# Ethanol and Overall Material Balance

Topics: Ethanol, Water, Petroleum Pages: 6 (1063 words) Published: December 28, 2012
ChE 101 Lab Exercise 405 December 2012
Materials Balance (Non-reacting Systems)THR/WRU

1. A 40-mol-% aqueous ethanol solution is to be distilled into two streams: a distillate with 98% ethanol and a side stream with 60%. 98% of the alcohol in the feed is to be recovered in these two products; only 2% goes into the bottoms. If the bottoms product contains 2% ethanol, calculate the amount of the different streams per 1000 kmol of feed stream.

Given: Recovery = 98% of ethanol in feed to distillate and side stream

Feed, F
xF = 0.40
Side stream, S
xS = 0.60
Distillate, D
xD = 0.98
Bottoms, B
xB = 0.02

Required: D, S and B

Solution:

2% of the ethanol in the feed goes to the bottoms

Overall material balance: →
Component material balance: →

System of Linear Equations with D and S unknown

2. Shown below is a flowchart of a process in which acetic acid (A) is extracted from a mixture of acetic acid and water (B) into 1-hexanol (C), a liquid immiscible with water. R xA = 0.005
xB = 0.995
F, 400 g/min
zA = 0.115
zB = 0.885
S
pureC
E
yA = 0.096
yC = 0.904

Figure 1

Calculate E, R, and S based on the given mixture feed rate F.

Given: Figure 1

Required: E, R, S

Solution:

Basis: 400 g/min F

Tie component in F and R: water (B)

Component Material Balance for acetic acid (A):

Overall Material Balance: →

3. A simplified flowsheet for the manufacture of sugar is shown in Figure 1. Sugarcane is fed to a mill where a syrup is squeezed out, and the resulting “bagasse” contains 80% pulp. The syrup (E) containing finely divided pieces of pulp is fed to a screen which removes all the pulp and produces a clear syrup (H) containing 15% sugar and 85% water. The evaporator makes a “heavy” syrup and the crystallizer produces 1000 lb/hr of sugar crystals. a. Find the water removed in the evaporator, lb/hr.

b. Find the composition of the waste stream G.
c. Find the rate of feed of cane to the unit, lb/hr.
d. Of the sugar fed in the cane, what percentage is lost with the bagasse? Figure 2
K40% Sugar
H
15% Sugar
E
13% Sugar
14% Pulp
Mill
Screen
Evaporator
Crystallizer
F Cane
16% Sugar
25% Water
59% Pulp
D Bagasse
80% Pulp
Solids G Contains
95% Pulp
J
Water
L
Water
M Sugar,
1000 lb/hr

Given: Figure 2

Required: J,

Component balance around the crystallizer: →

The other component in K is water:

Sugar is a tie component between H and K:

Material balance around the evaporator:

Material balance around the screen: → (1)
Pulp balance around the screen: → (2)

Substitute the expression in (2) to (1):

Composition xG of sugar in G from sugar balance around the screen:

Overall Material Balance around the mill: →
Pulp Balance around the mill:

System of linear equations with F and D unknown

Sugar balance around the mill: →

Percentage lost in the bagasse

ChE 101 Lab Exercise 116 November 2012

I. Unit Conversion. Using dimensional analysis, expressed the given values in terms of the desired units.

1. 40 slug| =| 583.756 kg| | 11. 135 in2/min| =| 0.016 ft2/s| 2. 5 mile (U.S.)| =| 3.168×105 in| | 12. 250 cp| =| 50.398 lb/in hr| 3. 30 ft2| =| 4.32×103 in2| | 13. 15 Btu/hr ft2 (°F/ft)| =| 3.72 cal/min cm2 (K/cm)| 4. 12 gallons (U.S.)| =| 45.425 L| | 14. 65 Btu/hr ft2°F| =| 369.087 J/s m2°C| 5. 50 years| =| 2.63×107 min| | 15. 15 cal/g °C| =| 62.802 J/g K| 6. 75 lbf| =| 333.617 N| | 16. 28 atm cm3/mol K| =| 0.249 atm ft3/lb mole °F| 7. 25°C| =| 536.67°R| | 17. 96 Btu/lbmol R| =| 4.019×103bar cm3/ mol K| 8. 35 psi| =| 2.413×105 Pa| | 18. 0.24 W/m K| =| 0.139 Btu/hrft (°R)| 9. 100 lbm/ft3| =| 1.602 g/mL| | 19. 270 rad/s| =| 2.578×103rpm| 10. 125 ft-lbf| =| 1.695×103 cm3-bar| | 20. 1036 molecules| =| 1.661×1012mol...