Chemical Engineering Tools1. Mass Balance (Neraca Massa)

Massa yang masuk suatu sistem akan sama dengan massa yang keluar sistem.2. Energy balance (Neraca Panas)

Energi yang masuk sistem + energi yang dihasilkan - energi keluar/hilang - energi berubah bentuk = energi yang terakumulasi

3. Equilibrium (Keseimbangan) Laju transfer massa atau panas dari kedua sistem sama (keseimbangan fasa, keseimbangan reaksi kimia )

4. Rate Process (Laju Proses) Laju perpindahan massa (laju reaksi/kinetika) maupun laju perpindahan panas.

5. Economy (Ekonomi)Proses kimia di scale up dalam skala pabrik. Strategi Perancangan dan kelayakan ekonominya.

6. Humanity (Humanitas)Mempertimbangkan aspek sosial dan lingkungan

04/28/2023 1

Material Balance

Mass In = Mass Out + Accumulation

No Chemical Reaction Involved

A Slurry containing 25% by weight of solids is fed into a filter. The filter cake contains 90% solids and the filtrate contains 1% solids. Make a material balance around the filter for a slurry feed rate of 2000 kg/h. For that feed rate, what are the corresponding flow rates for the cake and the filtrate ?

Calculation Procedure1. Sketch the system, showing the available data, indicating the unknown, defining

the system boundary, and establishing the basis for the calculations2. Set up and solve the material balance equatios3. Check the results

C kg/h

slurry

2000 kg/h75% liquid25% solid

Filter cake

Filtrate1% Solid99% LiquidF kg/h

System boundary

Overall balance : Filtrate out + cake out = slurry inF + C = 2000 kg/h

Liquid balance : Liquid in filtrate + liquid in cake = liquid in slurry or(wt fraction liquid in filtrate)(mass of filtrate) + (wt fraction liquid in cake)(mass of cake) = (wt fraction liquid in slurry)(mass of slurry)(1.0-0.001)F + (1.0-0.90)C = (1.0-0.25)2000

Simultaneous solving of the two equations, F+C=2000 and 0.99F+0.1C=1500, gives F to be 1460.7 kg/h of filtrate and C to be 539.3 kg/h of cake

In the present process, let the basis be 1 hour.

Chemical reaction involved

• Natural gas consisting of 95% methane and 5% nitrogen by volume is burned in a furnace with 15% excess air. How much air at 289K and 101.3 kPa (14.7 psia) is required if the fuel consumption is 10 m3/s measured at 289K ?. Make an overall material balance and calculate the quantity and composition of the flue gas.

Natural gas10 m3/s

Flue gasF m3/s

System boundary

AirA m3/s

Furnace

Natural gas at 289K and 101.3 kPa = 10 m3/s95% CH4 MW = 165% N2 MW = 28

Air at 289K and 101.3 kPa = A m3/s R = 8.314 m3 kPa/(kg.mol)(K) ideal gas constant21% O2 (MW = 32 ); 79% N2 (MW = 28)

The ideal gas law can be usedThus, n = PV/RT where n is number of moles, P is pressure, V is volume, R is the gas constant, and T is

absolute temperature n CH4 = (101.3kPa)[(10 m3/s)(0.95)]/[8.314 m3kPa/kg.mol)(K)][(289K)] = 0.40 kg.mol/s n N2 = (0.05/0.95)(0.40) = 0.02 kg.mol/s

The combustion reaction for CH4 is CH4 + 2O2 CO2 + 2 H2O0.40 mol/s CH4 requires 2(0.4) mol/s O2 for stochiometric combustion.Since 15% excess air is specified, the number of moles of oxygen in the air is (1.15)(0.80) = 0.92

kg.mol/sThe amount of Nitrogen in with the air is [(0.79 mol N2/mol air)/(0.21 mol O2/mol air)[0.92 kg.mol/s O2]

= 3.46 kg.mol/sTotal moles in the incoming air are 0.92 + 3.46 = 4.38 kg.mol/sFinally, using the ideal gas law to convert to volumetric flow rate, V = nRT/P = (4.38)(8.314)(289)/101.3

= 103.9 m3/s of air

• Set up the material balance and calculate the composition and quantity of the flue gas• Convert to a mass basis. In the present problem, there is no accumulation.• The output (the flue gas) includes nitrogen from the air ang from the natural gas, plus the

15% excess oxygen, plus the reaction products, namely : 0,40 mol/s CO2 and 2(0,40) = 0,80 mol/s water.

• Select 1 is as the basis. The inputs and outputs are as follows :

Component Kilogrammole x kilogram/mole = Kilogram

Input

Natural Gas

• CH4 0,40 16 6,40

• N2 0,02 28 0,56

Air

• N2 3,46 28 96,88

• O2 0,92 32 29,44

Tota;l 133,28

• The accumulation is zero. The overall material balance is 133.28 = 133.28 + 0• The total quantity of flue gas, therefore is 133.28 kg/s (293 lb/s).

Component Kilogrammole x kilogram/mole = Kilogram

Output

Flue Gas

• N2 (3,46 + 0,02) 28 97,44

• O2 (0,92-0,80) 32 3,84

• CO2 0,40 44 17,60

• H2O 0,80 18 14,40

Total 133,28

Equilbrium

Phase Equilibrium

• Vapor-liquid Equilibrium Ratios (VLE) for ideal solution Behavior.• Design caclculation involving VLE, such as distillation, absorption, or stripping, are

usually based on vapor-liquid equilibrium ratios (K)• Ki = yi/xi ; yi = is the mole fraction of that species in the vapour phase and x i = is its

mole fraction in the liquid phase.• Sometimes the design calculations are based on relative volatility ᾳi,j, which equals

Ki/Kj. The subscripts i and j referring to two different species• In general, K values depend on temperature and pressure and the compositions of

both phases

Example : assuming ideal system behavior, calculate the K values and relative volatility for the benzene-toluene systems at 373K (212 ⁰F) and 101.3 kPa (1 atm).

• When a system obeys Raoult’s law and Dalton’s law, it is known as an “ideal system”. Then Ki = P i ͦ/P and ᾳi,j = P i ͦ / P ͦj , where P i ͦ is the vapor pressure of the pure ith component at the system temperature and P is the total pressure.

• One way to obtain the necessary vapor-pressure data for benzene and toluene is employ the Antoine equation :

• Log Po = A-B/(t+C)• P = is in mm of mercury and t = is temperature in degrees Celsius, the following are the values

for the constants A,B and C

• Then, at 373 K (i.e., 100 ⁰C), log Pobenzene = 6.90565 – 1211.033/(100+220.790) = 3.1305. and

Pobenzene therefore equals 1350.5 mmHg (180.05 kPa)

• Similary, log Potoluene = 6.95464-1424.255/(100+219.482) = 2.4966 ,so Po

toluene equals 313.8 mm Hg (41.48 kPa)

• Total pressure P is 1 atm, or 760 mmHg. Therefore, Kbenzene = 1350.5/760 = 1.777, and Ktoluene = 313.8/760 = 0.413

• Relative volatility of benzene with respect to toluene , ᾳbenzene-toluene = 1350.5/313.8 = 4.304

A B C

Benzene 6.90565 1211.033 220.790

Toluene 6.95464 1424.255 219.482