15 Mixing and Solution

SKF 2123 - Energy Balance

Instructor: Dr Azeman Mustafa

Balances on Nonreactive Processes JKG- FKKKSA Page 8 - 1

Mixing and Solution

When two different liquids are mixed or when a gas or solid is dissolved into a liquid, bonds are broken between neighboring molecules.Net release in energy will result when the bonds are broken and solution is form.Suppose we mix 1 mol of pure liquid of sulfuric acid with water at specified temperature and pressure

The energy balance for this constant pressure-process is given by:

The pure solute and solvent at the same T and P is the Heat of Solution at that temperature and pressure.

( )OHlSOHaqSOH HHHHQ24242 )()( +−=∆=


Heat of Mixing and Solutions

Heats of solution, ∆Ĥs(T,r) - enthalpy change when 1 mol of solute (gas or solid) is dissolved in r moles of a liquid solvent at constant temperature T

MgSO4 added to water has ∆H soln = -91.2 kJ/mol.NH4NO3 added to water has ∆Hsoln = + 26.4 kJ/mol.MgSO4 is used in hot packs and NH4NO3 is used in cold packs

∆Hsoln is positive for endothermic (heat absorbing) processes and negative for exothermic (heat generating) processesDissolution can be spontaneous in either caseUsually exothermic for salts with small, very positive cations (high ∆Hhyd)

CaCl2, -82.8 kJ/molKCl, +17.2 kJ/mol





Heats of mixing, ∆Ĥm(T,r) - enthalpy change when 1 mol of liquid solute is mixed with r moles of a liquid solvent at constant temperature T

Enthalpy of a solution (or mixing) containing r moles H2O/mole solute for reference states of pure solute and solvent at 25oC and 1 atm is

and for reference states of pure solvent and an infinitely dilute solution at 25oC and 1 atm

( )rH∆H sˆˆ =

∞→→

∞=∞ n as H∆H∆ hence

)(H∆-(r)H∆H∆

s

ss

ˆˆ

ˆˆˆ


Heats of solution or mixing for an ideal mixture / solution (e.g. gas mixtures or liquid mixtures of structurally similar compounds, e.g. Paraffins, aromatics) is usually negligible, the enthalpy of mixtures is approximately

Aqueous solution of strong acids or bases of certain gases (HCl) or solids (NaOH) heats of solution should be included in energy balance calculationsData of the heats of solution is given in Perry’s Chemical Engineering Handbook on pp 2-201- 2-204.Some values of the heat of solution at 25oC of HCl(g) and NaOH (s) and the heat of mixing at 25oC of H2SO4(l) are given in Table B.11, p.653.

∑≈ iimix Hn H∆ ˆˆ





The heat of solution and mixing of HCl(g) and NaOH(s) and the heat of mixing at 25oC of H2SO4(l)


Example 17 - Prob. 8.82

Use table B-11 to determine the specific enthalpy (kJ/mol HCl) of hydrochloric acid containing 1 mol HCl/5 mol H2O at 25oC relative to ;

i. HCl(g) and H2O(l) at 25oC

ii. H2O(l) and an infinitely dilute HCl solution at 25oC

HCl molkJ-64.055)(rH∆ 11;-B Table

HCl mol 1

OH 5molr

s

2

==

=

ˆ

( )HCl mol

kJ11.0975.1464.05)(H∆-5)(rH∆H∆

molkJ 75.14- )(H∆ and

molkJ64.05 - 5)(rH∆ ; 11-B Table

ss

ss

+=−−−=∞==

=∞==

ˆˆˆ

ˆˆ





Sodium hydroxide (NaOH) is dissolved in enough water to make up a 20% mole % solution. If the NaOH and water were initially at 77oF (25oC), how much heat (Btu/Ibm solution) must be removed for the solution to be at 77oF. Assume the process is carried out at constant pressure. Use table B-11 to evaluate ∆Ĥs

20 mole % NaOH (aq)25oCH2O(l)

25oC

NaOH (s)25oC Mixing

tank



Basis : 100 moles of 20% mole NaOH solution( 20 moles NaOH(s), 80 moles H2O (l)) both at 25oC

Thus, Heat must be removed to keep the solution at 25oC

NaOH molkJ-34.434)(rH∆ 11;-B Table

4 NaOH mol 20

OH mol 80r

s

2

==

==

ˆ




Since the inlet temperature of both substances is 25oC (i.e. ref. state) the corresponding enthalpies are zero

( )

( )

IbmBTU 132.5 removed

solution IbmBTUHeat

Ibm 4.9281000g

Ibm 2.2 OH molOH g 18x

solution molOH mol 0.8

NaOH molNaOH g 40x

solution molNaOH mol 0.2 solution moles 100 solution Ibm

BTU 653J

BTU 9.486x10kJ

1000JNaOH molkJ 34.43- NaOH moles 20 4)(HH∆Q

HnHnQ

2

22

4

s

out inininoutout

−=

=

+

=

−=

====

−=

−

∑ ∑

rˆnˆn

ˆˆ

NaOHNaOH

-34.43 kJ/mol NaOH

100

00

2080

NaOH(s), 25oCH2O(l), 25oCNaOH(aq), 25oC

Hout (kJ/mol)nout (mol)Hin (kJ/mol)nin (mol)Substance



Example 18 - Heat of mixing (H2SO4 solution)

What is the specific enthalpy of 80 wt % H2SO4(aq, 110oF) relative to pure H2SO4 at (77 oF) and pure water at 32oF.

Basis : 1 lbm of 80 wt % H2SO4(aq, 110oF) ( 0.8 lbm H2SO4 (l),77 oF ; 0.2 lbm H2O (l),32oF)

0.8 lbm H2SO4 (l),77 oF1 lbm of 80 wt % H2SO4(aq, 110oF)

0.2 lbm H2O (l),32oF




Solution to Example 18 – Manually calculated

( ) BTU 121.2 - kJ 10

BTU10x 9.486 SOH lbm 0.8 2.2lbm1000g

SOH 98gSOH mol

SOH molkJ34.43-1.36)(r∆H 11;-B Table

1.36 SOH

lbm 98molx lbm 0.8

OH lbm 18

molx lbm 0.2r

3-

4-

4242

42

42s

42

2

=

≅=

≅

=

Pure H2O (l), 32oF

80 wt% H2SO4 (l), 77oFs∆H 2∆H ∆H1

Pure H2SO4 (l), 77oFPure H2O (l), 77oF

80 wt% H2SO4 (l), 110oF

( ) ( ) BTU 9.0 F32-77 Flbm-

BTU 1.0 OH lbm 0.2 dTClbm 0.2∆H oo2

F77

F32 OHp,1

o

o 2=

== ∫

(Hypothetical heating, cooling & isothermal mixing steps)∆HQ =


( )

( )

( ) BTU 16.51 lbm 1

BTU9.916.6 lbm 1H∆

Flbm-BTU0.00038T0.34

0.80.2 0.8

Flbm-BTU1

0.80.2 0.2

)conversionunit (require 2B e......Tabl

CyCy CyC

dTClbm 1H∆

2

oo

p,SOHp,OHi

ip,i(aq)SOHp,

F110

F77 (aq)SOH wt%, 80p2

4SO2H42O2H242

o

o42

=

+=

+

++

+=

−

+==

=

∑∫

ˆ

ˆ

( )lbmBTU95.7

solution lbm 1BTU 16.51121.29

solution lbm 1∆H∆H∆HH∆ 2s1

−=+−

=

++=ˆ

Note : Heat capacity for the solution (or dilute solution can be reasonably estimated using the heat capacity of the pure solvent




H2SO4-H2O Enthalpy-Concentration Diagram

Table B-11 gives heat of mixing for aqueous sulfuric acid, and heats of solutions for aqueous solutions of hydrochloride and sodium hydroxide at 25oC and 1 atm

Fig. 8.5-1, p. 399, is an enthalpy-concentration chart for a binary system involving liquid mixtures of sulfuric acid (ref. states; 77oF (25oC) & 1 atmand water (ref. states; 32oF (0oC) . It

is equivalent to heat of mixing or heat of solution data can be used to solve energy balance problems for mixing or separation of aqueous sulfuric acid streams



Reference states

H2SO4(l), 77oF (25oC) & 1 atmH2O (l), 32oF (0oC)



Balances on Nonreactive Processes JKG- FKKKSA - 15


An H – x chart for aqueous solutions of sulfuric acid at several temperatures.

How to read the figure:What is the specific enthalpy (BTU/lbm) of a 40 wt% sulfuric acid solution at 120oF?

From the figure the reading is about 55 Btu/lbm.


Solution to example 18 using H2SO4-H2O Enthalpy-Concentration Diagram (Fig. 8.5-1)

Locate mass fraction of H2SO4 ~ 0.8 (x-axis), move vertically to ~ 110oF line and then move horizontallyto estimate ∆H (y-axis), hence

∆H~ - 96 BTU/lbm




Adiabatic Mixing

Adiabatic mixing is considered to have taken place if substances are mixed in a well-insulated system, thus assuming no heat loss to its surrounding (i.e. Q = 0). This is the condition in which the maximum temperature the product solution can achieve.

The concentration of product solution can be estimated as follows;


Example 19 - Adiabatic Mixing

Pure water at 32oF is used to dilute 100 mL of 90% wt H2SO4(aq, 32oF) in a well-insulated flask. Estimate the maximum temperature the product solution can achieve, the concentration of sulfuric acid (wt %), the specific enthalpy of this solution .

Water (32oF)

90 wt% H2SO4 (32oF)? wt% H2SO4 (? oF)Q = 0




Example 19 - Adiabatic Mixing

From Fig. 8.5-1 ;

The maximum temp. ~190oF, the corresponding concentration ~ 65 wt% H2SO4 solution and the specific enthalpy ~ - 66 BTU/lbm

190oF


Working session 16- Problem 8.95 (Modified)

An aqueous solution containing 85% wt H2SO4 at 60oF (specific gravity = 1.78) is diluted with pure liquid water at the same temperature. The feed solution volume is 350 mL. The mixing may be considered adiabatic and the pressure is constant at 1 atm.

a. The product solution is to contain 30% wt H2SO4. Calculate the volume (mL) of the water needed for dilution.

b. Use Fig. 8.5-1 to estimate the specific enthalpies (BTU/lbm) of the feed solution and the water & calculate the enthalpy of the product solution and as well as the product temperature

c. Use Fig. 8.5-1, estimate the maximum concentration (wt %) of the product solution and temperature of the product solution that could be achieved by mixing pure water with a 350 mL of 85 wt% H2SO4 solution (both at 60oF). How much water is needed.

d. How much heat (BTU) must be removed from the product solution calculated in part (c) to keep its final temperature in part (c) at 77oF?





85% wt H2SO4 (aq),60 oF m1 kg m2 kg350 mL 30 wt % H2SO4(aq, ToF)

H2O (l),60oF mw kg

a. The product solution is to contain 30% wt H2SO4. Calculate the volume (mL) of the water needed for dilution.

( )

( ) mL 1142 L

mL 1000 kg 1L kg 1.142 ~neededwater pure of Volume

1.142kgm and kg 1.765 m then , kg 0.6230.35L Lkg1.78 m Given

0.7mm 0.15m : bal. mass OH and 0.3m0.85m : bal. mass SOH

w21

2w122142

=

===

=

=+=


b. Use Fig. 8.5-1 to estimate the specific enthalpies (BTU/lbm) of the feed solution and the water & calculate the enthalpy of the product solution and as well as the product temperature

( )

lbmBTU 18.39- F)T (aq,H∆ and

lbmBTU27.5 kg

lbm 2.2x kg 1.142

lbmBTU102.5- kg

lbm 2.2x kg 0.623 F)T (aq,H∆kg2.2lbmx 1.765kg Then

0Hm - HmQ : mixing adiabaticfor balance EnergylbmBTU27.5 F)60 (l,H∆

lbmBTU102.5- F)60 (aq,H∆

solutionfeed of enthalpies Specific

oSOH wt% 30

oSOH wt% 30

out iniiii

oOH

oSOH wt% 85

42

42

242

=

+

=

==

==

∑ ∑

ˆ

ˆ

ˆˆ

ˆˆ

From Fig. 8.5-1, at 30 wt % H2SO4 & ∆H ~ - 18. 4 BTU/lbm, the product temperature is about 130oF





Alternatively, the enthalpy of the product solution containing 30% wt H2SO4and as well as the product temperature can be estimated directly from From Fig. 8.5-1 ;

The specific enthalpy ~ -19 BTU/lbm, the product temp. ~130oF

~ 130oF



c. Use Fig. 8.5-1, estimate the maximum temperature of the product solution that could be achieved by mixing pure water with a 350 mL of 85 wt% H2SO4 solution (both at 60oF) and the concentration (wt %) of the product solution. How much water is needed?

From Fig. 8.5-1 : The maximum temp. ~ 180oF, the corresponding concentration ~ 60 wt% H2SO4 solution and the specific enthalpy ~ - 64 BTU/lbm





Volume of pure water needed ?

d. How much heat (BTU) that must be removed to keep the final temperature of the product solution calculated in part (c) at 77oF??

( )( ) ( )( )

( ) water mL 260 L

1000mL kg 1L kg 0.2596 m

OH kg 0.2596 m ........ m0.623

m00.6230.850.60

w

2ww

w

=

=

=+

+=


( ) ( ) ( ) F,OHF,SOH%wtF,SOH%wt oooˆˆˆ

ˆˆ

6060857760 24242 iiiiii

inii

outii

HmHmHm

HmHmQ

−−=

=−= ∑∑


From figure 8.5.1,

Alternatively, the heat transfer can estimated by the difference between the enthalpy of the adiabatic solution and that of the solution at 77oF.


( )

( )

( )IbmBTU27.5~H

IbmBTU102.5~H

IbmBTU121~H

FO,60H

F,60SO85wt%H

F,77SO60wt%H

o2

o42

o42

ˆ

ˆ

ˆ

−

−

BTU 110.2~HmHmQin

iiout

ii −−= ∑∑ ˆˆ

( ) ( )BTU 110.7 - ~

HmHm HmHmQ F,77SO60wt%HiiF,180SO60wt%Hiiin

iiout

ii o42

o42

ˆˆˆˆ −=−= ∑∑




NH3-H2O Enthalpy-Concentration Diagram

Fig. 8.5-2, p. 403, is a specific enthalpy of binary mixtures for liquid-vapor equilibria in the ammonia + water system at 1 atmpressure. It

has saturated liquid and vapor curves plus tie lines in the two-phase regioncan be used to solve energy balance problems, such as the flash vaporization of ammonia + water mixtures


NH3-H2O Enthalpy-Concentration Diagram




Enthalpy Concentration Charts (Vapor – Liquid Equilibrium)

An H – x chart for ammonia-water system at 1 atm is shown at several temperatures.

A tie line is drawn on the enthalpy concentration chart from x = 0.25 on the liquid phase to the vapor-phase curve at y = 0.95 at 100oF.

The tie line is to determine the equilibrium composition and the specific enthalpy of each phase at specified temperature


Example 20 - prob. 8.98

An NH3-H2O mixture containing 60 wt % NH3 is brought to equilibrium in a closed container at 140oF. The total mass of the mixture is 250 g. Use Fig. 8.5-2 to determine the masses of ammonia and of water in each phase of the system

Liquid phase , L, (~12.9 wt % NH3, 87.1 wt% H2O)Vapour phase, V, (80 wt % NH3, 20 wt% H2O)

( )

lbm 0.164 L lbm 0.386V V L 0.55 : bal. Overall

0.8V 0.129L 0.550.60 : bal. Ammonia

lbm 0.55 1000g

lbm 2.2x g 250 mass Total

==+=

+=

==

35.164.9

140.49.65

VapourLiquid

H2O (gram)NH3 (gram)




Working session 17 - Equilibrium Flash Vaporization

A 30 wt% NH3 solution at 100 psia is fed at a rate of 100 lbm/h to a tank in which the pressure is 1 atm. The enthalpy of the feed solution relative to the reference conditions is 100 Btu/lbm. The vapor composition is to be 89 wt% NH3. Determine the temperature of the stream leaving the tank, the mass fraction of NH3 in the liquid product, the flow rate of the liquid and vapor product streams, and the rate at which heat must be transferred to the vaporizer.


From the figure

xV = 0.89 lbm NH3/lbm

T = 120 oF

xL = 0.185 lbm NH3/lbm

HV = 728 Btu/lbm

HL = 45 Btu/lbm

100 lbm/h

0.3 lbm NH3/lbmHF = 100 Btu/lbm

mv lbm/h

0.89 lbm NH3/lbm0.11 lbm H2O/lbmHV Btu/lbm

mL lbm/h

xL lbm NH3/lbm(1-xL) lbm H2O/lbmHL Btu/lbm

Working session 17




Working session 17

LV

FV

m/hlb 100 xxxxmL

−−

=&

product liquid lb 84 185.089.0

3.089.0/hlb 100 mm =

−−

=Lm&

( ) product vapor lb 16 /hlb 84- 100 mm ==Vm&

[ ] Btu/h 5400 Btu/h (100)(100)-(84)(45)(16)(728)

ˆ100ˆˆ

=+=−+=∆= FLLVV HHmHmHQ &&&&

15 Mixing and Solution

Documents

Transcript of 15 Mixing and Solution