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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analogy between Heat and Mass Transfer
14-1C ( a ) Temperature difference is the driving force for heat transfer, ( b ) voltage difference is the driving force for electric current flow, and ( c ) concentration difference is the driving force for mass transfer.
14-2C The concentration of a commodity is defined as the amount of that commodity per unit volume. The concentration gradient dC/dx is defined as the change in the concentration C of a commodity per unit length in the direction of flow x. The diffusion rate of the commodity is expressed as
dx
dC Q & =− k diff A
where A is the area normal to the direction of flow and k diff is the diffusion coefficient of the medium , which is a measure of how fast a commodity diffuses in the medium.
14-3C Examples of different kinds of diffusion processes:
(a) Liquid-to-gas : A gallon of gasoline left in an open area will eventually evaporate and diffuse into air.
(b) Solid-to-liquid : A spoon of sugar in a cup of tea will eventually dissolve and move up.
(c) Solid-to gas : A moth ball left in a closet will sublimate and diffuse into the air.
(d) Gas-to-liquid : Air dissolves in water.
14-4C Bulk fluid flow refers to the transportation of a fluid on a macroscopic level from one location to another in a flow section by a mover such as a fan or a pump. Mass flow requires the presence of two regions at different chemical compositions, and it refers to the movement of a chemical species from a high concentration region towards a lower concentration one relative to the other chemical species present in the medium. Mass transfer cannot occur in a homogeneous medium.
14-5C ( a ) Homogenous reactions in mass transfer represent the generation of a species within the medium. Such reactions are analogous to internal heat generation in heat transfer. ( b ) Heterogeneous reactions in mass transfer represent the generation of a species at the surface as a result of chemical reactions occurring at the surface. Such reactions are analogous to specified surface heat flux in heat transfer.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-14 The maximum mass fraction of calcium bicarbonate in water at 300 K is to be determined.
Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N 2 and O 2 only.
Properties The solubility of [Ca(HCO 3 ) 2 ] in 100 kg of water at 300 K is 16.75 kg (Table 14-5).
Analysis The maximum mass fraction is determined from
( 16. 75 100 )kg
(CaHCO3)2 (CaHCO3) (CaHCO3) total m m w
m m
m w
14-15 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents are to be determined.
Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N 2 and O 2 only.
Properties The molar masses of N 2 , O 2 , and H 2 O are 28.0, 32.0, and 18.0 kg/kmol, respectively (Table A-1)
Analysis The molar mass of moist air is determined to be
M = (^) ∑ yi Mi = 0. 78 × 28. 0 + 0. 20 × 32. 0 + 0. 02 × 18 = 28. 6 kg/kmol Moist air 78% N 2 20% O 2 2% H 2 O (Mole fractions)
Then the mass fractions of constituent gases are determined from Eq. 14-10 to be
N 2 N N M
w y
O 2 O O M
w y
HO 2 HO HO
2 (^2 2) M
w y
Therefore, the mass fractions of N 2 , O 2 , and H 2 O in dry air are 76.4%, 22.4%, and 1.2%, respectively.
14-16E The error involved in assuming the density of air to remain constant during a humidification process is to be determined.
Properties The density of moist air before and after the humidification process is determined from the psychrometric chart to be (we used EES for more accurate values)
, 1 3 and 1
(^1 0). 07238 lbm/ft 45 %
air
3 , 2 1
(^1 0). 07124 lbm/ft 90 %
air
Analysis The error involved as a result of assuming constant air density is then determined to be Air 80 °F 14.7 psia RH 1 =45% RH 2 =90%
( 0. 07238 0. 07124 )lbm/ft %Error (^1003)
3
air , 1
air
which is acceptable for most engineering purposes.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-17 The mole fractions of the constituents of a gas mixture are given. The mass of each gas and apparent gas constant of the mixture are to be determined.
Assumptions None.
Properties The molar masses of H 2 and N 2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1)
Analysis The mass of each gas is
H 2 : m H 2 = N H 2 M H 2 =( 10 kmol)×( 2 kg/kmol)= 20 kg
10 kmol H 2 2 kmol N 2
N 2 : m N 2 = N N 2 M N 2 =( 2 kmol)×( 28 kg/kmol)= 56 kg
The molar mass of the mixture and its apparent gas constant are determined to be
6.333 kg/kmol 10 2 kmol
m
m N
m M
= 1.313 kJ/kg ⋅ K
333 kg/kmol
314 kJ/kmol K M
R u
14-18 The mole numbers of the constituents of a gas mixture at a specified pressure and temperature are given. The mass fractions and the partial pressures of the constituents are to be determined.
Assumptions The gases behave as ideal gases.
Properties The molar masses of N 2 , O 2 and CO 2 are 28, 32, and 44 kg/kmol, respectively (Table A-1)
Analysis When the mole fractions of a gas mixture are known, the mass fractions can be determined from
m
i i m m
i i m
i i (^) M
y N M
m
m w = = =
65% N 2 20% O 2 15% CO 2 290 K 250 kPa
The apparent molar mass of the mixture is
M = (^) ∑ yi Mi = 0. 65 × 28. 0 + 0. 20 × 32. 0 + 0. 15 × 44. 0 = 31. 2 kg/kmol
Then the mass fractions of the gases are determined from
(or 58.3%)
N 2 N =^ N M = = 0.
w y
(or 20.5%)
O 2 O =^ O M = = 0.
w y
(or 21.2%)
CO 2 CO = CO = = 0. M m
w y
Noting that the total pressure of the mixture is 250 kPa and the pressure fractions in an ideal gas mixture are equal to the mole fractions, the partial pressures of the individual gases become
P N 2 = y N 2 P =( 0. 65 )( 250 kPa)=162.5 kPa
P O 2 = y O 2 P =( 0. 20 )( 250 kPa)= 50 kPa
P CO 2 = y CO 2 P =( 0. 15 )( 250 kPa)= 37.5 kPa
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-21E The masses of the constituents of a gas mixture are given. The mass fractions, mole fractions, and the molar mass of the mixture are to be determined.
Assumptions None.
Properties The molar masses of N 2 , O 2 , and CO 2 are 28, 32, and 44 lbm/lbmol, respectively (Table A-1E)
Analysis ( a ) The total mass of the gas mixture is determined to be
m = (^) ∑ mi = m O 2 + m N 2 + m CO 2 = 7 + 8 + 10 = 25 lbm
Then the mass fractions of constituent gases are determined to be
7 lbm O 2 8 lbm N 2 10 lbm CO 2
N 2 N m
m w
O 2 O m
m w
CO 2 CO m
m w
( b ) To find the mole fractions, we need to determine the mole numbers of each component first,
= = = 0.286 lbmol 28 lbm/lbmol
8 lbm N : 2
2 2 N
N 2 N M
m N
= = = 0.219 lbmol 32 lbm/lbmol
7 lbm O : 2
2 2 O
O 2 O M
m N
= = = 0.227 lbmol 44 lbm/lbmol
10 lbm CO : 2
2 2 CO
CO 2 CO M
m N
Thus,
N (^) m = (^) ∑ Ni = N N 2 + N O 2 + N CO 2 = 0. 286 + 0. 219 + 0. 227 = 0. 732 lbmol
Then the mole fraction of gases are determined to be
N 2 N N m
y
O 2 O N m
y
CO 2 CO N m
y
( c ) The molar mass of the mixture is determined from
= = = 34.2 lbm/lbmol 0.732lbmol
25 lbm m
m N
m M
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-22 The diffusion coefficient of hydrogen in steel is given as a function of temperature. The diffusion coefficients at various temperatures are to be determined.
Analysis The diffusion coefficient of hydrogen in steel is given as
D (^) AB = 1. 65 × 10 −^6 exp(− 4630 / T ) m^2 /s
Using this relation, the diffusion coefficients at various temperatures are determined to be
300 K: DAB = 1. 65 × 10 −^6 exp(− 4630 / 300 )= 3. 27 × 10 −^13 m^2 /s
500 K: DAB = 1. 65 × 10 −^6 exp(− 4630 / 500 )= 1. 57 × 10 −^10 m^2 /s
1000 K: DAB = 1. 65 × 10 −^6 exp(− 4630 / 1000 )= 1. 61 × 10 −^8 m^2 /s
1500 K: DAB = 1. 65 × 10 −^6 exp(− 4630 / 1500 )= 7. 53 × 10 −^8 m^2 /s
14-23 Prob. 14-22 is reconsidered. The diffusion coefficient as a function of the temperature is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"The diffusion coeffcient of hydrogen in steel as a function of temperature is given" "ANALYSIS" D_AB=1.65E-6*exp(-4630/T)
[m^2 /s] 200 1.457E- 250 1.494E- 300 3.272E- 350 2.967E- 400 1.551E- 450 5.611E- 500 1.570E- 550 3.643E- 600 7.348E- 650 1.330E- 700 2.213E- 750 3.439E- 800 5.058E- 850 7.110E- 900 9.622E- 950 1.261E- 1000 1.610E- 1050 2.007E- 1100 2.452E- 1150 2.944E- 1200 3.482E-
200 400 600 800 1000 1200
0.0x10 0
7.0x10 -
1.4x10 -
2.1x10 -
2.8x10 -
AB
(^2)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Boundary Conditions
14-25C Three boundary conditions for mass transfer (on mass basis) that correspond to specified temperature, specified heat flux, and convection boundary conditions in heat transfer are expressed as follows:
0
AB A x
dx
dw − D = =
A A AB ∞ =
= − = As − A x
,s h w w x
w j D
(mass convection - corresponds to heat convection)
14-26C An impermeable surface is a surface that does not allow any mass to pass through. Mathematically it is expressed (at x = 0) as
0
x =
A dx
dw
An impermeable surface in mass transfer corresponds to an insulated surface in heat transfer.
14-27C Temperature is necessarily a continuous function, but concentration, in general, is not. Therefore, the mole fraction of water vapor in air will, in general, be different from the mole fraction of water in the lake (which is nearly 1).
14-28C When prescribing a boundary condition for mass transfer at a solid-gas interface, we need to specify the side of the surface (whether the solid or the gas side). This is because concentration, in general, is not a continuous function, and there may be large differences in concentrations on the gas and solid sides of the boundary. We did not do this in heat transfer because temperature is a continuous function.
14-29C The mole fraction of the water vapor at the surface of a lake when the temperature of the lake surface and the atmospheric pressure are specified can be determined from
atm
vapor sat@T vapor (^) P
y = =
where P vapor is equal to the saturation pressure of water at the lake surface temperature.
14-30C Using solubility data of a solid in a specified liquid, the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from
solid liquid
solid m m
m w (^) A
where m solid is the maximum amount of solid dissolved in the liquid of mass m liquid at the specified temperature.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-31C Using Henry’s constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henry’s law expressed as
y
i, liquidside(^0 )= i,gasside
where H is Henry’s constant and P i, gas side(0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids).
14-32C The permeability is a measure of the ability of a gas to penetrate a solid. The permeability of a gas in a solid, P , is related to the solubility of the gas by P = S D AB where D AB is the diffusivity of the gas in the solid.
14-33 A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium.
Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable.
Properties The saturation pressure of water at 15°C is 1.7051 kPa (Table A-9). Henry’s constant for air dissolved in water at 15ºC (288 K) is given in Table 14-6 to be H = 59,600 bar (determined by extrapolation). Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1).
Analysis ( a ) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 15°C,
P vapor (^) = Psat @ 15 ° C = 1. 7051 kPa Air 15ºC 97 kPa RH=100%
Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be
97 kPa
vapor^1.^7051 kPa vapor (^) P
y
Evaporation
Water 15ºC
( b ) Noting that the total pressure is 97 kPa, the partial pressure of dry air is
Pdry air = P − Pvapor = 97 − 1. 7051 = 95. 3 kPa=0.953 bar
From Henry’s law, the mole fraction of air in the water is determined to be
59,600bar
dryair,gasside 0. 953 bar dry air,liquidside H
y
Discussion The amount of air dissolved in water is very small, as expected.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-36 Prob. 14-35 is reconsidered. The mole fraction of dry air at the surface of the lake as a function of the lake temperature is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
P_atm=100 [kPa]
"PROPERTIES" Fluid$='steam_IAPWS' P_sat=Pressure(Fluid$, T=T, x=1)
"ANALYSIS" P_vapor=P_sat P_dryair=P_atm-P_vapor y_dryair=P_dryair/P_atm
ydry air
5 7 9 11 13 15 17 19 21 23 25 27 29 30
5 10 15 20 25 30
y^ dryair
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-37 A rubber plate is exposed to nitrogen. The molar and mass density of nitrogen in the rubber at the interface is to be determined.
Assumptions Rubber and nitrogen are in thermodynamic equilibrium at the interface.
Properties The molar mass of nitrogen is M = 28.0 kg/kmol (Table A-1). The solubility of nitrogen in rubber at 298 K is 0.00156 kmol/m^3 ⋅bar (Table 14-7).
Rubber plate
Analysis Noting that 250 kPa = 2.5 bar, the molar density of nitrogen in the rubber at the interface is determined from Eq. 14-20 to be
250 kPa
ρ N2 =?
= 0.0039kmol/m^3
( 0. 00156 kmol/m.bar)( 2. 5 bar)
3
It corresponds to a mass density of
= 0.1092kg/m^3
=(0.0039kmol/m )( 28 kmol/kg)
3
That is, there will be 0.0039 kmol (or 0.1092 kg) of N 2 gas in each m^3 volume of rubber adjacent to the interface.
14-38 A rubber wall separates O 2 and N 2 gases. The molar concentrations of O 2 and N 2 in the wall are to be determined.
Assumptions The O 2 and N 2 gases are in phase equilibrium with the rubber wall.
Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/kmol, respectively (Table A-1). The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156kmol/m^3 ⋅bar, respectively (Table 14-7).
Analysis Noting that 750 kPa = 7.5 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined from Eq. 14-20 to be Rubber plate
= 0.0234kmol/m^3
( 0. 00312 kmol/m .bar)( 7. 5 bar)
3
750 kPa
750 kPa = 0.0117kmol/m^3
( 0. 00156 kmol/m.bar)( 7. 5 bar)
3
That is, there will be 0.0234 kmol of O 2 and 0.0117 kmol of N 2 gas in each m^3 volume of the rubber wall.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-40 A nickel wall separates H 2 gas at different pressures. The molar diffusion rate per unit area through the nickel wall is to be determined.
Assumptions 1 Mass diffusion is steady and one-dimensional. 2 There are no chemical reactions in the nickel wall that result in the generation or depletion of hydrogen.
Properties The binary diffusion coefficient for hydrogen in the nickel at 85°C = 358 K is DAB = 1.2×10−^12 m^2 /s (Table 14– 3b). The solubility of H 2 in nickel at 85°C = 358 K is 0.00901 kmol/m^3 ⋅bar (Table 14-7).
Analysis The molar density of H 2 (for 5 atm) in the nickel at the interface is determined using
3
3
, 1
3 , 1
( 0. 00901 kmol/m bar)( 5 atm)( 1. 01325 bar/atm)
( 0. 00901 kmol/m bar)
Then, the molar density of H 2 (for 3 atm) in the nickel at the interface is
3
3
, 2
3 , 2
( 0. 00901 kmol/m bar)( 3 atm)( 1. 01325 bar/atm)
( 0. 00901 kmol/m bar)
The molar diffusion rate per unit area of hydrogen through the nickel wall can readily be determined using
= 2.18 × 1010 kmol/s ⋅ m^2
−
− 0 .0001 m
( 0. 0456 0. 0274 )kmol/m ( 1. 2 10 m/s)
3 12 2
diff ,^1 ,^2 diff L
j A A AB
Discussion The molar mass of H 2 is M = 2.016 kg/kmol (Table A-1). Hence, the mass diffusion rate per unit area of hydrogen through the nickel wall is
10 2 10 2 diff =^ (^2 18 ×^10 kmol/s⋅m)(^2.^016 kg/kmol)=^4.^40 ×^10 kg/s⋅m j. − −
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-41 A dry wall separates air in a room with vapor pressure of 3 kPa from air with negligible vapor pressure in the insulation adjoining the wall. The mass diffusion rate of water vapor through the wall is to be determined.
Assumptions 1 Mass diffusion is steady and one-dimensional. 2 Constant properties. 3 Condensation in the wall is negligible.
Properties The molar mass of water vapor is M = 18.015 kg/kmol (Table A-1).
Analysis The molar density of water vapor in the dry wall at the interface is determined using
3
3
, 1
3 , 1
( 0. 007 kmol/m bar)( 3 kPa)( 0. 01 bar/kPa)
( 0. 007 kmol/m bar)
On the opposite side, the molar density of water vapor is zero, since the vapor pressure is negligible,
CA, 2 = 0
The molar diffusion rate of water vapor through the wall can be determined using
105 10 kmol/s
0 .012m
( 0. 00021 0 )kmol/m ( 0. 2 10 m/s)( 3 10 m) 10
3 9 2 2
, 1 , 2 diff
A A AB
−
−
Hence the mass diffusion rate of water vapor through the wall is
m &diff (^) = N &diff M =( 1_._ 05 × 10 −^10 kmol/s)( 18. 015 kg/kmol)= 1.89 × 10 −^9 kg/s
Discussion At 25°C, the saturation pressure of water is 3169 Pa (from Table 14-9). With the given vapor pressure inside the room being 3 kPa, the relative humidity of the air is
3169 Pa
3000 Pa P sat
Pv
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-43E Water is sprayed into air, and the falling water droplets are collected in a container. The mass and mole fractions of air dissolved in the water are to be determined.
Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since water is constantly sprayed into it. 3 Air is weakly soluble in water and thus Henry’s law is applicable.
Properties The saturation pressure of water at 80°F is 0.5073 psia (Table A-9E). Henry’s constant for air dissolved in water at 80ºF (300 K) is given in Table 14-6 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 lbm / lbmol, respectively (Table A-1E).
Water droplets in air
Analysis Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 80°F,
P vapor = P sat@80°F= 0. 5073 psia
Then the partial pressure of dry air becomes
P dry air= P − P vapor= 14. 3 − 0. 5073 = 13. 79 psia
From Henry’s law, the mole fraction of air in the water is determined to be
74,000bar(1atm/1.0132 5 bar)
dryair,gasside^13.^79 psia(^1 atm/^14.^696 psia) dry air,liquidside H
y
Water
which is very small, as expected. The mass and mole fractions of a mixture are related to each other by
m
i i m m
i i m
i i (^) M
y N M
m
m w = = =
where the apparent molar mass of the liquid water - air mixture is
1 29. 0 0 18. 0 29. 0 kg/kmol
liquidwater water dryair dryair ≅ × + × ≅
M (^) m = (^) ∑ yiMi = y M + y M
Then the mass fraction of dissolved air in liquid water becomes
dryair 5 5 dryair, liquidside dryair,liquidside 29
M m
w y
Discussion The mass and mole fractions of dissolved air in this case are identical because of the very small amount of air in water.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
14-44 A rubber membrane separates CO 2 gas at different pressures. ( a ) The molar densities of CO 2 in the membrane and ( b ) outside the membrane are to be determined.
Assumptions 1 Rubber and nitrogen are in thermodynamic equilibrium at the interface. 2 Carbon dioxide is an ideal gas.
Properties The molar mass of CO 2 is M = 44.01 kg/kmol (Table A-1). The solubility of CO 2 in rubber at 25ºC = 298 K is 0.04015 kmol/m^3 ⋅bar (Table 14-7).
Analysis ( a ) The molar density of CO 2 (for 2 atm) in the rubber at the interface is determined using
= 0.0814kmol/m^3
( 0. 04015 kmol/m bar)( 2 atm)( 1. 01325 bar/atm )
( 0. 04015 kmol/m bar) 3
CO,gasside
3 C CO 2 ,solidside P 2
Then, the molar density of CO 2 (for 1 atm) in the rubber at the interface is
= 0.0407kmol/m^3
( 0. 04015 kmol/m bar)( 1 atm)( 1. 01325 bar/atm )
( 0. 04015 kmol/m bar) 3
CO,gasside
3 C CO 2 ,solidside P 2
( b ) The molar density of CO 2 (for 2 atm) outside the rubber is determined using
= 0.0818kmol/m^3 ⋅ ⋅ +
( 8. 314 kPa m/kmolK)( 273 25 )K
( 2 atm)( 101. 325 kPa/atm) 3
CO CO 2 (^2) RT
u
Then, the molar density of CO 2 (for 1 atm) outside the rubber is
= 0.409kmol/m^3 ⋅ ⋅ +
( 8. 314 kPa m/kmolK)( 273 25 )K
( 1 atm)( 101. 325 kPa/atm) 3
CO CO
2 (^2) RT
u
Discussion Due to its relatively high solubility in rubber, in comparison to O 2 and N 2 (see Table 14-7), the molar concentrations of CO 2 in the solid side and the gas side are almost equal.
