# Solucionario Felder Cap. 4

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CHAPTER FOUR
4.1 a. b. Continuous, Transient Input – Output = Accumulation No reactions ⇒ Generation = 0, Consumption = 0

6.00
c.

dn kg kg dn kg − 3.00 = ⇒ = 3.00 dt dt s s s

t=

100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg

4.2

a. b. c.

k = 0 ⇒ C A = C A0

k = ∞ ⇒ CA = 0

Input – Output – Consumption = 0 Steady state ⇒ Accumulation = 0 A is a reactant ⇒ Generation = 0
3 3

V

FG m IJ C FG mol IJ = V FG m IJ C FG mol IJ + kVC FG mol IJ ⇒ C HsK H sK Hm K H sK Hm K
A0 3 A 3 A

A

=

CA0 kV 1+ V

4.3

a.
100 kg / h 0.550 kg B / kg 0.450 kg T / kg

mv kg / h

b

g
Input – Output = 0 Steady state ⇒ Accumulation = 0 No reaction ⇒ Generation = 0, Consumption = 0

0.850 kg
…ver más…

b400gb0115g FGH g CH OOH IJK = 0.005m min
3

bg b g

R

+ 0.096mE

FG g CH OOH IJ H min K
3

Overall Balance: mC + 400
c.

FG g IJ = m + m FG g IJ ⇒ m = 417 g min H min K H min K g g . b0115gb400g − b0.005gb356g FGH min IJK = b0.096gb461g FGH min IJK ⇒ 44 g min = 44 g min
R E C

⇒ mE = 461g min

4- 3

4.7 (cont’d) d.
H 2O some CH3COOH CH3COOH H 2O C 4 H9OH C4 H9OH CH3COOH

CH 3COOH

Extractor

Distillation Column

C4 H 9OH

4.8

a.
120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg

X-large: 25 broken eggs/min 35 45 unbroken eggs/min Large: n 1 broken eggs/min n 2 unbroken eggs/min

b.

120 = 25 + 45 + n1 + n2 ( eggs min ) ⇒ n1 + n2 = 50 ⎫ n1 = 11 ⎪ ⇒ ⎬ ( 0.30 )(120 ) = 25 + n1 ⎪ n2 = 39 ⎭

c.

n1 + n2 = 50 large eggs min

n1 large eggs broken/50 large eggs = 11 50 = 0.22
d.

b

22% of the large eggs (right hand) and 25 70 ⇒ 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. m1 lb m strawberries 015 lb m S / lb m . 0.85 lb m W / lb m m2 lb m S sugar

b

g

g

4.9

a.

b

g

m3 lb m W evaporated

b

g

c

h

1.00 lb m jam 0.667 lb m S / lb m 0.333 lb m W / lb m

b.

c.

3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF