴,𝑠𝑘𝑐ℎ$$T_\infty-T_s=\lambda M_Ac_{A,s}\frac{k_c}{h}$$

Substituting the latent heat, concentration and coefficient ratio gives

𝑇∞−𝑇𝑠=2.461×106(0.018)(0.80465)(9.023×10−4)=32.161 K$$T_\infty-T_s=2.461\times10^6(0.018)(0.80465)(9.023\times10^{-4}) =32.161\ \mathrm{K}$$

Adding this temperature difference to the wet-bulb reading gives

𝑇∞=290+32.161=322.16 K$$T_\infty=290+32.161=322.16\ \mathrm{K}$$

Velocity cancels because the same flow affects both coefficients.

This cancellation holds under the stated analogy and energy assumptions; it does not remove ventilation requirements or radiative errors of a real thermometer.

Answer

The estimated dry-air temperature is 𝑇∞ =322.2 K$T_\infty=322.2\ \mathrm{K}$, approximately 49.0 °C$49.0\,\degree\mathrm{C}$.

Heat and mass analogy

Water evaporates from a wetted surface of rectangular shape, with an air stream (1 atm, 35 °C$35\ \degree\mathrm{C}$) blown over the surface at a velocity of 15 m/s. Heat transfer measurements indicate that for air at 1 atm and 35 °C$35\ \degree\mathrm{C}$, the average heat transfer coefficient is given by the following empirical relation:

⟨ℎ⟩=21𝑉0.6∞$$\langle h\rangle=21V_\infty^{0.6}$$

where ⟨ℎ⟩$\langle h\rangle$ is in W/(m2$^2$ K) and 𝑉∞$V_\infty$ is in m/s. Estimate the mass transfer coefficient and evaporation rate for an area of 1.5 m2$^2$. Assume dry incoming air and a surface maintained at 35 °C$35\ \degree\mathrm C$ by external heating. Use 𝑝sat𝐴 =5624 Pa$p_A^{\mathrm{sat}}=5624\ \mathrm{Pa}$ at 𝑇𝑠 =308.15 K$T_s=308.15\ \mathrm K$ and the following rounded, stipulated transport properties: 𝜌 =1.128 kg/m3$\rho=1.128\ \mathrm{kg/m^3}$, 𝐶𝑝 =1007 J/(kg K)$C_p=1007\ \mathrm{J/(kg\,K)}$, 𝜈 =15.89 ×10−6 m2/s$\nu=15.89\times10^{-6}\ \mathrm{m^2/s}$, 𝐷𝐴𝐵 =2.6 ×10−5 m2/s$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2/s}$ and Pr =0.71$\mathrm{Pr}=0.71$.

Solution

Data

Quantity	Value
Air speed	𝑈 =15 m s−1$U=15\ \mathrm{m\,s^{-1}}$
Wet area	𝐴𝑠 =1.5 m2$A_s=1.5\ \mathrm{m^2}$
Supplied heat correlation	¯¯¯¯ℎ =21𝑈0.6$\bar h=21U^{0.6}$
Air density	𝜌 =1.128 kg m−3$\rho=1.128\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Kinematic viscosity	𝜈 =15.89 ×10−6 m2 s−1$\nu=15.89\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.6 ×10−5 m2 s−1$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Surface saturation pressure	𝑝sat𝐴 =5624 Pa$p_A^{\mathrm{sat}}=5624\ \mathrm{Pa}$
Surface temperature	𝑇𝑠 =308.15 K$T_s=308.15\ \mathrm K$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

The transport properties are rounded model inputs. The surface saturation pressure is consistent with 35 degrees Celsius using the NIST Antoine correlation.

Assumptions

• Steady transfer from a uniformly wet surface.
• Dry incoming air.
• The surface is maintained at the stated air temperature.
• Constant supplied properties.
• Heat and mass transfer share the same flow and surface conditions.
• Normal evaporation flow is small.

Calculations

The measurement gives

¯¯¯¯ℎ=21×150.6=106.63 W m−2 K−1.$$\bar h=21\times15^{0.6}=106.63\,\mathrm{W\,m^{-2}\,K^{-1}}.$$

Compute Sc =0.61115$\mathrm{Sc}=0.61115$.

The matching-flow Chilton-Colburn relation is

¯¯¯¯𝑘𝑐=¯¯¯¯ℎ𝜌𝐶𝑝(PrSc)2/3$$\bar k_c=\frac{\bar h}{\rho C_p}\left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3}$$

Substituting the coefficient and property values gives

¯¯¯¯𝑘𝑐=106.631.128×1007(0.710.61115)2/3¯¯¯¯𝑘𝑐=0.10374 m s−1$$\begin{gathered} \bar k_c=\frac{106.63}{1.128\times1007} \left(\frac{0.71}{0.61115}\right)^{2/3} \\ \bar k_c=0.10374\ \mathrm{m\,s^{-1}} \end{gathered}$$(15)

At 308.15 K,

𝑐𝐴,𝑠=56248.314×308.15=2.1952 mol m−3.$$c_{A,s}=\frac{5624}{8.314\times308.15}=2.1952\,\mathrm{mol\,m^{-3}}.$$

The evaporation rate is coefficient times area times concentration difference:

˙𝑛𝐴=¯¯¯¯𝑘𝑐𝐴𝑠(𝑐𝐴,𝑠−𝑐𝐴,∞)$$\dot n_A=\bar k_cA_s(c_{A,s}-c_{A,\infty})$$

With dry bulk air, substitution gives

˙𝑛𝐴=0.10374(1.5)(2.1952)=0.3416 mol s−1$$\dot n_A=0.10374(1.5)(2.1952)=0.3416\ \mathrm{mol\,s^{-1}}$$

Convert the molar rate to mass rate:

˙𝑚𝐴=𝑀𝐴˙𝑛𝐴=0.018(0.3416)=0.00615 kg s−1=22.1 kg h−1$$\dot m_A=M_A\dot n_A=0.018(0.3416) =0.00615\ \mathrm{kg\,s^{-1}}=22.1\ \mathrm{kg\,h^{-1}}$$

Dryness and maintained surface temperature are explicit additions to the source question.

At 50 percent relative humidity with equal surface and ambient temperature the driving force and rate halve.

The saturated surface fraction is about 0.0555, exceeding the source’s 0.05 rule of thumb; a blowing correction may matter.

Maintaining the warm surface requires heat input.

This is not an adiabatic wet-bulb calculation.

Answer

Using the supplied additional properties, ¯¯¯¯𝑘𝑐 ≃0.104 m s−1$\bar k_c\simeq0.104\ \mathrm{m\,s^{-1}}$ and

˙𝑚𝐴≃0.00615 kg s−1=22.1 kg h−1.$$\dot m_A\simeq0.00615\ \mathrm{kg\,s^{-1}}=22.1\ \mathrm{kg\,h^{-1}}.$$

These estimates require the stated dry-air and maintained-surface-temperature assumptions.

Annular transfer

The following correlation has been proposed to characterise the heat transfer of fluids flowing in the annular section between concentric tubes:

Nu𝑑1=ℎ𝑑1𝑘=𝐶Re0.8𝑑1Pr1/3(𝜇𝜇𝑠)0.14(𝑑2𝑑1)0.15$$\mathrm{Nu}_{d_1}=\frac{hd_1}{k} =C\mathrm{Re}_{d_1}^{0.8}\mathrm{Pr}^{1/3} \left(\frac{\mu}{\mu_s}\right)^{0.14} \left(\frac{d_2}{d_1}\right)^{0.15}$$(16)

Show how this equation can be used to predict the mass-transfer coefficient from a naphthalene rod that is used to form the inner tube of an annular duct. 𝑑1$d_1$ and 𝑑2$d_2$ are the diameters of the inner and outer tubes, respectively, whereas 𝜇$\mu$ and 𝜇𝑠$\mu_s$ are the viscosity of the fluid at the bulk and surface temperatures.

Solution

Data

• Inner diameter 𝑑1$d_1$
• Outer diameter 𝑑2$d_2$
• Reynolds number based on inner diameter Re𝑑1$\mathrm{Re}_{d_1}$
• Empirical correlation constant 𝐶$C$

No numerical geometry, velocity or value of the correlation constant is supplied. Here the symbol for the correlation constant does not denote concentration.

Assumptions

• The heat and mass problems have matching geometry and flow regime.
• Uniform surface concentration corresponds to the heat correlation’s wall condition.
• Low mass flux leaves the velocity field essentially unchanged.
• The gas is dilute and isothermal, so wall and bulk viscosities are approximately equal.
• The translated correlation is used within the applicable Reynolds and Schmidt ranges.

Calculations

Keep the same geometry, Reynolds definition and experimental range.

Replace Nusselt by Sherwood and Prandtl by Schmidt.

At isothermal dilute conditions 𝜇𝑠 ≈𝜇$\mu_s\approx\mu$, giving

Sh𝑑1=𝐶Re0.8𝑑1Sc1/3(𝑑2𝑑1)0.15$$\mathrm{Sh}_{d_1}=C\mathrm{Re}_{d_1}^{0.8}\mathrm{Sc}^{1/3}\left(\frac{d_2}{d_1}\right)^{0.15}$$

𝑘𝑐=𝐷𝐴𝐵𝑑1Sh𝑑1$$k_c=\frac{D_{AB}}{d_1}\mathrm{Sh}_{d_1}$$(17)

The viscosity factor can be retained if wall temperature or composition changes viscosity appreciably, but the heat correlation alone does not validate a mass-composition correction.

The Reynolds and diameter-ratio factors describe flow and geometry, so they remain unchanged. The viscosity correction becomes unity under the stated isothermal dilute approximation.

Answer

For the stated isothermal, dilute model,

𝑘𝑐=𝐷𝐴𝐵𝑑1𝐶Re0.8𝑑1Sc1/3(𝑑2𝑑1)0.15$$k_c=\frac{D_{AB}}{d_1}C\mathrm{Re}_{d_1}^{0.8}\mathrm{Sc}^{1/3} \left(\frac{d_2}{d_1}\right)^{0.15}$$

The answer remains symbolic because the required numerical data are not supplied.

7 Example: concentration development in a pipe

At a transferring pipe wall the concentration layer grows inward until it reaches the centre. Fully developed concentration means that the normalized profile no longer changes, even while the bulk concentration continues changing along the pipe. Hydrodynamic, thermal and concentration development have different lengths. In laminar flow the estimates are 𝐿ℎ ≈0.05Re 𝑑$L_h\approx0.05\mathrm{Re}\,d$, 𝐿𝑡 ≈Pr𝐿ℎ$L_t\approx\mathrm{Pr}L_h$ and 𝐿𝑐 ≈Sc𝐿ℎ$L_c\approx\mathrm{Sc}L_h$, with 𝑑$d$ the pipe diameter. At Re =1000$\mathrm{Re}=1000$, Sc =600$\mathrm{Sc}=600$ and 𝑑 =0.01$d=0.01$ m, 𝐿𝑐 ≈300$L_c\approx300$ m. A short pipe then needs an entrance-region correlation. In turbulent flow, development is often of order ten diameters; check it rather than inferring development solely from Reynolds number.

Figure 4: Water enters through a wet pipe wall at constant wall concentration. The separate schematic graph shows bulk concentration rising from its dry-inlet value toward the wall concentration as distance downstream increases.

Water containing a slowly diffusing solute flows laminarly through a pipe with Sc =600$\mathrm{Sc}=600$. Using the entry-length estimates above, how does the concentration entry length compare with the hydrodynamic entry length?

• It is approximately 600 times shorter.
• It is approximately the same length.
• It is approximately 600 times longer. ✓
• It is approximately 24 times longer.

Explanation

The laminar entry-length estimates give

𝐿𝑐𝐿ℎ≈Sc=600$$\frac{L_c}{L_h}\approx\mathrm{Sc}=600$$

The concentration profile needs about 600 times as much distance to develop as the velocity profile. A developed velocity profile therefore does not guarantee a developed concentration profile.

7.1 Translating a heat correlation

Use 𝛼 →𝐷𝐴𝐵$\alpha\to D_{AB}$, Pr →Sc$\mathrm{Pr}\to\mathrm{Sc}$ and Nu →Sh$\mathrm{Nu}\to\mathrm{Sh}$ with the same geometry, wall condition and flow regime. The transfer must not appreciably modify the velocity field. The carrier fluid’s properties approximate mixture properties only at dilution.

A heat correlation assumes uniform wall temperature and negligible normal injection. Which mass-transfer situation most closely matches its assumptions?

• Only parts of the surface are wet, and strong flow out of the surface is ignored.
• The wall concentration is uniform, flow through the wall is negligible, and the geometry and flow match. ✓
• Any reacting porous bed with unknown wetted area.
• Any surface, provided Nusselt is renamed Sherwood.

Explanation

The analogy requires matching geometry, flow and boundary conditions as well as corresponding transport properties. Uniform temperature maps to uniform concentration. Strong injection changes the velocity field and needs a model that includes it.

7.2 Natural convection and buoyancy

Mass Grashof number

The composition-buoyancy group

Gr𝐷=𝑔∣Δ𝜌𝐷∣𝐿3𝜌𝜈2,$$\mathrm{Gr}_D=\frac{g|\Delta\rho_D|L^3}{\rho\nu^2},$$

where the density difference is due only to composition at a common temperature and pressure. Use the correlation’s reference density and characteristic length.

Small Gr𝐷$\mathrm{Gr}_D$ means weak composition buoyancy on the viscous scale; large values mean stronger natural motion is possible. The water-pan example below calculates Gr𝐷$\mathrm{Gr}_D$ from density and uses 𝐿 =𝐴𝑠/𝑃𝑠$L=A_s/P_s$. Using pan diameter instead makes its value 64 times too large. As with thermal Grashof number, the cube of length makes that choice especially important. The sign and orientation of the density change decide whether buoyancy assists or opposes flow; a magnitude alone does not.

Mass Rayleigh number

The composition natural-convection group

Ra𝐷=Gr𝐷Sc=𝑔∣Δ𝜌𝐷∣𝐿3𝜌𝜈𝐷𝐴𝐵.$$\mathrm{Ra}_D=\mathrm{Gr}_D\mathrm{Sc} =\frac{g|\Delta\rho_D|L^3}{\rho\nu D_{AB}}.$$

It combines composition buoyancy, momentum diffusion and species diffusion, using the same reference state and length in both factors.

Small Ra𝐷$\mathrm{Ra}_D$ supports a diffusion-dominated picture on the model scale; larger values can support stronger convective transfer when the density arrangement drives motion. Check the geometry and stability before using a correlation. The water-pan example gives Ra𝐷 =9.96 ×104$\mathrm{Ra}_D=9.96\times10^4$, within its stated upward-facing surface range. Multiplying Gr𝐷$\mathrm{Gr}_D$ by Pr instead of Sc would give the wrong diffusion comparison. Thermal Rayleigh number uses Pr.

Density differences generate natural flow. Water vapour is lighter than air at the same temperature and pressure, so composition alone can make air above water buoyant. Thermal expansion can reinforce or oppose that change. The thermal and composition Grashof magnitudes are

Gr𝑇=𝑔𝛽∣𝑇𝑠−𝑇∞∣𝐿3𝜈2$$\mathrm{Gr}_T=\frac{g\beta|T_s-T_\infty|L^3}{\nu^2}$$

Gr𝐷=𝑔∣Δ𝜌𝐷∣𝐿3𝜌𝜈2$$\mathrm{Gr}_D=\frac{g|\Delta\rho_D|L^3}{\rho\nu^2}$$(18)

Here 𝑔$g$ is gravity in m s⁻², 𝛽$\beta$ the expansion coefficient in K−1$^{-1}$, and Δ𝜌𝐷$\Delta\rho_D$ the density difference caused only by composition, evaluated at a common temperature. Both groups are dimensionless and use 𝐿3$L^3$. A total density difference between states at different temperatures already includes thermal buoyancy and must not have a thermal group added again. For small density variations one can add signed buoyancy contributions referenced to the same state; opposing forces require a suitable coupled correlation. Heat and concentration profiles also have different diffusivities, so adding groups alone does not establish an exact combined Sherwood law. If a thermal natural-convection correlation supplies ℎ$h$, Eq. 11 gives a conditional mass estimate. If composition drives flow, use an appropriate Sherwood correlation directly.

A measured density difference already includes both temperature and composition changes between two states. How should it be used in buoyancy modelling?

• Add a thermal Grashof contribution to it automatically.
• Discard composition effects whenever temperature differs.
• Replace the density difference by the species diffusivity.
• Use the measured total density difference once, including both temperature and composition effects. ✓

Explanation

A total density difference already includes thermal buoyancy. Adding a separate thermal contribution would count it again. If separating mechanisms, reference them to the same state and retain their signs because they can reinforce or oppose each other.

7.3 Correlations used in classroom examples 6 and 7

For developed turbulent pipe flow with fixed wall concentration, the slide correlation is

Sh=0.023Re0.8Sc0.4,$$\mathrm{Sh}=0.023\mathrm{Re}^{0.8}\mathrm{Sc}^{0.4},$$

with diameter in Re and Sh, Re >104$\mathrm{Re}>10^4$ and 0.7 <Sc <160$0.7<\mathrm{Sc}<160$.

For the upward-facing pan with light fluid above the surface,

Sh=0.54(Gr𝐷Sc)1/4$$\mathrm{Sh}=0.54(\mathrm{Gr}_D\mathrm{Sc})^{1/4}$$

over 104 <Gr𝐷Sc <107$10^4<\mathrm{Gr}_D\mathrm{Sc}<10^7$. Use 𝐿 =𝐴𝑠/𝑃𝑠$L=A_s/P_s$, where 𝑃𝑠$P_s$ is the perimeter. The worked examples check these ranges.

8 Classroom example: internal pipe flow

Wet felt pipe

Dry air (310 K, 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$, 1.5 m/s) is flowing through a pipe lined with a felt material that is continuously saturated with water at 290 K. The pipe is 0.15 m in diameter and 6 m long. The saturated felt material has a roughness index, 𝑒/𝐷$e/D$, of 0.00001. Assuming that the temperature of the air and pipe wall is constant, determine the rate at which water must be added to keep the felt continuously saturated.

Solution

The inlet, outlet and wetted wall are shown in Figure 4. Concentration rises along the pipe, so the downstream driving force is smaller than the inlet driving force.

Data

Quantity	Value
Pipe diameter	𝑑 =0.15 m$d=0.15\ \mathrm m$
Pipe length	𝐿 =6 m$L=6\ \mathrm m$
Bulk air speed	𝑈 =1.5 m s−1$U=1.5\ \mathrm{m\,s^{-1}}$
Dry inlet temperature	𝑇𝑖 =310 K$T_i=310\ \mathrm K$
Wet wall temperature	𝑇𝑠 =290 K$T_s=290\ \mathrm K$
Surface vapour pressure	𝑝𝐴,𝑠 =1917 Pa$p_{A,s}=1917\ \mathrm{Pa}$
Kinematic viscosity	𝜈 =1.569 ×10−5 m2 s−1$\nu=1.569\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.626 ×10−5 m2 s−1$D_{AB}=2.626\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Fanning friction factor	𝐶𝑓 =0.007$C_f=0.007$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

The vapour pressure, transport properties and friction factor come from the lecturer’s notes.

Assumptions

• Steady flow and evaporation.
• Constant bulk velocity and supplied properties.
• Uniform saturated wall concentration.
• Each cross-section has a representative mixed bulk concentration.
• Axial diffusion and normal-flow effects are neglected.
• A developed-flow average coefficient approximates the pipe transfer.

Calculations

The surface concentration is

𝑐𝐴,𝑠=19178.314×290=0.79508 mol m−3,$$c_{A,s}=\frac{1917}{8.314\times290}=0.79508\,\mathrm{mol\,m^{-3}},$$

and 𝑐𝐴,𝑖 =0$c_{A,i}=0$.

The groups are

Re𝑑=1.5×0.15𝜈=14340$$\mathrm{Re}_d=\frac{1.5\times0.15}{\nu}=14340$$

and Sc =0.59749$\mathrm{Sc}=0.59749$.

The pipe spans 40 diameters; the roughly ten-diameter turbulent development estimate motivates a developed average approximation, not exact development at the inlet.

The concentration model treats the maintained wall and gas temperatures through the supplied surface concentration and constant properties. It does not solve the axial thermal field. The friction analogy gives the coefficient as

𝑘𝑐=𝐶𝑓𝑈2Sc2/3=0.0074007 m s−1$$k_c=\frac{C_f U}{2\mathrm{Sc}^{2/3}} =0.0074007\ \mathrm{m\,s^{-1}}$$(19)

The source table’s Dittus-Boelter-type mass row gives about 0.00688 m s⁻¹, but Sc =0.5975$\mathrm{Sc}=0.5975$ falls below its stated 0.7 limit.

That comparison is an extrapolation and is not the governing answer.

The primary friction estimate also sits at the approximate lower edge of the source’s Chilton-Colburn range; treat the result as an engineering estimate.

For a slice 𝑑𝑥$dx$, steady balance is inflow minus outflow plus wall input equals zero.

Inflow is 𝑈𝐴𝑐𝑐𝐴(𝑥)$U A_c c_A(x)$, outflow is 𝑈𝐴𝑐𝑐𝐴(𝑥 +𝑑𝑥)$U A_c c_A(x+dx)$, and wall input is

𝑘𝑐𝜋𝑑 𝑑𝑥(𝑐𝐴,𝑠−𝑐𝐴),$$k_c\pi d\,dx(c_{A,s}-c_A),$$

where 𝐴𝑐 =𝜋𝑑2/4$A_c=\pi d^2/4$.

Dividing and taking the limit yields

𝑈𝑑4𝑑𝑐𝐴𝑑𝑥=𝑘𝑐(𝑐𝐴,𝑠−𝑐𝐴)$$\frac{Ud}{4}\frac{dc_A}{dx}=k_c(c_{A,s}-c_A)$$(20)

Integrate from inlet to exit.

For this circular pipe the Number of transfer units is

𝒩=4𝑘𝑐𝐿𝑈𝑑=𝑘𝑐𝐴𝑠𝑄,$$\mathcal N=\frac{4k_cL}{Ud}=\frac{k_cA_s}{Q},$$

using 𝐴𝑠 =𝜋𝑑𝐿$A_s=\pi dL$ and 𝑄 =𝑈𝜋𝑑2/4$Q=U\pi d^2/4$. Small 𝒩$\mathcal N$ leaves the gas close to its inlet composition; large 𝒩$\mathcal N$ brings it towards the maintained wall value. For 𝑘𝑐 =0.01 m s−1$k_c=0.01\ \mathrm{m\,s^{-1}}$, 𝐿 =1 m$L=1\ \mathrm m$, 𝑈 =2 m s−1$U=2\ \mathrm{m\,s^{-1}}$ and 𝑑 =0.02 m$d=0.02\ \mathrm m$, 𝒩 =1$\mathcal N=1$ and the remaining concentration difference is 𝑒−1 =0.368$e^{-1}=0.368$ of its inlet value. Omitting the factor four would underpredict the contact effect.

Then

𝑐𝐴,𝑠−𝑐𝐴,𝑒𝑐𝐴,𝑠−𝑐𝐴,𝑖=𝑒−𝒩𝒩=0.78941𝑐𝐴,𝑒=0.4340 mol m−3$$\begin{gathered} \frac{c_{A,s}-c_{A,e}}{c_{A,s}-c_{A,i}}=e^{-\mathcal N}\\ \mathcal N=0.78941\\ c_{A,e}=0.4340\ \mathrm{mol\,m^{-3}} \end{gathered}$$(21)

For the dry inlet, the integrated balance can be evaluated directly:

𝒩=4(0.0074007)(6)1.5(0.15)=0.78941$$\mathcal N=\frac{4(0.0074007)(6)}{1.5(0.15)}=0.78941$$𝑐𝐴,𝑒=𝑐𝐴,𝑠−(𝑐𝐴,𝑠−𝑐𝐴,𝑖)𝑒−𝒩=0.79508(1−𝑒−0.78941)=0.4340 mol m−3$$c_{A,e}=c_{A,s}-(c_{A,s}-c_{A,i})e^{-\mathcal N} =0.79508(1-e^{-0.78941})=0.4340\ \mathrm{mol\,m^{-3}}$$

The exit is about 54.6 percent of surface saturation on this concentration basis.

Let Δ𝑐𝑖 =𝑐𝐴,𝑠 −𝑐𝐴,𝑖$\Delta c_i=c_{A,s}-c_{A,i}$ and Δ𝑐𝑒 =𝑐𝐴,𝑠 −𝑐𝐴,𝑒$\Delta c_e=c_{A,s}-c_{A,e}$.

The logarithmic mean is

Δ𝑐𝑙𝑚=Δ𝑐𝑖−Δ𝑐𝑒ln⁡(Δ𝑐𝑖Δ𝑐𝑒)Δ𝑐𝑙𝑚=0.54981 mol m−3$$\begin{gathered} \Delta c_{lm}=\frac{\Delta c_i-\Delta c_e}{\ln\left(\frac{\Delta c_i}{\Delta c_e}\right)}\\ \Delta c_{lm}=0.54981\ \mathrm{mol\,m^{-3}} \end{gathered}$$(22)

It is the exact mean driving force for this exponential model.

At equal end differences its limiting value is that common difference.

Apply the mean driving force over wall area 𝜋𝑑𝐿$\pi dL$:

˙𝑛𝐴=𝑘𝑐𝜋𝑑𝐿Δ𝑐𝑙𝑚=0.0074007𝜋(0.15)(6)(0.54981)=0.01150 mol s−1$$\dot n_A=k_c\pi dL\Delta c_{lm} =0.0074007\pi(0.15)(6)(0.54981) =0.01150\ \mathrm{mol\,s^{-1}}$$

An independent overall balance gives

˙𝑛𝐴=𝑈𝐴𝑐(𝑐𝐴,𝑒−𝑐𝐴,𝑖)=1.5𝜋(0.15)24(0.4340−0)=0.01150 mol s−1$$\dot n_A=UA_c(c_{A,e}-c_{A,i}) =1.5\frac{\pi(0.15)^2}{4}(0.4340-0) =0.01150\ \mathrm{mol\,s^{-1}}$$

Agreement confirms that the wall-transfer calculation and outlet concentration satisfy the same material balance.

Thus the final replenishment rate is

˙𝑚𝐴=0.018˙𝑛𝐴=2.07×10−4 kg s−1,$$\dot m_A=0.018\dot n_A=2.07\times10^{-4}\,\mathrm{kg\,s^{-1}},$$

or 0.745 kg h⁻¹.

Answer

The required water replenishment rate is approximately 2.07 ×10−4 kg s−1$2.07\times10^{-4}\ \mathrm{kg\,s^{-1}}$, or 0.745 kg h−1$0.745\ \mathrm{kg\,h^{-1}}$. The exit bulk concentration is 0.434 mol m−3$0.434\ \mathrm{mol\,m^{-3}}$.

9 Classroom example: natural convection

Warm water pan

A 225 mm diameter pan of water is maintained at 37 °C$37\ \degree\mathrm{C}$ is exposed to a dry, still atmosphere at 17 °C$17\ \degree\mathrm{C}$. Estimate the heat and mass transfer coefficients, as well as the evaporation rate from the pan,

a. Assuming the convection is driven by species transfer; b. Assuming that the convection is driven by temperature gradients.

Solution

Data

Quantity	Value
Pan diameter	𝑑 =0.225 m$d=0.225\ \mathrm m$
Rounded surface temperature	𝑇𝑠 =310 K$T_s=310\ \mathrm K$
Rounded air temperature	𝑇∞ =290 K$T_\infty=290\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Gravitational acceleration	𝑔 =9.81 m s−2$g=9.81\ \mathrm{m\,s^{-2}}$
Kinematic viscosity	𝜈 =15.89 ×10−6 m2 s−1$\nu=15.89\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.6 ×10−5 m2 s−1$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Prandtl number	Pr =0.707$\mathrm{Pr}=0.707$
Specific heat capacity	𝐶𝑝 =1006 J kg−1 K−1$C_p=1006\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Air conductivity	𝑘 =0.0263 W m−1 K−1$k=0.0263\ \mathrm{W\,m^{-1}\,K^{-1}}$
Film density	𝜌 =1.1769 kg m−3$\rho=1.1769\ \mathrm{kg\,m^{-3}}$
Expansion coefficient	𝛽 =1/300 K−1$\beta=1/300\ \mathrm{K^{-1}}$
Surface saturation pressure	𝑝sat𝐴 =6220 Pa$p_A^{\mathrm{sat}}=6220\ \mathrm{Pa}$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$
Air molar mass	𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$

Use the source’s rounded temperatures and the supplementary properties from the lecturer’s notes.

Assumptions

• Steady natural convection above an upward-facing wet surface.
• Dry bulk air and saturated gas at the water surface.
• Uniform maintained surface temperature.
• Constant supplied film properties and ideal gas behaviour.
• Low normal evaporation flux.
• Composition buoyancy and thermal buoyancy are considered separately, as requested.

Calculations

Area is

𝐴𝑠=𝜋(0.225)24=0.039761 m2$$A_s=\frac{\pi(0.225)^2}{4}=0.039761\,\mathrm{m^2}$$

and horizontal characteristic length is

𝐿=𝐴𝑠𝑃𝑠=𝑑4=0.05625 m.$$L=\frac{A_s}{P_s}=\frac{d}{4}=0.05625\,\mathrm{m}.$$

Using diameter instead gives a Grashof number 64 times too large.

The surface water mass concentration is

𝜌𝐴,𝑠=𝑝𝐴,𝑠𝑀𝐴𝑅𝑇𝑠=0.043440 kg m−3$$\rho_{A,s}=\frac{p_{A,s}M_A}{RT_s}=0.043440\,\mathrm{kg\,m^{-3}}$$

and bulk water concentration is zero.

The Schmidt number is 0.61115.

For composition alone compare dry and saturated air at the same 310 K.

The densities are

𝜌𝐵,310=𝑃𝑀𝐵𝑅𝑇𝑠=1.1401$$\rho_{B,310}=\frac{PM_B}{RT_s}=1.1401$$

and

𝜌𝑠,310=𝜌𝐴,𝑠+𝜌𝐵,𝑠$$\rho_{s,310}=\rho_{A,s}+\rho_{B,s}$$𝜌𝐴,𝑠=6220(0.018)8.314(310)$$\rho_{A,s}=\frac{6220(0.018)}{8.314(310)}$$𝜌𝐵,𝑠=(101325−6220)(0.029)8.314(310)$$\rho_{B,s}=\frac{(101325-6220)(0.029)}{8.314(310)}$$

𝜌𝑠,310=1.1136 kg m−3$$\rho_{s,310}=1.1136\ \mathrm{kg\,m^{-3}}$$(23)

Thus Δ𝜌𝐷 =0.02655$\Delta\rho_D=0.02655$ kg m⁻³.

Use their mean density 1.12685 kg m⁻³ in Eq. 18.

The composition Grashof number is

Gr𝐷=𝑔Δ𝜌𝐷𝐿3𝜌𝜈2$$\mathrm{Gr}_D=\frac{g\Delta\rho_DL^3}{\rho\nu^2}$$

Substituting the same-temperature density difference gives

Gr𝐷=9.81(0.02655)(0.05625)31.12685(15.89×10−6)2=1.6291×105$$\mathrm{Gr}_D=\frac{9.81(0.02655)(0.05625)^3} {1.12685(15.89\times10^{-6})^2} =1.6291\times10^5$$

The composition Rayleigh number is Ra𝐷 =Gr𝐷Sc =9.9562 ×104$\mathrm{Ra}_D=\mathrm{Gr}_D\mathrm{Sc}=9.9562\times10^4$, within the specified horizontal-plate range.

The correlation gives

Sh=0.54Ra1/4𝐷=0.54(9.9562×104)1/4=9.5922$$\mathrm{Sh}=0.54\mathrm{Ra}_D^{1/4}=0.54(9.9562\times10^4)^{1/4}=9.5922$$𝑘𝑐=Sh𝐷𝐴𝐵𝐿=9.5922(2.6×10−5)0.05625=0.0044337 m s−1$$k_c=\frac{\mathrm{Sh}D_{AB}}L=\frac{9.5922(2.6\times10^{-5})}{0.05625}=0.0044337\ \mathrm{m\,s^{-1}}$$˙𝑚𝐴=𝑘𝑐𝐴𝑠𝜌𝐴,𝑠=0.0044337(0.039761)(0.043440)$$\dot m_A=k_cA_s\rho_{A,s}=0.0044337(0.039761)(0.043440)$$

˙𝑚𝐴=7.6580×10−6 kg s−1$$\dot m_A=7.6580\times10^{-6}\ \mathrm{kg\,s^{-1}}$$(24)

The composition-only estimate is 27.6 g h⁻¹.

Under the matching-flow transfer analogy, the corresponding heat coefficient is

ℎ=𝜌𝐶𝑝𝑘𝑐(ScPr)2/3=1.1769(1006)(0.0044337)(0.611150.707)2/3≃4.76 W m−2 K−1$$h=\rho C_pk_c\left(\frac{\mathrm{Sc}}{\mathrm{Pr}}\right)^{2/3} =1.1769(1006)(0.0044337) \left(\frac{0.61115}{0.707}\right)^{2/3} \simeq4.76\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For thermal buoyancy alone, use the thermal form of Eq. 18:

Gr𝑇=𝑔𝛽(𝑇𝑠−𝑇∞)𝐿3𝜈2$$\mathrm{Gr}_T=\frac{g\beta(T_s-T_\infty)L^3}{\nu^2}$$

Substitution gives

Gr𝑇=9.81(1300)(310−290)(0.05625)3(15.89×10−6)2=4.6100×105$$\mathrm{Gr}_T=\frac{9.81\left(\frac{1}{300}\right)(310-290)(0.05625)^3} {(15.89\times10^{-6})^2}=4.6100\times10^5$$

The thermal Rayleigh number is Ra𝑇 =Gr𝑇Pr =3.2593 ×105$\mathrm{Ra}_T=\mathrm{Gr}_T\mathrm{Pr}=3.2593\times10^5$. The matching heat correlation gives

Nu=0.54Ra1/4𝑇=0.54(3.2593×105)1/4=12.903$$\mathrm{Nu}=0.54\mathrm{Ra}_T^{1/4} =0.54(3.2593\times10^5)^{1/4}=12.903$$

Recover the heat transfer coefficient using the thermal conductivity:

ℎ=Nu𝑘𝐿=12.903(0.0263)0.05625=6.033 W m−2 K−1$$h=\frac{\mathrm{Nu}k}{L} =\frac{12.903(0.0263)}{0.05625} =6.033\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For the thermally driven flow, convert ℎ$h$ to a mass coefficient:

𝑘𝑐=ℎ𝜌𝐶𝑝(PrSc)2/3=6.0331.1769(1006)(0.7070.61115)2/3=0.005615 m s−1$$k_c=\frac{h}{\rho C_p}\left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3} =\frac{6.033}{1.1769(1006)} \left(\frac{0.707}{0.61115}\right)^{2/3} =0.005615\ \mathrm{m\,s^{-1}}$$

The evaporation rate then follows from the surface-to-bulk mass-concentration difference:

˙𝑚𝐴=𝑘𝑐𝐴𝑠(𝜌𝐴,𝑠−𝜌𝐴,∞)=0.005615(0.039761)(0.043440)=9.698×10−6 kg s−1=34.91 g h−1$$\begin{aligned} \dot m_A&=k_cA_s(\rho_{A,s}-\rho_{A,\infty})\\ &=0.005615(0.039761)(0.043440)\\ &=9.698\times10^{-6}\ \mathrm{kg\,s^{-1}}=34.91\ \mathrm{g\,h^{-1}} \end{aligned}$$

These are the two requested isolated-mechanism answers.

For comparison, using dry-air density at 290 K and saturated-air density at 310 K gives a total density difference that includes both mechanisms.

The mean density 1.16616 and total difference 0.10515 kg m⁻³ give Gr =6.2366 ×105$\mathrm{Gr}=6.2366\times10^5$, Sh =13.419$\mathrm{Sh}=13.419$ and 𝑘𝑐 =0.006202$k_c=0.006202$ m s⁻¹.

The corresponding total-density estimate is ˙𝑚𝐴 =1.0712 ×10−5$\dot m_A=1.0712\times10^{-5}$ kg s⁻¹, about 38.6 g h⁻¹.

This approximately 39 g h⁻¹ estimate already includes thermal expansion. Adding a thermal Grashof contribution would count that expansion twice.

The real pan has reinforcing thermal and composition buoyancy.

The total-density route is a possible single-group engineering approximation; an exact coupled answer requires a validated heat-and-species natural-convection model.

The two isolated evaporation rates cannot be added to predict the combined rate, because the same velocity field transports both heat and species.

The saturation fraction is 0.0614, so low-flux and dry-air-property assumptions remain approximate.

The maintained water temperature implies external heat supply; no latent-heat constraint has been omitted from an adiabatic balance because this pan is not adiabatic.

Answer

For composition-driven convection alone, 𝑘𝑐 =0.00443 m s−1$k_c=0.00443\ \mathrm{m\,s^{-1}}$, ℎ ≃4.76 W m−2 K−1$h\simeq4.76\ \mathrm{W\,m^{-2}\,K^{-1}}$ and ˙𝑚𝐴 ≃27.6 g h−1$\dot m_A\simeq27.6\ \mathrm{g\,h^{-1}}$.

For temperature-driven convection alone, ℎ =6.03 W m−2 K−1$h=6.03\ \mathrm{W\,m^{-2}\,K^{-1}}$, 𝑘𝑐 =0.00562 m s−1$k_c=0.00562\ \mathrm{m\,s^{-1}}$ and ˙𝑚𝐴 ≃34.9 g h−1$\dot m_A\simeq34.9\ \mathrm{g\,h^{-1}}$.

10 Additional examples

Evaporation from a tray and an outlet-humidity check

A water tray is flush with the floor of a rectangular duct. The tray is 0.60 m long along the air flow and 0.25 m wide. The duct is 0.25 m wide and 0.12 m high. Initially dry air enters at 2.0 m s⁻¹, 30 °C$30\,\degree\mathrm{C}$ and 101325 Pa. A heater maintains the water at 30 °C$30\,\degree\mathrm{C}$. Assume a uniform air stream over the tray and a concentration boundary layer beginning at its leading edge. Neglect upstream velocity-profile effects.

Use 𝜈 =1.60 ×10−5$\nu=1.60\times10^{-5}$ m² s⁻¹, 𝐷𝐴𝐵 =2.60 ×10−5$D_{AB}=2.60\times10^{-5}$ m² s⁻¹, water saturation pressure 4240 Pa, 𝑀𝐴 =0.0180$M_A=0.0180$ kg mol⁻¹ and 𝑅 =8.314$R=8.314$ J mol⁻¹ K⁻¹. Calculate the average mass transfer coefficient and evaporation rate. Estimate the outlet relative humidity and use it to check the assumption of dry bulk air over the entire tray. Use the dilute-transfer flat-plate model without a Stefan-flow correction.

Solution

Data

Quantity	Value
Tray length	𝐿 =0.60 m$L=0.60\ \mathrm m$
Tray width	𝑏 =0.25 m$b=0.25\ \mathrm m$
Duct height	𝐻 =0.12 m$H=0.12\ \mathrm m$
Air speed	𝑈 =2.0 m s−1$U=2.0\ \mathrm{m\,s^{-1}}$
Temperature	𝑇 =303.15 K$T=303.15\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Kinematic viscosity	𝜈 =1.60 ×10−5 m2 s−1$\nu=1.60\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.60 ×10−5 m2 s−1$D_{AB}=2.60\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface saturation pressure	𝑝sat𝐴 =4240 Pa$p_A^{\mathrm{sat}}=4240\ \mathrm{Pa}$
Water molar mass	𝑀𝐴 =0.0180 kg mol−1$M_A=0.0180\ \mathrm{kg\,mol^{-1}}$

Assumptions

• Steady external transfer with a boundary layer starting at the tray leading edge.
• Uniform incoming velocity and negligible upstream profile effects.
• Maintained saturated surface at the air temperature.
• Constant properties and dilute transfer without a Stefan-flow correction.
• Dry bulk air is used for the initial rate estimate and then checked.

Calculations

The concentration boundary layer grows along the tray, so use its length 𝐿 =0.60$L=0.60$ m for external transfer. The duct hydraulic diameter is not the length in this flat-plate correlation:

Re𝐿=2.0(0.60)1.60×10−5=75000Sc=1.60×10−52.60×10−5=0.6154$$\mathrm{Re}_L=\frac{2.0(0.60)}{1.60\times10^{-5}}=75000 \qquad \mathrm{Sc}=\frac{1.60\times10^{-5}}{2.60\times10^{-5}}=0.6154$$

The layer is laminar, and Sc exceeds 0.5. The average flat-plate correlation and coefficient definition are

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3$$\overline{\mathrm{Sh}}_L=0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3}$$¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L}$$

Substitution gives

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3=154.7$$\overline{\mathrm{Sh}}_L=0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3} =154.7$$¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿=0.006702 m s−1$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}L =0.006702\ \mathrm{m\,s^{-1}}$$

At the interface the vapour is saturated. With 𝑇 =303.15$T=303.15$ K,

𝑐𝐴𝑠=42408.314𝑇=1.682 mol m−3.$$c_{As}=\frac{4240}{8.314T}=1.682\,\mathrm{mol\,m^{-3}}.$$

Approximate the bulk concentration as zero and use transfer area

𝐴𝑠=0.60(0.25)=0.150 m2:$$A_s=0.60(0.25)=0.150\,\mathrm{m^2}:$$˙𝑛𝐴=¯¯¯¯𝑘𝑐𝐴𝑠(𝑐𝐴𝑠−𝑐𝐴,∞)=0.006702(0.150)(1.682−0)=1.691×10−3 mol s−1˙𝑚𝐴=𝑀𝐴˙𝑛𝐴=0.1096 kg h−1$$\begin{aligned} \dot n_A&=\bar k_cA_s(c_{As}-c_{A,\infty})=0.006702(0.150)(1.682-0)=1.691\times10^{-3}\ \mathrm{mol\,s^{-1}}\\ \dot m_A&=M_A\dot n_A=0.1096\ \mathrm{kg\,h^{-1}} \end{aligned}$$

For the outlet balance use the duct flow area,

𝐴𝑑𝑢𝑐𝑡=0.25(0.12)=0.0300 m2.$$A_{duct}=0.25(0.12)=0.0300\,\mathrm{m^2}.$$

The dry-air molar flow is

˙𝑛𝑎𝑖𝑟=𝑢𝐴𝑑𝑢𝑐𝑡𝑃𝑅𝑇=2.412 mol s−1$$\dot n_{air}=uA_{duct}\frac P{RT}=2.412\ \mathrm{mol\,s^{-1}}$$

The outlet contains both dry air and the added vapour:

𝑦𝐴,𝑜𝑢𝑡=˙𝑛𝐴˙𝑛𝑎𝑖𝑟+˙𝑛𝐴=7.007×10−4$$y_{A,out}=\frac{\dot n_A}{\dot n_{air}+\dot n_A} =7.007\times10^{-4}$$RH𝑜𝑢𝑡=𝑦𝐴,𝑜𝑢𝑡𝑃𝑝𝑠𝑎𝑡𝐴=0.01674=1.67 percent$$\mathrm{RH}_{out}=\frac{y_{A,out}P}{p_A^{sat}} =0.01674=1.67\text{ percent}$$

Under the assumed mixed-bulk description, the bulk concentration reaches only 1.67 percent of the surface saturation concentration. The driving force therefore falls by less than 2 percent from inlet to outlet. This supports using dry bulk air in the first evaporation estimate. A much wetter outlet would require coupling local transfer to the changing bulk humidity.

This humidity check does not remove all model approximations. The surface mole fraction is

4240101325=0.04185,$$\frac{4240}{101325}=0.04185,$$

so a Stefan-flow correction can also change the dilute estimate by a few percent. The imposed heater avoids an evaporative-cooling calculation; without it, the surface temperature and vapour pressure would need a coupled energy balance.

Answer

The average mass transfer coefficient is ¯¯¯¯𝑘𝑐 =0.00670 m s−1$\bar k_c=0.00670\ \mathrm{m\,s^{-1}}$ and the evaporation rate is 0.110 kg h−1$0.110\ \mathrm{kg\,h^{-1}}$. The estimated outlet relative humidity is 1.67 percent, supporting the dry-bulk approximation.

Drying a thin coating on a spherical pellet

A rigid spherical pellet of diameter 0.010 m has a thin wet coating containing 0.0020 g of volatile solvent per cm² of pellet surface. Solvent-free air flows past it at 0.10 m s⁻¹. Both the air and pellet surface are maintained at 300 K and 101325 Pa. External temperature control supplies the evaporation heat.

Use air properties 𝜌 =1.18$\rho=1.18$ kg m⁻³, 𝜇 =1.85 ×10−5$\mu=1.85\times10^{-5}$ Pa s, 𝑘 =0.0263$k=0.0263$ W m⁻¹ K⁻¹ and Pr =0.707$\mathrm{Pr}=0.707$. The solvent has 𝐷𝐴𝐵 =9.62 ×10−6$D_{AB}=9.62\times10^{-6}$ m² s⁻¹, saturation pressure 1270 Pa at 300 K, molar mass 0.0780 kg mol⁻¹ and liquid density 876.5 kg m⁻³. For this low-Re sphere, adopt the matching correlations

Nu𝐷=2+0.6Re1/2𝐷Pr1/3Sh𝐷=2+0.6Re1/2𝐷Sc1/3$$\begin{aligned} \mathrm{Nu}_D&=2+0.6\mathrm{Re}_D^{1/2}\mathrm{Pr}^{1/3}\\ \mathrm{Sh}_D&=2+0.6\mathrm{Re}_D^{1/2}\mathrm{Sc}^{1/3} \end{aligned}$$

Estimate the heat and mass transfer coefficients and the drying time. Assume the exposed area stays saturated at the stated vapour pressure until the solvent is exhausted, coating resistance is negligible, and normal evaporation flow is small. Check the thin-coating assumption.

Solution

Figure 5: A thin solvent coating surrounds the pellet. Its surface concentration is fixed by vapour equilibrium, and solvent transfers into the surrounding air. The diagram shows the external gas film; the model tracks coating loss rather than diffusion inside the pellet.

Data

Quantity	Value
Pellet diameter	𝐷 =0.010 m$D=0.010\ \mathrm m$
Air speed	𝑢 =0.10 m s−1$u=0.10\ \mathrm{m\,s^{-1}}$
Initial solvent loading	𝑚𝐴/𝐴𝑠 =0.0200 kg m−2$m_A/A_s=0.0200\ \mathrm{kg\,m^{-2}}$
Solvent diffusivity	𝐷𝐴𝐵 =9.62 ×10−6 m2 s−1$D_{AB}=9.62\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Solvent saturation pressure	𝑝sat𝐴 =1270 Pa$p_A^{\mathrm{sat}}=1270\ \mathrm{Pa}$
Solvent molar mass	𝑀𝐴 =0.0780 kg mol−1$M_A=0.0780\ \mathrm{kg\,mol^{-1}}$
Liquid density	𝜌𝐿 =876.5 kg m−3$\rho_L=876.5\ \mathrm{kg\,m^{-3}}$
Air and surface temperature	𝑇 =𝑇𝑠 =300 K$T=T_s=300\ \mathrm K$

Use the supplied air properties.

Assumptions

• Quasi steady external transfer and constant supplied properties.
• Negligible solvent in the bulk air.
• The exposed coating remains saturated until the solvent is exhausted.
• Negligible coating resistance and normal evaporation flow.
• Surface temperature is maintained by external heat supply.
• Fixed pellet diameter is used and the coating thickness is checked.

Calculations

Use pellet diameter for both the flow Reynolds number and the transfer groups:

Re𝐷=𝜌𝑢𝐷𝜇=1.18(0.10)(0.010)1.85×10−5=63.78$$\mathrm{Re}_D=\frac{\rho uD}{\mu}=\frac{1.18(0.10)(0.010)}{1.85\times10^{-5}}=63.78$$Sc=𝜇𝜌𝐷𝐴𝐵=1.85×10−51.18(9.62×10−6)=1.630$$\mathrm{Sc}=\frac{\mu}{\rho D_{AB}}=\frac{1.85\times10^{-5}}{1.18(9.62\times10^{-6})}=1.630$$Nu𝐷=2+0.6(63.78)1/2(0.707)1/3=6.269$$\mathrm{Nu}_D=2+0.6(63.78)^{1/2}(0.707)^{1/3}=6.269$$Sh𝐷=2+0.6(63.78)1/2(1.630)1/3=7.639$$\mathrm{Sh}_D=2+0.6(63.78)^{1/2}(1.630)^{1/3}=7.639$$

This Reynolds number is within the low-Re use described for the sphere correlation in the external forced-convection notes. The two transfer groups differ because Schmidt and Prandtl numbers differ. Their additive two is the stagnant-sphere limit and must be retained.

Recover the dimensional coefficients with the corresponding transport property:

ℎ=Nu𝐷𝑘𝐷=16.49 W m−2 K−1𝑘𝑐=Sh𝐷𝐷𝐴𝐵𝐷=0.007349 m s−1$$\begin{aligned} h&=\frac{\mathrm{Nu}_Dk}{D}=16.49\ \mathrm{W\,m^{-2}\,K^{-1}}\\ k_c&=\frac{\mathrm{Sh}_DD_{AB}}D=0.007349\ \mathrm{m\,s^{-1}} \end{aligned}$$

Since temperature is controlled, the heat coefficient need not determine the evaporation rate. The supplied vapour pressure fixes

𝑐𝐴𝑠=12708.314(300)=0.5092 mol m−3.$$c_{As}=\frac{1270}{8.314(300)}=0.5092\,\mathrm{mol\,m^{-3}}.$$

With dry bulk air,

𝑁𝐴=𝑘𝑐(𝑐𝐴𝑠−𝑐𝐴,∞)=0.007349(0.5092−0)=0.003742 mol m−2 s−1$$N_A=k_c(c_{As}-c_{A,\infty})=0.007349(0.5092-0)=0.003742\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

The spherical area is

𝐴𝑠=𝜋𝐷2=3.1416×10−4 m2.$$A_s=\pi D^2=3.1416\times10^{-4}\,\mathrm{m^2}.$$

The loading converts as 0.0020$0.0020$ g cm⁻² =0.0200$=0.0200$ kg m⁻², so the initial solvent mass is 𝑚𝐴0 =6.283 ×10−6$m_{A0}=6.283\times10^{-6}$ kg. The constant-rate estimate is

𝑡𝑑𝑟𝑦=𝑚𝐴0𝑀𝐴𝑁𝐴𝐴𝑠=0.02000.0780(0.003742)=68.5 s$$t_{dry}=\frac{m_{A0}}{M_AN_AA_s} =\frac{0.0200}{0.0780(0.003742)}=68.5\ \mathrm s$$

The area cancels because both inventory and evaporation rate scale with it. Diameter still affects drying time through the transfer coefficient.

The solvent-equivalent coating thickness is

0.0200876.5=2.28×10−5 m,$$\frac{0.0200}{876.5}=2.28\times10^{-5}\ \mathrm m,$$

only 0.46 percent of the pellet radius. Treating the diameter as fixed is consistent with a thin coating provided the nonvolatile coating is also thin. The surface mole fraction is 0.01253, supporting the dilute approximation. Actual paint can dry more slowly as solvent activity falls or diffusion through the coating becomes limiting; the calculated time uses the stated constant-activity model.

Answer

The heat transfer coefficient is ℎ =16.5 W m−2 K−1$h=16.5\ \mathrm{W\,m^{-2}\,K^{-1}}$, the mass transfer coefficient is 𝑘𝑐 =0.00735 m s−1$k_c=0.00735\ \mathrm{m\,s^{-1}}$ and the constant-rate drying time is approximately 68.5 s. The solvent-equivalent thickness is 22.8 𝜇m$22.8\ \mathrm{\mu m}$, only 0.46 percent of the pellet radius.

11 What you should now be able to calculate

Recover a mass-transfer coefficient, combine phase resistances, and calculate evaporation, drying or stream enrichment with the stated interfacial and bulk conditions.

Practice: Workshop 11 for coating removal and tube dissolution.

12 Report an error

Email Ranjeet Utikar with the title Convective mass transfer, the URL /mass-transport/convective-mass-transfer/, and the equation or passage to check.

References

Bird, R. Byron, Warren E. Stewart, and Edwin N. Lightfoot. 2006. Transport Phenomena. Revised 2nd. John Wiley & Sons. https://www.wiley-vch.de/en/areas-interest/engineering/transport-phenomena-978-0-470-11539-8.

Çengel, Yunus A., and Afshin J. Ghajar. 2015. Heat and Mass Transfer: Fundamentals and Applications. 5th ed. McGraw-Hill Education. https://glencoe.mheducation.com/sites/0073398187/information_center_view0/application_chapters.html.

Geankoplis, Christie J. 2003. Transport Processes and Separation Process Principles. 4th ed. Prentice Hall.

Shah, Milinkumar. n.d. “CHEN3001-TP: Mass Transport 4.” Curtin University. Unpublished.

Welty, James R., Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer. 2001. Fundamentals of Momentum, Heat, and Mass Transfer. 4th ed. John Wiley & Sons.