SOLUTION TO TUTORIAL (9)CHE 302 - COMPUTERIZED MATERIAL AND ENERGY BALANCES

PROBLEM 8.25 PAGE 530 (G.V.REKLAITIS,” Introduction to Material and EnergyBalances”, John Wiley and Sons.

Sulfur dioxide is converted to sulfur trioxide by oxidation at atmospheric pressureaccording to data shown in process flow diagram below. Sulfur dioxide and oxygen arefed to the process in stoichiometric ratio. Conversion is 96%. Ten kcal per mole ofcombined feed to reactor is removed by external medium. Assume no heat losses to theenvironment.

Calculate all unknown temperatures and flows. Given C p (cal/ mol O C).

O 2 SO 2 SO 3

Avg. C p 25 - 300 O C 7.6 10.4 15.3

Avg. C p 300 - 900 O C 7.8 11.8 17.2

SOLUTION

Basis: 1.0 mole of sulfur dioxide feed in stream (1)The reaction is:SO 2 + ½ O 2 SO 3

Material balances around heat exchangers are trivial. Therefore flows through heatexchangers are unchanged.Flow rates of species in streams will be as follows:Stream (1) and (2): 1.0 mole of SO 2

Streams (3), (4), (5), 96): N2

3SO , N

3

3SO , N

2

3O

Streams (7) and (8): N 7Sat Water

Streams (9) and (10): N2

9O

Degree of Freedom Table:

Stream VariablesTQ

6 + 1104

Material Balances (Round Reactor only)Energy BalancesConversionStoichiometric relationGiven Temperatures ♣Given heat loadsBasis

3411741

Degree of freedom Zero♣ Temperatues of saturated water and steam can be read from steam tables at 300 psia.

Table of Independent Unknown Variables:

Stream Number (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) TotalNumber ofUnknownVariables

- - 3 - - - 1 - 1 -

SelectedUnknownIndependentVariables

- - N2

3SO ,

N3

3SO ,

N2

3O

- - - N 7Sat Water

- N2

9O

- 5 +1♠

Unknowntemperatures

- - 1 1 1 - - - - - 3

♠ There is one reaction.

Number of independent unknown variables = 9Number of independent equations = 3 (MB) + 4(EB) + 2 (relations) = 9

EZ Solve Computer Code

/*Symbols: Sulfur Dioxide = S, Sulfur Trioxide = ST, Oxygen = O, Water = WRate of reaction =R *///Material Balances Round ReactorS3 = 1.0 – R // S3 is flow of sulfur dioxide in stream 3ST3 = RO3 = O9 – 0.5*R//Conversion relation1.0 – S3 = 0.96//Stoichiometric relationO9 = 0.5

//Energy Balance round reactor , Tref = 25HEAT_REACTION = (-94.47 – (-70.95))*1000*4.182 // 1.0 Calorie = 4.182 Joules-10000 = R* HEAT_REACTION + (S3*11.8 + SD*17.2 + O3*7.8)*(T4-25) – 11.8*(425– 25) – O9*7.8(425 – 25)

//Energy Balance round heat exchanger 1, Tref = 4250.0 = (S3*11.8 + SD*17.2 + O3*7.8)*(T4-425) – 10.4*(300-425) - (S3*11.8 + SD*17.2

+ O3*7.8)*(T3-425)

//Energy Balance round heat exchanger 2, Tref = 4250.0 = (S3*11.8 + SD*17.2 + O3*7.8)*(T5-425) – O9*7.6*(25-425) - (S3*11.8 +SD*17.2 + O3*7.8)*(T4-425)

//Energy Balance round boiler/*Tref = 400 for S, SD and OTref = Steam tables reference conditions (Triple point)W7 = Flow rate of water in stream (7) in grams to maintain consistent units in equation*/H_SATWATER = 908.6 kJ/kg //Enthaply of sat water from steam tables at 300 psiaH_SATSTEAM = 2707.2 kJ/kg //Enthaply of sat steam from steam tables at 300 psia0.0 = W7*H_SATSTEAM - (S3*11.8 + SD*17.2 + O3*7.8)*(T5-425) –W7*H_SATWATER