principios elementales de los procesos quimicos solucionario - felder

s-1 d Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inad

Calculate the total cost to travel miles.

6 3

3 5

plane h 1 d 1 yr 220.83 imp gal 1 cm 1000 g 1000 kg

tonne kerosene 1.188 10

plane yr

⋅

⋅9

5

4.02 10 tonne crude oil 1 tonne kerosene plane yr

yr 7 tonne crude oil 1.188 10 tonne kerosene

3 m

ρ f

ρ c

m m

y= –1

y= –1+h

dA

(a) (i) On the earth:

2 m 2

2 m 2

(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor

temperature (failure of reactor control system), problems with the color measurement system, operator carelessness

(c) Beginning with Run 11, the process has been near or well over the upper quality assurance

limit An overhaul would have been reasonable after Run 12

2 exact

ρ µ

N s / m

N s / m m)

2

3 2

2

(b) The diameter of the particles is not uniform, the conditions of the system used to model the

equation may differ significantly from the conditions in the reactor (out of the range of

empirical data), all of the other variables are subject to measurement or estimation error

crystals 0.050 in (25.4) mm min mm in

238 crysta ls / min 238 crystals 1 min

(b) The t in the exponent has a coefficient of s-1

2.28 (a) 3 00 mol / L, 2.00 min-1

exact

For C=0.10 mol/L: t

t

int exact

min

2.00ln

C3.00= -

1

2ln

0.103.00= 1.70 min

t (min)

(t=0.6, C=1.4) (t=1.12, C=0.10)

DO 1 I = 1, 6 READ (5, *) TD(I), PD(I)

2

FORMAT (10X, F5.1, 10X, F5.1) CONTINUE

END

2.29 (cont’d)

SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6)

2.31 (b) Plot y2 vs on rectangular axes Slopex3 =m,Intcpt= −n

L / s

L / s

L / s

(c) The estimate for T=175°C is probably closest to the real value, because the value of

temperature is in the range of the data originally taken to fit the line The value of T=290°C

is probably the least likely to be correct, because it is farthest away from the date range

the plot is a straight line

- 2 -1.5

- 1 -0.5

2.35 (a) ft and h , respectively3 -2

(b) ln(V) vs t2 in rectangular coordinates, slope=2 and intercept= ln( 3 53 10× −2); or

V(logarithmic axis) vs t2 in semilog coordinates, slope=2, intercept= 3 53 10 × −2

m

L L m

.483

C=475 ppm is well beyond the range of the data

2.38 (a) For runs 2, 3 and 4:

a = 86.7 volts kPa1.46/ (L / s)0.678

(b) When P is constant (runs 1 to 4), plot lnZ vs lnV& Slope=b, Intercept= lna+clnp

lnZ = 0.5199lnV + 1.0035

0 0.5 1 1.5 2

1 1.5

Plot Z vs &V P b c Slope=a (no intercept)

a=slope=311 volt kPa / (L / s)⋅ .52

The results in part (b) are more reliable, because more data were used to obtain them

i n

1 2 3 4

(d) t=0 and C=0.01 are out of the range of the experimental data

(e) The concentration of the hazardous substance could be enough to cause damage to the

biotic resources in the river; the treatment requires an extremely large period of time; some

of the hazardous substances might remain in the tank instead of being converted; the

decomposition products might not be harmless

2.41 (a) and (c)

1 10

-4 -2 0

-2 0

t

(c) Disregarding the value of k that is very different from the other three, k is estimated with

the average of the calculated k’s k=0 0063 s-1

(d) Errors in measurement of concentration, poor temperature control, errors in time

measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor

i i i

i

i i

i i i

n

i i n

4FORMAT (1H0, 'SLOPEb bAb =', F6.3, 3X 'INTERCEPTb b8b =', F7.3/)

RESIDUALS SSQ = 0.0

DO 200J = 1, N

YC = A * X(J) + B

RES = Y(J) - YC

WRITE (6, 5) X(J), Y(J), YC, RES

5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X

3 3

(b) – No buildup of mass in unit

– ρ B and ρ H at inlet stream condit ions are equal to their tabulated values (which are strictly valid at 20oC and 1 atm.)

– Volumes of benzene and hexane are additive

– Densitometer gives correct reading

3.8 Buoyant force up = Weight of block downb g b g

⇒Wdisplaced liquid =Wblock⇒(ρ Vg)disp Liq =(ρ Vg)block

Let ρ w = density of water Note: ρ A >ρ w (object sinks)

Volume displaced: V d1= A h b si = A h bd p1−h b1i (1)

Subst (1) for V d 1 , solve for hd p1−h b1i

h

A B w

b h

V h

W W

Neglected the weight of the bag itself and of the air in the filled bag

(c) The limestone would fall short of filling three bags, because

– the powder would pack tighter than the original particles

– you could never recover 100% of what you fed to the mill

(

b

m V

b b f

b b m

f m b f mnf mb x f b

f f

f nf

122 51

0 9

111

(c) The density of H2O increases as T decreases, therefore the density was higher than it should have been to use the calibration formula The valve of r and hence the Ile mass flow rate calculated in part (b) would be too high

0.987 0.989 0.991 0.993 0.995 0.997

Density (g/cm3)

(b)

Rotameter

Reading

Collection Time (min)

Collected Volume (cm3)

Mass Flow Rate (kg/min)

Difference Duplicate (Di)

There is roughly a 95% probability that the true flow rate is between 0.592 kg / min and 0.628 kg / min

3.15 (a) &m=175 m 1000 L 0.866 kg 1 h =2526

(b) &n= 2526 kg 1000 mol 1 min=457

(c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm

3.16 (a) 200 0. kg mix 0150. kg CH OH kmol CH OH 1000 mol 936

(c) 150 g CaCO3/500 g suspension=0 300 g CaCO3 g suspension

3.19 Assume 100 mol mix

3.20 (a)

Crystallizer Form solid gypsum particles from a solution

Filter Separate particles from solution

kmolmin

mol CSAmol FB

=

=

U V

She was wrong

The mixer would come to a grinding halt and the motor would overheat

3.22 (a) 150 mol EtOH 46.07 g EtOH 6910

6910 g EtO

10365

H 0.600 g H O0.400 g Et OH g H O

2 264 kmol CH 0.91 kmol air

4

4

=

5% CH 2.264 kmol CH 0.95 kmol air

h 0.05 kmol CH 43.01 kmol air h4

4

4

Dilution air required: 43.01 - 22.89 kmol air

h 1 kmol mol air h

kg Okg

i

i i i

i i

i i

i i i

3.26 (a)

Area Fraction Fraction

MOLT = MOLT + MOL(J)

MASST = MASST + MASS(J)

1 2838

(b) Hexane recovery = & × = × =

&

.

n n

3 1

3.31 (a) kt is dimensionless ⇒ k (min-1)

0 06243 C′ =A 1334 expb−0 419 t′ 60gdrop primes⇒ C Ab g=214 expb−0 00693 tg

t (min)

P h g ρ

3.34 (cont’d)

(b)

,

b b

— Structural flaw in the tank

— Tank strength inadequate for that much force

— Molasses corroded tank wall

Since the pressure at every point on the door is greater than

P a, Since the pressure at every point on the door is greater than P a, F > ρghAdoor

(b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill

(ii) If the door holds, it will take

V

fill room

(b) Air in the line (lowers average density of the water.)

(c) The line could be clogged, or there could be a leak between the junction and the tap

2 m

1 m

Use mercury, because the water manometer would have to be too tall

(b) If the manometer were simply filled with toluene, the level in the glass tube would be at the

level in the tank

Advantages of using mercury: smaller manometer; less evaporation

(c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen,

minimizing the risk of combustion

(b) — Nonreactive with the vapor in the apparatus

— Lighter than and immiscible with mercury

— Low rate of evaporation (low volatility)

3.47 (a) Let ρf = manometer fluid density c110 g cm3h , ρ ac = acetone density c0 791 g cm3h

Differential manometer formula: ∆P=dρ f −ρ acigh

ln( P)

C

1000 - 100 L

CL ; b=15 0 0311 100− × =119 o C

⇒ T C) ( o = 0 0311 T ( o L) + 11 9 and T ( o L) = 32 15 T ( o C) − 382 6

(d) Tbp = − 88 6 o C ⇒ 184.6 K ⇒ 332.3 R o ⇒ -127.4 F o ⇒ − 9851 o FB ⇒ − 3232 o L

(e) ∆ T = 50 0 o L ⇒ 1.56 C o ⇒ 16 6 o FB ⇒ 156 K ⇒ 2 8 o F ⇒ 2 8 o R

mV C mV

3.54 REAL, MW, T, SLOPE, INTCPT, KO, E

REAL TIME (100), CA (100), TK (100), X (100), Y(100)

3 FORMAT (' K (L/MOL – MIN): ', F5.3, //)

5 FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4) END

SUBROUTINE LS (X, Y, N, SLOPE, INTCPT)

REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J

CHAPTER FOUR 4.1 a Continuous, Transient

b Input – Output = Accumulation

No reactions ⇒ Generation = 0, Consumption = 0

s

kg s

kg s

dt

dn dt

s kg

3 3

4.2 a Continuous, Steady State

c Input – Output – Consumption = 0

Steady state ⇒ Accumulation = 0

A

m s

mol m

m s

mol m

mol s

01

&m l kg / h 0.106kg B / kg 0.894kg T / kg

Input – Output = 0 Steady state ⇒ Accumulation = 0

No reaction ⇒ Generation = 0, Consumption = 0

(1) Total Mass Balance: 100 0 kg / h=m&v+m&l

(2) Benzene Balance: 0 550 100 0 × kg B / h=0 850 m&v+0106 m&l

Solve (1) & (2) simultaneously ⇒ m &v = 59 7 kg h, m &l = 40 3 kg h

b The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg) The balance equations are also identical (initial input = final output)

c Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors

4.4 b n (mol)

2 4

lbg

lb - mole C H

lb C H

sh

&

&

=+ F

&

n n

Basis: 1000 lbm C3H8 / h fresh feed

(Could also take 1 h operation as basis

-flow chart would be as below except

that all / h would be deleted.)

Note: the compressor and the off gas from

the absorber are not mentioned explicitly

in the process description, but their presence

should be inferred.

b Overall objective : To produce C3H6 from C3H8

Preheater function: Raise temperature of the reactants to raise the reaction rate

Reactor function: Convert C3H8 to C3H6

Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other components

Stripping tower function: Recover the C3H8 and C3H6 from the solvent

Distillation column function: Separate the C3H5 from the C3H8

4.6 a 3 independent balances (one for each species)

b 7 unknowns (m m m x y y z& , & , & , , , ,1 3 5 2 2 4 4)

– 3 balances

– 2 mole fraction summations

2 unknowns must be specified

c y2 = − 1 x2

h

kg Ah

kgh

B Balance: 0 03 m&1+5300x2 F 1200y4 0 60 m&5

kg Bh

kg Bh

z4 = − 1 0 70 − y4

g H O

b gAcetic Acid Balance: b gb g400 0115 g CH OOH 0 005 & 0 096 &

min

g CH OOHmin

H O someCH COOH

2 3

C H OH

CH COOH

4 9 3

C H OH4 9

Distillation Column

1

1 2

U V|

eggs min

1+ 2 =50 large eggs min

n1 large eggs broken/50 large eggs = b 11 50 g = 0 22

d 22% of the large eggs (right hand) and b25 70g⇒36% of the extra-large eggs (left hand)

are broken Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right Therefore, Fred is right-handed

4 unknowns (m m V1, 2, 40,m3) – 2 balances

3

n1

0 0403

0 9597

mol / smol C H / molmol air / mol

3 8

&

n2

0 21

0 79

mol air / smol O / molmol N / mol2 2

&

n3

0 0205

0 9795

mol / smol C H / molmol air / mol

b Propane feed rate: 0 0403 n &1 = 150 ⇒ = n &1 3722 b mol / s g

Propane balance: 0 0403 n &1= 0 0205 n &3⇒ n &3 = 7317 b mol / s g

Overall balance: 3722 + n &2 = 7317 ⇒ n &2 = 3600 b mol / s g

c > The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly

&

m kg / h

kg CH OH / kg

kg H O / kg

3 2

x x

kg CH OHkg

gkg

molCH OH

kgh

kg H Okg

gkg

beyond analyzer calibration data is risky recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter

lb- mole HO/ h

ft / h

2 3

d i

&

n3

01000900

lb- mole/ h

lb -mole H O / lb- mole lb- mole DA/ lb - mole

2

4 unknowns (n n n v& , & , & , &1 2 3 ) – 2 balances – 1 density – 1 meter reading = 0 DF

Assume linear relationship: &v aR b= +

lb - mol

3

m 3

m

2

Solve (1) & (2) simultaneously ⇒ =n&1 5890 lb - moles / h , n&3=6480 lb - moles / h

b Bad calibration data, not at steady state, leaks, 7% value is wrong, &v−Rrelationship is not linear, extrapolation of analyzer correlation leads to error

&

&m x x

x x

E S

kg / s

kg E / kgkgS / kg

kg H SO / kg

kg H O / kg SG

2 4 2

b g

.

=

v

m

2 2

0 600

0 400

1 498

L kg

kg H SO / kg

kg H O / k g SG

2 4 2

b g

b g

.

=

v m

3 3

0 323

0 677

1 213

Lkg

kg H SO / kg

kg H O / kgSG

2 4 2

b g

b g

=

5 unknowns (v v v m m1, 2, 3, 2, 3) – 2 balances

100

44 4144

2 3

100017

0 83

Solve (1) and (2) simultaneously ⇒m1 =0 385 kg 25% paint,m2=0 615 kg12% paint

Cost of blend: 0 385 b $18 00 g + 0 615 b $10 00 g = $13 08 per kg

Selling price:110 b$13.08g=$14.39per kg

kg H O / kg2

m m

2 3

day

tons DSton WS

3 3

– 1 volume balance – 3 specific gravities

0 DF

b Specific gravity of coal < 1.48 < Specific gravity of slate

c The suspension begins to settle Stir the suspension 1.00 < Specific gravity of coal < 1.48

2

x x

2

1

mol / hmol H O / molmol DA / mol

2

97% adsorbed: 156 =0 97 0 04 b n&1g⇒ =n&1 401 mol / h

Total mole balance: n&1 =n&2+n&3⇒n&2 =401 1556 − =38 54 mol / h

20.040 40.1 = 1.556 + x 38.54 ⇒ = x 1.2 10 × − molH O/mol

b The calcium chloride pellets have reached their saturation limit Eventually the mole fraction will reach that of the inlet stream, i.e 4%

H2SO4 balance: 0 55 300 b g + 0 90 m &B = 0 75 m &C (2)

Solve (1) and (2) simultaneously ⇒m&B =400lb / hm ,m&C=700lb / hm

2 2

2 2

2 2

&

kmol

Solve (1) and (2) simultaneously ⇒n&A =3 29 kmol / h , n&B =110 kmol / h

Dialyzing fluid

ml / min1500

ml / min

mg urea / ml

&

v c

a Water removal rate: 200 0 195 0. − . =5 0. ml / min

Urea removal rate: 1 90 200 0 b g−1 75 195 0 b g =38 8 mg urea / min

b v&=1500 5 0+ =1505 ml/ min

38.8mgurea/min

0.0258mgurea/ml1505ml/min

Solve (1) and (2) simultaneously ⇒n&2 =55 6 kmol / min,n&3=561 kmol / min

1

m / minkmol / minkmol SO / kmolkmol A / kmol

1

kmol / minkmol SO / kmolkmol A / kmol

4 4 4

8 unknowns (n n v m m x y y& , & , & , & , & , , ,1 3 1 2 4 4 1 3)

b Orifice meter calibration:

A log plot of vs is a line through the points V& h dh1=100, &V1=142i d and h2=400, &V2=290i.

||

| W

kg

2 2

3 3 3 31

kmol / minkmol SO / kmolkmol A / kmol2

4 4 41

kg / minkgSO kg

kg B / kg2

– 3 analyzer and orifice meter readings

– 1 gas density formula (relates n&1andV&1)

– 1 specific gravity (relates m&2andV&2)

& & kg

h

mkg

3

For a given SO2 removal rate (y 3), a higher solvent feed rate ( &V2) tends to a more dilute

SO2 solution at the outlet (lower x 4)

d Answers are the same as in part c

4.28 Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3

Overall mass balance ⇒ &m3

Mass balance - Unit 1 ⇒ &m1

A balance - Unit 1 ⇒x1

Mass balance - mixing point ⇒ &m2

A balance - mixing point ⇒ x2

C balance - mixing point ⇒ y2