Stoic-solution to Problems

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 1

1-8Given: Reading of main meter: 250 m3; Consumptions in m3: tenant 1, 40; tenant 2, 35.5; tenant 3,54.3; tenant 4, 4.36; and tenant 5, 55.Required: the amount of leakage.Solution: Arithmetic solution (algebraic solution also acceptable)Overall mass balance:Σmass inputs = Σmass outputs

Total watery supplied = supply to tenant 1 + supply to tenant 2 + supply to tenant 3 + supply totenant 4 + supply to tenant 5 + leakage (all in m3)

Leakage = 250 - (40 + 35.5 + 54.3 + 36 + 55) = 29.2 m3

1-10. As an irrigation system delivers water from the source to the point of usage, the following losses areencountered: 9% to evaporation, 7% to run-off, and 15% to infiltration. If 4,000 liters was finally usedwhat was the volume of the water at the source?

evaporation��

��

��

��

��

��

��

��

��

��

��

��

�

�

runoff waterdelivered

Infiltration

watersource

Given: Losses to evaporation, run-off, and infiltration with 4,000 liters finally usedRequired: The volume of the water at the sourceSolution:

Let x = the volume of the water at the source

0.09x + 0.07x + 0.15x + 4000 = xx - 0.31x = 4000

x = 4000/0.69 = 5,797 L

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piping dist.system

leakage

1 2 3

5

4

DRAFT


2-1

a. 8 cp% p100 cp %

gcm $ sp %

kg1000 g %

100 cmm % N

kg $ms2

% PaNm2

= 0.8 Pa$s

b. 150 ft$lbf %

ggc % lblbf % 12 inft % 2.54 cmin % m

100 cm %kg

2.2 lb % 9.8ms2 %

Nkg ms2

= 203.7 J

c. 0.1 N1000 dynes %

g $ cms2dyne %

kg1000 g %

m100 cm % N

kg $ms2

=

f. 0.01634 pints%qts

2 pints %gal4 qts %

ft37.48 gal %

(12 in)3

ft3%(2.54 cm)3

in3% m3

(100 cm)3 =

g. 0.344 m321,000 in3 %(2.54 cm)3

in3% m3

(100 cm)3 =

h. 900 fl. ounce%0.05 pintfl. ounce %

gal8 pints %

ft37.48 gal %

(12 in)3

ft3%(2.54 cm)3

in3% m3

(100 cm)3 = 0.021 m3

i. = 3.371x 104 kg/m32100 lbmft3

%kg

2.2 lbm % ft3(12 in)3 %

in3(2.54 cm)3 %

(100cm)3m3

j. 7.532 x 106 W7150 Btus % 252 calBtu % 4.18 Jcal =

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DRAFT

2.4 Find out which of the following formulas is correct:

or ØCvØV T

= T Ø2PØT2 V

T ØCvØV T

= Ø2PØT2 V

C, temperature, θP, pressure, F/L2

V, specific volume, L3/MCv, heat capacity, FL/MT

Solution:Substituting the dimensions;

Eq. 1: Left side Right side FLMTL3M

= FTL2

T F/L2

T2= FTL2

Eq. 2 Left side Right side TFL/MTL3/M = FTL2

F2/L4T2

= F2L4T2

The first equation is correct since it is dimensionally consistent.

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DRAFT

2-5 Find the value of the following dimensionless groups for the given data:

a. The Reynolds No., NRe =Dv

where D = diameter, Lv = velocity L/θρ = density, M/L3

µ = viscosity, M/Lθ

for a case where D = 16 in, v = 6 m/s, ρ = 4.0 kg/gal and µ = 300 #ft $ h

Solution:

NRe =Dv

=(16 in % 2.54 cmin )(6ms % 100 cmm )(4

kggal %

7.48 galft3

% ft3123 in3

% in32.543cm3 )

300 lbft $ h %

kg2.2 lb %

ft12 in %

in2.54 cm % h

3600 s

NRe = = 2.073 % 104

b. The Schmidt No., NSc = Dvwhere µ = viscosity, M/L

ρ = density, M/L3

Dv = diffusivity, L2/θ

for a case where µ = 12 centipoise, ρ = 30 kg/cu ft, Dv = 745 cm2/h.

Solution:

NSc = Dv =12 cp %

p100 cp %

gcm $ sp

30kgft3

%1000 gkg % ft3

123 in3% in32.543 cm3 % 745

cm2

h % h3600s

0.5473NSc =

c. The Froude No., NFr = v2

Lgwhere v = velocity, L/θ

L = length, Lg = acceleration due to gravity

for a case where L = 6 ft, v = 20 cm/s

NFr = vLg =

(20 cms )2 in2(2.54 cm)2

% ft2(12 in)2

6 ft % 32.2 fts2= 2.229 % 10−3

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DRAFT

2.7 The viscosity of a liquid at the critical point is given by formula

c = 61.6MTc

(Vc)2/3

where µ = critical viscosity in micropoise M = molecular weight

Tc = critical temperature, K Vc = critical volume, cm3/mol

For a similar formula using µc in Pa⋅s, and Vc in m3/kmol, and Tc in K, find the value of the constantcorresponding to 61.6.The constant with the corresponding units is

16 p( cm

3

mol )2/3

K

Converting to the desired units

new constant = 16 P( cm

3

mol %m3

(1000 cm)3 %1000 molkmol )2/3

K % P106 P %

0.1 Pa $ sP

= Pa$s( m

3

kmol )2/3

K

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DRAFT

2-9 is changed to by conversion of units0.086ft ( lb

ft3)0.4

(cP)2(dynecm )3

cm (gcm3 )0.4

( lbft $ h )2(dynecm )3

=0.086 ft % 12 inft % 2.54 cmin ( lbm

ft3%454 glbm % ft3

(12 % 2.54 cm)3 )0.4

(cP %poise100 cP %

1g/cm $ spoise % lb

454 g %12 % 2.54 cm

ft % 3600 sh )2

% 1

(dynecm %

g $ cm/sdyne %

kg1000 g %

m100 cm % n

kg $m/s2% 100 cmm )3

= 8.59 %107c,(g/cm3)0.4

( lbmft $h )2 % ( Nm )3

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DRAFT

2-10. The temperature of a cold water being heated varies linearly with the distance from one end of the heatexchanger. From the following data find the suitable equation representing the variation graphically:

4.112.992.071.130.240Distance, m 32.2225.0019.4411.675.564.44Temperature, °C

Solution:

The equation is T = (slope) + mx T, temp x, distancePlot Temperature on the ordinate and x on the abcissa

Intercept: 4.2751052333Slope: 6.8955757685

Equation: T = 4.275 + 6.896

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Prob-2-10

X , distance in m

0 1 2 3 4 5

T, te

mpe

ratu

re d

eg, C

elsi

us

0

5

10

15

20

25

30

35

DRAFT

2.11 The following table gives the relationship between dissolved methane in a solution and the partial pressureof methane above the solution. (Chem. Eng. Prog. symp. series, Volume 65, 73 (1969).

3447275820681379689138Partial pressure of CH4, kPa56045034022511520Dissolved CH4, ppm

If the variation is linear, find the suitable relationship.

2-14

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Dissolved CH4 , ppm

0 100 200 300 400 500 600

Parti

al p

ress

ure

of C

H4,

kPa

0

1000

2000

3000

4000

DRAFT

5.263.82.51.25x1.953.866.7512.6y

Solution:Plot log y vs x on a rectangular coordinate graph or plot y vs x on semilog graph. We use a semilog plot.

intercept = 1.3437

slope = - 0.2005

equation: log y = -0.2005 x + 1.3437

y = 10−0.2005x + 1.3437

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Problem 2-14

X Data

1 2 3 4 5 6

Y D

ata

1

10

Y vs XPlot 1 Regr

DRAFT

2-18 Convert the following temperature readingsa. 180K to °C, °R, and °F

Rdg in °C = 180 - 273 = -93°CRdg in °R = 180× 1.8 = 324°RRdg in °F = (180× 1.8) - 460 = 136°F

b. 800°F to K, °C, and °R

Rdg in K = (800 - 32)/1.8 + 273 = 699.7 KRdg in °C = (800 -32)/1.8 = 426.7°CRdg in °R = 800 + 460 = 1260°R

c. 100,000°C to K, °R, and °F.

Rdg in K = 100,000 -273 = 99,723 KRdg in °F = (100,000)1.8 + 32 = 180032°FRdg in °R = (100,000)1.8 + 32 -460 = 179572°R

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DRAFT

2-26 Convert the following gage pressures to absolute values if the barometric pressure is 757 mm Hg:

a. 200 psigb. 4 cm of Hg vacuumc. 29 mm of H20 headd. 15 mm of H20 draft

a. Pabs = Pbar +PgagePabs = 757 mm Hg %

14.7 psi760 mm Hg + 200 = 214.6 psi

b. Pabs = Pbar −PvacPabs = 757 mm Hg %

76 cm Hg760 mm Hg −4 cm Hg= 71.7 cm Hg

c. Pabs = Pbar +headPabs = 757 mm Hg + 29 mm H2O%

1 ft H2O12 % 2.54 in %

760 mm Hg33.9 ft H2O = 778.3 mm Hg

d. Pabs = Pbar −draftPabs = 757 mm Hg −15 mm H2O%

1 ft H2O12 % 2.54 in %

760 mm Hg33.9 ft H2O = 746 mm Hg

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DRAFT

2-27 solution

Convert the following pressures to the indicated absolute pressures:a.a. 20.0 psig to atm abs with barometric pressure = 740 mm Hgb.c.Pabs = 740 mm Hg % 1 atm

760 mm Hg + 20 psi % 1 atm14.7 psi = atm abs

d.e.b. 28" Hg vacuum to mm Hg absolute with barometric pressure = 760 mm Hgf.g.Pabs = 740 mm Hg − 28 in Hg %

760 mm Hg29.92 in Hg = mm Hg

h.i.c. 4" H2O draft to mm Hg abs with barometric pressure = 760 mm Hgj.k.Pabs = 760 mm Hg − 4 in H2O % 1 ft

12 in %760 mm Hg33.9 ft H2O

= mm Hg

l.m.d. 5" H2O head to mm Hg abs with barometric pressure = 750 mm Hg.a.b. Pabs = 750 mm Hg +5 in H2O% 1 ft

12 in %760 mm Hg33.9 ft H2O = mm Hg

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DRAFT

2-28 A flue gas has the following composition on a molar basis:

12% CO2 77% N2

2% CO 2% H2O 7% O2

What is the composition in mass percent? What is the average molecular weight of the mixture?

Solution:Basis: 100 mols mixture

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3,0001001.236182H2O

71.872,1562877N2

7.47224327O2

1.8756282CO17.65284412CO2

% massmassMWmolComp

DRAFT

2-29 Some ionic components of sea water are given in the following table:

10.0001F-10.01Sr+2

0.0065Br-10.04K+1

0.01HCO3-10.04Ca+2

0.26SO4-20.13Mg+2

1.90Cl-11.06Na+l

mass %Anionmass %Cation

If the specific gravity of sea water is 1.03, how much NaC1 could be recovered from a cubic km of sea water?How much Mg(OH)2? MW Na = 23 MW Cl = 35.45

Solution:Basis: 1 km3 of sea water

kg1 km3 % 10003 m3

km3 % 1030kgm3 %0.0106

kg Na+kg sea water %

57.45 kg NaCl23 kg Na+ = 2.727 % 1010

1 km3 % 10003 m3

km3 %1030kgm3 %0.0013

kg Mg+2kg sea water %

58.33 kg Mg(OH)224.31 kg Mg+2

= 3.213 %109 kg

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DRAFT

2-33 5.7 kg of sucrose is dissolved in a 20-liter can containing 18 liters of water. What is the sugar concentration ofthe solution formed in

a. mass fraction?b. mole fraction?c. molality?d. molarity? Note: To find molarity,

1.383 at 25°C.

Solution:Basis: 5.7 kg of sucrose

a. the mass fractionkg sucrose = 5.7 kg water = 18 L x 1 kg/L = 18 kgTotal mass = 5.7 + 18 = 23.7 kg

mass fraction sucrose = 5.7/23.7 = mass fraction water = 18/23.7 =

b. the mol fraction (molecular formula of sucrose = C12H22O12

MW sucrose = 12(12) + 22(1) + 12(16) = 342 MW H2O = 18

kmol sucrose = 5.7/342 = 0.0167 kmol water = 18/18 = 1 kmol solution = 1.0167

mol fraction sucrose = 0.0167/1.0167 = mol fraction water = 1/1.0167 =

c. molality

kmol sucrose = 5.7/342 = 0.0167 g H2O = 18,000 g

molality = gmol sucrose per 1000 g H20

gmol sucrose = 16.7 no. of 1000 g H2O = 18therefore, per 1000 g H2O, the gmol of sucrose = 167/18 = and the molality is molal

d. the molarityFor this problem, we need the specific gravity of the sucrose solution, which we can obtain from handbooksor other references.

Percent sucrose by mass = 76%the specific gravity = 1.383 from referencetotal mass = 57,000 + 18,000 = 75,000 gVolume = 75,000 g % ml

1.383 g %1 L

1000 ml = 54.23

molarity = 143/54.23 = 2.64 M

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DRAFT

2.35 Ethane is flowing through a 1 m I.D. pipe at the rate of 6530 kg/hr. If the density of ethane is 1.3567 g/L whatis

a. the flow velocityb. the mass velocityc. the volumetric flow rate?

a. Flow velocity in m/sv =

QA = w

A w = 6530 kg/h 1.3567g/L A = (1)2 /4 m2

v =6530

kgh % 1 h

3600 s

1.3567gL %

kg1000 g %

1000 Lm3 % (1)2 % 4m

2= 1.702 m/s

b. mass velocity in kg/m2⋅s

G = wA =6530

kgh % 1 h

3600 s(1)2 % 4m

2= 2.31 kg/m2 $s

c. vol flow rate in m3/s

Q = w =6530

kgh % 1 h

3600 s

1.3567gL %

kg1000 g %

1000 Lm3

= 1.337 m3/s

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DRAFT

2-36 A lube oil (sp. gr. = 0.85) is pumped to a header at the rate of 4,000 liters/hr. At the header, the flow branchesin two lines. One pipe has an inside diameter of 7 cm while the other, 15 cm. Assuming that the mass flowrate is directly proportional to the cross sectional area of flow, calculate for both pipes

a. the flow velocity in m/sb. the mass velocity in kg/hr·m2

c. the volumetric flow rate in m3/hr

Solution:Basis: 4,000 L/hcross sectional area of 7 cm dia (0.07 m dia)

A = D24 = 0.072

4 = 0.003848

cross sectional area of 15 cm dia (0.15 m dia)

A = D24 = 0.152

4 = 0.01767

Let x = flow through 7 cm dia pipe4000 - x = flow through the 15 dia m pipe

x4000 − x =

D724D1524

= 0.072

0.152 = 49225

225x = 49(4000−x)x = 715 kg/h4,000 - x = 3285 kg/h

a. flow velocitythrough 7 cm dia

v = 715kgh % h

3600 s %m3

850 kg %1

0.003848 m2 = 0.061 m/s

through 15 cm dia

v = 3285kgh % h

3600 s %m3

850 kg %1

0.01767 m2 = 0.061

b. The mass velocity, GG = wA

for the 7 cm dia pipe kg/(s⋅m2)G = wA = 7150.003848 = 1.846 % 106

for the 15 cm dia pipe kg/(s⋅m2)G = wA = 32850.017678 = 1.859 % 105

c. See above.

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DRAFT

2.39 Butane gas is stored in a cylinder at a pressure of 100 psig and at a temperature of 30°C. Some of the butanegas was used and after some time, the pressure had gone down to 50 psig. If the temperature is 25°C, whatfraction of the butane had been used?

Solution:

The pressure is high. We must use the real gas law.

Mass balance

Butane used = Initial butane content - final butane content

= P1V1/z1RT1 - P2V2/z2RT2

fraction used: = =

P1V1z1RT1 − P2V2

z2RT2P1V1z1RT1

=

P1z1T1 − P2

z2T2P1z1T1

=

114.7z1(30 + 273) −

64.7z2(25 + 273)

114.7z1(30 + 273)

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DRAFT

2.45 The flue gas from a furnace goes out with the following composition (dry basis): 12% CO2, 3% O2, 2% CO,and 83% N2. The partial pressure of water is 10 mm Hg. The total pressure is 4 cm H20 draft. The barometricpressure is 759 mm Hg. Calculate the composition on a wet basis (including H20).

Solution:

Basis: 100 mols of flue gas (dry basis

It is best to use tabulation. PH2O = 10 mm Hg PT = 4 cm H2O draft Pbar = 759 mm Hg

99.99101.34100Total1.321.34H2O 81.98383N2

1.9722CO2.9633O2

11.841212CO2

mol % (wet basismol (wet basis)mol (dry basis)Components

Patm = 759 mm Hg

PH2O = 10 mm Hg

PT = Patm - Pdraft = 756.1 mm Hg759 −4 cm H2O%ft H2O

(2.54 $ 12) cm H2O %760 mm Hg33.9 ft H2O =

Pdry gas + PH2O = 756.1 mm Hg

Pdry gas = 756.1 - 10 = 746.1 mm Hg

Mol water =100 mol dry gas% 10 mol H2O746.1mol dry gas = 1.34 mols

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DRAFT


3-7 A producer gas consisting of 8.4% CO2, 0.8% O2, 21.2% CO, 7.9% H2, and 61.7% N2 at a temperature of 30°Cand 770 mm Hg total pressure, with Pwater = 25 mm Hg, flowing at a rate of 500 m3 per minute is mixed with anatural gas consisting of 81.11% CH4, 6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2 at a totalpressure of 2 atm and a temperature of 25°C and flowing at a rate of 100 m3 per minute. Its water vapor content hasa partial pressure of 30 mm Hg. What is the composition of the resulting mixture on a dry basis? What is the watervapor content in grams per kmol of dry gas?Given:

Producer gas: 500 m3 per minute at 30° C and 770 mm total Press PH2O = 25 mm Hg8.4% CO2, 0.8% O2, 21.2% CO, 7.9% H2, and 61.7% N2

Natural gas: 100 m3/min at 25°C and total pres = 2 atm PH2O = 30 mm Hg81.11% CH4, 6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2

Required: Composition of the resulting mixture on a dry basis Water vapor content of the mixture, g/kmol dry gas

Basis: One minute operationMixing of producer gas and natural gas

kmol producer gas = 500 m3

min %0 +27330 + 273 %

770760 %

kmol22.4 m3 = 20.376 kmol (with H2O)

kmol H2O in prod. gas = 20.376 kmol wetb gas % 25770 = 0.684 kmol

kmol dry producer gas = 20.376 - 0.684 = 19.629kmol natural gas = 100 m

3

min % 0 +27325 +273 %

21 %

kmol22.4 m3 = 8.18

kmol H2O in nat. gas = 0.136 kmol dry gas % 302(760) − 30 = 0.165 kmol

kmol dry natural gas = 8.18 - 0.165 = 8.015For the composition of the mixture

kmol dry gas = 28.558

kmol H2O = 0.684 + 0.165 = 0.849

g H2O = (0.849)(18)(1000) = 1.528 x 104

g water vapor per kmol dry gas = (1.528 x 104)/28.558 = 535

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99.97827.734Total0.2120.0590.0074 (8.018) = 0.061C4H10

0.6020.1720.021 (8.018) = 0.168C3H8

1.840.5160.0644 (8.018) = 0.527C2H6

23.236.5030.8111 (8.018) = 6.635CH4

5.641.5570.079 (19.714) = 1.61H2

15.134.1790.212(19.714) = 4.32CO0.5710.1580.008 (19.714) = 0.163O2

46.6612.90.0919(8.018) = 0.7520.617 (19.714) = 12.572N2

6.0931.690.0042(8.018) = 0.0340.084(19.714) = 1.656CO2

mol %kmolNat. gasProducer gasComp.

DRAFT

3-8 Moist air at 40°C and at a total pressure of 765 mm Hg is flowing through a pipeline. The partial pressure ofwater is 20 mm Hg. In order to measure the rate of flow, pure carbon dioxide gas is bled into the gas at rate of 20kg/min. At a point downstream where complete mixing has occurred, it is found that the air contains 5% CO2 byvolume. Find the volumetric flow rate of the moist air in m3/min. The % CO2 is on a dry basis.

Given:

mixed gases5% CO2 vol

moist air40oCPt = 765 mm HgPH2O = 20 mm Hg

pure CO2 gas20 kg/min

Required: Vol flow rate of moist air in m3/min

Basis: 1 minute operation

Solution:

Vol flow rate of moist air =

20kg CO2min % kmol44 kg %

0.95 kmol dry air0.05 kmol CO2 % 765 kmol moist air

(765 −20) kmol dry air %22.4 m3

kmol % 40 +2730 +273 % 760765

= 226.3 m3/min

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DRAFT

3-9

Basis: 1000 kg wet sand

Required: water evaporated

A. Algebraic method

Let x, water evap; y, dried sand

OMB: 1000 = x + y Eq. 1

BDS balance: (1-0.3)(1000) = (1- 0.09)y Eq. 2

Subst y = 1000 - x in Eq. 2

700 = 0.91(1000 - x)

x = 910 - 700 = 210/0.91 = 230.8 kg H2O evap

Arithmetic solution:

Basis: 1000 kg feed

Required: water evaporated

kg product =1000 kg feed %(1 − .3) kg BDS

kg feed %kg product

(1 − 0.09) kg BDS =

769.2 kg prod

OMB:

kg H2O evap = 1000 - 769.2 = 230.8 kg

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DRAFT

3-10 Wood chips are to be dried in a rotary drier from 35 to 12% moisture. If 10 kg of wet wood is charged perminute, how much product can be obtained over an 8-hr period? Find the ratio kg H2O evaporated/kg wet woodcharged

Drier

Water evaporated

10 kg wet wood/min

Driedproduct

12%moisture

36 %moisture

Required:Product for 8 hr periodratio kg water evap./kg wet wood charged.

Solution:We can identify a key component, which is the bone-dry solids. Therefore, the arithmetic method is

recommended.

Basis:1 min (10 kg wet wood charged)

kg dry prod/min =10 kg wet wood(1 − 0.35) kg BDS1 kg wet wood %

1 kg product(1 − 0.12) kg BDS = 7.386 kg

Overall mass balance:

For the 8 hr period:kg dry prod = 7.386 kg dry prod/min% 60min/hr % 8 hr= 3545 kg

kg water evap/kg wet wood charged = 2.614/10 = 0.261

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DRAFT

3-11 A drier is being used to reduce the moisture content of bagasse to be used for fiberboard manufacture. Itcontains 50% moisture. 5.75 kg of wet bagasse is charged into the drier. After 3 hours, it was found that theweight has gone down to 3.25 kg. What is the moisture content of the final product?

Using the arithmetic method:

Given: 5.75 kg wet bagasse with 50% moisture

weight of product = 3.25 kg

Required: Moisture content of the final product

Basis: 5.75 kg of wet bagasse

BDS balance:

BDS in feed = BDS in product

BDS in feed = 0.5(5.75) = 2.875 kg = BDS in product =

moisture in product = 3.25 - 2.875 = 0.375 kg

% moisture content = (0.375)(100)/3.25 = 11.54%

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DRAFT

3-14 It is desired to concentrate a 6% KN03 solution (in water) to 20% KN03 solution. 7000 kg of product liquor isto be needed per hour. How much solution should be charged? How much water is evaporated?

Given: input solution = 6% KNO2

7,000 kg/h product containing 20% KNO3

Required: solution charged; kg water evaporated.

Basis: 1 hour operation

Solution:

kg KNO3 = 0.2(7000) = 1400 kg

KNO3 balance:

KNO3 in product = KNO3 in feed = 1400 kg

kg feed input = = 23,300 kg1400 kg %100 kg solution6 kg KNO3

OMB

kg water evaporated

= kg input solution - kg product = 23300- 7000 = 16,300 kg

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DRAFT

3-19 An evaporator is concentrating solutions coming from two different sources. The solution from the firstsource containing 10% NaCl and 10% NaOH8% NaC1 and 12% NaOH flows at the rate of 70 kg per minute. The two streams are fed directly to the evaporator.If 50% of the total water is to be evaporated, calculate the composition and the flow rate of the product.Given:

Evaporator

10% NaCl10% NaOH

water evap =50% of total

H2O50 kg/min

Product70 kg/min8% NaCl

12% NaOH

Required: Composition and flow rate of productBasis: 1 min operationSolution: We can use the aritmetic solution.

Feed Stream 1: Feed Stream 2kg NaOH = 0.1(50) = 5 kg kg NaOH = 0.12(70) = 8.4kg NaCl = 0.1(50) = 5 kg kg NaCl = 0.08(70) = 5.6kg H2O = 0.8(50) = 40 kg H2O = 0.8 (70) = 56

Product composition:kg NaOH = 5 + 84 = 13.4 % NaOH = (13.4)(100)/72 = 18.6%kg NaCl = 5 + 5.6 = 10.6 % NaCl = (10.6)(100)/72 = 14.7%kg H2O = 0.5(40+56) = 48 kgkg product = 13.4 + 10.6 + 48 = 72 kg product flow rate = 72 kg/min

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DRAFT

3-20 A mixture containing 70% methanol and 30% water is to be distilled. If the distillate product is to contain99.9% methanol and the bottoms product 0.004% methanol, how much distillate and bottoms product areobtained per 100 kg of feed distilled?

Given:

Bottoms, B

Distillate , D

70% methanol30% water

Feed = 100 kg

xm = 0.999

xM = .00004

Required: kg distillate and kg bottomsBasis: 100 kg feedSolution: We cannot identify a tie component.We have to use the algebraic solution.

Let D = kg distillate B = kg bottoms

Overall mass balance: Eq. 1100 =D +B

Methanol balanceEq. 20.7(100) = 0.999D +0.00004B

from Eq 1, B = 100−D

Substiting in Eq. 2, 70 = 0.999D+0.00004(100−D)

Solving for D, D = 70.07 kg B = 29.93 kg

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3-21 100 kmol of a benzene-toluene mixture in equimolal amounts is distilled. 96% of the benzene is obtained inthe distillate product while 90% of the toluene is in the bottom. Calculate the composition of the product streams.Given:

Bottoms, B

Distillate , D

benzene-toluenesolution, equimolar

Feed,F = 100 kmol

90% of thetoluene goes tothe bottoms

96% of thebenzene goesto the distillate

Required: Composition of the distillate and the bottomsBasis: 100 kmol feedSolution: While we cannot identify a tie component in this problem, we can analyze the problem and see that thegiven quantities make the problem amenable to arithmetic solution and we do not have to use the algebraic solution.

From the given quantities, benzene in the feed = 50 kmol

96% of the benzene in the feed goes to the distillate = 0.96(50) = 48 kmolthe difference goes to the bottoms: 50 - 48 = 2 kmol

90% of the toluene in the feed goes to the bottoms = 0.9(50) = 45 kmolthe difference goes to the bottoms: 50 - 45 = 5 kmol

Therefore the distillate contains 48 kmol benzene and 5 kmol tolueneThe composition, mole fraction benzene = 48/(48 + 5) = 0.906

The bottoms contains 45 kmol toluene and 2 kmol tolueneThe composition, mole fraction benzene = 2/(45 + 2) = 0.043

While in general, problems without tie components are usually solved by algebraic methods, someproblems are so structured that they can be solved by arith metic method.

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DRAFT

3-23 A mixture consisting of isobutane, n-butane, and n-pentane is to be distilled in batch mode. The followingtabulation gives the composition of the streams in mole fraction:

z0.0240.45n-pentane0.3y0.32isopentane0.129x0.23n-butanebottomsdistillatefeedComponent

Per 100 moles of feed, find the amount of the distillate and the bottom streams. Complete the tabulation ofthe composition.

Solution:

Given: tabulation of the composition with unknowns

Basis: 100 moles of feed

Examining the table, we see that in the bottoms column, z can be solved from the equation,

z + 0.129 + 0.3 = 1 and z = 0.571

The unknown compositions are x and y

Let D = mol distillate and B mol bottoms

We have four unknowns and we need four equations

O.M. Balance:

D + B = 100 (1)

n-butane balance:

0.23(100) = xD + 0.129B (2)

isopentane balance:

0.32(100) = yD + 0.3B (3)

n-pentane balance:

0.45(100) = 0.024D + 0.571B (4)

Substituting the value of D (from Eq. 1) in Eq. 4

45 = 0.024(100 - B) + 0.571B

B = 77.88 mol D = 22.12 mol

Substituting the value of B and D in (2)

0.23(100) = x(22.12) + 0.129(77.88)

x = (23 - 0.129(77.88))/22.12 = 0.586

Substituting the value of B and D in (3)

0.32(100) = y(22.12) + 0.3(77.88)

y = (32 - 0.3(77.88))/22.12 = 0.390

As a check, the sum of the composition of the distillate = 0.390 + 0.586 + 0.024 = 1

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3-24 Ethane is to be removed in an absorber from an ethane-hydrogen mixture by using a mixture ofhexane containing 2% ethane. The feed contains 40% ethane and 60% hydrogen. If 99% of the ethane isremoved, what is the composition of the effluent at the top and at the bottom? The liquid to gas molar ratio is2.

Given:

Required: Composition of outlet gas and liquid

Basis: 100 kmol gas 200 kmol of liquid

Hexane in: 0.98(200) = 196 kmol

Ethane in with hexane = 0.02(200) = 4 kmol

Ethane carried by hydrogen in: 0.4(100) = 40kmol

Ethane absorbed = 0.99(40) = 39.6 kmol

Hydrogen in = 0.6(100) = 60 kmol

Liquid out = 196 kmol hexane + 39.6 kmolsethane + 4

Composition of outgoing liquid = 18.2 %43.643.6 + 196 % 100 =

Gas out = 60 + 0.4 = 60.4

Composition of outgoing gas = 0.40.4 + 196 % 100 = 0.662%

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40% ethane60% hydrogen

Absorber

Hexane +absorbedethane

hydrogen andethane

2

Hexane2% ethane

99% of theethane isremoved

1

DRAFT

3-26 In order to recover the benzene from a benzene-air mixture, the mixture is scrubbed withkerosene in a packed tower. 8000 m3/h of a benzene-air mixture containing 2 mole % benzene (T = 25°C,P = 765 mm Hg) is treated with 10,000 kg/h of kerosene. If 98% of the benzene is recovered, find thecomposition of the gas stream and liquid stream leaving the tower. What is the volumetric flow rate ofthe outgoing gas stream (T = 25°C, P = 759 mm Hg)?

Solution:

Basis: 1 hour operation

kmol air-benzene in = 8000m3

h % 0 +27325 +273 %

765760 %

kmol22.4 m3 = 325 kmol

kmol air = (0.98)(325) = 318.5 kmol

kmol benzene = 0.98(325) = 6.5 kmol

kmol benzene in gas out = 0.02(6.5) = 0.13 kmol

kmol benzene in liquid out = 0.98(6.5) = 6.37 kmol = 6.37(80) = 509.6 kg

composition of air-benzene out = 0.13(100)318.5 + 0.13 = 0.041% mol benzene

composition of benzene-kerosene mixture = 509.6(100)

10,000 +509.6 = 4.85%

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25oC 759 mm Hg

8000 m3/hAir-benzene mixture2% mol benzene25oC 765 mm Hg

Absorber

kerosene +absorbed benzene

Outlet gas

2

Kerosene

10,000 kg/h

98% of the benzeneis recovered

1

DRAFT

3-34 Eight hundred kg/h of halibut livers containing 23% oil is extracted with 570 kg/h of pure ether. The extractedlivers is analyzed and is found to contain 1.12% oil, 32.96% ether, and 65.92% oil-free livers. Find the compositionand the weight of the extract, and the percentage recovery of the oil.

Given:

Extract

Extracted liversExtractor

800 kg/hr23% oil

1.12 % oil32.96% ether

65.92% oil-free livers

570 kg/hr ether

Halibut livers

Required: Composition and mass of the extractBasis: 800 kg of halibut liversSolution: The oil-free livers is a tie component. We can use the arithmetic method.

kg livers = 800 kgkg oil = 0.23(800) = 184 kgkg oil-free livers = 800 - 184 = 616 kgOil-free livers balance:Oil-free livers in feed = oil-free livers in extracted livers = 616 kg

Oil in extracted livers = 616 kg oil-free livers %1.12 kg oil

65.92 kg oil-free livers = 10.466 kg

Ether in extracted livers = 616 kg oil-free livers %32.96 kg ether

65.92 kg oil-free livers = 308 kg

Oil balance:Oil in extract = oil in charge - oil in extracted livers = 184 - 10.466 = 173.53 kg

Ether balance:Ether in extract = ether input - ether in extracted livers = 570 - 308 = 262 kg

kg extract = ether + extracted oil = 173.53 + 262 = 435.53 kg% oil = 173.53(100)/435.53 = 39.84%% oil recovery = 173.53(100)/184 = 94.31%

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3-40 A calcium carbonate slurry is to be filtered in a plate-and-frame filter press with 15 frames measuring 90 cm ×120 cm × 2.5 cm. The solids in the slurry is 5 mass percent. The filter cake is 32% porous. Assuming that thefilter cake is incompressible, how much volume of the slurry is required to fill the filter press? The specificgravity of the solids in the slurry is 2.63. The density of the liquid is 1.0. Assume that the specific gravity ofthe slurry is almost equal to 1.

Given: filter press with 15 frames 90 cm x 120 cm x 2.5 cm (assume the inside volume)material to be filtered: 5% (mass) calcium carbonate slurryfilter cake: 32% porous; incompressibleSp. g of solids 2.63 sp. g. of liquid and slurry = 1

Required: Volume of slurry required to fill the filter press

Solution: arithmetic method

Total volume of the frames: = volume of the cake0.9m % 1.2 m % 0.025 m = 4.05 m3

Volume of the solids in the filter cake = (1 - 0.32)(4.05) = 2.754 m3

density of the solid = 2630 kg/m3

kg solids in filter cake = 2.754 m3 % 2630 kg/m3 = 7245 kg

liquid in the slurry = kg slurry7243 kg solids %1.0 kg slurry0.05 kg solids = 1.449 % 105

volume of the slurry = 144900 kg slurry % 1 m3

1000 kg = 144.9 m3

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3-47 A double-effect evaporator is to concentrate 1,000,000 kg/day of a liquor containing 5% solids to 40% solids.Assuming equal evaporations are obtained from each effect, calculate the composition of the solution fromthe first effect (if forward feeding is used) and the flow rate of the product in kg/h. How much evaporationfrom each effect is obtained?

Given:

E1 = E2

Required:

a. Composition of the product from the first effect

b. The flow rate of the product in kg/h

c. Evaporation from each effect

Solution:

Basis: One-day operation

Use combination of arithmetic and algebraic methods

Let E1 equals the evaporation from Effect No. 1

E2 equals the evaporation from Effect No. 2

Overall Mass Balance around the whole system, solution side

Eq. 11,000,000 = P2 +E2 +C2Eq. 2C2 =E1Eq. 3E1 =E2

Solids balance around the whole system, solution side

Eq. 40.05(1,000,000) = 0.4P2

P2 =0.05(1,000,000)

0.4 = 125,000kgday %

day24 h = 5208 kg/h

Subst Eq 2 and Eq 3 in Eq1

1,000,000 = 125,000+E1 +E1E1 = (1,000,000−125,000)/2 = 875,000 kg/day = 36,460 kg/h

E2 =E1 = 36,460 kg/h

Overall Mass Balance around effect 1

1,000,000 =E1 +P1

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E1.

F = 106 kg/day

E2

P2P1

C2C1

S

Effect 1 Effect 2xF = 0.05 xP1 xP2 = 0.4

DRAFT

= 23,440 kg/hP1 = 1,000,000−437,500 = 562,500 kg/day

xP1 =0.05(1,000,000)

562,500 = 0.089

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DRAFT

3-48. It is desired to produce 7% NaNO3 solution continuously. The water line (NaNO3-free) is split into two. 500 kg per hour issent to a tank where NaNO3 is added. The mixtureis stirred well to form a saturated solution ofNaNO3 (47.9%). The other line bypasses the tankand is mixed with the 47.9% solution. What is theflow rate of the bypass stream and the finalproduct?

Solution:Basis: 1 hr operationRequired: Flow rate of bypass stream and the finalproductBalance around saturator inside the bypass loop

Water balancewater in 47.9% NaNO3 solution = 500 kg

NaNO3 balance:

NaNO3 input = 500 kg water %47.9 kg NaNO3

(100 − 47.9) kg H2O= 459.693 kg

Balance around saturator outside the bypass loopNaNO3 balance:

NaNO3 input = 459.693 kg = NaNO3 in 7% NaNO3 solutionWater balance:

Water in 7% NaNO3 solution = 459.693 kg NaNO3 %93 kg H2O7 kg NaNO3

= 6107.35 kg

Balance around point of splitting of bypass streambypass stream = 6107.35 - 500 = 5607.36 kg/h

product stream = 459.693+6107.35 = 6567.04 kg/h

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Water

Bypass stream

Saturator500kg/hr 47.9%NaNO3

7%NaNO3

NaNO3

DRAFT

3-49 A process needs an air supply which should contain 0.12 mol H2O/ mol dry air exactly. 1500 m3/min of air at 25°C and101.3 kPa is to be treated. Part of this air goes to a spray chamber where the air picks up water and goes out with 0.3 molH2O/mol dry air. The other part of the air feed bypasses the water spray and is mixed with the humidified air to produce amixture containing 0.12 mol H2O per mol dry air. What is the water consumption? What is the ratio of the flowmeter readingsof the bypass stream and the stream to the water spray?


Using the arithmetic method

kmol bone-dry air = 1500 m3

min % 0 + 27325 + 273 %

kmol22.4 m3 = 63.35 kmol BDA

BDA balance around spray chamber outside the bypass loopBDA in feed = BDA in process air

= 61.35 kmol BDAkmol H2O in process air = 61.35 kmol BDA % 0.12 kmol H2O

kmol BDA = 7.36 kmol H2O

H2O balance around spray chamber outside the recycle loopwater spray = H2O in process air = 7.36 kmol = 7.36 x 18 = 132.14 kmol H2O/ min

H2O balance around the spray chamber within the bypass loop

water in stream of air going out of the chamber = 7.36 kmol

BDA balance around the spray chamber within the bypass loop

BDA in net feed = BDA in air out or chamber

24.53 kmolBDA = 7.36 kmol H2O % kmol BDA0.3 kmol H2O

=

2.52bypass stream

stream to spray chamber = 61.3524.53 =

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Dry Air1500 m3

25oC101.5 kPa

Bypass stream

Spray chamber

0.12 kmol H2O/kmol dry airNet Feed

0.3 kmol H2O/kmol dry air

water

DRAFT

3-50 An air conditioning system supplies 1000 m3/min of air containing 0.01 mol H2O/mol dry air. It is at 20°Cand 1 atm. To conserve energy, part of the exhaust air containing 0.08 mol H2O/mol dry air is recycled andmixed with the fresh air from the air conditioner to produce a gross air feed to the room containing 0.035mol H20/mol dry air. How many kg of water is picked up by the air per minute? What is the volumetric flowrate of the recycle stream? (27°C and 99 kPa)

Given:

Required: a. kg water picked up by the air per minute

b. volumetric flow rate of the recycle stream


Bone-dry air balance around whole system

bone-dry in = bone-dry out = 1000 m3

min %0 +27320 +273 %

kmol22.4 m3 = 41.6 kmol

Water in air in = 41.6 kmolmin % 0.01 kmol H2Okmol bda = 0.416 kmol

Water in air out = 41.6 kmolmin % 0.08 kmol H2Okmol bda = 3.328 kmol

Water picked up by the air per minute

Using water balance,

Water picked up by the air =

(3.328 kmol water outkmol bda − 0.416 kmol water inkmol bda ) % 18kgkmol = 52.42 kg

Balance around point of mixing

Let R = kmol recycle (dry basis)

water balance

41.6(0.01) +0.08R = 0.035(R+41.6)Solving for R,

R =41.6(0.035 − 0.01)0.08 − 0.035 = 23.11 kmol

Volumetric flow rate of recycle = 23.11 kmol %22.4 m3

kmol %27 + 2730 + 273 % 101.32599 = 582.2 m3

room inside recycle loop:

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0.035 mol H20per mol dry air

1000 m3 air20oC 1 atm

Recycle stream

0.01 kmol H2O/kmol dry air

27oC and 99 kPa

Room

0.08 mol H20 permol dry air

DRAFT

Water picked up by the air = 52.42 kg

Let x = kmol dry air entering the room = kmol dry air leaving the room

Water balance:

0.08x−0.035x = 52.42/18

x = 52.42(18)(0.025) = 64.72 kmol

Balance around point of splitting,

Kmol recycle = 64.72 - 41.6 = 23.12 kmol

Volumetric flow rate of recycle = 23.12 kmol %22.4 m3

kmol %27 + 2730 + 273 % 101.32599 = 582.5 m3

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DRAFT


4-3For each of the following reactions, calculate the amount of the compounds asked for: (Note: The equations are not balanced.)

a) Benzene + Oxygen → Carbon dioxide + water per 120 kg of benzene burned, how many kilograms CO2 and kilograms H2O are produced? How much oxygen is

needed?b) NaCl + H2O + SO3 → Na2SO4 + HC1

Per 300 kg NaC1, how much Na2SO4 and HC1 are formed?Solution:a) C6H6 + 7.5O2 → 6CO2 + 3H2OBasis: 120 kg benzene

kgkg CO2 = 120 kg C6H6 %6MWCO2MWC6H6

= 120 %6(44.01)78.114 = 405.7

kgkg H2O = 120 kg C6H6 %3MWH2OMWC6H6

= 120 %3(18.015)78.114 = 83.03

368.7 kgkg O2 = 120 kg C6H6 %7.5MWO2MWC6H6

= 120 %7.5(31.999)78.114 =

b) 2NaCl + H2O + SO3 → Na2SO4 + 2HC1Basis: 300 kg NaCl

kgkg Na2SO4 = 300 kg NaCl % MWNa2SO42MWNaCl = 300 % 142.04

2(58.44) = 364.6

kgkg HCl = 300 kg NaCl % 2MWHCl2MWNaCl = 300 %

2(46.46)2(58.44) = 187.2

c) Ca3(PO4)2 + 3SiO2 + C → 3CaSiO3 + 2P + CO252

52

Per 100 tons Ca3(PO4)2 how much of each of the products are formed?Basis: 100 tons Ca3(PO4)2

tons CaSiO2 = 100 tons Ca3(PO4)2 %3MWCaSiO3MWCa3(PO4)2 = 100 %

3(116.16)(310.18) = 112.4 tons

tons P = 100 tons Ca3(PO4)2 % 2MWPMWCa3(PO4)2 = 100 %

2(30.97)(310.18) = 19.97 tons

tons CO2 = 100 tons Ca3(PO4)2 %(5/2)MWCO2MWCa3(PO4)2 = 100 %

(5/2)(44.01)(310.18) = 35.47 tons

d) C2H6 + (7/2)O2 → 2CO2 + 3H2OHow many kg CO2 and H2O can be produced from 13 kg of ethane?

kgkg CO2 = 13 kg C2H6 %2MWCO2MWC2H6

= 13 %2(44.01)32.07 = 38.05

kgkg CO2 = 13 kg C2H6 %3MWH2OMWC2H6

= 13 %3(18.015)32.07 = 32.35

e) C12H22O11 (sucrose) + H2O → C6H12O6

C6H12O6 → C2H5OH (ethyl alcohol) + CO2

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DRAFT

4-6 Basis: 1,000 kg of fertilizer

MW (NaNO3)= 85.0kg N2 = 0.02(1000) = 20 kg MW Ca3(PO4)2 =310.2kg P2O5 = 0.12(1000) = 120 kg MW KCl = 74.55kg K2O = 0.06(1000) = 60 kg MW P2O5 = 141.94 MW K2O = 94.18

kg 98% NaNO3 needed =

20 %2(MW NaNO3)

MW N2 %1 kg 98% NaNO30.98 kg NaNO3

= 20 %2(85)28 % 1

0.98 = 123.8 kg

kg Ca3(PO4)2 = 120 %MW Ca3(PO4)2MW(P2O5) = 120 % 310.2

141.94 = 262.5 kg

kg 97% KCl =

60 %2(MW KCl)MW K2O %

1 kg 97% KCl0.97 kg KCl = 60 %

2(74.55)94.18 % 1

0.97 = 97.93 kg

Overall Mass Balance:kg inerts = kg mixed fertilizer - (kg 98% NaNO3 + kg 97% KCl + kg Ca3(PO4)2)kg inerts = 1,000 - (123.8 + 262.5 + 97.93) = 515.8 kg

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DRAFT

4-7 Basis: 1,000 kg of fertilizer 2% N2, 12% P2O5, 6% K2O

Calcium nitrate, purePhosphoric acid KNO3

Inert fillers.

kg N2 = 0.02(1000) = 20 kg MW Ca(NO3)2 = 226.1 kg P2O5 = 0.12(1000) = 120 kg MW H3PO4 = 98 kg K2O = 0.06(1000) = 60 kg MW KNO3 = 101.1

MW P2O5 = 141.94 MW K2O = 94.18

kg KNO3 needed = 60 %2(MW KNO3)MW K2O = 60 %

2(101.1)94.18 = 128.8 kg

N2 from KNO3 = 17.84128.8 kg KNO3 %MWN2

2MW KNO3%= 128.8 % 28

2(101.1) =

N2 balance

N2 from Ca(NO3)3 = total N2 - N2 from KNO3 = 20 - 17.84 = 2.16

g Ca(NO3)3 = 2.16 kg N2 %2MW Ca(NO3)33MW(N2) = 2.16 %

2(226.1)3(28) = 11.63 kg

kg 97% H3PO4 =

120 %2(MW H3PO4)MW P2O5

= 1202(98)141.9 = 165.8 kg

Overall Mass Balance:kg inerts = kg mixed fertilizer - ( kg KNO3 + kg Ca (NO3)3 KCl + kg H3PO4)2)kg inerts = 1,000 - (128.8 + 11.63 + 165.8) = 693.8 kg

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DRAFT

4-8 To make 5,000 kg of mixed fertilizer containing 6% N2, 12% P2O5 and 12% K2O, find the amount of thefollowing ingredients necessary:

(NH4)3PO4

Ca3(PO4)2

KNO3

Inerts fillers

Solution:Basis: 5,000 kg mixed fertilizer

MW (NH4)3PO4 = 149kg N2 = 0.06(5000) = 300 kg MW Ca3(PO4)2 = 214.24kg P2O5 = 0.12(5000) = 600 kg MW KNO3 = 101.09kg K2O = 0.12(5000) = 600 kg MW P2O5 = 141.94 MW K2O = 94.18

kg KNO3 needed = 600 %2(MW KNO3)MW K2O = 600 %

2(101.09)94.18 = 1288 kg

N2 balance:N2 in mixed fertilizer = N2 from KNO3 + N2 from (NH4)3PO4

N2 from KNO3 = 1288 % 2(14)101.09 = 173.376

N2 from (NH4)3PO4 = 300 - 173.376 = 121.624

kg (NH4)3PO4 = 600 %2(MW (NH4)3PO4)

3(MW N2) = 600 %2(149)3(28) = 431.176 kg

kg P2O5 from (NH4)3PO4 = 431.476 % MW P2O52(MW (NH4)3PO4)

= 431.476 % 141.942(149) = 205.516

P2O5 balance:P2O5 in mixed fertilizer = P2O5 from (NH4)3PO4 + P2O5 from Ca3(PO4)2

P2O5 from Ca3(PO4)2 = 600 - 205.516 = 394.484 kg

kg Ca3(PO4)2 = 394.484 %MW Ca3(PO4)2MW(P2O5

= 394.484 % 212.24141.94 = 589.864 kg

Overall Mass Balance:kg inerts = kg mixed fertilizer - (kg KNO3 + kg (NH4)3PO4 + kg Ca3(PO4)2)kg inerts = 5,000 - (1288 + 431.1746 + 589.864) = 2691 kg

Answer:kg inerts = 2691 kgkg KNO3 = 1288 kgkg (NH4)3PO4 = 431 kgkg Ca3(PO4)2 = 590 kg

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DRAFT

4-10 Solution:

Basis: 1,000 kg of solution at 30°C

We have no tie component and we have to use the algebraic methodLet x = Na2SO4⋅10H2O produced

1,000 - x = kg solution at 15°C (Using an overall mass balance)

Na2SO4 balance: Na2SO4 in feed = Na2SO4 in Na2SO4⋅10H2O + Na2SO4 in final solution

1000 %40 kg Na2SO4

(100 +40)kg solution = x MW Na2SO4MW Na2SO4 $ 10H2O + (1000 − x) %

13.5 kg Na2SO4(100 +13.5) kg solution

1000 % 40140 = x 142

322.2 + (1000 − x) 13.5113.5

Solving for x,

= 518.3 kg Na2SO4⋅10H2Ox =40000140 − 13500113.5142322.2 − 13.5

113.5

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DRAFT

4-13 For the following reactions, find the composition the final product. If the second reactant used is 10% in excess of thetheoretical amount or complete reaction. (The first is the limiting reactant) The reaction is 100% complete and the reactantsare pure.a. 2KCl + H2SO4 → K2SO4 + 2HCl

b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2 Solution:

a. 2KCl (s) + H2SO4(l) → K2SO4(s) + 2HCl(g)MW Cl: 35.45 H: 1 S: 32 O: 16 K: 39basis: 100 kg KClkg theoretical H2SO4 needed = kg100 kg KCl % MWH2SO4

2MWKCl = 982 % 74.45 = 65.82

excess H2SO4 needed = 0.1( 65.82 ) = 6.58 kg

kg K2SO4 produced = kg100 kg KCl % MWK2SO42MWKCl = 174

2 % 74.45 = 116.9

kg HCL produced = kg100 kg KCl % MWHClMWKCl =36.4574.45 = 48.96

product composition: HCl is a gas and normally will not mix with the other substances. The product mixture consist ofH2SO4 and K2SO4.

%H2SO4 = 6.586.58 + 116.9 % 100 = 5.33%

%K2SO4 = 116.96.58 + 116.9 % 100 = 94.67%

b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2

Basis: 100 kg Na2CO3

kg theoretical Br2 needed = 100 kg Na2CO3 %MWBr2

MWNa2CO3= 159.8106 = 150.8 kg

excess Br2 needed = 0.1(150.8) = 15.08 kg

kg NaBr produced = 100 kg Na2CO3 % 5MWNaBr3MWNa2CO3

=5(102.8)3(106) = 32.33 kg

kg NaBrO3 produced = 100 kg Na2CO3 %MWNaBrO33MWNa2CO3

= 342.63(106) = 107.7 kg

kg product = 15.08 + 32.33 + 107.7 = 155.1 kg

% Br2 = 15.08155.1 %100 = 9.722%

% NaBr = 32.33155.1 %100 = 20.84%

% NaBrO3 = 107.7155.1 %100 = 69.44%

CO2 is a gas and will not be in the solid and liquid substances.

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DRAFT

4.14a For the following, the reaction goes only to 95% completion. Pure reactants and theoretical amounts areused. For every 500 kg of the impure products, how much of each of the reactants are used?a. KCl +NaNO3 → KNO3 + NaCl

Basis: 100 kg KCl

kg NaNO3 needed = 100 kg KCl %MWNaNO3MWKCl = (100)(84.995)

74.551 = 114.009

kg KNO3 produced = 100 kg KCl %MWKNO3MWKCl % 0.95 = (100)(101.103)(0.95

74.551 = 128.835

kg CaCl2 produced = 100 kg KCl %MWNaClMWKCl % 0.95 =(100)(58.442)(0.95)

74.551 = 74.473

The product consists of KNO3, NaCl2, and unreacted KCl and NaNO3. unreacted KCl = 5 kg unreacted NaNO3 = 0.05(114.009) = 5.7 kg

kg product = 5 + 5.7 + 128.835 + 74.473 = 214.008 kg

By ratio and proportion, we can obtain the reactants needed to form 500 kg product. kg KCl : 500 = 100 : 214.008

kg KCl = 500 % 100214.008 = 233.636 kg

kg NaNO3 : 500 = 114.009 : 214.008

kg NaNO3 = 500 % 114.009214.008 = 266.366 kg

4.14b For the following, the reactions go only to 95% completion. Pure reactants and theoretical amounts areused. For every 500 kg of the impure products, how much of each of the reactants are used?

MgO + CaCl2 + CO2 → MgCl2 +CaCO3

Basis: 100 kg MgO

kg CO2 needed = 100 kg MgO %MWCO2MWMgO =

(100)(44.01)(40.3) = 109.2 kg

kg CaCl2 needed = 100 kg MgO %MWCaCl2MWMgO =

(100)(111)(40.3) = 275.4 kg

kg MgCl2 produced = 100 kg MgO%MWMgCl2MWMgO % 0.95 =

(100)(95.21)(0.95)40.3 = 224.4 kg

kg CaCO3 produced = 100 kg MgO %MWCaCO3MWMgO % 0.95 =

(100)(100.1)(0.95)40.3 = 236 kg

The product consists of MgCl2, CaCO3, and unreacted MgO and CaCl2 unreacted MgO = 5 kg unreacted CaCl2 = 0.05(275.4) = 13.77kg

kg product = 5 + 13.77 + 224.4 + 236 = 479.2 kg

By ratio and proportion, we can obtain the reactants needed to form 500 kg product.

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kg MgO : 500 = 100 : 479.2 kg MgO = 500479.2 % 100 = 104.3 kg

kg CaCl2 : 500 = 275.4 : 479.2 kg CaCl2 = 500479.2 % 275.4 = 287.4 kg

kg CO2 : 500 = 109.2 : 479.2 kg CO2 = 500479.2 % 109.2 = 113.9 kg

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4-18 100 kg of pure carbon is burned. For each of the following cases, calculate the composition of thecombustion gases:

a. Theoretical amount of pure O2 used; complete combustionb. Theoretical amount of air used; complete combustionc. 30% excess air is used: 90% of the C burns to CO2 and 10% to COd. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid.

Basis: 100 kg pure carbon

a. Theoretical pure O2 used; complete combustionReaction: C + O2 → CO2

katom C = 100/12 = 8.333 katomkmol CO2 formed = 8.333 kmolO2 needed = 8.333 kmolcombustion gas = 100% CO2

b. Theoretical air used; complete combustionkmol CO2 formed = 8.333 kmolTheo O2 needed = 8.333 kmolN2 from air = 8.333(79/21) = 31.348 kmolTotal mols combustion gas = 8.333 + 31.348 = 39.681 kmol% CO2 = 8.333(100)/39.681 = 21%% N2 = 31.348(100)/39.681 = 79%

c. 30% excess air is used: 90% of the C burns to CO2 and 10% to COTheo O2 needed = 8.333excess O2 = 0.3(8.333) = 2.5 kmoltotal O2 needed = 1.3(8.333) = 10.833 kmolN2 from air = kmol10.833(7921 ) = 40.753CO2 formed = 0.9(8.333) = 7.5 kmolCO formed = 0.1(8.333) = 0.833 kmolO2 left due to incomplete reaction to CO = 0.833/2 = 0.416 kmolO2 in combustion gas = 2.5 + 0.416 = 2.92 kmol

10052Total78.3640.75N2

1.60.83CO5.622.92O2

14.427.5CO2

%kmolComponent

d. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid.

Theo O2 needed = 8.333excess O2 = 0.2(8.333) = 1.667 kmoltotal O2 needed = 1.2(8.333) = 10 kmol

N2 from air = kmol10(7921 ) = 37.619CO2 formed = 0.9(8.333) = 7.5 kmolO2 left due to incomplete reaction = 0.833 kmolO2 in combustion gas = 1.667 + 0.833 = 2.5 kmol

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10047.62Total7937.62N2

5.252.5O2

15.757.5CO2

%kmolComp

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4.19e The following pure compound (CO) is burned with 25% excess of air required over the theoretical.

Basis: 100 kmol CO

Reaction: CO + 1/2 O2 → CO2

Theo O2 required = 100/2 = 50 kmolExcess O2 required = 0.25(50) = 12.5 kmolTotal O2 required = 50+12.5 = 62.5 kmolN2 from air = 62.5(79/21) = 235.1CO2 formed = 100 kmol

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100.01347.6total67.64235.1N2

3.612.5O2

28.77100CO2

%kmolComponent

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4-34 500 m3 per min of a fuel gas at 30°C and 800 mm Hg consisting of 15% CH4, 40% C2H6, 15% C3H8, 10%C6H6, and 20% N2 is burned with 20% excess air. All the C burns to CO2 and H2 to H2O. The air enters at atemperature of 60°C and a total pressure of 760 mm Hg wing PH2O = 12 mm Hg. Calculate:

a. the flue gas analysis (dry basis)b. the partial pressure of water in the flue gasc. m3 humid air used/m3 fuel at the given conditions

PT of flue gas = 755 mm Hg.

Given:Required: a. flue gas analysis

b. partial pressure of water in the flue gasc. m3 humid air used/m3


fuel input = 500 m3

min %0 +27330 + 273 %

800760 %

1 kmol22.4 m3 = 21.17 kmol

50.8542.3821.18100total50.854.2320N2

6.3612.722.1210C6H6

12.729.543.1815C3H8

25.4116.948.4740C2H6

6.363.183.1815CH4

kmol H2katoms Ckmol%Comp

Theo O2 required = 42.38 + 50.852 = 67.805 kmol

Excess O2 = 0.02(67.805) = 13.56 kmol

total O2 required = 81.37 kmol

N2 from air = 81.37(79/21) = 306.09 kmol

Air required = 81.37 + 306.09 = 387.46 kmol

H2O in air = 387.46 % 12760 −12

(a) Products of combustion

(b) Partial pressure of the flue gas

total H2O in flue gas = 50.85 + 6.22 = 57.07 kmol

PH2O = 57.07363.03 + 57.07 % 760 = 102.68 mmHg

(c) m3 humid airm3 fuel =

(387.46 + 6.22)(22.4) % 50 + 2730 + 273500 = 21.50

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84.43306.09N23.7413.56O2

11.8342.38CO2%kmolComp

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5-2 One kg of iron is initially at 35°C. It absorbs 500 joules of heat? What will be its final temperature?Solution:

Basis: 1 kg of iron at 35°C that absorbs 500 joules of heatRequired: Final temperature of the iron

Let Tf as the final temperature of the iron

The sensible heat gain of the iron is 500 J.

m ¶TiTf CpdT = 500

m = 1 kg Ti = 273 + 35 = 315 K Cp = 4.13 + 0.00635T Cp, cal/(mol)(K) T, K

(1000) ¶315Tf (4.13 + 0.00635T)dT = 500

(1000) 4.13T + 0.00635 T2

2 315

Tf= (1000) 4.13(Tf − 315) + 0.00635

Tf2 − 31522 = (500)(4.18)

4.13Tf −(4.13)(315) +(0.00635/2)Tf2 −(0.00635/2)(315)2 =4.18/2

Simplifying,

Tf = 315.3 K = 42.3 °C

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SOME SOLUTION TO PROBLEMS IN CHAPTER 7

7-1 Find the degrees of freedom for the following systems in equilibrium:

Formula: F = C - P + 2 a. salt and sugar (with undissolved quantities) in water.

F = 2 - 2 + 2 = 2b. hexane, heptane and octane in equilibrium with the vapor.

F = 3 - 2 + 2 = 3c. oil and water in a closed vessel at 100°C.

F = 2 - 2 + 2 - 1= 1d. carbon, iron and silicon.

F = 3 - 1 + 2 = 4e. coffee, milk, and sugar.

F = 3 - 1 + 2 = 4

7-2 Calculate the boiling point ofa. benzene at 30 in Hg.

30 in Hg %760 mm Hg29.92 in Hg = 762.0 mm Hg

From Cox Chart, BP at 762.0 mm Hg = 81°CFrom Antoine Equation,

t = 1211.0336.90565 − log 762.0 − 220.790 = 80.2°C

b. water at 10 mm Hg.

From Cox Chart, BP at 10 mm Hg = 12°C

c. toluene at 10 atm.

10 atm %760 mm Hg

1 atm = 7600 mm Hg

From Cox Chart, BP at 7600 mm Hg = 220°CFrom Antoine equation,

t = 1344.8006.95464 − log 7600 − 219.482 = 218°C

d. propane at 50 psig.

50 psig %760 mm Hg

14.7 psi + 760 mm Hg = 3345 mm Hg

From Cox Chart, BP at 3345 mm Hg = −4°CFrom Antoine equation,

t = 813.206.82973 − log 3345 − 248.00 = −2.0°C

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7-3 What is the vapor pressure ofa. water at 80°F?

(80-32)/1.8 = 26.7°CFrom Cox Chart, Pv at 26.7°C = 27 mm Hg

b. n-octane at 250°C?

From Cox Chart, Pv at 250°C = 9000 mm Hg

c. n-pentane at 125°C?

From Cox Chart, Pv at 125°C = 7400 mm HgFrom Antoine equation,

= 7414 mm HgPv = anti log 6.85221 − 1064.63232.000 + 125

d. mercury at 200°C?

From Cox Chart, Pv at 200°C = 16 mm Hg

7-4 Given that glycerol at mm Hg boils at 167.2°C and at 760 mm Hg, 290°C, calculate the constants Aand B in the equation

Log PV = A - BT + C

where PV is vapor pressure in mm HgT, temperature in K, and C = 230What is the vapor pressure of glycerol at 200°C?

(1)log 10 = A − B167.2 + 273 + 230

(2)log 760 = A − B290 + 273 + 230

Solving (1) and (2) simultaneously, A = 13.1456 B = 8140.01

Pv = anti log 13.1456 − 8140.01200 + 273 + 230 = 36.9 mm Hg

7-5 A flue gas on a wet basis contains 10% CO2 , 1% CO, 5% O2, 5% H2O and 79% N2. It is at 200°Cand 758 mm Hg. To what temperature should it be cooled to make it saturated with water vapor,keeping the total pressure constant at 758 mm Hg?

Given: Flue gas composition on a wet basis: 10% CO2, 1% CO, 5% O2, 79% N2, and 5%H2OT = 200°C P = 758 mm Hg

Required: Temperature to which it should be cooled to make it saturated with water vapor

PH2O = 0.05(758) = 37.9 mm Hg

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The equilibrium temperature corresponding to the an equilibrium vapor pressure of 37.9 mm Hg iswhat we want. We can find this from Appendix 9 to Appendix 7.

From Appendix 9 (Cox Chart) the temperature corresponding to P = 37.9 mm Hg for water is33°C. From the Appendix 7 (Steam Table), the pressure is 5.053 kPa. By interpolation, thecorresponding equilibrium temperature is

T = 32 + 2 (5.053 - 4.759)/(5.324 - 4.759) = 32 + 0.98 = 33°C

7-6 In a room measuring 4m × 5m × 4m was left an open tank containing ethyl alcohol. Considering thatthe room is sealed from the outside, calculate the amount of alcohol that has evaporated into theroom if the temperature is 30°C and the atmospheric pressure is 755 mm Hg. Assume that the roombecomes completely saturated with alcohol.

PvEtOH, 30oC = anti log 8.16290 − 1623.22228.98 + 30 = 78.552 mm Hg

YEtOH,sat. =PvEtOH

PT − PvEtOH = 78.522755 − 78.552 = 0.11612 kmol EtOH

kmol dry air

Vol. of room = 4 m % 5 m % 4 m = 80 m3 = vol. of dry air

80 m3 dry air %(755 − 78.552) mm Hg

760 mm Hg % 273 K30 + 273 K % kmol

22.4 m3 %0.11612 kmol EtOH

kmol dry air %46.07 kg

kmol

= 15.3 kg EtOH

7-7 The atmospheric pressure varies with elevation according to the equation P = 29.92 − 0.00111h

where P is pressure in in Hg and h is the altitude in ft.What is the elevation where water boils at 95°C?

At 95°C, from Cox chart, Pvwater = 650 mm Hg = 25.59 in. Hgh = 29.92 − 25.59

0.00111 = 3900 ft.

7-8 A liquid mixture at 0°C consists of 30% ethane, 40% propane, and 30% n-butane. Find thecomposition of the vapor mixture in equilibrium with this liquid. What is the total equilibriumpressure?

From Antoine equation,

Pvethane = anti log 6.80266 − 656.40256.00 = 17322 mm Hg

Pvpropane = anti log 6.82973 − 813.20248.00 = 3553.8 mm Hg

Pvbu tan e = anti log 6.83029 − 945.90240.00 = 774.53 mm Hg

PT = 0.30(17322) + 0.40(3553.8) + 0.30(774.53) = 6850 mm Hg

yethane =0.30(17322)

6850 = 0.7586

ypropane =0.40(3553.8)

6850 = 0.2075

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ybu tan e =0.30(774.53)

6850 = 0.0339

7-9 A gaseous mixture consists of 61% n-hexane, 26% n-octane and 13% n-decane at a temperature of50°C. Find the equilibrium pressure and the composition of the liquid in equilibrium with thismixture.

At 50°C, from Cox chart,Pvhexane = 390 mm HgPvoctane = 50 mm HgPvdecane = 8 mm Hg

xhexane =PT

Pvhexane $ yhexane =PT390 (0.61)

xoc tan e =PT

Pvoc tan e$ yoc tan e =

PT50 (0.26)

xdecane =PT

Pvdecane $ ydecane =PT8 (0.13)

xhexane + xoc tan e + xdecane = 1

0.61 PT390 + 0.26PT50 + 0.13PT8 = 1

PT = 10.61390 + 0.26

50 + 0.138

= 43.45 mm Hg

xhexane = 0.6796xoctane = 0.2259xdecane = 0.7061

7-10 Find the dew point of a gaseous mixture of 40% ethyl alcohol and 60% water at a total pressure of 2atm. What is the composition of the first liquid that appears?

PT = 2 atm = 1520 mm Hg

yEtOH = 0.40 = PvEtOHPT $ xEtOH xEtOH =1520(0.4)PvEtOH = 608

PvEtOH

ywater = 0.60 = PvwaterPT $ xEtOH xwater =1520(0.6)Pvwater = 912

Pvwater

At the correct T: xEtOH + xwater = 1

608PvEtOH + 912

Pvwater = 1

Assume T = 125°C

PvEtOH = 3750 mm Hg Pvwater = 1700 mm Hg

6083750 + 912

1700?= 1 0.162 + .536 = 0.698 ! 1

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T = 117°C


6083000 + 912

1300?= 1 0.203 + 0.702 = 0.905 ! 1

T = 114°C


6082200 + 912

1000?= 1 0.276 + 0.912 = 1.188 ! 1

T = 115°C

PvEtOH = 2500 mm Hg Pvwater = 1150 mm Hg6082500 + 912

1150 = 0.243 + 0.793 = 1.036 ! 1

T = 116°C

PvEtOH = 2530 mm Hg Pvwater = 1200 mm Hg6082530 + 912

1200 = 0.24 + 0.76 = 1

7-11 Find the bubble point of a mixture of 50% n-pentane and 50% n-butane at a pressure of 4 atm.What is the composition of the first vapors formed?

PT = 4 atm = 3040 mm Hgxpentane = 0.50 ypen tan e =

PvpentanePT $ xpentane = 0.50

3040Pvpentane

xbutane = 0.50 ybu tan e =Pvbu tan ePT $ xbu tan e = 0.50

3040Pvbu tan e

At the correct T: ypentane + ybutane = 1

Assume T = 60°C

Pvpentane = 1700 mm Hg Pvbutane = 4750 mm Hg0.50(1700)

3040 +0.50(4750)

3040?= 1 0.2796 + 0.7813 = 1.0609 l 1

â Bubble point = 60°C

Using Antoine equation:

(1)Pvpen tan e = anti log 6.85221 − 1064.63232.000 + T

(2)Pvbu tan e = anti log 6.83029 − 945.90240.00 + T

(3)0.503040Pvpen tan e + 0.50

3040Pvbu tan e = 1

Solving simultaneously (1), (2) and (3), bubble point = 58.2°C

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7-12 The composition of a liquefied petroleum gas (LPG) is 0.5% ethane, 0.1% ethylene, 16.4%propane, 2.1% propylene, 74.0% n-butane, and 6.9% n-butene. If the room temperature is 30°C,what is the pressure inside the tank and the composition of the gas that first issues from themixture?

At T = 30°C

Pvethane = anti log 6.80266 − 656.40256.00 + 30 = 32177.7 mm Hg

Pvethylene = anti log 6.74756 − 585.00255.00 + 30 = 49536.9 mm Hg

Pvpropane = anti log 6.82973 − 813.20248.00 + 30 = 8026.94 mm Hg

Pvpropylene = anti log 6.81960 − 785.00247.00 + 30 = 9675.31 mm Hg

Pvn−bu tan e = anti log 6.83029 − 945.90240.00 + 30 = 2123.03 mm Hg

Pvn−butene = anti log 6.84290 − 926.10240.00 + 30 = 2587.62 mm Hg

PT = 0.005(32177.7) + 0.001(49536.9) + 0.164(8026.94) + 0.021(9675.31) + 0.74(2123.03)

+0.069(2587.62) = 3479.62

yethane =0.005(32177.7)

3479.62 = 0.04624

yethylene =0.001(49536.9)

3479.62 = 0.01424

ypropane =0.164(8026.94)

3479.62 = 0.37832

ypropylene =0.021(9675.31)

3479.62 = 0.05839

yn−bu tan e =0.74(2123.03)

3479.62 = 0.45150

yn−butene =0.069(2587.62)

3479.62 = 0.05131

7-16 A helium-toluene mixture contains 30% mole toluene at 100°C and 760 mm Hg. Calculate thepercentage saturation, relative saturation, dew point and the molar humidity of the mixture.

T = 100°CPT = 760 mm HgPtoluene = 0.30(760) = 228 mm HgPvtoluene = anti log 6.95464 − 1344.800

219.482 + 100 = 556.3 mm Hg

a) % saturation =

PtoluenePT − PtoluenePvtoluene

PT − Pvtoluene

% 100 =228

760 − 228556.3

760 − 556.3% 100 = 15.69 %

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b) rel. saturation = PtoluenePvtoluene % 100 = 228

556.3 % 100 = 40.99 %

c) dew point: t = 1344.8006.95464 − log 228 − 219.482 = 73.1°C

d) molar humidity, Y = 0.301 − 0.30 = 0.429

7-17 200 grams of benzene is allowed to evaporate in a 10 m3 tank containing CO2 gas at 40°C and 750mm Hg. If the mixture is brought to 60°C and 1.2 atm, what is the percentage saturation and therelative saturation of the mixture?

At 60°C

Pvbenzene = anti log 6.90565 − 1211.033220.790 + 60 = 391.5 mm Hg

PT = 1.2 atm %760 mm Hg

1 atm = 912 mm Hg

10 m3 CO2 %kmol

22.4 m3 %750760 %

273 K273 + 40 K = 0.384 kmol CO2

0.2 kg benzene % kmol78.12 kg = 0.00256 kmol benzene

Pbenzene = ybenzene $PT = 0.002560.00256 + 0.38425 (912) = 6.04

percent saturation =

PbenzenePT − PbenzenePvbenzene

PT − Pvbenzene

% 100 =6.04

912 − 6.04391.5

912 − 391.5% 100 = 0.886 %

relative saturation = PbenzenePvbenzene % 100 = 6.04

391.5 % 100 = 1.54 %

7-18 A nitrogen-acetone mixture containing 20% acetone and 80% nitrogen is at a temperature of 30°Cand a pressure of 755 mm Hg. To what temperature should the mixture be chilled so that whenheated back to 30°C at the same pressure, the percentage saturation will be 40%?

At 30°C, Pvacetone = anti log 7.23157 − 1277.03237.23 + 30 = 283.7 mm Hg


PacetonePT − PacetonePvacetone

PT − Pvacetone

% 100 =

Pacetone755 − Pacetone

283.7755 − 283.7

% 100 = 40

â Pacetone = 146.5 mm Hg

At chilling temperature, Pacetone = Pvacetone

t = 1277.037.23157 − log 146.5 − 237.23 = 14.86°C

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7-19 A mixture of benzene-nitrogen contains 15% benzene and is at 40°C and 780 mm Hg. What willhappen if the mixture is compressed until the final pressure becomes 1500 mm Hg with temperaturekept at 40°C? What is the final percentage saturation and concentration, (kg benzene)/(kgbenzene-free nitrogen)?

Pvbenzene = anti log 6.90565 − 1211.033220.790 + 40 = 182.8 mm Hg


PbenzenePT − PbenzenePvbenzene

PT − Pvbenzene

% 100 =

0.15(780)1500 − 0.15(780)

182.81500 − 182.8

% 100 =

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Stoic-solution to Problems

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