MATERIAL BALANCE

MATERIAL BALANCE


GIVEN:

TO DESIGN A 100 TPD CAPACITY H2SO4 ACID PLANT

BASIS:
1 HOUR OF OPERATION.

PURITY:
Product which is to be manufactured is assumed to have strength of 98% acid.


100 TPD implies that we have 100 x 103 / 24 = 4166.67 Kg/hr of Acid

With 98% purity, the acid that is produced per hour = ( 98 x 4166.67) / 100
= 4083.34 Kg/Hr

Kmoles of Sulphuric acid to be produced = 4083.34 / 98
= 41.667 Kmoles/Hr

It’s assumed that overall absorption of the acid is 100 %

Then, SO3 required = 41.6667 / 1.0
= 41.6667 Kmoles/Hr

Also its assumed that the overall conversion of SO2 to SO3 in the reactor is 99.8%

Then SO2 required = 41.6667 / 0.998
= 41.75 Kmoles/Hr
= 2672 Kg/ hr

Assuming 100% combustion of Sulfur,

Then S required = 41.75 Kmoles/Hr
= 1336.0 Kg/ hr

Amount of oxygen required to convert 1Kmole of S to SO3 = 1.5 Kmoles

Then, O2 required = 41.75 x 1.5
= 62.625 Kmoles/ hr


As cited in the literature that some amount of excess oxygen must be used,
Using 40% excess,
O2 required = 62.625 x 1.4
= 87.675 Kmoles/ hr
= 2805.6 Kg/ hr

From this the total dry air that is coming in can be calculated as.

Dry air in = ( 87.675 x 100) / 21
= 417.5 Kmoles/Hr

At 300C, assuming 65% Relative Humidity,

Humidity as calculated from the psychometric chart is,

Humidity = 0.0165 Kg water/ Kg dry air

Then, water entering with dry air = 417.5 x 28.84 x 0.0165
= 198.672 Kg/Hr
= 11.037 Kmoles/Hr

Total weight of entering air = 417.5 x 28.84 + 11.037 x 18
= 12239.336 Kg/hr

Then,
Weight of sulphur required = 1336.0 Kg/ hr
Weight of air entering = 12239.366 Kg/ hr


DRYING TOWER:
















Making a Mass balance around the Drying Tower

P + R = Q + S

As water is being removed from the incoming air to make it dry, the 98% acid that is being recycled to the tower, decreases in concentration and let this concentration be assumed as 97%, then we can write,

0.02 x R + 198.672 = S x 0.03 (1)

H2SO4 Balance will give,
R x 0.98 = S x 0.97 (2)

Solving the above equations

Amount of 98% H2SO4 entering R = 19271.184 Kg/hr
Amount of 97% H2SO4 leaving S = 19469.856 Kg/hr

SULFUR BURNER:











The combustion reaction takes place inside the burner, where Sulphur is oxidized to Sulphur Dioxide
Moles of Sulfur coming in = 41.75 Kmoles/ hr = 1336.0 Kg / hr
Moles of Oxygen coming in = 87.675 Kmoles/ hr = 2805.6 Kg / hr

As mentioned before we have assumed 100% combustion of sulphur,

Sulfur Dioxide formed = 41.75 Kmoles/ hr = 2672.0 Kg/ hr

Oxygen leaving = 2805.6 – 1336.0
= 1469.6 Kg/ hr

Nitrogen leaving = (87.675 x 79) / 21
= 329.825 Kmoles/ hr = 9235.1 Kg / hr


REACTOR:























As cited in the reference by author NORMAN SHREVE et al Pg 337, the temperature and conversions in each stage of a Monsanto Converter is given as follows:


Location Temperature 0C Equivalent conversion (%)
Gas entering I pass
Gas leaving I pass
Rise in temperature 410.0
601.8
191.8
74.0
Gas entering II pass
Gas leaving II pass
Rise in temperature 438.0
485.0
47.30
18.4
Gas entering III pass
Gas leaving III pass
Rise in temperature 432.0
444.0
12.0
4.3
Gas entering IV pass
Gas leaving IV pass
Rise in temperature 427.0
430.3
3.30
1.30

Total rise
254.4
98.0

As we see from the table that the overall conversion in the reactor is 98% but to validate our assumption that was made earlier, we assume that the conversion in the last stage of the reactor is 3.1% instead of 1.3% so that the assumption of 99.8% as overall conversion remains unaffected and thus temperature for the gas leaving the fourth pass is then assumed to be 4370C.

Component
I stage II stage III stage

S 74 %
conversion 18.4%
conversion 4.3%
conversion



For II stage:

Sample calculation for the 2nd stage is shown as follows:

Components:

SO2 INLET = 694.72 Kg/ hr
SO2 OUTLET = 694.72 – 41.75 x 0.184 x 64
= 203.072 Kg/ hr

N2 INLET = 9235.1 Kg/ hr
N2 OUTLET = 9235.1 Kg/ hr

O2 INLET = 975.28 Kg/ hr
O2OUTLET = 975.28 – 41.75 x 0.5 x 0.184 x 32
= 852.368 Kg/ hr

SO3 INLET = 2471.6 Kg/ hr
SO3 OUTLET = 2471.6 + 41.75 x 0.184 x 80
= 3086.16 Kg/ hr




components I stage
(Kg/ hr) II stage
(Kg/ hr) III stage
(Kg/ hr)

Inlet
Outlet Inlet Outlet Inlet Outlet
N2 9235.1 9235.1 9235.1 9235.1 9235.1 9235.1
SO2 2672.0 694.72 694.72 203.072 203.072 88.176
SO3 0.0 2471.6 2471.6 3086.16 3086.16 3229.78
O2 1469.6 975.28 975.28 852.368 852.368 823.644
Total (Kg/ hr) 13376.7 13376.7 13376.7 13376.7 13376.7 13376.7
After the passage from the 3 stages or after the first contact, the gases are let into the interpass absorber where the absorption of the SO3 takes place. After the contact with H2SO4 in the tower, the gases are returned back to the 4thstage for the second contact.

We can write from the reaction for sulfur dioxide oxidation to give sulfur trioxide that,

KP = (PSO3) / (PSO2) (PO2)1/2 (A)

And available in the reference by author NORMAN SHREVE et al Pg 333, the equilibrium constants for the Sulfur Dioxide Oxidation are given at different temperatures.

Now for KP at the entering temperature of 4th stage i.e. 4270C,

We have KP = 270.2


INTERPASS ABSORBER:

From the equation (A), SO3 present are calculated by the following

Let
X1 = Moles of SO2
X2 = Initial moles of SO2 entering the reactor
X3 = Moles of O2
X4 = Moles of SO3

Then
KP = (X4 + 0.031 x X2) x (X1 + X3 + X4)0.5 (B)
(X1 - 0.031 x X2) x (X3 – 0.05 x 0.031 x X2)0.5

The above equation is once again mentioned in the literature. Calculating the value of the unknown X4, we have, with KP = 270.2, we get

X4 = 15.8 Kmoles/ hr (or) 1264 Kg/ hr


Then Moles of SO3 removed in the interpass absorber is given as

(40.373 – 15.8) = 24.573 Kmoles

SO3 + H2O H2SO4

As from the stoichiometric coefficients of the reaction given, we can find out the weight of sulfuric acid to be absorbed as

( 24.573 x 98 ) = 2408.154 Kg/Hr
Also mentioned in the literature that “its required to take the strength of the solvent H2SO4 for absorption of SO3 not to increase by more than 1-2%, and the best absorption will occur when the absorbing acid has the strength between the range 97.5 to 99%”.
























SO3 absorbed in this stage = 40.373 – 15.801
= 24.572 Kmoles
= 1965.76 Kg/ hr

From stoichiometry

H2SO4 required for absorption of 1965.76 Kg of SO3 = 1965.76 x 98 / 80
= 2408.056 Kg/ hr

The recycle stream contains 98% H2SO4

0.98 X = 2408.056 Kg/ hr
X = 2457.2 Kg/ hr

Weight of the recycle stream = 2457.2 Kg/ hr
Weight of H2S2O7 produced = 1965.76 x 178 / 80
= 4373.816 Kg/ hr

Weight of H2O accompanying = 491.44 Kg/ hr
DILUTION TANK I:















Weight of H2O required to dilute 4373.816 Kg of H2S2O7 = 4373.816 x 18 / 178
= 442.296 Kg/ hr

H2SO4 produced from H2S2O7 = 4373.816 x 196 / 178
= 4816.112 Kg/ hr

The above H2SO4 contributes 98% of X, then
0.98 X = 4816.112
X = 4914.4 Kg/ hr

H2O actually to be added = 4914.4 x 0.02 + 442.292 – 49.144
= 491.44Kg/ hr


REACTOR:
















For the second contact we have the following details,

Components
N2 SO2 SO3 O2 Total

Inlet (Kg/ hr)

9235.24
88.1792
1264.08
823.648
11411.21

Outlet (Kg/ hr)

9235.24
5.344
1367.68
802.944
11411.21


FINAL ABSORPTION TOWER:

SO3 + H2SO4 H2S2O7

SO3 entering = 1367.68 Kg/ hr
H2SO4 required =1367.68 x 98 / 80
= 1675.408 Kg/ hr

0.98 X = 1675.408 Kg/ hr

Recycle stream, X = 1709.6 Kg/ hr

In the exit stream,

H2S2O7 produced = 1675.408 x 178 / 98
= 3043.088 Kg/ hr

H2O leaving = 34.192 Kg/ hr


DILUTION TANK II:

H2S2O7 + H2O 2 H2SO4

H2O required to dilute H2S2O7 = 3043.088 x 18 / 178
= 307.78 Kg/ hr

H2SO4 produced = 3043.088 + 307.78
= 3350.816 Kg/ hr

This contributes 98% of the acid
0.98 X = 3350.816

X = 3419.2 Kg/ hr

Then, H2O to be added = 3419.2 – 3043.088
= 376.112 Kg/ hr

Then, H2O added from dryer = 19469.856 – 19271.184
= 198.67 Kg/ hr

H2O added from final absorption tower = 34.91 Kg/ hr

H2O to be added = 376.112 – 198.67 – 34.91
= 143.24 Kg/ hr

H2SO4 produced = 3419.2 – 1709.6
= 1709.6 Kg/ hr


STORAGE TANK:












H2SO4 from dilution tank I = 2457.2 Kg/ hr
H2SO4 from dilution tank II = 1709.6 Kg/ hr
Total sulphuric acid produced = 4166.8 Kg/ hr
= 100 TPD