1-1(a) degree of freedom(080407)

8/3/2019 1-1(a) Degree of Freedom(080407)

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Unit Operation (III)

part I(1)a

Degree of Freedom


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Degree of freedom analysis

Phase rule Process

System equilibrium state process streams

Variable intensive variables process variables(temp., pressure, (flowrate, composition, temp.,composition) energy transfer rate)


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Degree of freedom analysis for a process(Reklaitis, Introduction to Material & EnergyBalances)

In order to make a complete material (streams andcomposition) and energy (energy required and system

temperature) balances, number of unknown should beequal to number of independent equation.

To know a problem is correctly specified, it is tediousto set up the simultaneous equations and count thenumber of unknown; instead, you can do the degree of

freedom analysis, which is a more convenient approach.Besides it can tell you where the starting place to solve aproblem is in a manual calculation.


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1. For material balance only

degree of freedom= total no. of stream variables (flow and composition)

- total no. of mass balance (no. of component no. of unit)- total no. of subsidiary relation (such as equilibrium relation)- total no. of specified stream variables

= 0 correctly specified (1)

+ underspecified (2)- overspecified (3)

2. Special unit splitter

If a flow contains S components and splits into N branches,composition relations imposed are (N-1)(S-1), which are

XjI = Xj

II= = XjN, j = 1, 2, 3, , S (4)


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Non-active system


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3. Reactor

For material balanceNs

out = Nsin + esr, s = 1, 2, 3, , S (5)

Where es is stoichiometry coefficient, which is zero for inert species.

Therefore, around a reactor, the number of independent massbalance is equal to the number of species plus one additionalvariable r (Rs/es), in which Rs is the molar production rate,

Rs = Nsout Ns

in (6)

4. Combined material and energy balances

1. One energy balance for one unit

2. Variables should include Tv, dQ/dt and dW/dt (if work=involved)


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Reactive system


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Exp. 1

Process overall I II III

Streamvariable

11+1=12 6+1=7 9 7+1=8 8

Mass balance 4+4=8 4 4 4 4

Specified flow 1 1 0 1 0

composition 2 2 2 0 2

subsidiary 1

(conversion)

0 0 1 1

Degree offreedom

0 0 3 2 1

This problem is correctly specified, start with over-all boundary.


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Exp. 2


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Exp. 2


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Exp. 2

condenser

reactor splitter

E: 50oC

HClC2H4C2H5ClC2H6

C: 0oCHCl

C2H4C2H61.5%A1.5% of 0.93B

F:C2H5Cl

D: 0oCC

2

H5

Cl1600kg/h

=24800mol/hG: 0oCC2H5Cl

A: 0oCHCl

B: 0oC93% C2H4

7% C2H6

1-1(a)

1-1(a) degree of freedom(080407)

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