A gas mixture containing 60% ethanol (C_2H_5OH) and 40 % benzene (C_6H_6) by vol

Question

A gas mixture containing 60% ethanol (C_2H_5OH) and 40 % benzene (C_6H_6) by volume flows to a furnace at a rate of 3,000 ft^3/hr at 200degreeF and 20 psig, where it is burned with 20% excess air. Calculate the required flow rate of air in SCFH (standard cubic feet per hour) assuming 90% completion of the burning process and 20% of the burned gas from CO. Humid air is bubbled through 100 moles of warm liquid water at a rate of 40 liters/min (under 1 atm, 20 degree C and 50% relative humidity). The air leaving the liquid is at 40 degree C, 1 atm and 80% relative humidity. Determine: the flow rate of humid air leaving the liquid in L/min; and the time required for all the liquid water to be carried away by the humid air. Pure elemental sulfur can be produced through the following reaction (nor balanced yet): If 100 Kg/hr of sulfur are to be produced, determine the input rates of H_2S and SO_2 in Kg/hr assuming SO_2 is 30% in excess and the fractional conversion of H_2S is 70%. Also, determine the dry and wet compositions of the product gas and the corresponding total molar flow rates. Liquid benzoic alcohol (C_7H_8O) flowing at 12 L/min is burned with air. Calculate the flue gas flow rate (in m^3/min under 50 mm Hg (gauge) and 1500 degree C) and its wet composition under the following conditions: 10% excess air. 90% conversion of benzoic alcohol, and 8% of the burned benzoic alcohol form CO. Also, determine the dew point of the flue gas.

Explanation / Answer

Flow rate of gas mixture V= 3000 ft3/hr, P = 20Psig= 20+14.7= 34.7 Psia, =

T= 200+460 R= 660 OR, R= 10.731 ft3.psis lbmole.R

Hence n= no of moles of gas mixture = PV/RT= 34.7*3000/(10.731* 660)= 14.7 lb moles/hr

They contains ethanol at 60% , ethanol moles/hr= 0.6*14.7= 8.82 and benzene= 14.7-8.82=5.88

The combustion reactions are C2H5OH + 3O2 ------à2CO2 + 3H2O and C6H6+7.5O2-à 6CO2 + 3H2O

Moles of oxygen required = 3*8.82( 1 moles of C2H5OH requires 3 moles of oxygen) + 7.5*5.88( 1 mole of C6H6 requires 7.5 moles of oxygen)=70.56 lb moles/hr of oxygen

Air contains 21% O2 and 79% N2, moles of air = 70.56/0.21 =336

Air supplied is 20% excess, moles of air supplied =1.2*336= 403 lb moles/hr

Hence n= 403 lb moles/hr, T= 0deg.c= 32deg.F+460= 492OR, P= 14.7 psia

V= nRT/P= 403*492*0.7302/14.7 =9849 ft3/hr

2.

At 20 deg.c, vapor pressure of water = 17.5 mm Hg and 40 deg.c= 55.3 mm Hg

Relative humidity = 100*(partial pressure of water vapor/vapor pressure of water )

At inlet conditions, 0.5= partial pressure of water vapor/17.5, partial pressure of water vapor= 17.5*0.5= 8.75 mm Hg

Partial pressure of water vapor/partial pressure of dry air = moles of water vapor/moles of dry air

8.75/(760-8.75) =0.012 = x/(40-x), x= mole of water vapor

Hence x= 0.012*(40-x), x= 0.012*40-0.012x, x= 0.012*40/1.012 =0.47 moles of water vapor

Moles of dry air = 40-0.47= 39.53

At the outlet, partial pressure of water vapor = 55.3*0.8 mm Hg=44.24 mm Hg

Again, partial pressure of water vapor/moles of dry air = partial pressure of water vapor/moles of dry air pressure , 44.24/(760-44.72)= moles of water vapor/39.53

Moles of water vapor= 39.53*0.062= 2.45 moles

Moles of wet air leaving =39.53+2.45 =41.98 moles

Volume of wet air leaving = 41.98*0.0821*313/1 L=1079 L

Flow rate of water = 40 L/min

Time taken =1079/40 =27 min

3.

Molar masses : S= 32, H2S= 34, SO2= 64, H2O=18

The reaction is H2S+ SO2--à2S + H2O

Moles of S to be produced= 100/32= 3.125 kg moles/hr

From the stoichiometry, moles of H2S required= 3.125/2 = 1.56 kg moles/hr

However, the reaction is only 70% complete. Hence moles of H2S required= 1.56/0.7= 2.23 kg moles/hr, flow rate in kg /hr= molar flow rate* molar mass =flow rate in Kg/hr= 2.23* 34=76 kg/hr.

SO2 required = 2.23 kg moles/hr, SO2 supplied is 30% excess, moles of SO2= 2.23*1.3=2.9 kgmoles/hr, mass of SO2 = 2.9*64=186 kg/hr

Product ( dry basis ) : 3.125 kg moles/hr S, H2S= 2.23-1.56= 0.67 kg moles/hr, SO2= 2.9-1.56= 1.34 kg moles/hr, total moles = 3.125+ 0.67+1.34= 5.135 kg moles/hr Composition ( dry basis) : S= 3.125/5.135= 0.61, H2S= 0.13, SO2=1-0.61-0.13=0.26

Wet basis : moles of water additionally will be there which is = 3.125 kg moles/hr

Total moles =5.135+3.125= 8.26 kg moles/hr

Composition ( wet basis ) : S= 3.125/8.26= 0.38 = H2O, SO2= 1.34/8.26= 0.16, H2S= 1-0.16-0.76=0.08

4.

The reaction is C7H8O + 8.5O2-----à7CO2 + 4H2O (1)

Flow rate of benzoic alcohol= 12 L/min density = 1040 g/L, mass flow rate= 12*1040= 12480 g/min

Molar mass = 7*12+8+16= 108

Moles of benzoic alcohol= 12480/108= 693 gmoles/min

Moles of oxygen required= 8.5* moles of benzoic alcohol= 8.5*693= 5899 moles/min

Moles of air to be supplied ( since air contains 21% O2 and 79% N2), moles of air =5899/0.21= 28090 moles/min, Air supplied = 10%, ait actually supplied= 1.1*28090=30899 moles/min

The combustion is 90% complete. So moles of benzoic alcohol consumed= 693*0.9= 624 gmoles/min

92% of this got converted to CO2 via reaction -1

The second combustion reaction for formation of CO is

C7H8O+ 5O2----à7CO + 4H2O (2)

Hence moles of benzoic alcohol converted to CO2= 624*0.92= 574 gmoles/min

Remaining Benzoic alcohol= 624-574= 50 gmoles/min

Products :

Moles of CO2 formed = 7*574= 4018 gmoles/min

Moles of CO formed from reaction -2 = 7*50= 350 gmoles/min

H2O formed from reaction-1 and reaction-2 = 574*4+50*4= 2496 gmoles/min

O2 remaining= O2 supplied-( O2 consumed for reaction- 1 + oxygen consumed for reaction-2)

=30899*0.21-(8.5*574+5*50)= 1360 gmoles/min

N2= 30899*0.79= 24410moles/min

Products :

N2= 24410 moles/min, CO2= 4018 gmoles/min, CO= 350 gmoles/min, H2O= 2496 gmo.es/min and O2= 1360 gmoles/min

Total moles : 24410+4018+350+2496+1360= 32634 moles/min

Composition : N2= 24410/32634=0.75, CO2= 4018/32634=0.12, H2O= 0.076, O2= 0.042

Partial pressure of water = mole fraction* total pressure

Pressure = 50mm (g)= 50+760= 810mm Hg

Partial pressure = 0.042*760= 31.92 mm Hg. This corresponds to vapor pressure of water at 30 deg.c which is the dew point since at dew point, partial pressure = vapor pressure

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