2. Quiz
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(System) (System Boundary)
Antimony (Sb) is obtained by heating pulverized stibnite (Sb2S3) with scrap iron and drawing off the molten antimony from the bottom of the reaction vessel.
Sb2S3 + Fe 2 Sb + 3 FeS
Sppose that 0.600 kg of stibnite and 0.250 kg of iron turnings are heated together to give 0.200 kg of Sb metal. Determine:
Q, W
Q, W
+
+


mi
me
Q + W = DU + DKE + DPE
D = final state initial state
Q + W = DH + DKE + DPE
D = Out In
..2551
50,000 75,000 22,000 19,000
()
=
+ 
= 50,000 + 22,000 75,000 19,000 /
=  22,000 /
22,000 / Ans.
.. 2552
=
+
1.
2.
3.
(Component balance)
4.
5.
6.
7. ()
Unsteady State Process Transient Process
Steady State Process ( )
Steady State
Steady State
= (1)
= (2)
(1) (2)
(Thickener) Steady State 100 kg
Basis 100 kg
Steady State
=
100 kg = 70 kg
+ W kg
W = 100 70 = 30 kg
= 30 kg
Ans
1. streamline
2. Basis
3. Set
4.
5.
Differential balance
Integral balance
Differential balance (/)
Integral balance 2 kg
Generation= 0 Consumption= 0
Steady  state
Accumulation= 0
300 lb C 24.0 lb reactor 600F Reactor
1. C, O2 reactor
2. C O reactor
3. reactor
O2 = 21%
N2 = 79%
Basis: 300 lb ( C 24.0 lb)
=2.17 + 8.17=10.34 lb mol
C + O2 CO2
(lb mol)2.002.170
(lb mol)00.172.00
1. N2 Inert (Impurities) Reactor
2. Reactor Reactor Reactor
3. O O2 , H H2
3. O O2 , H H2
C = C
= 2.00 lb mol
O = O
= 4.34 lb mol
C = C
= 24.0 lb
O = O
= 138.9 = 128 + 10.9
Accumulation = In Out + Generation  Consumption
C balance
0=C in C out + Generation Consumption
0=2.00 lb mol C out +0 2.00 lb mol
C out= 2.002.00 = 0
O2 balance
0=O2 in O2 out + Generation Consumption
0=2.17 lb mol O2 out +0 2.00 lb mol
O2 out= 2.172.00 lb mol = 0.17 lb mol
Accumulation = In Out + Generation  Consumption
C balance
0=C in C out + Generation Consumption
0=2.00 lb mol C out +0 2.00 lb mol
C out= 2.002.00 = 0
O2 balance
0=O2 in O2 out + Generation Consumption
0=2.17 lb mol O2 out +0 2.00 lb mol
O2 out= 2.172.00 lb mol = 0.17 lb mol
CO2 balance
0=CO2 in O2 out + Generation Consumption
0= 0 CO2 out + 2.00 lb mol 0
CO2 out = 2.00lb mol
N2 balance
N2 in = N2 out
1000 kg/h Benzene (B) Toluene (T) B 50 % T 50 % 2 (Top Stream) B 450 kg/h 475 kg T/h Steadystate flowrate
Basis 1000 kg / h 1 hr.
Overall balanceF=D + B
Benzene balance
0.50 F =450 kg/hr + q2
0.50(1000) = 450 + q2
q2= 50 kg B/hr
Toluene balance
0.50 F =q1 + 475 kg/hr
0.50(1000) = q1 + 475
q2= 25 kg B/hr.
In=Out
1000=(450 + 25) + (475 + 50)
1000=1000
1000 kg/h Benzene (B) Toluene (T) B 50 % T 50 % 2 (Top Stream) B 450 kg/h 475 kg T/h Steadystate flowrate
Batch mixing process
Basis A = 200 g, B = 150 g
x mass fraction MeOH C
Overall balanceC =200 + 150 = 350 g
CH3OH balance
200(0.4) + 0.7(150)= x (350)
80 + 105= x (350)
x= 185 / 350
= 0.5286
H2O balance
In=Out
0.6(200) + 0.3(150)=(10.5286)(350)
165=165
1. flowchart
Flowchart Box Symbol Process Unit (Reactor, Mixer, Dryer, ) Input Output Stream Box Stream Stream F Feed , T
(Database)
2. Basis
flowrate Process Basis Stream Basis
flowrate 1 100
3. Flowchart
Component balance
=
( L)
2. Basis
flowrate Process Basis Stream Basis
flowrate 1 100
3. Flowchart
Component balance
=
( L)
4.
5. Redundant equation ( total mass balance)
*
4
4 3 unknown
3 unknown 4 1 3
Determine the Number of Degree of Freedom
D
Degree of Freedom 100 %
Alcohol Alcohol Bottom Product
BasisF = 1000 kg
EtOH, H2O B B, 2 2
overall balanceF=D + B
1000 kg=100 kg + B
B=900 kg
B=F D
Composition B
Total Balance B = 900 kg
EtOH 40 kg B = 900 40 kg
= 860 kg
7.12
7.13
7.14