hydrocarbons assignment jinny

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CME 8022 Assignment - Ethane-Propane Separation

Submitted By :Jinny Varkey (B0916817)

Introduction

A process to separate ethane and propane using distillation is shown in Figure 1. In thisprocess a feed containing ethane and propane is compressed from 101.3 kPa to 850 kPa(compress or isentropic eff iciency = 85%). The compr essed feed is then cooled in three stages,firstly using the ethane and propane produc t streams, second ly using a refriger ated pre-cooler and finally by passing it through the reboiler of the distillation column. The compressed cooled

feed is then let down t hrough a valve t o a pressure of 101.3 kPa.

After the valve the vapour and liquid phases are in equilibrium. They are separated and theliquid is fed to the top plate of the distillation column. The vapour from the separator and thevapour from the top of the column are combined to form the ethane product stream. Thepropane product stream is taken from the reboiler. The two product streams are heated usingenergy from pre-co oler 1. Each product str eam receives half of the heat load of Pre-cooler 1.

The Aim of this course work is to

1. Using the data provided and assuming that the liquid phase is incompressible and that the gas/vapour phase isan ideal gas:

(a) Find the mass flow rates, compositions, pressures and temperatures of all the process streams in Figure 1.

(b) Calculate the work required to compress the feed.

(c) Calculate the number of stages required in the column.

This is accomplished in theNumerical approach section below.

2. Model the process using HYSYS process simulation software and compare the simulation results with thosefrom part 1.

1. Numerical Approach

Data available to us are as follows:

Ethane:

Propane:



3

No: of moles of ethane = =

= 48.86 mol s-1

No: of moles of propane = =

= 34.01 mol s-1

Mole fraction of ethane, = = 0.59

Mole fraction of propane, = = 0.41

Therefore, molar fraction of propane and ethane will be and respectively.

We can then calculate the molar mass of the mic

ture

1.1

d ompre

e e

or Ie

entropic work

The compression is assumed to be isentropic, which applies there is no change in entropy,

�� = [Si, t ± R ln

f

i] (for liquid mif

ture)= [Si, t ± R ln { ] (for gas/vapor mi

f

ture)

��

��

��

, therefore:

�� .�� . ��

°



4

To estimate the isentropic work done by the compressor, the isentropic enthalpy change across the compressor

is required.

Enthalpy at the inlet of the compressor,

Enthalpy at the inlet of the compressor,

I

g

entropic Work done by the compreg g

or ig

Real Work done by the compre

h h or i

h calculated dividin

i the i

h entropic work by the compre

h h or

ep p

iciency.

=

=

= 593604.16 J s

-1

We can now calculate the real enthalpy chan

q e



5

°

2. Pre-cooler r Ess

im t

s

ion

u iv

:w

Px y

�

� � �

ly x

�

+

+



6

1.2 Pre-cooler 2

Si� � �

,

�

�

�

i� �

�

�

t�

�

�

t�

�

�

7 � K

� i

�

�

: P

l

�

Diff �

t

�

tl

t t � �

t�

j

k

� j j

� �

�

,k

l

tl

li� �

l �

�

ij

�

l

�

l�

t

�

� �

�

� �

�

t

tl

�

im

�

m

�

l�

t

m

�

li�

�

t

tl

ij

� j j

� �

ti �

.� l

t�

n

l

n

l

k

j

l

k

j

tl

� j j

� �

�

t � �

t�

j

� �

�

tl

j �

�

i� �

l

�

t

t � � j

f

m

�

l�

j

�

l

�

l�

t

�

:

�

� �

t�

[Kk

]

Q@� o

[K]

o �

.�

�

� 7.�

Q@� o �

[K]

�

.

� o

.

Q@� o o

[K]

�

7�

.o �

o

.o �

Q@� o

[K]

� 7

.� �

�

.7o

IN�

E

POLA�

ION @

�

T4 [K] 267.4

t

z

7.{

K,|

} ~

f

it }

ti

}

t|

i

t

�

l

�

i�

~

t}

�

f � �

KW

�

t

x

i} � }

f

t

}

�

}

~

~

t

}

t

�

KP}

}

z

7.{

K

The method of calculating Tout:

Vapour composition at 268K f rom Txy diagram at 850 Pa: Ethane = 0.57, Propane = 0.43

Liquid composition at 268K f rom Txy diagram at 850 Pa: Ethane = 0.21, Propane = 0.79

S ~

t

} ~ ~

}

l}

:

�

V}

+ Li

i

�

V + L

�

f

L

�

V

} ~ ~

}

l}

f

t

}

:

X

�

Y

+ XLL

X

�

Y

+ XL

�

� V�



7

V =

Assuming vapour fraction (266° K)

Feed = Vapour �

� iquid

F = V � � = 3

Therefore�

= 3 ± V , Therefore � = 0.56 Kg/s

Ener � y Balance:

FHF � �

= VHv � H

Datum point chosen

2

1, therefore HF = 0�

= VHv � H



8

Vapour Enthalpy

Hmi = Ye(v) [He(vap)] Yp(v) [Hp(vap)]

Hethane(vap) = e(vap) dT = = 1

41

(-3.6) = - 626

.6 J/Kg

Hpropane(vap) =

p(vap) dT =

= 1

15

(-3.6) = - 61

4 J/Kg

1.

.1 iª

uid Enthalpy

iª

uid Ethane

Li«

uid Propane

Hmi¬ = 2.5 [ 0.5 (-626 .6) ® 0.43 (-61 4) ] = - 1 ¯ °

±

.3±

J/² ³

Usin

Ener

y balance



9

1.3 Re-boiler

Fig 5: Re-boiler

Flow composition, from Tµ

y diagram at 850 KPa and 254.¶

K

Vapour mass fraction of ethane = 0.¶

9Vapour mass fraction of propane = 0.21

· iquid mass fraction of ethane = 0.39

· iquid mass fraction of propane = 0.61

V =

Vapour fraction of mass flow , V = = 0.

¹ º

»

¼

/s

½ iquid fraction of mass flow ,

½ = 2.1

¾

º

»

¼

/s

1.4 Valve

Flow composition at 203 K and 101.2 kPa :

Vapour mass fraction of ethane = 0.¿

9Vapour mass fraction of propane = 0.21

½ iquid mass fraction of ethane = 0.2

¿

½ iquid mass fraction of propane = 0.

¿

3

Vapour fraction of mass flow, V = = 1.32

À

Á /s,

Â iquid fraction of mass flow, Â = 1.Ã

Ä

À

Á /s

Fig 6: Valve



10

1.Å

Separator

MiÆ

ture entering the separator:

Vapour fraction of mass flow, V = 1.32 Kg/s

Ethane mass fraction in vapour = 1.043 Kg/s

Propane mass fraction in vapour = 0.2

Ç Ç

Kg/s

FigÇ

: Separator

È iquid fraction of mass flow, È = 1.6Ç

Kg/s

Ethane mass fraction in liquid = 0.45 Kg/sPropane mass fraction in liquid = 1.22 Kg/s

Separator outlet:

Top vapour line

Mass flow rate = 1.32 É Ê

/s

Ethane = 1.043 É Ê /s Propane = 0.277 É Ê /s

Bottom liquid line

Ë iquid flow rate = 1.

Ì

7 É Ê

/s

Ethane = 0.4Í

É Ê /s

Propane = 1.22 É Ê

/s

1.Ì

DistillationÎ

olumn

Mass balance of whole system:

F = 3Kg/sD = ethane product line

B = propane product line

F =D Ï

B

Mass balance of ethane:

F =D Ï

B

3Ð

0.5 = 0.Ñ

5D Ï

0.005B

1.5 = 0.Ñ

5D Ï 0.005B

Fig 8: Multistage Separator

Substituting the equation of whole system into mass balanceEquation of ethane B and D are found to be,

B = 1.01 Kg/s (propane product line mass flow)

D = 1.99 Kg/s (ethane product line mass flow)

Vapour leaving the reboiler from TÐ

y diagram at 0.5% ethane is 231 K

Temperature of ethane product line from TÐ

y diagram atÑ

5% ethane is 205 K

Following McCabe Thiele principles which states that liquid and vapour flow rate through the column are

constant, thereforeË iquid into reboiler = liquid into column = 1.6

Ñ

kg/s



11

The vapour stream out of top of column can therefore be calculated as the Mass flow rate of ethane product ±

vapour mass flow rate from separator, 1.99 ± 1.32 = 0.6Ò

Kg/s

This applies that the vapour out of reboiler into separator is 0.6Ò

Kg/s

Ethane product line containsÒ

5% ethane, 1.99Ó

0.Ò

5 = 1.49 kg/s ethane

Therefore ethane vapour coming from column = De(product line) ± De(first separator)

= 1.49 ± 1.043 = 0.449 Kg/s

Propane vapour coming from column = 0.6Ò

± 0.449 = 0.221 Kg/s

The temperature of the vapour coming out of the column can be read from the TÓ

y diagram using the mass

fraction of ethane, which is calculated as follows,

0.6Ò

Ó

mass fraction = 0.449, mass fraction = 0.6Ò

Therefore temperature of ethane is 209 Ô

Number oÕ

staÖ

es re×

uired:

To plot the operation line on theØ

y diagram the following calculations are required,

Ù iquid feed in moles, n =

Ethane: n = = 14.69 moles

Propane: n = 2

Ú

.64 moles

Mole fraction of ethane = =

= 0.343, theÛ

-aÛ

is coordinate of the operating line on

theÛ

y diagram for ethane propane system at 101.3 kPa

Vapour, ethane product line in moles:


Propane: n = 5.01 moles

Therefore mole fraction of ethane in ethane product line = = 0.

Ú

45, the y-aÛ

is coordinate of the operating

line on theÛ

y diagram for ethane propane system at 101.3 kPa

Vapour, propane product line in moles:


Propane: n = 22.69 moles

Therefore mole fraction of ethane in propane product line = = 0.00Ú

1, the second coordinate of the y-aÛ

is of the operating line on the

Û

y diagram for ethane propane system at 101.3 kPa. This low value effectively means

that the operating line is going through the origin of the graph.

Figure 9 below shows the operating line plotted using the data calculated above:



12

Fig 9:Ü

y Diagram for Ethane- Propane system at 101.3kPa

From the above diagram it can be derived at that a separator with 6 stages is required with 100% efficiency.

1.7 Pre-cooler 1 (HeatinÝ )

To calculate the output temperature of both product lines an energy balance is carried out;

Therefore,

EP = Ethane product line

PP = Propane product line



13

And,

Þ

1 = 1.99 [0.ß

5à

1ß

41 (Tout ± 205)á

0.25à

1ß

15 (Tout ± 205)]á

1.006 [0.005à

1ß

41 (Tout -231)á

oooo0.995à

1ß

15 (Tout -231) ]

ß

33260 = 511ß

.0ß

Tout ± 110ß

161.46

Therefore, Tout = 3â 9.6

ã

Tout is equal for both product lines.

2. HYSYS Modellinä

The process was modelled in HYSYS and simulated. The results obtained showed slight variations from the

numerical solutions but there was some degree of adherence too. The process flow diagram and the report

including the various parameters of the streams are included.

Fig 10: HYSYS PFD Screen Shot



14

Conclusion

Some deviations were observed in the results obtained from the HYSYS simulation and the numerical method.

This can be accounted to two main reasons. They are as follows:

� Specific heat capacity (Cp) was assumed to be constant value in the numerical method , which inn reality is a

function of temperature .This assumption is overridden by the temperature calculation in HYSYS calculation.

� The gas was assumed it to be an ideal gas in the numerical method, which is not the case in HYSYS. Peng

Robinson Equation was used in HYSYS modelling.



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