CHAPTER 5 PROBLEMS
CHAPTER 5 PROBLEMS
1. Raw Sulfur which is 75% pure is burned in excess air supplied at the rate of 4.713 m³/kg of
raw Sulfur at 25ºC, 745 mm Hg with 80% RH. If 87% of the sulfur charged burns to SO2 and
the rest to SO3, calculate:
a)
% excess air (S to SO2) (65%)
b)
% excess air (S to SO3) (10%)
c)
Complete analysis of the burner gas (10.88% SO2, 7.32% O2, 77.62% N2, 1.63% SO3,
2.565% H2O)
60
CHAPTER 5 PROBLEMS
2. The burner gas from a sulphur burner analyzes 9.2% SO2, 7.13% O2 and 83.67% N2. The raw
Sulfur charged contains 82% pure Sulfur and analysis of the cinder shows 20% unburned
sulfur. Calculate:
a)
% excess air (S to SO2)
(60%)
b)
% excess air (S to SO3)
(6.67%)
c)
m³ of saturated air (28ºC, 750 torrs)/kg raw S (5.076)
d)
m³ of burner gas (300ºC, 730 torrs)/kg raw S (9.75)
61
CHAPTER 5 PROBLEMS
62
CHAPTER 5 PROBLEMS
3.
Pyrite analyzing 78% FeS2 and 22% gangue is burned at the rate of 1000 kg/hr. Analysis
of the cinder shows 7.22% S as unburned FeS2 and SO3 absorbed by Fe2O3. Air supplied is 70% in
excess based on conversion of FeS2 to SO2. The ratio of SO2 to SO3 in the burner gas is 3.48:1.
Calculate:
a)
% excess air (FeS2 to SO3) (24.67%)
b)
% of the FeS2 charged lost in the cinder (12%)
c)
Complete analysis of the burner gas (6.26% SO2, 9.57% O2, 82.35% N2, 1.82% SO3)
63
CHAPTER 5 PROBLEMS
64
CHAPTER 5 PROBLEMS
65
CHAPTER 5 PROBLEMS
4.
In the burning of pyrite containing 92% FeS2 and 8% gangue, 13% of the FeS2 charged is
lost in the cinder. A partial analysis of the cinder also shows 5.31% SO3. The orsat analysis of the
burner gas shows 6.75% SO2, 6.83% O2 and 86.38% N2. Air supplied is at 23ºC, 743 mm Hg and
88% RH. Calculate:
a)
% excess air (FeS2 to SO2)
(40%)
b)
% excess air (FeS2 to SO3)
(2.676%)
c)
m³ air/kg pyrite (3.58)
d)
m³ burner gas (250ºC, 750 mm Hg)/kg pyrite (5.926)
66
CHAPTER 5 PROBLEMS
67
CHAPTER 5 PROBLEMS
5.
Raw Sulfur containing 83% pure S is burned together with 80% excess air (S to
SO2). An analysis of the cinder shows 20% unburned sulphur and 80% inerts. Air is
supplied saturated at 30ºC and 750 mm Hg. The gases from the burner enter a converter
where catalytic oxidation of SO2 to SO3 takes place. A partial orsat analysis of the
converter shows 1.37% SO2. The gases from the converter enter an absorber where after
absorption in acid solution forms a waste gas analyzing 0.55% SO 2, 11.99% O2 and
87.46% N2. Calculate per 100 kg raw sulfur:
a)
Complete analysis of the burner gas (9.59% SO2, 9.03% O2, 76.07% N2, 1.07%
SO3, 4.25% H2O)
b)
% of the SO2 entering the converter that is converted to SO3 (88.3%)
c)
Weight of a 60% dilute H2SO4 needed to produce an 87% H2SO4 (248.04)
d)
If the absorbing acid is 94% H2SO4, what weight of a 14% oleum is formed
(587.61)
Solution: Basis: 100 kg Raw Sulfur
Sulfuric acid sol’n
Primary Air 80% x’ss, 30⁰C, 750 mm Hg,
Saturated
BURNER
a) 60% H2SO4
Burner Gas Catalytic
Converter
Raw Sulfur
1.87% SO2
83% pure S
17% inerts
Total at S =
Cinder
83
32
= 2.5938
Theo O2 (S to SO2) = 2.5938
68
Gas
Absorber
b)94% H2SO4
CHAPTER 5 PROBLEMS
O2 from air = 1.8 (2.5938) = 4.6688
N2 from air = 4.6688 (
79
) = 17.5636
21
At 30ºC, VP = 31.6869 mm Hg
(4.6688+17.5636)(31.6869)
Moles H2O from air =
= 0.9807
750−31.6869
Inert balance:
0.8 x Wt of cinder = 17
Wt of cinder = 21.25
0.30(21.25)
S gasified = 2.5938 –
= 2.4610
0.30
S converted to SO2 = 0.90(2.4610) = 2.2149
S converted to SO3 = 0.10(2.4610) = 0.2461
3
Free O2 = 4.6688- 2.2149- 0.2461( ) = 2.0848
2
a)
Complete Analysis of the Burner Gas
GAS
n
%
SO2
2.2149
9.59
O2
2.0848
9.03
N2
17.5636
76.07
SO3
0.2461
1.07
H2O
0.9807
4.25
23.0901
b)
Converter Analysis
Let n = moles SO2 + SO3
In the reaction
1
SO2 + 2 O2
n
GAS
SO2
O2
N2
SO3
0.5n
n
n
2.2149 - n
2.2149 – 0.5n
17.5636
21.8683 – 1.5n
Converter Gas
0.2893
1.107
17.8636
69
CHAPTER 5 PROBLEMS
For SO2 :
2.2149−n
21.8683−1.5n
= 0.0137
n= 1.9556
% SO2 to SO3 =
1.9556
2.2149
x 100 = 88.29% ANS
70
CHAPTER 5 PROBLEMS
6.
Pyrite containing 80% FeS2 and 20% gangue is burned in excess air to produce a
gas with a complete analysis of 7.78% SO2, 1.39% O2, 82.06% N2, 4.92% SO3 and 3.85%
H2O. Analysis of the cinder shows a total sulfur content of 6.46% due to the presence of
unburned FeS2 and SO3 absorbed by Fe2O3. Air is supplied is at 27ºC, 745 mm Hg and
saturated with vapour. The burner gases then enter a converter together with 30%
excess secondary air (supplied at the same conditions as primary air) based on the
complete conversion of all SO2 to SO3. 75% of the SO2 actually burns to SO3. The converter
gases enter an absorber and absorbed in acid solution. The waste gases formed has a
partial orsat analysis of 0.7% SO2. Calculate per kg of pyrite:
a)
% excess air (S to SO2) (10.17)
b)
Complete analysis of the converter gas (1.79% SO2, 1.58% O2, 83.77% N2, 9.55%
SO3, 3.42% H2O)
c)
Kg of an 80% H2SO4 acid formed from a 40% acid charged (1.94)
d)
Kg of a 75% H2SO4 needed to produce 10% oleum (0.662)
Solution: Basis: 100 Moles SO3 Free BG
O2 from Air = 82.06 (21/79) = 21.8134
For the Reaction:
4 FeS2 + 11 O2
→ 2 Fe2O3 + 8SO2
3.89 10.6975
1.945
7.78
O2 Conversion of FeS2
→
SO3
= 21.8134 - 10.6975 – 1.39
= 9.7259
For the Reaction
4 FeS2 + 15 O2
→
2 Fe2O3 + 8 SO3
2.5936 9.7259
1.2968 5.1871
SO3 in the order = 5.1871 - 4.92 = 0.2671
Cinder Analysis:
Wt.
Wt.%
FeS2
120W
6.46
Fe2O3
(1.2968 + 1.945) (160)
SO3
0.26 (80)
6.46
Gangue
(3.89 + 2.5936 + W)
120 (20/8)
Total Wt. Cinder = 120 W + 518.688 + 21.36 + 194.508 +30 W
Equation 1
0.0646 (Wt. cinder) = 120 W
Equation 2
Wt. Cinder = 120 W + 518.688 + 21.368 + 194.508 + 30 W
120W
0.0646
W
= 120 W + 518.688 + 21.368 + 194.508 + 30W
= 0.4302
71
CHAPTER 5 PROBLEMS
Wt. pyrite
Total FeS2
Theo O2
a)
=
=
(3.89 + 2.5936+ 0.4302)
0.8
= 1037.07
= 6.9138
= 19.0130
% excess O2 O2 (SO2)
21.8134−19.0130
19.0130
x 100 = 10.17%
b)
Theo O2 for the complete conversion
½ (7.78) = 1.39 = 2.5
O2 from Air = 2.5 x 1.3 = 3.25
N2 from Air = 12.2262
SO2 →
SO3 = 7.78 x 0.75 = 5.835
log VP
= 7.9681 –
1668.21
228+20
VP = 26.5956
nH2O from Air
= 0.5729
The Reaction is:
SO2 + 1/2O2
5.835 2.9175
→
SO3
5.835
SO2 unreacted = 7.78 - 5.835 = 1.945
O2 after reaction = 1.7225
N2 after reaction = 94.2862
SO3 after reaction = 10.755
GAS
SO2
SO3
O2
N2
H2O
n
1.945
10.755
1.7225
94.2862
4.4229
113.1316
72
%
1.79
9.55
1.58
83.77
3.42
CHAPTER 5 PROBLEMS
c.) kg of an 80% H2SO4 acid formed from a 40% acid charged
let x = kg of 80% H2SO4 acid formed
x – 10.755 (80) = x – 860.4 = wt. 40% acid charged
H2SO4 BAL:
0.4(x – 860.4) + 860.4 (98/80) = 0.80x
X = 1774.575 kg/100 moles SO3 free BG
1774.575 – 860.4 = wt. Of 40% acid charged914.175 = wt. Of 40%
acid charged
1774.575/914.175 = 1.94 kg of an 80% H2SO4 acid formed from a 40% acid
charged
d.) kg of 75% H2SO4 needed to produce 10% oleum
let y = wt. Of 75% H2SO4
y + 10.755(0.75) = 7 + 806.625 = wt. Of 10% oleum
0.75 y + 806.625 (98/80) = (0.10)(y + 806.625)(98/80) + 0.75(y +
806.625)
Y = 2321.104
Wt. Of 10 % oleum = 2321.104 + 806.625 = 3122.729
7.
The burning of raw S consisting a 95% S and 5% inerts produces a gas
whose orsat analysis shows 11.39% SO2, 7.76% O2 and 80.84% N2. Ten % of
the total sulfur charged is lost in the cinder. The burner gases are cooled and
absorbed in milk of lime obtained by slaking a lime consisting of 58% CaO,
32% MgO and 10% inerts with water. The bisulfite liquor formed contains
12% SO2 of which 2% is free and the rest present as bisulfites. Orsat analysis
of the waste gas shows that it contains 7.39% O2 and 92.61% N2. Calculate:
a)
Kg bisulfite liquor/kg raw sulfur
(9.65)
b)
Kg lime consumed/ kg raw sulfur
(0.8832)
c)
Kg of water used for slaking/kg raw sulfur
(6.92)
73
CHAPTER 5 PROBLEMS
Waste Gas
7.39% O2
92.61% N2
Slaker
Air
Burn
er
Raw S
95% S
5% inerts
Burner Gas
Gas
Absorber
Cooler
11.39% SO2
7.76% O2
80.84% N2
Bisulfite Liquor
12% SO2
-2% free
-10% bisulfites
Cinder
10% of total S
21.4891 – 11.39 – 7.76 = 2.3391
S converted to SO3 = 2.3391 (2/3) = 1.5594
Mole S in the cinder
(1.5594 + 11.39) (.10/.90) = 1.4388
wt. of raw S
(1.5594+11.39+1.4388)×32
0.95
= 484.6522
Mole of waste gas (N2 balance)
80.84 = 0.9261 x
X= 87.2908
mol O2 in waste gas
87.2908 (0.0739) = 6.4508
O2 in burner gas ≠ O2 in waste gas
O2 used = 7.76 – 6.4508 = 1.3092
SO2 +
2.6184
½ O2 =
1.3092
SO3
2.6184
Let y = wt. of the bisulfite liquor
0.12y = (11.39 – 2.6184) 64
y= 4678.18667
74
Lime
58% CaO
32% MgO
10% inerts
CHAPTER 5 PROBLEMS
a)
kg bisulfite liquor
kg raw sulfur
4678.18667
=
484.6552
= 9.6526
Moles CaO and MgO converted to bisulfites
4678.18667 (0.10)
64
1
× 2 = 3.6548
Mole CaO and MgO converted to sulfate = moles SO3
= 1.5594 + 2.6184 = 4.1778
Total moles CaO and MgO supplied
3.6548 + 4.1778 = 7.8326
For 100 kg lime
Moles CaO and MgO =
wt.lime
58
56
+
32
40.3
100
= 1.8298
100 mol burner gas
b)
kg lime consumed
kg raw sulfur
=
= = 1.8298
× 7.8326 = 428.0577
428.0577
484.6552
= 0.8832
wt. of burner gas + wt. lime + wt. H2O for slaking = wt. waste gas +
wt. bisulfite liquor
SO2 + SO3 + O2 + N2 + lime + H2O = O2 + N2 + BL
11.39(64) + 1.5594(80) + 7.76(32) + 428.0577 + H2O = 6.4508(32)
+ 4678.1867
H2O = 3354.5226
c)
kg of H2 O used for slaking
kg raw sulfur
=
3354.5226
484.6552
75
= 6.9215
ANS
CHAPTER 5 PROBLEMS
8.
The roasting of pyrites analyzing 85% FeS2 and 15% gangue utilizes
40% excess air (FeS2 to SO2) supplied at the rate of 358 m³/hr at 23ºC, 743
mm Hg and 88% RH. A partial analysis of the cinder showed 25.92% FeS2
and 17.83% gangue. Only 65% of the FeS2 gasified is converted to SO2 and
the rest to SO3. The burner gases are cooled and charged to a converter
together with slaked lime from a dolomitic lime containing 75% CaO, 25%
MgO. If 850 kg/hr of bisulfite liquor are produced with no oxidation of SO2
to SO3 taking place. Assume all SO2 converted to bisulfite. Calculate:
a)
Kg/hr of lime (33.724)
b)
Kg/hr of water for slaking (704.77)
c)
Complete analysis of the burner gas (5.35% SO2, 8.83% O2, 81.06%
N2, 2.15% SO3 and 2.61% H2O)
76
CHAPTER 5 PROBLEMS
77
CHAPTER 5 PROBLEMS
78