Stoic-solution to Problems

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 1 1-8

Given: Reading of main meter: 250 m3_{; Consumptions in m}3_{: 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 to tenant 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 are encountered: 9% to evaporation, 7% to run-off, and 15% to infiltration. If 4,000 liters was finally used what was the volume of the water at the source?

evaporation runoff water delivered Infiltration water source

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

Solution:

Let x = the volume of the water at the source 0.09x + 0.07x + 0.15x + 4000 = x x - 0.31x = 4000 x = 4000/0.69 = 5,797 L piping dist. system leakage 1 2 3 5 4

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 2 2-1 a. 8 cp%_{100 cp %}p g cm $ s p % _{1000 g %}kg 100 cmm % _{kg $ m}N s2 % Pa_N m2 = 0.8 Pa$s b. 150 ft$lbf % g g_c %lb lbf % 12 inft % 2.54 cmin % 100 cm %m 2.2 lb % 9.8kg sm2 % _{kg m}N s2 = 203.7 J c. 1000 dynes % 0.1 N g $ cm s2 dyne % 1000 g %kg 100 cm %m kg $ mN s2 =

f. 34 pints%_{2 pints %}qts _{4 qts %}gal _{7.48 gal %}ft3 (12 in)3 0.016 ft3 % (2.54 cm) 3 in3 % m 3 (100 cm)3 = g. _{21, 000 in}3_% (2.54 cm)3 0.344 m3 in3 % m 3 (100 cm)3 =

h. 900 fl. ounce% 0.05 pint_{fl. ounce %8 pints %}gal _{7.48 gal %}ft3 (12 in)3

ft3 % (2.54 cm) 3 in3 % m 3 (100 cm)3 = 0.021 m3 i. _{2100 lbm} = 3.371x 104 _kg/m3 ft3 % kg 2.2 lbm % ft 3 (12 in)3 % in 3 (2.54 cm)3 % (100 cm) 3 m3 j. _{7150 Btu} 7.532 x 106 _W

DRAFT

2.4 Find out which of the following formulas is correct: or ØCv ØV T = T Ø 2_P ØT2 V T ØØVCv T = Ø 2_P ØT2 V C, temperature, θ P, pressure, F/L2 V, specific volume, L3_/M Cv, heat capacity, FL/MT Solution:

Substituting the dimensions;

Eq. 1: Left side Right side

FL MT L3 M = F TL2 T F/L 2 T2 = F_TL2

Eq. 2 Left side T FL/MT Right side

L3_{/M = TL}F2 F

2_/L4

T2 = F 2

L4_T2

DRAFT

2-5 Find the value of the following dimensionless groups for the given data: a. The Reynolds No., NRe= Dv!

where D = diameter, L

v = velocity L/θ ρ = 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 )(6 ms % 100 cm m )(4_{gal %}kg 7.48 gal_ft3 % ft 3 123 _in3 % in3 2.543_cm3) 300 lb_{ft $ h % 2.2 lb %}kg _{12 in %}ft _{2.54 cm %}in _{3600 s}h NRe = = 2.073 % 104

b. The Schmidt No., NSc= _!D

v

where 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= _D! v = 12 cp % _{100 cp %}p g cm $ s p 30kg ft3 % 1000 g kg % ft 3 123_in3 % in3 2.543_cm3 %745 cm 2 h % 3600sh 0.5473 NSc =

c. The Froude No., NFr= v_Lg2 where v = velocity, L/θ

L = length, L

g = acceleration due to gravity

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

NFr= v_{Lg =} (20 cm_{s )}2 in2 (2.54 cm)2 % ft 2 (12 in)2 6 ft % 32.2 ft_s2 = 2.229 % 10 −3

DRAFT

2.7 The viscosity of a liquid at the critical point is given by formula c= 61.6 _(VMTc

c)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 constant corresponding 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 % m 3 (1000 cm)3 % 1000 mol_kmol )2/3 K % 10P6_P % 0.1 Pa $ s_P =Pa$s ( m 3 kmol )2/3 K

DRAFT

2-9 0.086 ft( lbft3) is changed to by conversion of units 0.4 (cP)2₍dyne_{cm )}3  cm( g cm3)0.4 ( lb_{ft $ h )}2(dyne_{cm )}3 = 0.086 ft % 12 inft % 2.54 cmin ( lbmft3 % 454 g lbm % ft 3 (12 % 2.54 cm)3)0.4 (cP % _{100 cP %}poise 1g/cm $ s_poise % _{454 g %}lb 12 % 2.54 cm_ft % 3600 s_h )2 % 1 (dyne_{cm %} g $ cm/s_{dyne % 1000 g %}kg _{100 cm %}m _{kg $ m/s}n 2% 100 cmm )3 = 8.59 % 107 c,(g/cm3)0.4 ( lbm_{ft $ h )}2_{%( Nm )}3

DRAFT

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

4.11 2.99 2.07 1.13 0.24 0 Distance, m 32.22 25.00 19.44 11.67 5.56 4.44 Temperature, °C Solution:

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

Intercept:

4.2751052333

Slope:

6.8955757685

Equation:

T = 4.275 + 6.896

Prob-2-10

X , distance in m 0 1 2 3 4 5 T, temper

ature deg, Cels

ius 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 pressure of methane above the solution. (Chem. Eng. Prog. symp. series, Volume 65, 73 (1969).

3447 2758 2068 1379 689 138 Partial pressure of CH4, kPa

560 450 340 225 115 20 Dissolved CH4, ppm

If the variation is linear, find the suitable relationship.

2-14 Dissolved CH₄ , ppm 0 100 200 300 400 500 600 Partial pressure o f CH 4 , k P a 0 1000 2000 3000 4000

DRAFT

5.26 3.8 2.5 1.25 x 1.95 3.86 6.75 12.6 y 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

Problem 2-14

X Data 1 2 3 4 5 6 Y D ata 1 10 Y vs X Plot 1 Regr

DRAFT

2-18

Convert the following temperature readings a. 180K to °C, °R, and °F Rdg in °C = 180 - 273 = -93°C Rdg in °R = 180× 1.8 = 324°R Rdg 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 K Rdg in °C = (800 -32)/1.8 = 426.7°C Rdg in °R = 800 + 460 = 1260°R c. 100,000°C to K, °R, and °F. Rdg in K = 100,000 -273 = 99,723 K Rdg in °F = (100,000)1.8 + 32 = 180032°F Rdg in °R = (100,000)1.8 + 32 -460 = 179572°R

DRAFT

2-26 Convert the following gage pressures to absolute values if the barometric pressure is 757 mm Hg: a. 200 psig

b. 4 cm of Hg vacuum c. 29 mm of H20 head

d. 15 mm of H20 draft

a.

Pabs= Pbar+ Pgage

Pabs= 757 mm Hg %_{760 mm Hg + 200 = 214.6 psi}14.7 psi

b.

Pabs= Pbar− Pvac

Pabs= 757 mm Hg %_{760 mm Hg − 4 cm Hg= 71.7 cm Hg}76 cm Hg

c.

Pabs= Pbar+ head

Pabs= 757 mm Hg + 29 mm H2O%_{12 % 2.54 in %}1 ft H2O 760 mm Hg_{33.9 ft H}

2O = 778.3 mm Hg

d.

Pabs= Pbar− draft

Pabs= 757 mm Hg −15 mm H2O%_{12 % 2.54 in %}1 ft H2O 760 mm Hg_{33.9 ft H}

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 Hg b.

c.Pabs= 740 mm Hg % _{760 mm Hg + 20 psi %}1 atm _{14.7 psi = atm abs}1 atm

d.

e.b. 28" Hg vacuum to mm Hg absolute with barometric pressure = 760 mm Hg f.

g.Pabs= 740 mm Hg − 28 in Hg % 760 mm Hg_{29.92 in Hg = mm Hg}

h.

i.c. 4" H2O draft to mm Hg abs with barometric pressure = 760 mm Hg

j.

k.Pabs= 760 mm Hg − 4 in H2O % 1 ft_{12 in %} 760 mm Hg_{33.9 ft H}

2O = 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 Hg_{33.9 ft H}

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

3,000

100

1.2

36

18

2

H

2

O

71.87

2,156

28

77

N

2

7.47

224

32

7

O

2

1.87

56

28

2

CO

17.6

528

44

12

CO

2

% mass

mass

MW

mol

Comp

DRAFT

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

10.0001 F-1 0.01 Sr+2 0.0065 Br-1 0.04 K+1 0.01 HCO3-1 0.04 Ca+2 0.26 SO4-2 0.13 Mg+2 1.90 Cl-1 1.06 Na+l mass % Anion mass % 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}

kg 1 km3_{% 1000}3m3 km3 %1030 kg m3 %0.0106 kg Na+

kg sea water % 57.45 kg NaCl23 kg Na+ = 2.727 % 1010

1 km3_{% 1000}3m3 km3 %1030 kg m3 %0.0013 kg Mg+2 kg sea water % 58.33 kg Mg(OH)₂ 24.31 kg Mg+2 = 3.213 % 109kg

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 of the 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 fraction

kg sucrose = 5.7 kg water = 18 L x 1 kg/L = 18 kg Total 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 = 18

therefore, per 1000 g H2O, the gmol of sucrose = 167/18 =

and the molality is molal d. the molarity

For this problem, we need the specific gravity of the sucrose solution, which we can obtain from handbooks or other references.

Percent sucrose by mass = 76%

the specific gravity = 1.383 from reference total mass = 57,000 + 18,000 = 75,000 g

Volume = 75, 000 g %_{1.383 g %}ml _{1000 ml = 54.23}1 L molarity = 143/54.23 = 2.64 M

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 what is

a. the flow velocity b. the mass velocity c. the volumetric flow rate? a. Flow velocity in m/s v= Q_{A = !A}w w= 6530 kg/h ! 1.3567 g/L A= (1)2_{/4 m}2 v= 6530 kg h % 3600 s1 h 1.3567_{L %}g _{1000 g %}kg 1000 L_m3 %(1)2_% 4 m2 = 1.702 m/s b. mass velocity in kg/m2_⋅s G= w_{A =} 6530 kg h %3600 s1 h (1)2_% 4 m2 = 2.31 kg/m2_$s

c. vol flow rate in m3_/s

Q= w_{! =} 6530 kg

h % 3600 s1 h 1.3567_{L %}g _{1000 g %}kg 1000 L_m3

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 branches in two lines. One pipe has an inside diameter of 7 cm while the other, 15 cm. Assuming that the mass flow rate is directly proportional to the cross sectional area of flow, calculate for both pipes

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

c. the volumetric flow rate in m3_/hr

Solution:

Basis: 4,000 L/h

cross sectional area of 7 cm dia (0.07 m dia) A = D_{4 =}2 0.07₄ 2 = 0.003848 cross sectional area of 15 cm dia (0.15 m dia)

A = D_{4 =}2 0.15₄ 2 = 0.01767 Let x = flow through 7 cm dia pipe

4000 - x = flow through the 15 dia m pipe x 4000− x = D7 2 4 D15 2 4 = 0.07_0.152₂ = 49₂₂₅ 225x= 49(4000 − x) x = 715 kg/h 4,000 - x = 3285 kg/h a. flow velocity through 7 cm dia v= 715kg_{h % 3600 s %}h _{850 kg %}m3 _{0.003848 m}1 ₂ = 0.061 m/s through 15 cm dia v= 3285kg_{h % 3600 s %}h _{850 kg %}m3 _{0.01767 m}1 ₂ = 0.061 b. The mass velocity, G

G= w_A

for the 7 cm dia pipe

_{G= w}

kg/(s⋅m

2

₎

A = 0.003848 = 1.846 % 10715 6

for the 15 cm dia pipe

_{G= w}

kg/(s⋅m

2

₎

A = 0.017678 = 1.859 % 103285 5

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 butane gas was used and after some time, the pressure had gone down to 50 psig. If the temperature is 25°C, what fraction 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: = = P1V1 z1RT1 − P 2V2 z2RT2 P1V1 z1RT1 = P1 z1T1 − P 2 z2T2 P1 z1T1 = 114.7 z1(30 + 273) − 64.7 z2(25 + 273) 114.7 z1(30 + 273)

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 barometric

pressure 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

99.99

101.34

100

Total

1.32

1.34

H

2

O

81.9

83

83

N

2

1.97

2

2

CO

2.96

3

3

O

2

11.84

12

12

CO

2

mol % (wet basis

mol (wet basis)

mol (dry basis)

Components

Patm = 759 mm Hg PH2O = 10 mm Hg PT = Patm - Pdraft = 759− 4 cm H2O%_{(2.54 $ 12) cm H}ft H2O 756.1 mm Hg 2O % 760 mm Hg 33.9 ft H2O =

Pdry gas + PH2O = 756.1 mm Hg Pdry gas = 756.1 - 10 = 746.1 mm Hg

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 3

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°C

and 770 mm Hg total pressure, with Pwater = 25 mm Hg, flowing at a rate of 500 m3 per minute is mixed with a

natural gas consisting of 81.11% CH4, 6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2 at a total

pressure of 2 atm and a temperature of 25°C and flowing at a rate of 100 m3 _{per minute. Its water vapor content has}

a partial pressure of 30 mm Hg. What is the composition of the resulting mixture on a dry basis? What is the water vapor content in grams per kmol of dry gas?

Given:

Producer gas: 500 m3 _{per minute at 30° C and 770 mm total Press P}

H2O = 25 mm Hg

8.4% 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 P}

H2O = 30 mm Hg

81.11% 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 operation

Mixing of producer gas and natural gas

kmol producer gas = 500 m_{min %}3 ₃₀0+ 273_{+ 273 %} 770_{760 % 22.4 m}kmol₃ = 20.376 kmol (with H2O)

kmol H2O in prod. gas = 20.376 kmol wetb gas % 25_{770 = 0.684 kmol}

kmol dry producer gas = 20.376 - 0.684 = 19.629

kmol natural gas = 100 m_min 3 % 0 + 273_{25+ 273 %} 2_{1 %22.4 m}kmol₃ = 8.18

kmol H2O in nat. gas = 0.136 kmol dry gas % _{2(760) − 30 = 0.165 kmol}30

kmol dry natural gas = 8.18 - 0.165 = 8.015 For 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}

99.978 27.734 Total 0.212 0.059 0.0074 (8.018) = 0.061 C4H10 0.602 0.172 0.021 (8.018) = 0.168 C3H8 1.84 0.516 0.0644 (8.018) = 0.527 C2H6 23.23 6.503 0.8111 (8.018) = 6.635 CH4 5.64 1.557 0.079 (19.714) = 1.61 H2 15.13 4.179 0.212(19.714) = 4.32 CO 0.571 0.158 0.008 (19.714) = 0.163 O2 46.66 12.9 0.0919(8.018) = 0.752 0.617 (19.714) = 12.572 N2 6.093 1.69 0.0042(8.018) = 0.034 0.084(19.714) = 1.656 CO2 mol % kmol

Nat. gas

Producer gas

Comp.

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 of water is 20 mm Hg. In order to measure the rate of flow, pure carbon dioxide gas is bled into the gas at rate of 20 kg/min. At a point downstream where complete mixing has occurred, it is found that the air contains 5% CO2 by

volume. Find the volumetric flow rate of the moist air in m3_{/min. The % CO}

2 is on a dry basis. Given: mixed gases 5% CO₂ vol moist air 40o_C P_t = 765 mm Hg P_H2O = 20 mm Hg pure CO₂ gas 20 kg/min

Required: Vol flow rate of moist air in m3_/min

Basis: 1 minute operation Solution:

Vol flow rate of moist air =

20kg CO_min2 % kmol_{44 kg %} 0.95 kmol dry air_{0.05 kmol CO}

2 %

765 kmol moist air

(765 − 20) kmol dry air % 22.4 m

3

kmol % 400+ 273 %+ 273 760765 = 226.3 m3_/min

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} % _{(1 − 0.09) kg BDS =}kg product 769.2 kg prod

OMB:

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 per minute, how much product can be obtained over an 8-hr period? Find the ratio kg H2O evaporated/kg wet wood

charged Drier Water evaporated 10 kg wet wood/ min Dried product 12% moisture 36 % moisture Required:

Product for 8 hr period

ratio 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 BDS_{1 kg wet wood} % _{(1 − 0.12) kg BDS = 7.386 kg}1 kg product Overall mass balance:

For the 8 hr period:

kg dry prod = 7.386 kg dry prod/min% 60 min/hr % 8 hr= 3545 kg kg water evap/kg wet wood charged = 2.614/10 = 0.261

DRAFT

3-11 A drier is being used to reduce the moisture content of bagasse to be used for fiberboard manufacture. It contains 50% moisture. 5.75 kg of wet bagasse is charged into the drier. After 3 hours, it was found that the weight 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

DRAFT

3-14 It is desired to concentrate a 6% KN03 solution (in water) to 20% KN03 solution. 7000 kg of product liquor is

to 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 = 1400 kg % 100 kg solution_{6 kg KNO} = 23,300 kg

3

OMB

kg water evaporated

DRAFT

3-19

An evaporator is concentrating solutions coming from two different sources. The solution from the first source containing 10% NaCl and 10% NaOH

8% 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% NaCl 10% NaOH water evap = 50% of total H2O 50 kg/min Product 70 kg/min 8% NaCl 12% NaOH

Required: Composition and flow rate of product Basis: 1 min operation

Solution: We can use the aritmetic solution.

Feed Stream 1: Feed Stream 2

kg NaOH = 0.1(50) = 5 kg kg NaOH = 0.12(70) = 8.4 kg NaCl = 0.1(50) = 5 kg kg NaCl = 0.08(70) = 5.6 kg 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 kg

DRAFT

3-20 A mixture containing 70% methanol and 30% water is to be distilled. If the distillate product is to contain 99.9% methanol and the bottoms product 0.004% methanol, how much distillate and bottoms product are obtained per 100 kg of feed distilled?

Given:

Bottoms, B Distillate , D 70% methanol 30% water Feed = 100 kg x_m = 0.999 x_M = .00004 Required: kg distillate and kg bottoms

Basis: 100 kg feed

Solution: We cannot identify a tie component. We have to use the algebraic solution.

Let D = kg distillate B = kg bottoms Overall mass balance:

Eq. 100= D + B 1 Methanol balance Eq. 2 0.7(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

DRAFT

3-21100 kmol of a benzene-toluene mixture in equimolal amounts is distilled. 96% of the benzene is obtained in the distillate product while 90% of the toluene is in the bottom. Calculate the composition of the product streams. Given: Bottoms, B Distillate , D benzene-toluene solution, equimolar Feed, F = 100 kmol 90% of the toluene goes to the bottoms 96% of the benzene goes to the distillate

Required: Composition of the distillate and the bottoms Basis: 100 kmol feed

Solution: While we cannot identify a tie component in this problem, we can analyze the problem and see that the given 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 kmol the difference goes to the bottoms: 50 - 48 = 2 kmol

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

Therefore the distillate contains 48 kmol benzene and 5 kmol toluene The composition, mole fraction benzene = 48/(48 + 5) = 0.906 The bottoms contains 45 kmol toluene and 2 kmol toluene The composition, mole fraction benzene = 2/(45 + 2) = 0.043

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

DRAFT

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

z 0.024 0.45 n-pentane 0.3 y 0.32 isopentane 0.129 x 0.23 n-butane bottoms distillate feed Component

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

DRAFT

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

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) = 40 kmol

Ethane absorbed = 0.99(40) = 39.6 kmol Hydrogen in = 0.6(100) = 60 kmol

Liquid out = 196 kmol hexane + 39.6 kmols ethane + 4

Composition of outgoing liquid = _43.643.6_{+ 196 % 100 =}18.2 % Gas out = 60 + 0.4 = 60.4

Composition of outgoing gas = _0.40.4_{+ 196 % 100 = 0.662%} 40% ethane 60% hydrogen Absorber Hexane + absorbed ethane hydrogen and ethane 2 Hexane 2% ethane 99% of the ethane is removed 1

DRAFT

3-26 In order to recover the benzene from a benzene-air mixture, the mixture is scrubbed with kerosene 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 the

composition of the gas stream and liquid stream leaving the tower. What is the volumetric flow rate of the outgoing gas stream (T = 25°C, P = 759 mm Hg)?

Solution:

Basis: 1 hour operation

kmol air-benzene in = 8000 m_{h %}3 ₂₅0+ 273_{+ 273 %}765_{760 % 22.4 m}kmol₃ = 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 = _318.50.13(100)_{+ 0.13 = 0.041% mol benzene} composition of benzene-kerosene mixture = _{10, 000}509.6(100)_{+ 509.6 = 4.85%}

25o_{C 759 mm Hg} 8000 m3_/h Air-benzene mixture 2% mol benzene 25o_{C 765 mm Hg} Absorber kerosene + absorbed benzene Outlet gas 2 Kerosene 10,000 kg/h 98% of the benzene is 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 extracted livers is analyzed and is found to contain 1.12% oil, 32.96% ether, and 65.92% oil-free livers. Find the composition and the weight of the extract, and the percentage recovery of the oil.

Given: Extract Extracted livers Extractor 800 kg/hr 23% oil 1.12 % oil 32.96% ether 65.92% oil-free livers 570 kg/hr ether Halibut livers

Required: Composition and mass of the extract Basis: 800 kg of halibut livers

Solution: The oil-free livers is a tie component. We can use the arithmetic method. kg livers = 800 kg

kg oil = 0.23(800) = 184 kg

kg oil-free livers = 800 - 184 = 616 kg Oil-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 % _{65.92 kg oil-free livers = 10.466 kg}1.12 kg oil Ether in extracted livers = 616 kg oil-free livers % _{65.92 kg oil-free livers = 308 kg}32.96 kg ether 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%

DRAFT

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 the filter cake is incompressible, how much volume of the slurry is required to fill the filter press? The specific gravity of the solids in the slurry is 2.63. The density of the liquid is 1.0. Assume that the specific gravity of the 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 slurry

filter cake: 32% porous; incompressible

Sp. 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:0.9 m % 1.2 m % 0.025 m = 4.05 m3 = volume of the cake 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/m}3 _{= 7245 kg}

liquid in the slurry = 7243 kg solids % _{0.05 kg solids = 1.449 % 10}1.0 kg slurry kg slurry5

DRAFT

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 from the first effect (if forward feeding is used) and the flow rate of the product in kg/h. How much evaporation from 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. 1 1,000, 000 = P2+ E2+ C2 Eq. 2 C2= E1 Eq. 3 E1= E2

Solids balance around the whole system, solution side

Eq. 4 0.05(1,000,000) = 0.4P2

P2= 0.05(1, 000, 000)_0.4 = 125, 000_{day %}kg _{24 h = 5208 kg/h}day

Subst Eq 2 and Eq 3 in Eq1

1,000,000= 125,000 + E1+ E1

E1= (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 E₁

.

F = 106 _kg/day E₂ P₂ P₁ C₂ C₁ S Effect 1 Effect 2 x_F = 0.05 x_P1 x_P2 = 0.4

DRAFT

= 23,440 kg/h

P1= 1,000,000 − 437,500 = 562,500 kg/day

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 is

sent to a tank where NaNO3 is added. The mixture

is stirred well to form a saturated solution of NaNO3 (47.9%). The other line bypasses the tank

and is mixed with the 47.9% solution. What is the flow rate of the bypass stream and the final product?

Solution:

Basis: 1 hr operation

Required: Flow rate of bypass stream and the final product

Balance around saturator inside the bypass loop Water balance

water 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 loop NaNO3 balance:

NaNO3 input = 459.693 kg = NaNO3 in 7% NaNO3 solution

Water balance:

Water in 7% NaNO3 solution = 459.693 kg NaNO3% _{7 kg NaNO}93 kg H2O

3 = 6107.35 kg

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

=

459.693+ 6107.35 = 6567.04 kg/h Water Bypass stream Saturator 500kg/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 and

101.3 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 mol H2O/mol dry air. The other part of the air feed bypasses the water spray and is mixed with the humidified air to produce a

mixture containing 0.12 mol H2O per mol dry air. What is the water consumption? What is the ratio of the flowmeter readings

of the bypass stream and the stream to the water spray? Basis: 1 minute operation

Using the arithmetic method

kmol bone-dry air = 1500 m_{min %}3 ₂₅0+ 273_{+ 273 % 22.4 m}kmol₃ = 63.35 kmol BDA BDA balance around spray chamber outside the bypass loop

BDA in feed = BDA in process air = 61.35 kmol BDA

kmol H2O in process air = 61.35 kmol BDA %0.12 kmol H_{kmol BDA}2O = 7.36 kmol H2O

H2O balance around spray chamber outside the recycle loop

water 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 kmol BDA = 7.36 kmol H2O % kmol BDA_{0.3 kmol H}

2O =

2.52 bypass stream

stream to spray chamber = 61.3524.53 = Dry Air 1500 m3 25o_C 101.5 kPa Bypass stream Spray chamber 0.12 kmol H₂O/ kmol dry air Net 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 H}

2O/mol dry air. It is at 20°C

and 1 atm. To conserve energy, part of the exhaust air containing 0.08 mol H2O/mol dry air is recycled and

mixed with the fresh air from the air conditioner to produce a gross air feed to the room containing 0.035 mol H20/mol dry air. How many kg of water is picked up by the air per minute? What is the volumetric flow

rate 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 Basis: 1 minute operation

Bone-dry air balance around whole system

bone-dry in = bone-dry out = 1000 m_{min %}3 ₂₀0+ 273_{+ 273 % 22.4 m}kmol₃ = 41.6 kmol Water in air in = 41.6kmol_{min % 0.01}kmol H_{kmol bda = 0.416 kmol}2O

Water in air out = 41.6kmol_{min % 0.08}kmol H_{kmol bda = 3.328 kmol}2O Water picked up by the air per minute

Using water balance, Water picked up by the air =

(3.328 kmol water out_{kmol bda} − 0.416 kmol water in_{kmol bda} ) % 18_{kmol = 52.42 kg}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 m_{kmol %}3 27_{0+ 273 %}+ 273 101.325₉₉ = 582.2 m3

room inside recycle loop:

0.035 mol H₂0 per mol dry air 1000 m3 _air

20o_{C 1 atm}

Recycle stream 0.01 kmol H₂O/

kmol dry air

27o_{C and 99 kPa}

Room

0.08 mol H₂0 per mol 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= _{(18)(0.025) = 64.72 kmol}52.42 Balance around point of splitting,

Kmol recycle = 64.72 - 41.6 = 23.12 kmol

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 4

4-3

For 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 + 3H2O Basis: 120 kg benzene kg kg CO2= 120 kg C6H6% 6MWCO_MWC 2 6H6 = 120 % 6(44.01) 78.114 = 405.7 kg kg H2O= 120 kg C6H6% 3MWH_MWC 2O 6H6 = 120 % 3(18.015) 78.114 = 83.03 368.7 kg kg O2= 120 kg C6H6% 7.5MWO_MWC 2 6H6 = 120 % 7.5(31.999) 78.114 = b) 2NaCl + H2O + SO3 → Na2SO4 + 2HC1 Basis: 300 kg NaCl kg kg Na2SO4 = 300 kg NaCl % MWNa_{2MWNaCl = 300 %}2SO4 _{2(58.44) = 364.6}142.04

kg kg HCl = 300 kg NaCl % 2MWHCl_{2MWNaCl = 300 %}2(46.46)_{2(58.44) = 187.2} c) Ca3(PO4)2 + 3SiO2 + C → 3CaSiO5₂ 3 + 2P + CO5₂ 2

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% 3MWCaSiO_MWCa 3

3(PO4)2 = 100 %

3(116.16)

(310.18) = 112.4 tons tons P= 100 tons Ca3(PO4)2% _MWCa2MWP

3(PO4)2 = 100 %

2(30.97)

(310.18) = 19.97 tons tons CO2= 100 tons Ca3(PO4)2% _MWCa(5/2)MWCO2

3(PO4)2 = 100 %

(5/2)(44.01)

(310.18) = 35.47 tons d) C2H6 + (7/2)O2 → 2CO2 + 3H2O

How many kg CO2 and H2O can be produced from 13 kg of ethane?

kg kg CO2= 13 kg C2H6% 2MWCO_MWC 2 2H6 = 13 % 2(44.01) 32.07 = 38.05 kg kg CO2= 13 kg C2H6% 3MWH_MWC 2O 2H6 = 13 % 3(18.015) 32.07 = 32.35 e) C12H22O11 (sucrose) + H2O → C6H12O6 C6H12O6 → C2H5OH (ethyl alcohol) + CO2

DRAFT

4-6 Basis: 1,000 kg of fertilizer MW (NaNO3)= 85.0 kg N2 = 0.02(1000) = 20 kg MW Ca3(PO4)2 =310.2 kg P2O5 = 0.12(1000) = 120 kg MW KCl = 74.55 kg K2O = 0.06(1000) = 60 kg MW P2O5 = 141.94 MW K2O = 94.18 kg 98% NaNO3 needed = 20 % 2(MW NaNO_{MW N} 3) 2 % 1 kg 98% NaNO₃ 0.98 kg NaNO₃ = 20 % 2(85)28 % 0.98 = 123.8 kg1 kg Ca3(PO4)2 = 120 % MW Ca_MW(P3(PO4)2 2O5) = 120 % 310.2141.94 = 262.5 kg kg 97% KCl = 60 % 2(MW KCl)_{MW K} 2O % 1 kg 97% KCl 0.97 kg KCl = 60 % 2(74.55)94.18 % 0.97 = 97.93 kg1 Overall Mass Balance:

kg inerts = kg mixed fertilizer - (kg 98% NaNO3 + kg 97% KCl + kg Ca3(PO4)2)

DRAFT

4-7

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

Calcium nitrate, pure Phosphoric 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 KNO_{MW K} 3)

2O = 60 %

2(101.1)

94.18 = 128.8 kg

N2 from KNO3 = 128.8 kg KNO3% _{2MW KNO}MWN2 17.84 3 %= 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(NO_3MW(N 3)3 2) = 2.16 % 2(226.1) 3(28) = 11.63 kg kg 97% H3PO4 = 120 % 2(MW H_{MW P}3PO4) 2O5 = 120 2(98) 141.9 = 165.8 kg Overall Mass Balance:

kg inerts = kg mixed fertilizer - ( kg KNO3 + kg Ca (NO3)3 KCl + kg H3PO4)2)

DRAFT

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

following ingredients necessary: (NH4)3PO4

Ca3(PO4)2

KNO3

Inerts fillers Solution:

Basis: 5,000 kg mixed fertilizer

MW (NH4)3PO4 = 149

kg N2 = 0.06(5000) = 300 kg MW Ca3(PO4)2 = 214.24

kg P2O5 = 0.12(5000) = 600 kg MW KNO3 = 101.09

kg K2O = 0.12(5000) = 600 kg MW P2O5 = 141.94 MW K2O = 94.18

kg KNO3 needed = 600 % 2(MW KNO_{MW K} 3)

2O = 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 % _{101.09 = 173.376}2(14) 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 % _{2(MW (NH}MW P2O5 4)3PO4) = 431.476 % 141.94 2(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 Ca_MW(P3(PO4)2

2O5 = 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 kg kg KNO3 = 1288 kg kg (NH4)3PO4 = 431 kg kg Ca3(PO4)2 = 590 kg

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 method Let 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 % _{(100 + 40)kg solution = x}40 kg Na2SO4 _{MW Na}MW Na2SO4 2SO4$10H2O + (1000 − x) % 13.5 kg Na₂SO4 (100 + 13.5) kg solution 1000 % 40_{140 = x 322.2 + (1000 − x)}142 _113.513.5 Solving for x, = 518.3 kg Na2SO4⋅10H2O x= 40000 140 − 13500113.5 142 322.2 − 113.513.5

DRAFT

4-13 For the following reactions, find the composition the final product. If the second reactant used is 10% in excess of the theoretical amount or complete reaction. (The first is the limiting reactant) The reaction is 100% complete and the reactants are 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: 39 basis: 100 kg KCl kg theoretical H2SO4 needed = 100 kg KCl % MWH_{2MWKCl =}2SO4 _{2 % 74.45 = 65.82}98 kg excess H2SO4 needed = 0.1( 65.82 ) = 6.58 kg kg K2SO4 produced = 100 kg KCl % MWK_{2MWKCl =}2SO4 _{2 % 74.45 = 116.9}174 kg kg HCL produced = 100 kg KCl % MWHCl_{MWKCl =} 36.45_{74.45 = 48.96} kg

product composition: HCl is a gas and normally will not mix with the other substances. The product mixture consist of H2SO4 and K2SO4.

% H2SO4= _6.586.58_{+ 116.9 % 100 = 5.33%}

% K2SO4= _6.58116.9_{+ 116.9 % 100 = 94.67%}

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

Basis: 100 kg Na2CO3

kg theoretical Br2 needed = 100 kg Na2CO3 % _MWNaMWBr2

2CO3 = 159.8106 = 150.8 kg

excess Br2 needed = 0.1(150.8) = 15.08 kg

kg NaBr produced = 100 kg Na2CO3 % 5MWNaBr_3MWNa 2CO3 =

5(102.8)

3(106) = 32.33 kg kg NaBrO3 produced = 100 kg Na2CO3 % MWNaBrO_3MWNa 3

2CO3 = 342.63(106) = 107.7 kg

kg product = 15.08 + 32.33 + 107.7 = 155.1 kg % Br2 = 15.08_{155.1 % 100 = 9.722%}

% NaBr = 32.33_{155.1 % 100 = 20.84%} % NaBrO3 = 107.7_{155.1 % 100 = 69.44%}

DRAFT

4.14a For the following, the reaction goes only to 95% completion. Pure reactants and theoretical amounts are used. 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 % MWNaNO_MWKCl3 = (100)(84.995)_74.551 = 114.009

kg KNO3 produced = 100 kg KCl % MWKNO_{MWKCl % 0.95 =}3 (100)(101.103)(0.95_74.551 = 128.835

kg CaCl2 produced = 100 kg KCl % MWNaCl_{MWKCl % 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 % _{214.008 = 233.636 kg}100 kg NaNO3 : 500 = 114.009 : 214.008

kg NaNO3 = 500 % 114.009_{214.008 = 266.366 kg}

4.14b For the following, the reactions go only to 95% completion. Pure reactants and theoretical amounts are used. 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 % MWCO_{MWMgO =}2 (100)(44.01)_(40.3) = 109.2 kg

kg CaCl2 needed = 100 kg MgO % MWCaCl_{MWMgO =}2 (100)(111)_(40.3) = 275.4 kg

kg MgCl2 produced = 100 kg MgO%

MWMgCl₂

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

kg CaCO3 produced = 100 kg MgO %MWCaCO_{MWMgO % 0.95 =}3 (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

DRAFT

kg MgO : 500 = 100 : 479.2 kg MgO = _{479.2 % 100 = 104.3 kg}500 kg CaCl2 : 500 = 275.4 : 479.2 kg CaCl2 = _{479.2 % 275.4 = 287.4 kg}500

DRAFT

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

a. Theoretical amount of pure O2 used; complete combustion b. Theoretical amount of air used; complete combustion

c. 30% excess air is used: 90% of the C burns to CO2 and 10% to CO

d. 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 combustion Reaction: C + O2 → CO2

katom C = 100/12 = 8.333 katom kmol CO2 formed = 8.333 kmol O2 needed = 8.333 kmol combustion gas = 100% CO2

b. Theoretical air used; complete combustion kmol CO2 formed = 8.333 kmol

Theo O2 needed = 8.333 kmol

N2 from air = 8.333(79/21) = 31.348 kmol

Total 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 CO Theo O2 needed = 8.333

excess O2 = 0.3(8.333) = 2.5 kmol

total O2 needed = 1.3(8.333) = 10.833 kmol N2 from air = 10.833( 79_{21 ) = 40.753}kmol CO2 formed = 0.9(8.333) = 7.5 kmol CO formed = 0.1(8.333) = 0.833 kmol

O2 left due to incomplete reaction to CO = 0.833/2 = 0.416 kmol O2 in combustion gas = 2.5 + 0.416 = 2.92 kmol

100 52 Total 78.36 40.75 N2 1.6 0.83 CO 5.62 2.92 O2 14.42 7.5 CO2 % kmol Component

d. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid. Theo O2 needed = 8.333

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

N2 from air = 10( 79_{21 ) = 37.619}kmol CO2 formed = 0.9(8.333) = 7.5 kmol

O2 left due to incomplete reaction = 0.833 kmol O2 in combustion gas = 1.667 + 0.833 = 2.5 kmol

DRAFT

100 47.62 Total 79 37.62 N2 5.25 2.5 O2 15.75 7.5 CO2 % kmol Comp

DRAFT

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 kmol Excess O2 required = 0.25(50) = 12.5 kmol Total O2 required = 50+12.5 = 62.5 kmol N2 from air = 62.5(79/21) = 235.1 CO2 formed = 100 kmol 100.01 347.6 total 67.64 235.1 N2 3.6 12.5 O2 28.77 100 CO2 % kmol Component

DRAFT

4-34 500 m3 _{per min of a fuel gas at 30°C and 800 mm Hg consisting of 15% CH}

4, 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 a

temperature 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 gas c. m3 _{humid air used/m}3 _{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 gas c. m3 _{humid air used/m}3

Basis: 1 minute operation

fuel input = 500 m_{min %}3 ₃₀0+ 273_{+ 273 %} 800_{760 %22.4 m}1 kmol₃ = 21.17 kmol

50.85 42.38 21.18 100 total 50.85 4.23 20 N2 6.36 12.72 2.12 10 C6H6 12.72 9.54 3.18 15 C3H8 25.41 16.94 8.47 40 C2H6 6.36 3.18 3.18 15 CH4 kmol H2 katoms C kmol % 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 % ₇₆₀12_{− 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= 363.0357.07+ 57.07 % 760 = 102.68 mm Hg

(c) m3_mhumid air_{3 fuel} = (387.46 + 6.22)(22.4) % 50 + 273₅₀₀ 0+ 273 = 21.50 84.43 306.09 N2 3.74 13.56 O2 11.83 42.38 CO2 % kmol Comp

DRAFT

SOLUTION TO SOME PROBLEMS IN CHAPTER 5

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 heat Required: 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

¶

_T i Tf 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 T₂2 315 Tf = (1000) 4.13(Tf− 315) + 0.00635Tf 2_{− 315}2 2 = (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

DRAFT

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 = 2

b. hexane, heptane and octane in equilibrium with the vapor. F = 3 - 2 + 2 = 3

c. oil and water in a closed vessel at 100°C. F = 2 - 2 + 2 - 1= 1

d. carbon, iron and silicon. F = 3 - 1 + 2 = 4 e. coffee, milk, and sugar.

F = 3 - 1 + 2 = 4 7-2 Calculate the boiling point of

a. benzene at 30 in Hg.

30 in Hg %760 mm Hg_{29.92 in Hg = 762.0 mm Hg} From Cox Chart, BP at 762.0 mm Hg = 81°C From Antoine Equation,

t= _6.905651211.033_{− 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°C From Antoine equation,

t= _6.954641344.800_{− 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°C From Antoine equation,

DRAFT

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

(80-32)/1.8 = 26.7°C

From 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 Hg From Antoine equation,

= 7414 mm Hg

Pv= anti log 6.85221 − _232.0001064.63_{+ 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 A and B in the equation

Log PV= A - _{T+ C}B

where PV is vapor pressure in mm Hg

T, temperature in K, and C = 230

What is the vapor pressure of glycerol at 200°C? (1) log 10= A − _{167.2+ 273 + 230}B

(2) log 760= A − _{290+ 273 + 230}B

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

Pv = anti log 13.1456− ₂₀₀8140.01_{+ 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°C and 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%H2O

T = 200°C P = 758 mm Hg

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

DRAFT

The equilibrium temperature corresponding to the an equilibrium vapor pressure of 37.9 mm Hg is what 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 is 33°C. From the Appendix 7 (Steam Table), the pressure is 5.053 kPa. By interpolation, the corresponding 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 that the room is sealed from the outside, calculate the amount of alcohol that has evaporated into the room if the temperature is 30°C and the atmospheric pressure is 755 mm Hg. Assume that the room becomes completely saturated with alcohol.

PvEtOH, 30o_C= anti log 8.16290 − 1623.22

228.98+ 30 = 78.552 mm Hg

YEtOH,sat.= _P PvEtOH

T− PvEtOH =

78.522

755− 78.552 = 0.11612kmol dry airkmol EtOH 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 % 30273 K+ 273 K % 22.4 mkmol3 % 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. Hg

h= 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 the composition of the vapor mixture in equilibrium with this liquid. What is the total equilibrium pressure?

From Antoine equation,

Pvethane= anti log 6.80266 − 656.40_{256.00 = 17322 mm Hg}

Pvpropane= anti log 6.82973 − 813.20_{248.00 = 3553.8 mm Hg}

Pvbu tan e= anti log 6.83029 − 945.90_{240.00 = 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)

DRAFT

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 of 50°C. Find the equilibrium pressure and the composition of the liquid in equilibrium with this mixture.

At 50°C, from Cox chart, Pvhexane = 390 mm Hg Pvoctane = 50 mm Hg Pvdecane = 8 mm Hg xhexane= _PvPT hexane $ yhexane= PT 390 (0.61) xoc tan e= _PvPT oc tan e $ yoc tan e= PT 50 (0.26) xdecane= _PvPT decane $ ydecane= PT 8 (0.13)

xhexane+ xoc tan e+ xdecane = 1

0.61PT 390 + 0.26P50 + 0.13T P8 = 1T PT= _0.61 1 390 + 0.2650 +0.138 = 43.45 mm Hg xhexane = 0.6796 xoctane = 0.2259 xdecane = 0.7061

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

PT= 2 atm = 1520 mm Hg yEtOH= 0.40 = Pv_PEtOH T $ xEtOH xEtOH = 1520(0.4) PvEtOH = 608PvEtOH ywater= 0.60 = Pv_Pwater T $ xEtOH xwater= 1520(0.6) Pvwater = 912Pvwater

At the correct T: xEtOH+ xwater= 1

608_Pv EtOH + 912Pvwater = 1 Assume T = 125°C PvEtOH = 3750 mm Hg Pvwater = 1700 mm Hg 608 3750 + 1700912 = 1? 0.162+ .536 = 0.698 ! 1

DRAFT

T = 117°C PvEtOH = 3000 mm Hg Pvwater = 1300 mm Hg 608 3000 + 1300912 = 1? 0.203+ 0.702 = 0.905 ! 1 T = 114°C PvEtOH = 2200 mm Hg Pvwater = 1000 mm Hg 608 2200 + 1000912 = 1? 0.276+ 0.912 = 1.188 ! 1 T = 115°C PvEtOH = 2500 mm Hg Pvwater = 1150 mm Hg 608 2500 + 1150 = 0.243 + 0.793 = 1.036 ! 1912 T = 116°C PvEtOH = 2530 mm Hg Pvwater = 1200 mm Hg 608 2530 + 1200 = 0.24 + 0.76 = 1912

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 Hg xpentane = 0.50 ypen tan e=

Pvpentane

PT $ xpentane= 0.503040 Pvpentane

xbutane = 0.50 ybu tan e= Pv_Pbu tan e

T $ xbu tan e= 0.503040 Pvbu tan e

At the correct T: ypentane + ybutane = 1 Assume T = 60°C Pvpentane = 1700 mm Hg Pvbutane = 4750 mm Hg 0.50(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 − 1064.63_{232.000+ T}

(2)

Pvbu tan e= anti log 6.83029 − 945.90_{240.00+ T}

(3) 0.50

3040 Pvpen tan e+ 0.50_{3040 Pv}bu tan e= 1

DRAFT

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 the mixture?

At T = 30°C

Pvethane= anti log 6.80266− _256.00656.40_{+ 30 = 32177.7 mm Hg}

Pvethylene= anti log 6.74756 − _255.00585.00_{+ 30 = 49536.9 mm Hg}

Pvpropane= anti log 6.82973 − _248.00813.20_{+ 30 = 8026.94 mm Hg}

Pvpropylene= anti log 6.81960 − _247.00785.00_{+ 30 = 9675.31 mm Hg}

Pvn−bu tan e= anti log 6.83029 − _240.00945.90_{+ 30 = 2123.03 mm Hg}

Pvn−butene= anti log 6.84290 − _240.00926.10_{+ 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 the percentage saturation, relative saturation, dew point and the molar humidity of the mixture. T = 100°C

PT = 760 mm Hg

Ptoluene = 0.30(760) = 228 mm Hg

Pvtoluene= anti log 6.95464 − _219.4821344.800_{+ 100 = 556.3 mm Hg}

a) % saturation = Ptoluene PT− Ptoluene Pvtoluene PT− Pvtoluene % 100= 228 760− 228 556.3 760− 556.3 % 100= 15.69 %

DRAFT

b) rel. saturation = _PvPtoluene

toluene % 100= 228556.3 % 100 = 40.99 %

c) dew point: t = _6.954641344.800_{− log 228 − 219.482 = 73.1°C} d) molar humidity, Y = 0.30_{1− 0.30 = 0.429}

7-17 200 grams of benzene is allowed to evaporate in a 10 m3 _{tank containing CO}

2 gas at 40°C and 750 mm Hg. If the mixture is brought to 60°C and 1.2 atm, what is the percentage saturation and the relative saturation of the mixture?

At 60°C

Pvbenzene= anti log 6.90565 − 1211.033_{220.790+ 60 = 391.5 mm Hg}

PT= 1.2 atm %

760 mm Hg

1 atm = 912 mm Hg 10 m3_CO

2% kmol_{22.4 m3} % 750_{760 % 273}273 K_{+ 40 K = 0.384 kmol CO}2

0.2 kg benzene % kmol_{78.12 kg = 0.00256 kmol benzene}

Pbenzene= ybenzene$PT= _0.002560.00256_{+ 0.38425 (912) = 6.04} percent saturation = Pbenzene PT− Pbenzene Pvbenzene PT− Pvbenzene % 100= 6.04 912− 6.04 391.5 912− 391.5 % 100= 0.886 % relative saturation = _PvPbenzene

benzene % 100= 6.04391.5 % 100 = 1.54 %

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

At 30°C, Pvacetone= anti log 7.23157 − 1277.03_{237.23+ 30 = 283.7 mm Hg}

percent saturation = Pacetone PT− Pacetone Pvacetone PT− Pvacetone % 100= Pacetone 755− Pacetone 283.7 755− 283.7 % 100= 40 â Pacetone = 146.5 mm Hg

At chilling temperature, Pacetone = Pvacetone

DRAFT

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

benzene-free nitrogen)?

Pvbenzene= anti log 6.90565 − 1211.033_{220.790+ 40 = 182.8 mm Hg}

percent saturation = Pbenzene PT− Pbenzene Pvbenzene PT− Pvbenzene % 100= 0.15(780) 1500− 0.15(780) 182.8 1500− 182.8 % 100=