Solution manual Chemical Process : Design and Integration, 2nd Edition, by Robin Smith (COMPLETE)

Chapter 2 Solutions - 1

1
Chapter 2 Solutions
Process Economics
1. A new agitated reactor with a new external shell-and-tube heat exchanger and new centrifugal pump are to
be installed in an existing facility. The agitated reactor is to be glass-lined, which can be assumed to have
an equipment cost of three times the cost of a carbon steel vessel. The heat exchanger, pump and
associated piping are all high-grade stainless steel. The equipment is rated for moderate pressure. The
reactor has a volume of 9 m3, the heat exchanger an area of 50 m2 and the pump has a power of 5 kW. No
significant investment is required in utilities, off-sites, buildings, site preparation or working capital. Using
Equation 2.1 and Table 2.1 (extrapolating beyond the range of the correlation if necessary), estimate the
cost of the project (CE Index of Equipment = 680.0).


Solution
The installed cost can be estimated from Equation 2.6:
C F    f M f P f T ( 1  f PIP )i C Ei
i
  f ER  f INST  f ELEC  f UTIL  f OS  f BUILD  f DEC  f CONT  fWS  C E,i
i

First estimate the equipment cost from Table 2.1 using Equation 2.1:
M
 Q 
C E  C B  
 QB 
Reactor cost:
0.45
9
C E  1.15 10 4  
1
= $30,911
Correcting for equipment cost index:
680.0
= 30,911 
435.8
= $48,232
Heat exchanger cost: Extrapolating beyond the range of the cost correlation:
0.68
 50 
C E  3.28 10 4  
 80 
= $23,827
Correcting for equipment cost index:
680.0
= 23,827 
435.8


Chemical Process Design and Integration, Second Edition. Robin Smith. © 2016 John Wiley & Sons, Ltd.
Published 2016 by John Wiley & Sons, Ltd. Companion website: www.wiley.com/go/smith/chemical2e

, Chapter 2 Solutions - 2

= $37,178
Centrifugal pump: Assuming a small pump with motor in high grade stainless steel:
0.35
5
C E  1.97 10 3   
1
= $3, 460
Correcting for equipment cost index:
680.0
= 3,460 
435.8
= $5,399
For equipment cost factors take the following:
fm fm fT

Reactor 3 1 1
Heat exchanger 3.4 1 1
Centrifugal pump 1 1 1


For contributions to the overall installation factor as appropriate from Table 2.7 for fluid processing plant:
fER = 0.4
fPIP = 0.7
fINST = 0.2
fELEC = 0.1
fUTIL =0
fOS =0
fBUILD =0
fSP =0
fDEC = 1.0
fCONT = 0.4
fWC =0
Substituting in Equation 2.6:
C F = 3 11 1 0.7 48,232  3.4 11 1 0.7 37,178 111 1  0.7 5,399

0.4  0.2  0.1  0  0  0  0  1.0  0.4  0  48,232  37,178  5,399

 4.70  105 1.91 105

 $6.61 105


2. Steam is distributed on a site via high-pressure and low-pressure steam mains. The high-pressure main is at
40 bar and 350ºC. The low-pressure main is at 4 bar. The high-pressure steam is generated in boilers. The
overall efficiency of steam generation and distribution is 75%. The low-pressure steam is generated by
expanding the high-pressure stream through steam turbines with an isentropic efficiency of 80%. The cost
of fuel in the boilers is 3.5 $· GJ–1, and the cost of electricity is $0.05 kW–1·h–1. The boiler feedwater is

, Chapter 2 Solutions - 3

available at 100ºC with a heat capacity of 4.2 kJ·kg–1·K–1. Estimate the cost of the high-pressure and low-
pressure steam.


Solution
Cost of 40 bar steam. From steam tables, for 40 bar steam at 3500C:
Enthalpy = 3095 kJ∙kg−1
For boiler feedwater:
Enthalpy = 4.2 (100 − 0) (relative to water at 0ºC)
−1
= 420 kJ∙kg
To generate 40 bar steam at 3500C:
Heat duty = 3095 − 420 = 2675 kJ∙kg−1
For 40 bar steam:
1
Cost  3.5  10 6  2675 
0.75
= 0.01248 $∙kg−1
= 12.48 $∙t−1
Cost of 4 bar steam. Here 40 bar steam is now expanded to 4 bar in a steam turbine.
From steam tables, steam turbine inlet conditions at 40 bar and 3500C are:
H1 = 3095 kJ∙kg−1
S1 = 6.587 kJ∙kg−1K−1
Turbine outlet conditions for isentropic expansion to 4 bar are:
H2 = 2610 kJkg−1
S2 = 6.587 kJ∙kg−1K−1
For a single−stage expansion with isentropic efficiency of 80%:
H'2 = H1 − I (H1 − H2)
= 3095 − 0.8 (3095 − 2610)
= 2707 kJ∙kg−1
Power generation
= 3095 – 2707
= 388 kJ∙kg−1
Value of power generation
0.05
 388 
3600
= 0.00539 $∙kg−1
Cost of 4 bar steam
= 0.01248 − 0.00539
= 0.00709 $∙kg−1
= 7.09 $∙t−1


3. A refrigerated distillation condenser has a cooling duty of 0.75 MW. The condensing stream has a

, Chapter 2 Solutions - 4

temperature of –10ºC. The heat from a refrigeration circuit can be rejected to cooling water at a
temperature of 30ºC. Assuming a temperature difference in the distillation condenser of 5ºC and a
temperature difference for heat rejection from refrigeration to cooling water of 10ºC, estimate the power
requirements for the refrigeration.


Solution
The power requirements for refrigeration can be estimated from Equation 2.10:

QC  TH  TC 
W   
0.6  TC 

TH = 273 + 30 + 10
= 313 K
TC = 273 – 10 – 5
= 258 K
QC = 0.75 MW
Substituting in Equation 2.10:
0.75  313  258 
W=  
0.6  285 
= 0.27 MW


4. Acetone is to be produced by the dehydrogenation of an aqueous solution of isopropanol according to the
reaction:
(CH3 )2 CHOH  CH3COCH3  H2
Isopropanol Acetone Hydrogen
The effluent from the reactor enters a phase separator that separates vapor from liquid. The liquid contains
the bulk of the product, and the vapor is a waste stream. The vapor stream is at a temperature of 30ºC and
an absolute pressure of 1.1 bar. The component flowrates in the vapor stream are given in Table 2.13,
together with their raw material values and fuel values. Three options are to be considered:
a) Burn the vapor in a furnace
b) Recover the acetone by absorption in water recycled from elsewhere in the process with the tail gas
being burnt in a furnace. It is expected that 99% will be recovered by this method at a cost of 1.8 $·
kmol–1 acetone recovered.
c) Recover the acetone by condensation using refrigerated coolant with the tail gas being burnt in a
furnace. It is anticipated that a temperature of –10ºC will need to be achieved in the condenser. It can
be assumed that the hydrogen is an inert gas that will not dissolve in the liquid acetone. The vapor
pressure of acetone is given by:
2940.5
ln P  10.031 
T  35.93
where P = pressure (bara)
T = absolute temperature (K)
The cost of refrigerant is 11.5 $∙GJ–1, the mean molal heat capacity of the vapor is 40 kJ·kmol–1·K–1, and

, Chapter 2 Solutions - 5

the latent heat of acetone is 29,100 kJ·kmol–1. Calculate the economic potential of each option given the
data in Table 2.13.
Table 2.13 Data for Exercise 4.
Component Flowrate in vapor Raw material value Fuel value
–1 –1
(kmol·h ) ($· kmol ) ($· kmol–1)
Hydrogen 51.1 0 0.99
Acetone 13.5 34.8 6.85


Solution
a) First option to burn the vapour in a furnace to provide useful heat. Take the value of the useful heat as
revenue:
Economic potential = fuel value of vapor stream 
= 51.1×0.99 + 13.5×6.85
= 143.1 $∙h−1

b) Second option recovers acetone in an absorption column at a cost of 1.8 $∙kmol 1 acetone recovered.
Assume hydrogen is insoluble in the water:
Economic potential = value of acetone recovered 

+ fuel value of remaining vapor stream 

− operating cost of separation 

= 13.5  0.99  34.8 + 51.1 0.99  13.5  0.01 6.85

− 13.5  0.99 1.8
= 492.6 $∙h−1
c) Third option recovers acetone by refrigerated condensation. Acetone will condense when its partial
pressure equals its vapor pressure. Assume hydrogen is insoluble in the condensed acetone. At −10°C
(263K) the vapour pressure of acetone is given by:
 2940.5 
P SAT  exp 10.031 
 263  35.93 
= 0.054 bar
For a partial pressure of 0.054 bar the mole fraction of acetone at a pressure of 1.1 bar is given by:

0.054
yCOND 
1.1
= 0.049
Mole fraction of acetone in feed:

13.5
yFEED 
51.1 13.5
= 0.209
Fractional recovery of acetone
0.209  0.049

0.209