Question

Your boss has told you that you need to process a full 55 gallon drum of 1-butene that is stored at 15 atm and ambient temperature. You are going to combust this material with the stoichiometric amount of air at 1 atm and the product gasses will emerge at 579 oF with a conversion of 97%. You have a metal gas sphere that is 55 feet in diameter at ambient temperature that contains the air you are feeding in. You want to make sure you have enough air on hand. The 1-butene enters through a pipe that is ½ inch in diameter, while the air comes in through a 1 ½ in diameter pipe. The flue gases exit through a stack that is 6 inches in diameter. If you use another resource for physical properties, remember to cite it. Answer the following questions:

1) What does the pressure gauge on the tank need to read in order for you to make sure you have enough air, in psig?

Answer 1

The stoichiometric reaction for the combustion of 1-butene in the air can be given as

C4H8 + 6 ( O2 + 3.76 N2 ) ----------> 4 CO2 + 4 H2O + 22.56 N2

( as in air 1 mol of oxygen is with 3.76 mol of nitrogen )

Now we have

the volume of butene to burn = 55 gallon = 0.2081 m3

stored at pressure = 15 atm and temperature ( ambient ) = 293 K

Assuming ideal gas law applies

Number of moles of butene = ( PV / RT )

here

P = 15 atm

V = 0.2081 m3

R = gas constant =  0.082058 m3. atm / kmol. K

T = 293 K

we get

Number of moles of butene = ( 15 atm * 0.2081 m3 ) / ( 0.082058 m3. atm / kmol. K * 293 K )

= 0.12982 Kmol = 129.829 mol

From stoichiometry for combustion of 1mol of butene , ( 6 * 4.76) mol of air is required

so for combustion of 129.829 mol of butene air required = ( 6 * 4.76) * 129.829 = 3707.91 mol = 3.70791 Kmol

Now

We have gas sphere with diameter = 55 ft = 16.764 m

Radius of sphere ( r)  = 16.764 m / 2 =8.382 m

The volume of sphere = ( 4/3) ( r3 ) = ( 4/3) ( 3.14 * 8.3823 ) = 2465.53 m3

Assuming ideal gas law applies

we have

P = ( nRT ) / V

here

P = pressure atm

n= number of mooles of air = 3.70794 Kmol

R = gas constant =  0.082058 m3. atm / kmol. K

T = 293 K ( ambient )

V = 2465.53 m3

we get

P = ( 3.70791 * 0.082058 * 293 ) / 2465.53 = 0.0361 atm = 0.5305 psi

Here pressure is absolute

As absolute pressure is less than atmospheric pressure Guage pressure = Abs. Pressure - atmospheric pressure

= 0.5305 psi - 14.6959 psi = -14.1654 psig

Which is almost equals to atmospheric pressure so, pressure gauge reading on spherical tank = 0 psig