Material balance
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Material Balance

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Material balance

Material Balance ()

:

(System) (System Boundary)


Quiz 6

Quiz # 6

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:

  • the limiting reactant

  • the percentage of excess reactant

  • the degree of completion

  • the percen conversion based on

  • the yield in kg Sb produced / kg Sb2S3 fed to the reactor.


Material balance

Q, W

Q, W

+

+

-

-

mi

me

Q + W = DU + DKE + DPE

D = final state initial state

Q + W = DH + DKE + DPE

D = Out In


Material balance

  • /


Material balance

..2551

50,000 75,000 22,000 19,000

()


Material balance

=

+ -

= 50,000 + 22,000 75,000 -19,000 /

= - 22,000 /

22,000 / Ans.


Material balance

.. 2552


Material balance

=

+


Material balance

()

1.

2.

3.

(Component balance)

4.

5.

6.

7. ()


Material balance

Unsteady State Process Transient Process

Steady State Process ( )

Steady State


Material balance

Steady State

= (1)

= (2)

(1) (2)


Material balance

(Thickener) Steady State 100 kg


Material balance

Basis 100 kg

Steady State

=

100 kg = 70 kg

+ W kg

W = 100 70 = 30 kg

= 30 kg

Ans


Material balance

1. streamline

2. Basis

3. Set

4.

5.


Material balance

Differential balance

Integral balance


Material balance

Differential balance (/)

Integral balance 2 kg

Generation= 0 Consumption= 0

Steady - state

Accumulation= 0


Material balance

300 lb C 24.0 lb reactor 600F Reactor

1. C, O2 reactor

2. C O reactor

3. reactor

O2 = 21%

N2 = 79%


Material balance

Basis: 300 lb ( C 24.0 lb)

=2.17 + 8.17=10.34 lb mol


Material balance

C + O2 CO2

(lb mol)2.002.170

(lb mol)00.172.00


Material balance

1. N2 Inert (Impurities) Reactor

2. Reactor Reactor Reactor


Material balance

3. O O2 , H H2


Material balance

3. O O2 , H H2


Material balance

C = C

= 2.00 lb mol

O = O

= 4.34 lb mol


Material balance

C = C

= 24.0 lb

O = O

= 138.9 = 128 + 10.9


Material balance

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= 2.00-2.00 = 0

O2 balance

0=O2 in O2 out + Generation Consumption

0=2.17 lb mol O2 out +0 2.00 lb mol

O2 out= 2.17-2.00 lb mol = 0.17 lb mol


Material balance

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= 2.00-2.00 = 0

O2 balance

0=O2 in O2 out + Generation Consumption

0=2.17 lb mol O2 out +0 2.00 lb mol

O2 out= 2.17-2.00 lb mol = 0.17 lb mol


Material balance

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


Material balance

1000 kg/h Benzene (B) Toluene (T) B 50 % T 50 % 2 (Top Stream) B 450 kg/h 475 kg T/h Steadystate flowrate


Material balance

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


Material balance

1000 kg/h Benzene (B) Toluene (T) B 50 % T 50 % 2 (Top Stream) B 450 kg/h 475 kg T/h Steadystate flowrate


Material balance

Batch mixing process


Material balance

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)=(1-0.5286)(350)

165=165


Material balance

1. flowchart

Flowchart Box Symbol Process Unit (Reactor, Mixer, Dryer, ) Input Output Stream Box Stream Stream F Feed , T

(Data-base)


Material balance

2. Basis

flowrate Process Basis Stream Basis

flowrate 1 100

3. Flowchart

Component balance

=

( L)


Material balance

2. Basis

flowrate Process Basis Stream Basis

flowrate 1 100

3. Flowchart

Component balance

=

( L)


Material balance

4.

5. Redundant equation ( total mass balance)

*


Material balance

4

4 3 unknown

3 unknown 4 1 3


Material balance

Determine the Number of Degree of Freedom

D


Material balance

Degree of Freedom 100 %


Material balance

Alcohol Alcohol Bottom Product


Material balance

BasisF = 1000 kg

EtOH, H2O B B, 2 2

overall balanceF=D + B

1000 kg=100 kg + B

B=900 kg

B=F D


Material balance

Composition B

Total Balance B = 900 kg

EtOH 40 kg B = 900 40 kg

= 860 kg


Homework

Homework

7.12

7.13

7.14


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