Exchanging Substances Facilitated Diffusion, Osmosis and Active Transport.
MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion...
Transcript of MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion...
MASS TRANSFER
DiffusionConvectionExamples
Fick’s Diffusion Law
.
x
A Am CDA x
Q T ukA x y
τ µ∂ ∂ = − = ∂ ∂∂−∂
=
D – diffusion coefficient (m2/s)m – mass flux per unit time (kg/s)
CA - mass concentration of component A (kg/m3)
In blue:Heat and momentum transfer analogies
MASS TRANSFER
Diffusion in Gases
( ).
3/ 2
1/3 1/3
. .
1 1436A A
A BA B
A AA A A
A
A BA A B BAB BA
m C TD DA x M Mp V V
MR p Mp RT R CM MRT
m M dp m M dpD DA MRT dx A MRT dx
ρ ρ
∂= − = +∂ +
= = = =
= − = −
D – diffusion coefficient (cm2/s)Gilliand semiempirical coefficient of diffusion
m – mass flux per unit time (kg/s)CA - mass concentration of component A (kg/m3)
T – temperature (K)V – molecular volume of constituents (m3)
M – molecular mass (kg/kmol)
MASS TRANSFER
Example 1. Diffusion Coefficient for CO2
Calculate the diffusion coefficient for CO2 in air at atmospheric pressure and 25 oC and compare this value with that in the Table
Atmospheric pressure: p=1 atm =101320 PaVCO2=34.0, Vair=29.9, MCO2=44 kmol/kg, Mair=28.9 kmol/kg
( )
( )
3/ 2
1/3 1/3
1.5
1/3 1/3
2
1 1436
436 298 1 144 29.9101320 34 29.9
0.132 /
A BA B
TDM Mp V V
cm s
= + =+
= + =+
=
Tabular value is D=0.164 cm2/s
MASS TRANSFER
Diffusion in Gases, Equimolal Counterdiffusion. .
.
2 1
A BA BA AB B BA
A B
A BA B A B
A AAB BA AB BA
A A A A
m A dp m A dpN D N DM MRT dx M MRT dx
dp dpN N Since p p p Constdx dx
A dp A dpD D D D DMRT dx MRT dx
m M p pIntegrate DA MRT x
= = − = = −
= − = + = = −
− = − = =
−= −∆
Isothermal Evaporation
.
.
.
,
.
,
A A A AAA
A
A
w w w ww w
w w w w Aw total
A
A BA B
w ww total
w
M A dp p MAir down m D Aw AwMRT dx MRT
D dpwp dx
M A dp p MWater up m D Aw AwMRT dx MRT
Total mass transportM A dp p M D dpm D AMRT dx MRT p dx
dp dpSince p p pdx dx
M A dppm D StMRT p p dx
ρ
ρ
= − − = −
=
= − =
= − +
= + = −
= −−
( ) ( ).
2 2,
2 1 1 2 1 1
'
ln lnw w w Aw total
w A
efan s law
pM A p p pM A pm D DMRT x x p p MRT x x p
−= − =− − −
MASS TRANSFER
Example 2. Diffusion of Water in a Tube
Estimate the diffusion rate of water from the bottom of a test tube 100 mm in diameter and 15 cm long into a dry atmospheric air at 25 oC
Atmospheric pressure p=101320 PaFor t= 25 oC, the partial pressure of water at water surface at the pipe bottom is: pw1=3125 Pa, at the pipe top air is dry, pw1=0 PaDiffusion coefficient for water at 25 oC is D=0.256 cm2/s
( ).
2,
2 1 1
4 5 2
10
ln
0.256 10 1.0132 10 18 0.005 101320ln8314 298 0.15 101320 3125
3.131 10 /
w ww total
w
pM A p pm DMRT x x p p
kg s
π−
−
−= − =− −
−= =−
=
MASS TRANSFER
Diffusion in Liquids and Solids
Fick’s law describes diffusion in liquids and solids too,
Diffusion coefficient is difficult to estimate, Approximate or experimental values,
smaller than in gases due to high intermolecular forces
MASS TRANSFERThe Mass Transfer Coefficient
(Like the heat transfer coefficient)
.
1 1( )A A Am KA C C= −
– mass flux of component A, kg/s
K - mass transfer coefficient, m/s
CA1, CA1 - concentrations through which diffusion occurs, kg/m3
A - surface area m2
.
Am
MASS TRANSFER
Steady state diffusion across a layer of thickness ∆∆∆∆x
.
1 1 1 1( ) ( )A A A A ADm A C C KA C Cx
DKx
= − − = −∆
= −∆
Water evaporation
( )( )
1 1 1 1
2
2 1 1 2 1
( )
ln
ww w w w
w
w w w
MC C p pRT
Thereforep pDpK
x x p p p p
− = −
−= −− − −
MASS TRANSFER
Phenomenological laws - Dimensional AnalysisSimilar to Momentum and Heat Transfer
For example, a boundary layer flow 2
2
2
2
2
2A A A
u u uu v Momentumx y xT T Tu v a Energyx y xC C Cu v D Concentrationx y x
ν∂ ∂ ∂+ =∂ ∂ ∂∂ ∂ ∂+ =∂ ∂ ∂∂ ∂ ∂+ =∂ ∂ ∂
u, v -velocity components, m/s, νννν – viscosity, m2/s, a - thermal diffusivity a=k/ρρρρcp, m2/s
k – coefficient of thermal conductivity,W/mKρ −−−− density, kg/m3, cp - specific heat, J/kgK
D – mass diffusion coefficient, m2/s
MASS TRANSFER
Dimensional AnalysisConcentration and velocity profiles will have the same shape if
νννν=D, or Schmidt number Sc=νννν/D=µ/µ/µ/µ/ρρρρD=1, a role similar to Prandtl number
Pr=νννν/a in heat-momentum analogyIn mass transfer - heat transfer analogy,
concentration and temperature profiles will be similar ifLewis number, Le=a/D=Sc/Pr
In heat transfer, Nusselt number, Nu=hx/k=f(Re,Pr) defines a heat transfer coefficient h, W/m2K,
while in mass transfer, Sherwood number, Sh=Kx/D=f(Re,Sc)defines a mass transfer coefficient,
Empirical correlations should be similar for mass and heat transfer,therefore, for example, in smooth tubes
Sh=Kd/D=0.023Re 0.83 Sc 0.44
MASS TRANSFER
Mass transfer coefficient in terms of the friction factor2/3 / 8
m
K Sc fu
=
Analogy with heat transfer:2/3Pr / 8
m p
h fu c ρ
=
Combined2/3 2 /3
2 /3
Prp p ph Sc ac c c LeK D
ρ ρ ρ = = = Reynolds analogy
1/ 22 /3
1/52 /3
/ 2 0.332Re
/ 2 0.0296Re
f x
f x
KLam Sc CuKTurb Sc Cu
−
∞
−
∞
= =
= =
MASS TRANSFER
Example 3. Wet Bulb Temperature
Dry air at atmospheric pressure blows across a thermometer which is enclosed in a dampened cover, a classical wet bulb thermometer. TheThermometer reads 18.3 oC. What is the temperature of the dry air.
Heat to evaporate the water comes from the air. Energy balance is:hA(Tw2-Tw1)=mwhfg
h - heat transfer coefficient, mw - the mass of water evaporated, mw = KA(Cw -C)
Therefore hA(Tw2-Tw1)= KA(Cw1 -Cw2) hfg
Since:
It follows: ( ) ( )
2/3 2 /32 /3
2 /3
2 1 1 2
Prp p p
p w w w w fg
h Sc ac c c LeK D
ac T T C C hD
ρ ρ ρ
ρ
= = =
− = −
From the steam tables, partial pressure of water is pw= 2107 Pa for 18.3 oC
311 1
1
2 2
3
2107 18 0.01566 /8314 291.3
0:
101320 28.9 1.212 /8314 291.3
w ww w
w
w w
p MC kg mMRT
CFurthermore for air
pMC kg mMRT
ρ
ρ
ρ
= = = =
= =
= = = =
hfg=2456 kJ/kg, cp=1.004 kJ/kgK, Le=a/D=Sc/Pr=0.845
( ) ( )1 12 1 2/3 2 /3
2
0.0156 0 245635.4
1.212 1.004 0.845
53.7
w w fg ow w
p
ow
C C hT T C
c Le
T C
ρ− −
− = = =
=
Since the estimated temperature differs from the guessed one, a repeated calculation gives: ρρρρ=1.143 kg/m3 and Tw2=55.8 oC
MASS TRANSFER
Example 4. Relative Humidity of Air Stream If the air stream from the Example 3 is at Tw2=32.2 oC, while the wet bulb remains the same Tw1=18.3 oC, calculate the relative humidity of the air stream.
Relative humidity is the ratio of concentration (or density, or partial pressure) of vapour to concentration of air stream.
Vapour concentration is:
( ) ( ) ( )
( )
( )
2/3
2 1 1 2 1 2
2/3
2 1
2 1
2/3
3
1.212 1.004 0.845 32.2 18.30.0156 0.0095
2456
p w w w w fg w w fg
p w w
w wfg
ac T T C C h hD
ac T TD
h
kgm
ρ ρ ρ
ρρ ρ
− = − = −
− = − =
−= − =
The saturation pressure at Tw2=32.2 oC is from steam tables pw2= 4825 Pa,Therefore the saturation concentration of the stream is,
Cw2sat=ρρρρw2sat= pw2 Mw/MRTw2=(4825 18)/(8314 305.4)=0.0342 kg/m3
The relative humidity is:
φφφφw2= ρρρρw2/ρ/ρ/ρ/ρw2sat=0.0095/0.0342=0.278=27.8%
MASS TRANSFER
Evaporation Process in the Stagnant Atmosphere
Significant effect to the everyday life
Evaporation from the horizontal surface.
.
,
18, 0.62228.9
0.622
w w ww
w
w w w ww w
A
M dpm DA MRT dz
MRTTotal pressure pM
MM essentially airM
M dp dpm D DA M p dz p dz
ρ
ρ ρ
= −
=
= =
= − = −
Integrate with boundary conditions
pw = pw1= psat at z=0pw = pw2 at z=z1
.
2
1
0.622w sat ww
p pm DA p z
ρ −=
Since evaporation includes moving atmospherewhich may be turbulent, convection is usually modelled,
for example by using a ‘standard land pan’
( )( )0.881 185 1.29p sat wE u p p= + −
E1p - Land pan evaporation mm/day u – daily wind movement km/day
psat , pw saturation and actual vapour pressureunder temperature and humidity conditions 1.5 m above the water surface, bar (105 Pa)
MASS TRANSFER
Example 5. Water Evaporation Rate A standard land pan is used to measure the evaporation rate in atmospheric air at T=37.7 oC and φφφφ=30 % relative humidity. The wind Speed is 16 km/h. What is the evaporation rate on the land under these conditions.
Standard land pan dimensions are: 1.22 m diameter and 0.254 m height
( )( )0.881 185 1.29p sat wE u p p= + −
psat= 6545 Pa, φφφφ= pw/ psat=0.3, pw=0.3 6545=1964 Pau=16 m/h = 384 km/day
E1p=(185+1.29 384)(0.06545-0.01964)0.88= 45.1 mm/day
.
12 2
0.04511000 38.60.61
w pw
Em kgA A m day
ρπ
= = =
To compare this result with the molecular diffusion rate, let us take z1 = 1.5 m
( )
.
2
1
4
72 2
0.622
6545 19640.256 10 180.6228315 311 1.5
3.38 10 0.029
w w w sat wD M p pmA MRT z
kg kgm s m day
−
−
−= =
−= =
= =
This is negligible compared with the convective mass transfer.
MASS TRANSFER
Humid Air
Content of water x=mw/ma (kg of water per 1 kg of dry air) 1 kg of dry air contains x kg of moisture, 1+x kg of humid air. Moisture can be vapour, liquid or ice. For vapour:
( ) ( ) ( ) ( )18 0.622
28.9
w
w w w w w w
wa w a w w
a
p Vm R T p M p px
p p Vm p p M p p p pR T
= = = = =− − − −
( )0.622 ss
s
pxp p
=−
For saturated air:
For temperatures higher than saturation temperatures any moisture content.For temperatures lower than saturation temperature not more thanmaximum moisture content
MASS TRANSFER
Humid Air
Saturation ratio (percentage): ψψψψ=xs/x (=100 xs/x %)Relative humidity (percentage): φφφφ=pw/ps (=100 pw/ps %)
( )
( )
( )
0.622
0.622
1
w
w s
ss s
s
s
pp p p px
px p pp p
pp p
ψ φφ
φ ψψ
− −= = =−
−
=− −
For practical (not too high) temperatures, implying low saturation pressure compared with total pressure: ψ=φψ=φψ=φψ=φ
MASS TRANSFERPsychrometric Chart (Carrier diagram)
h-x diagram is a standard engineering tool in air conditioning. It is drawn in inclined h-x coordinates, where h (kJ/kg of dry air) stands for enthalpy of humid air, x stands for moisture content per 1 kg of dry air (humidity ratio), with the wet and dry bulb temperatures, relative humidity or percentage of saturation as parameters.
Enthalpy of humid air:
h = ha + hw = cpaT + x (hfg+ cpwT) kJ/kg of dry air
hfg = 2500 kJ/kg enthalpy of water evaporation cpw = 1.930 kJ/kgK specific heat at constant pressure for water vapour cpa = 1.005 kJ/kgK specific heat at constant pressure for air
Enthalpy of air saturated with vapour:
h = cpaT + xs (hfg+ cpwT) kJ/kg
MASS TRANSFER
Psychrometric Chart (Carrier diagram)
Enthalpy of air containing water vapour, liquid water, fog and solid water, ice:
h = cpaT + x (hfg+ cpwT) + cw,lT + xw,i (hif+ cw,iT) kJ/kg
cw,l = 4.18 kJ/kgK specific heat for liquid waterhfg = 334 kJ/kg enthalpy of water solidification cw,i = 2.09 kJ/kgK specific heat for solid water
Third and fourth term can exist together only at 0 oC
MASS TRANSFER
Psychrometric Chart (Carrier diagram)
In h-x coordinates these equations are plotted for each temperaturetogether with curves of constant relative humidity or percentage of saturation. Also saturation values are plotted. Since in humid and Saturated area the line slopes are:
respectively. In the saturation area the isotherms are positioned almost vertically. Diagram area of interest is small. Therefore, the abscissa is inclined to utilize t=0 oC vertically in the humid area.
,
2500 1.93
4.18
fg pwT
w lT
h h c T T andxh c T Tx
∂ = + = + ∂ ∂ = = ∂
MASS TRANSFER
Psychrometric Chart (Carrier diagram)
Heating and Cooling of Air Streams
Enthalpy changes from h1 to h2 Content of moisture (absolute humidity) remains constantTemperature changes from t1 to t2Relative humidity changes from φφφφ1 to φφφφ2Process is represented in x-h diagram by a (straight) line of constant enthalpyHeat exchanged is q12= h2 - h1 (kJ/kg)The heating will increase temperature and decrease relative humidityThe cooling will decrease temperature and increase relative humidity. If undergoes to the saturated region the cooling might be associated with extraction of liquid water
Psychrometric Chart (Carrier diagram)
Mixing of Air Streams
If there is no work or heat exchange with surrounding, mass and energy balance for dry air and moisture and heat are respectively:
1 1
1 1 1 2
1 1 1 2
m m mmx m x m xmh m h m h
= += += +
Therefore:1 1 1 2 1 1 1 2
1 1 1 1
2 1
2 1
m x m x m h m hx hm m m m
h h h h hx x x x x
+ += =+ +
− − ∆= =− − ∆
That implies that the mixing point will be on the line 1-2
Psychrometric Chart (Carrier diagram)
Mixing of Air Streams
Furthermore if: m1/m=r1and m2/m=r2 , r1 + r2 =1
2 11 1
2 1 2 1
,x x x xr rx x x x
− −= =− −
Therefore the mixing point divides 1-2 in the ratio of r1/ r2
The mixing can be combined with heatingHeat is added to the stream 1 or stream 2 or to the mixture,the effect is the same
Psychrometric Chart (Carrier diagram)
Mixing of Air and Water Streams
If water is added to air:
( ) ( )1 1
11
1
,
,
w w w
ww
m m x x m h m h hmh h h h x x
x x x m
= − = −− ∆= = = −− ∆
Therefore the slope of the mixing line is hwThis is usually given or it can be plotted in the diagram
Psychrometric Chart (Carrier diagram)
Drying in Air Stream
A combination of heating and adding water. Water and energy balances are:
( )( ) ( )
( )
2 1 1 1 2
2 1
2 1
2 1
2 1
2 1
,
:
w s s in w w s s out
s s out in w w in
s ss s out in w in
w w
s
w
m m x x m c t m c t Q mh m c t mh
Q m h h m c t t m c tm cQ h h q q t t c t
m x x mSince q is smallQ h h hm x x x
= − + + + = +
= − + − +−= + = − −−
− ∆= =− ∆
Psychrometric Chart (Carrier diagram)
Estimation of Humidity by a Psychrometer(Dry and Wet Bulb Thermometer)
Wet bulb thermometer shows lower temperature if air is not saturated. Energy balance for air immediately above the interface surface is:
Energy balance for air immediately below the interface surface is:
( ) ( )( )
w w cw w
w w cw w
mh dm h h t t dA m h dh
mdh dm h h t t dA
+ + − = +
= + −
( ) ( )( )
( )
w v cv v
w v cv v
w v
mh dm h h t t dA m h dh
mdh dm h h t t dA
dm K x x dA
+ + − = +
= + −
= −
Psychrometric Chart (Carrier diagram)
Dry and Wet Bulb Thermometer
Combination of these equations gives:
This helps to construct a line in h-x diagram
( ) ( ) ( )
( ) ( ) ( )
v fg cw w v cv v
fg v w cw v
pp v v v
v
K x x h h t t h t th h h h h
Kcc t t x x Le x x
h
− = − − −
= −
− = − = −
!
Mass Transfer
Psychrometric Chart (Carrier diagram)
Question 1
20,000 m3/h of air, t1,dry=3.5 oC, t1,wet=1 oC is conditioned to t4=21 oC, φφφφ4=0.65 by heating to t2=21 oC, humidification by injection of water h=1400 kJ/kg and by repeated heating.Show the process in h-x diagram and calculate heater powers and mass of water injected. Would it be possible to use the steam injection only and not the heaters to reach the same condition. What are the water enthalpy and water flow in such a case.
Question 1
( ) ( )
1 2 3 4
1 2
4
3 2
3 2
3 2 3 2
1
0.003 / 0.010 /11.2 / 32.6 /46.2 /
32.6 1400 0.010 0.003 42.3 /500
w
w
w
x x kg kg x x kg kgh kJ kg h kJ kgh kJ kg
h hhSince hx x x
h h h x x kJ kgp Pa
= = = == ==
−∆= =∆ −
= + − = + − ==
Question 1
( ) ( )
( ) ( )( ) ( )
( ) ( )
1
1
1,2 2 1
3,4 4 3
3 2
4 11
4 1
101320 50 20,000287 276.5
25,495 / 7.057 /
7.057 32.6 11.2 151.0
7.057 46.2 42.3 27.5
7.057 0.01 0.003 0.049 / 178 /
46.2 11.20.01 0.003
w
w
w
p p Vm
RTkg h kg s
Q m h h kW
Q m h h kW
m m x x k s kg h
h hhx x
− −= = =
= =
= − = − =
= − = − =
= − = − = =
− −= =− −
( ) ( )1 4 1
5000 /
7.057 0.01 0.003 0.049 / 178 /w
kJ kg
m m x x k s kg h
=
= − = − = =
Psychrometric Chart (Carrier diagram)
Question 2
A two-stage dryer with a full air recirculation is used for drying of sensitive material in sterile conditions. Maximum and minimum temperatures are 30 and 14 oC respectively. Humidity at the dryer exits should not excess φφφφ=0.9. Draw the process in h-x diagram and calculate the mass flow of air and heater powers if mw=130 kg/h of moisture is extracted.
Question 2
( ) ( )( ) ( )
1 2 3 4
5
1 2 3
4 5
3 4
1,2 2 1
3,4 4 3
5
0.0100 / 0.0136 /0.0164 /39.0 / 55.5 /
65.0 /
130 46,429 / 12.9 /0.0164 0.0136
12.9 55.5 39 212.8
12.9 65 55.5 122.6
w
x x kg kg x x kg kgx kg kgh kJ kg h h kJ kgh h kJ kg
mm kg h kg sx x
Q m h h kW
Q m h h kW
Q
= = = === = == =
= = = =− −
= − = − =
= − = − =
( ) ( ),1 5 1 12.9 65 39 335.4m h h kW= − = − =