Computer Applications

Distillation Alcohol-water solution is fed into a distilling still at the rate of 1000 kg per hour. The solution analyzes 30% ethyl alcohol by weight, the distillate contains 90% ethyl alcohol and the bottom 8% ethyl alcohol, calculate (a) kg of distillate per hour (b)kg bottom per hour (c) % recovery of alcohol. Solution: OMB: D + B = 1000 B= 1000-D Ethyl Alcohol Balance: 300= 0.90D +0.080B 300=0.90D+0.080(1000-D) (a) D = 268.29 kg/hr (b) B= 731.7 kg/hr (c) % recovery alcohol = (alcohol in D/alcohol in F) x 100 = (0.90 x268.290)/300 = 0.8049 x 100 = 89.49% A distilling column is fed with a solution containing 0.45 mass fractions of benzene and 0.55 mass fraction toluene. If 95% of benzene must appear in the overhead product, while 92% of toluene appears in the residue, calculate the composition of distillate and bottoms. Solution: Basis: 100 kg feed OMB: 100= D + B Benzene balance: Benzene in D= 0.95(0.45)(100) Benzene in D = 42.75 kg Benzene in B= 0.45(0.08)(100) Benzene in B=3.6 kg

Toluene Balance:

Toluene in B= 0.92(0.55)(100) Toluene in B= 50.6 kg Toluene in D= 4.4 kg Wt. Benzene

Wt. Toluene

% Benzene

% Toluene

D

42.75

4.4

90.67

9.33

B

3.6

50.6

6.64

93.36

D = 47. 15kg B = 52.85 kg Dilute alcohol from fermenting vats contains 11.15% ethanol. The distillate contains 95% alcohol while the bottom contains 0.5% alcohol. As distillation proceeds, a side stream containing 40% alcohol is removed. The total recovery of alcohol in the distillate and in the side stream is 96.5%. On the basis of 1000 kg of feed, calculate the amounts of the bottom, distillate and the side stream. Solution: OMB: 1000= D + S + B Alcohol balance: Alcohol in B =0.035 (alcohol in F) 0.005B = 0.035(1000)(.1115) B= 780.5 D + S = 1000 – 780.5 D + S = 219.5 0.1115(1000) = 0.95D + 0.40S+ 0.005(780.5) 107.6= 0.95D + 0.40S D = 36 kg S = 183.5kg

Flow of Fluids Aniline is to be cooled from 200 to 150 degrees Fahrenheit in a double-pipe heat exchanger having a total outside area of 70 ft². For cooling a stream of toluene amounting to 8600 lb/h at a temperature of 100 degrees Fahrenheit is available. The exchanger consists of Schedule 40 ,1 and ¼ inches pipe and Schedule 40,2-inches pipe. The aniline flow rate is10,000 lb/h. if flow is countercurrent, what are the toluene outlet temperature, the LMTD and the overall heat transfer coefficient? Cp of Aniline=0.545 BTU/lb-°F and Cp for toluene= 0.438 BTU/lb-°F. Solution: Aniline : Q= mCpΔT Q= 10,000 (0.545)(200-150) Q=27 x 10⁴ BTU Toluene: Q= 8600 (0.438)(T- 100) Q= 3.77 x 10³ (T-100) Solving simultaneously to get T: 3.77 x 10³ (T-100) = 27 x 10⁴ BTU T = 170 °F ΔTLM = (ΔT2-ΔTi)/ Ln (ΔT2/ΔT1) ΔTLM = (200-170)-(150-100)/ Ln ((200-170)/(150-100)) ΔTLM = 39°F Over-all Heat Transfer Coefficient: 10000(0.545)(200-150)/(70 x 39) = 100 BTU/ ft²-hr -°F

A simple U-tube manometer is installed across an orifice meter. The manometer is filled with mercury( specific gravity= 13.6), and the liquid above the mercury is carbon tetrachloride (specific gravity of 1.6). The manometer reads 200mm. What is the pressure difference over the manometer in Newton per meter square? Solution: Pa-Pb= (g/gc)Rm (ρA-ρB) 1 kg wt. = 9.8 N ΔP = (9.8 m/s²/9.8N) (0.2 m) (13.6 x 10³ -1.6 x 10³) ΔP = 2.4 x 10³ N/m²

Crude oil,specific gravity 60°F/60°F = 0.887, flows through the piping with Pipe A (2-in Schedule 40), Pipe B (3-in Schedule 40) and Pipe C (1 and ½ in Schedule 40). An equal quantity of liquid flows through pipe C. The flow through pipe A and B is 30 gal/min. Calculate the mass flow rate in each pipe. The dimension and cross-sectional areas of standard pipes needed are 0.0233 ft² for a 2-in pipe, 0.0513 ft² for a 3-in pipe and 0.01414 ft² for a 1 and 1/2 pipe.

Solution: Density of the fluid = 0.887 x 62.37 =55.3 lb/ft³ Since there are 7.48 gal in 1 ft³, the total volumetric flow rate is: q=30*607.48=240.7 ft³/hr

The mass flow rate for pipes A and B is the product of the density and the volumetric flow rate: Mass flow rate = 240.7 x 55.3 = 13,300 lb/h Mass flow rate in pipe C: 13,300/ 2 = 6650 lb/hr

Filtration

Slurry of talc and water contains 75% water by weight. If 80% of the original water is removed and 600 kg of filter cake is obtained. Calculate: (a) Original weight of the slurry (b) Water content of the filter cake

Solution: OMB: F = W + Fc F = W + 600 Water balance: W = 0.80 (0.75)F W = 0.60 F F = 0.60F + 600 F = 1500 kg W = 900 kg Water in Fc= water in F – W = 0.75(1500) – 900 = 225 kg

Saturated solution of a salt is made by agitating 600 kg of salt in 1200 kg of water. The salt is soluble to the extent of 0.25 kg per kg of water. On filtration, 0.40 kg of solution adheres to every kilogram of undissolved salt. Calculate: (a) Weight of the filtrate (b) Kg of dry salt recovered on drying the wet filter cake

Solution: Wt. of dissolved salt = 1200 kg water x (0.25 kg salt/1 kg water) = 300 kg

Wt. of saturated solution = 1200 kg water x (1.25 kg solution/1 kg water) = 1500 kg Wt. of undissolved salt = 600 - 300 = 300 kg Wt. of adhering sol’n = 300 kg undis. salt x (0.4 adhering sol’n/ 1 kg undissolved salt)= 120 kg Filter cake = Adhering solution + undissolved solution Fc = 120 + 300 = 420 kg FL = saturated solution – adhering solution FL = 1500 -120 FL = 1380 kg (filtrate) Balance at the Drier: Fc = W + P 420 = W + P Water balance: Water in Fc = W 120 adhering sol’n x (1 kg water/ 1.25 adhering solution) = W W = 96 kg water P = 324 kg salt

Screening The data below were obtained on the operation of a 6-mesh (square) hummer screen at the tipple of a coal mine. The screening was done to separate a very fine refuse from a fine coal stream so

that it could be reprocessed. Calculate (a) the recovery and rejection of each size fraction and (b) the screen effectiveness .Feed is 131tons/hr and approximately 5% moisture. Size

Sample Weight

+1/4 in

3825 grams

¼ x6 mesh

1006

6 x 14

750

14 x 28

303

28 x48

219

48 x 0

807

Overflow from screen

Underflow from screen (9.8 Tons/hr) Dry Solids

Size

sample Wt.

Size

% Sample

+1’4 in

2905 grams

1/4x 6

11.3

¼ x 6 mesh

767

6x8

7.8

14 x 28

117

8 x14

6.9

28 x 48

68

14 x 48

8.6

48 x 0

278

28 x 48

3.3

48 x 0

62.1

Solution: Very fine particles (coal) will be reprocess to fine coal, therefore Oversize= P Since 5% of F is moisture, 95% of F is dry F dry = (131 tons/hr)(0.95) = 124.45 tons/hr P = 124.45 – 9.8 = 114.65 tons/hr F

Xi

P

Xi

3825

0.55

2905

0.640

1006

0.146

767

0.169

750

0.109

405

0.089

303

0.044

117

0.026

219

0.032

68

0.015

807

0.117

278

0.061

Desired size in F: 3825 to 1006 at Xi = 0.55 to 0.146 Desired size in P: 2905 to 767 at Xi = 0.640 to 0.169

What is the power requirement to crush 100 tons per hour of limestone if 80% of the feed passes a 2-inch screen and 80% of the product passes a 1/8-inch screen? Use Bond’s law. Wi =12.74 kW-hr/ton.

Solution: Dp1 = 2 in (25.4mm/1 in) = 50.8 mm Dp2 = 1/8 in (25.4 mm/ 1 in) = 3.175 mm P/m = 0.3162Wi [(1/√Dp2) – (1/√Dp1)] P = 100 tons/hr (0.3162 x 12.74) [(1/√3.175) – (1/√50.8)] P = 169.6 kW or 227 HP