MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion...

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MASS TRANSFER Diffusion Convection Examples Fick’s Diffusion Law . x A A m C D A x Q T u k A x y τ µ =− = = D – diffusion coefficient (m 2 /s) m – mass flux per unit time (kg/s) C A - mass concentration of component A (kg/m 3 ) In blue: Heat and momentum transfer analogies

Transcript of MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion...

Page 1: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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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

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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)

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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

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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

= = − = = −

= − = + = = −

− = − = =

−= −∆

Page 5: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric
Page 6: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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

−= − =− − −

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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−

−= − =− −

−= =−

=

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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

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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

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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

− = −

−= −− − −

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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

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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

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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

= =

= =

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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

= = =

− = −

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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

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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

− = − = −

− = − =

−= − =

Page 17: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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%

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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

= −

=

= =

= − = −

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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)

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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

= = =

Page 21: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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.

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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

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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

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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

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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

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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

∂ = + = + ∂ ∂ = = ∂

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Page 28: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric
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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

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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

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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

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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

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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

= − + + + = +

= − + − +−= + = − −−

− ∆= =− ∆

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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

+ + − = +

= + −

= −

Page 35: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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

− = − − −

= −

− = − = −

!

Page 36: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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.

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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

= = = == ==

−∆= =∆ −

= + − = + − ==

Page 38: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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

=

= − = − = =

Page 39: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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.

Page 40: MASS TRANSFER Diffusion Convection Examples …sj376/mt1.pdf · MASS TRANSFER Example 1. Diffusion Coefficient for CO2 Calculate the diffusion coefficient for CO2 in air at atmospheric

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= − = − =