Soluciones cap.3 felder y r.w. rosseau “ principios elementales de los procesos químicos “ 3°ed.

CHAPTER THREE
3.1 (a) m =

16 × 6 × 2 m3 1000 kg ≈ 2 × 10 5 2 103 ≈ 2 × 105 kg 3 m

b

gb gb gd i
4 × 106

(b) m =

8 oz 2s

1 qt

106 cm3

1g
3

32 oz 1056.68 qt cm

≈

b3 × 10gd10 i
3

≈ 1 × 102 g / s

(c) Weight of a boxer ≈ 220 lb m 12 × 220 lb m 1 stone Wmax ≥ ≈ 220 stones 14 lb m (d)

dictionary

V= ≈

πD 2 L
4

=

314 4.5 ft . 4

2

2

800miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 3 42 gal

3 × 4 × 5 × 8 × 10 2 × 5 × 10 3 × 7 4 × 4 × 10

d

i d

i

≈ 1 × 10 7 barrels

(e) (i) V ≈

6 ft × 1 ft × 0.5 ft 28,317 cm3 ≈ 3 × 3 × 104 ≈ 1 × 105 cm3 3 1 ft

(ii) V ≈

150 lb m

1 ft 3 62.4 lb m

28,317 cm3 1 ft 3

≈

150 × 3 × 104 ≈ 1 × 105 cm3 60

(f) SG ≈ 105 .

3.2

(a) (i)

995 kg 1 lb m 0.028317 m3 =62.12 lb m / ft 3 3 3 m 0.45359 kg 1 ft 995 kg / m3 62.43 lb m / ft 3 1000 kg / m3 = 62.12 lb m / ft 3

(ii)

(b) ρ = ρ H2 O × SG = 62.43 lb m / ft 3 × 5.7 = 360 lb m / ft 3 3.3 (a)

50 L

0.70 × 103 kg

1 m3

m3 103 L 1150 kg min 10 gal

= 35 kg

(b)

m3 1000 L 1 min = 27 L s 0.7 × 1000 kg 1 m3 60 s 1 ft 3 0.70 × 62.43 lb m 1 ft 3 ≅ 29 lb m / min

(c)

2 min 7.481 gal

3-13.3 (cont’d) (d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline

Vg cm3gasoline ⇒ 0.70Vg g gasoline
3

i d i 1dcm kerosenei ⇒ 0.82dg kerosenei d0.70V + 0.82idg blendi = 0.78 ⇒ V SG = V + 1dcm blend i
g

d

3

g

=

g

0.82 − 0.78 3 = 0.5 0 cm 0.78 − 0.70

Volumetric ratio =
50.0 kg

Vgasoline 0.50 cm3 3 3 = 3 = 0.50 cm gasoline / cm kerosene Vkerosene 1 cm3.4

In France: In U.S.:

L 5 Fr $1 = $68.42 0.7 × 10 kg 1L 5.22 Fr . 50.0 kg L 1 gal $1.20 = $22.64 0.70 × 10 kg 3.7854 L 1 gal .

3.5

V B ( ft 3 / h ), m B ( lb m / h ) V ( ft 3 / h), SG = 0.850
V H ( ft 3 / h ), m H ( lb m / h )

700 lb m / h

(a) V =

700 lb m h
VB ft
3

ft 3 0.850 × 62.43 lb m

= 1319 ft 3 / h .

mB = mH

d i 0.879 × 62.43 lb = 54.88V bkg / hg ftb hg = dV hb0.659 × 62.43g = 4114V b kg / hg .
m 3 B H H 3

VB + VH = 1319 ft / h .
mB + mH = 54.88VB + 4114VH = 700 lb m .
⇒
VB = 114 ft 3 / h ⇒ mB = 628 lb m / h benzene .

VH = 1.74 ft 3 / h ⇒ mH = 71.6 lb m / h hexane
(b) – No buildup of mass in unit. – ρ B and ρ H at inlet stream conditions are equal to their tabulated values (which are

strictly valid at 20 C and 1 atm.) – Volumesof benzene and hexane are additive. – Densitometer gives correct reading.

o

3-2

3.6

(a) V = (b)

195.5 kg H 2SO 4

1 kg solution

L

0.35kg H 2SO 4 12563 × 1000 kg . .
195.5 kg H 2 SO 4

= 445 L

Videal =

L 18255 × 1.00 kg . 195.5 kg H 2 SO 4 0.65 kg H 2 O L + = 470 L 0.35 kg H 2 SO 4 1.000 kg 470 − 445 % error = × 100% = 5.6% 445

3.7

Buoyant force up = Weightof block down

b gE
2

b

g

Mass of oil displaced + Mass of water displaced = Mass of block

ρ oil 0.542 V + ρ H O 1 − 0.542 V = ρ c V
. From Table B.1: ρ c = 2.26 g / cm3 , ρ w = 100 g / cm3 ⇒ ρ oil = 3.325 g / cm3 moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm3 = 117.4 g moil + flask = 117.4 g + 124.8 g = 242 g
3.8

b

g

b

g

Buoyant force up = Weight of block down

b gb

g
15 . 2

⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp. Liq = ( ρVg ) block Expt. 1: ρ w 15 A g = ρ B 2 A g ⇒ ρ B = ρ w × .
ρ w =1.00 g/cm3

b g

b g

ρ B = 0.75 g / cm3 ⇒ SG

b g

B

= 0.75

Expt. 2: ρ soln A g = ρ B 2 A g ⇒ ρ soln = 2 ρ B = 15 g / cm3 ⇒ SG .
3.9
hs 1 hρ1

bg

b g

b g

soln

= 15 .

Let ρ w = density of water. Note: ρ A > ρ w (object sinks)WA + WB hb1

Volume displaced: Vd 1 = Ab hsi = Ab hp1 − hb1 Archimedes ⇒

d

i

(1)

ρ wVd 1 g
weight of displaced water

= WA + WB

Before object is jettisoned

Subst. (1) for Vd 1 , solve for h p1 − hb1

d

i

h p1 − hb1 =

WA + WB pw gAb

(2)

Volume of pond water: Vw = Ap h p1 − Vd 1 ⇒Vw = Ap h p1 − Ab h p1 − hb1
subst. 2

bi g

bg

d

i

for b p 1 −...