Ch E 416 - Assignment 3 Solutions

Question 1

(1a)  

 

(1b)  Unit (c)

         I am deriving Nv, NE and ND for each of the elements. You were only required to derive the ND for the overall units (c) and (g).

For a single non – adiabatic stage:

 

Four streams with C + 3 variables plus the heat transfer rate.

 

Nv = 4(C + 3) + 1 = 4C + 13

 

Number of equations:

 

4     Composition summations

1     Heat balance

1     Equality of temperatures for equilibrium streams

1     Equality of pressures for equilibrium streams

C    Component mass balance

C    Equality of fugacities for equilibrium streams

 

NE = 2C + 7

 

Hence:

 

ND = 2C + 6

 

For the feed stage:

 

The only difference for this stage is the additional input

 

One additional stream with C + 3 variables.

 

Nv = 4C + 13 + C + 3 = 5C + 16

 

Number of additional equations:

 

1          Composition summation

 

NE = 2C + 8

 

Hence:

 

ND = 3C + 8

 

For the partial condenser:

 

Three streams with C + 3 variables plus the heat transfer rate.

 

Nv = 3(C + 3) + 1 = 3C + 10

 

Number of equations:

 

3     Composition summations

1     Heat balance

1     Equality of temperatures for equilibrium streams

1     Equality of pressures for equilibrium streams

C    Component mass balance

C    Equality of fugacities for equilibrium streams

 

NE = 2C + 6

 

Hence:

 

ND = C + 4

 

The partial reboiler information comes from part 1a.

 

ND = C + 4

 

For the enriching section stages:

 

Number of stages: (f – 1 )

ND = (f – 1) (2C + 6)

 

For the stripping section stages:

 

Number of stages: (N – f – 1 )

ND = (N – f – 1) (2C + 6)

 

Total of ND = C + 6N + 2CN + 4

 

Redundant streams:

 

Condenser:

 

2: The vapour entering the condenser and the liquid leaving it, which is the liquid returning to the column.

 

Enriching section:

 

Stage 1 has been counted doing the analysis on the condenser and the splitter

From stage 2 to stage f – 1 = f – 1 – 1 = f – 2. Two streams are counted twice between stages, therefore:

 

2(f – 2)

 

Feed stage:

 

4: The streams leaving the stage

 

Stripping section:

 

From stage f + 1 to stage N – 1 = N – 1 – f – 1= N – f – 2. Two streams are counted twice between stages, therefore:

 

2(N – f – 2)

Reboiler::

 

2: The liquid entering and the vapour leaving it.

 

Total of NR = 2N

 

Each stream adds (C + 2) variables

 

(C + 2)NR = 2CN + 4N

 

Additional variables:

 

2: the location of the feed stage + the number of stages N

 

Finally:

 

(ND)unit = NV – (C + 2)NR + NA = C + 2N + 6

 

 

Unit (g)

 

The only addition to this unit is the stream divider:

 

For the stream divider:

 

Three streams with C + 3 variables plus the heat transfer rate.

 

Nv = 3(C + 3) + 1 = 3C + 10

 

Number of equations:

 

3     Composition summations

1     Equality of temperatures for equilibrium streams

1     Heat Balance

1     Equality of pressures for equilibrium streams

C    Component mass balance

(C – 1)    Equality of compositions for leaving streams

 

NE = 2C + 5

 

Hence:

 

ND = C + 5

 

Total of ND = C + 6N + 2CN + 4 + C + 5 = 2C + 6N + 2CN + 9

 

 

Redundant streams:

 

Condenser:

 

2: The vapour entering the condenser and the liquid leaving it.

 

Stream divider::

 

1: The liquid returning to the column.

 

One additional variable is added, the liquid leaving the condenser. The rest of the analysis is identical to that of unit (c).

Total of NR = 2N + 1

Finally:

 

(ND)unit = NV – (C + 2)NR + NA = C + 2N + 9

 

 

If unit (g) has an additional feed stream: As explained in the last paragraph in chapter 5 in the text book, for every additional stream, C + 3 degrees of freedom are added to the unit D.O.F. Therefore,

 

(ND)unit = C + 2N + 9 + C + 3 = 2(C + N + 6)

 

Question 2

1.

Water = 10 litres/s = 555.56 mol/s = L

Ammonia = 291.09 mol/s = G

The operating line equation is given by:

 

The slope for the line is

Taking this slope from the intercept , clearly shows that the degree of removal of ammonia will not be possible with such a flow rate of water (The initial Y1 required to reach the degree of removal should be between 0.6 and 0.7 ). See figure1.

 

Figure 1. Low flow rate of water

 

2.

The minimum water flow rate allows reaching the desired separation at the expense of infinite separation stages. The slope is estimated according to the operating line equation , where XN is obtained from the equilibrium line (point YN+1,XN). Lmin=745 mol/s=13.4 l/s. See figure 2

 

Figure 2. Minimum water flow rate

 

3.

With a molar flow rate of 745 moles/s (16.1 l/s), the new slope of the operating line is , 3 stages are needed to reach the required separation. See figure 3.

 

 

 

Posted October 21th, 2006

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