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Overview of Mass balance

MASS
BALANCE
CIVL270-Maraqa


Objective
Develop a mass balance expression under different
case scenarios and design a simple system to meet
desired needs

Conservation of Mass
Mass cannot be created nor destroyed.

Implication
Atoms are conserved but molecules may change to
other forms.

CIVL270-Maraqa


Types of Reactors

• Batch Reactors
V

• Completely Stirred Tank Reactors (CSTR)

Q,
Cin

Q,
Cout
V

• Plug-Flow Reactors

Q
Q
Q

CIVL270-Maraqa

Q
to
t1
t2

Q
Q


• Dispersed Plug-Flow Reactors

Q

Q
to

Q

Q
t1


Q

Q
t2

• Packed bed Reactors

• Fluidized Packed bed Reactors

Q,
Cin

Q, Cout

Q,
Cout

CIVL270-Maraqa

Q,
Cin


Completely-Mixed Tank Reactor (CSTR)
Accumulation
Input,
Cin
System boundary

Output,
Cout

Decay

Mass Balance- General Equation
Input
rate

CIVL270-Maraqa

=

Ouput
rate

+

Decay +
rate

Accumulation
rate


Steady-State Conservative System
Steady state
Conservative

Accumulation = 0
Decay = 0

Input = Q C +Q C
s s
w w
rate
Ouput = QmCm
rate

Cs Qs + C w Q w
Cm =
Qm

CIVL270-Maraqa

}

Input =
rate

Stream

Ouput
rate

Mixture

Qs, Cs

CSTR
Waste
Qw, Cw

Qm, Cm


Steady-State with Nononservative Pollutants
Steady state

Accumulation = 0

Nonconservative
Input =
rate

Ouput + Decay
rate
rate

Input
QsCs+QwCw
rate =
Ouput = QmCm
rate
Decay = kCmV
rate

Cs Qs + C w Q w
Cm =
Q m + kV
CIVL270-Maraqa

Decay rate (1st-order) = kCV = kCmV
CSTR
Stream

V

Mixture

Qs, Cs

k

Qm, Cm

Waste
Qw, Cw


Reaction Term
Sedimentation
Radioactive decay
Volatilization
Biodegradation
Bio uptake
Hydrolysis
Photodegradation
CIVL270-Maraqa


Example

Stream
Qs=5 m3/s

Well-mixed lake

Cs =10 mg/L

V=10x106 m3

Waste

k=0.2/day

Qw 0.5 m3/s

Solution

Cw =100mg/L

Qm= Qs+ Qw = 5.5 m3/s

Cm =

Cs Qs + C w Q w
Q m + kV

mg m 3
mg
m3
(10 )(5 ) + (100 )(0.5 )
mg
L
s
L
s
Cm =
=
3
.
5
m3
1
day
L
6 3
(5.5 ) + (0.2
×
)(10 × 10 m )
s
day 24 × 60 × 60s
If there was no reaction in the reactor, Cm will be 18.2 mg/L
CIVL270-Maraqa

Outgoing
Qm=?
Cm=?


Example

Stream
Qs=5 m3/s

V=5x106 m3

V=5x106 m3

Cs =10 mg/L
k=0.2/day

Waste

Cm1

k=0.2/day

Cm2=?

Qw 0.5 m3/s

Solution

Cw =100mg/L

Qm= Qs+ Qw = 5.5 m3/s

Cm =

Cs Qs + C w Q w
Q m + kV

Cm1=5.86 mg/L

Cm1Qm=Cm2Qm+kCmV

Cm2

C m1Q m
=
Q m + kV

Cm2=1.9 mg/L
CIVL270-Maraqa



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