Compressor, Pompa, Reactor

8/11/2019 Compressor, Pompa, Reactor

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Compressor

A. Compressor C-101

Function : To increase pressure of CO2from the storage tank

Type : Centrifugal Compressor

Gas Properties

- k = 1.316

- Z = 0.966

Operating Condition;

- P inlet = 6 Bar

- P outlet = 12 Bar

- T Inlet = 306.9 K

- Mass Flow = 16 ton/day

Since we use 12 batch processes so, the mass flow would be 1333.33 kg/batch. Since the

charging process on each batch process need only 30 min or 0.5h. so, the flow rate have

to be 2666.7 kg/h for 1 batch process.

Discharge Temperature Calculation

To calculate discharge temperature, we use the following equation;

Compressor Head Calculation

To calculate Compressor Head, we use the following equation;

()

[

]


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BHP power Calculation

() (

) (

)

()

( ) ( ) () [ ] B. Compressor C-102

Function : To increase pressure of CO2from the storage tank

Type : Centrifugal Compressor

Gas Properties- k = 1.286

- Z = 0.990

Operating Condition;

- P inlet = 2 Bar

- P outlet = 6 Bar

- T Inlet = 318 K

- Mass Flow = 172.5 kg/h

Since we use 12 batch processes so, the mass flow would be 1333.33 kg/batch. Since the

charging process on each batch process needs only 30 min or 0.5 h. so, the flow rates

have to be 690 kg/h for 1 batch process

Discharge Temperature Calculation

To calculate discharge temperature, we use the following equation;

Compressor Head Calculation

To calculate Compressor Head, we use the following equation;


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()

[

]

BHP power Calculation

() ( ) () ()

( ) ( ) () [ ]


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Reactor

A. Precipitation Reactor

Operating Condition

- Pressure = 11.5 bar

- Temp = 318 K

- Flow Rate

o Black Liquor = 26000 kg/h

o CO2 = 685.4 kg/h

- Since we donthave enough information of kinetic reaction we cannot determine

total volume by using Fogler formula for batch reactor process. In this case, we get

reaction time information about 45 min (Per Tomani, 2001).

Step 1: Determine Batch Process per Day

Process Time (min)

Charging 30

Reaction time 45

Emptying 30

Cleaning Time 15

Total time needed for completing 1 batch process is 120 min or 2h. Based on this result,

total batch process per day can be calculated by using the following equation:

So, the capacity of 1 batch process is:

- Liquid = 624000 kg/day x 1 day/12 batch = 52000 kg/batch

- CO2= 16450 kg/day x 1 day/12 batch = 1370.8 kg/batch


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Step 2: Determine Volume Tank (basis: 1 batch process)

a. Volume Liquid

b. Volume Gas

Gas holdup calculation

Gas holdup in bubble column can be estimated by empirical equation (Hikita &

Kikukaw, 1974)

( ) ( ) (Per Tomani, 2001) (Walas,1988) After gas hold up fraction known, we can use the following equation (Doran, 1995)

to determine the volume of the gas phase on the tank:

c. Total Volume

After we evaluated the volume of the liquid and gas on the feed of the reactor, we

are able to determine total working volume of the reactor. The total volume is thesum of the volume of liquid and gas. On the other hand the vessel also consists of

vapor space. Vapor space take 20% of column volume (Perry, 1999).


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Step 3: Determine Size of Tank

a. Body (Shell)

Based on the rule of thumb on the Walas (1988), Brownell (1969), typical dimension for

reactor are about;

So,

b.

Head

Since this vessel are working on the high pressure around 11.5 bar, we choose

Ellipsoidal with major/minor = 2:1 , since this head are able to working on the pressure

11.5 bar or 170 psi (Brownell, 1969), so

So total volume of the tank can be evaluated with the following equation:

Based on this equation, we can do a modification to get diameter of the tank as

following equation;

So, the height of the tank can be evaluated by using the following equation;


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While for height of ellipsoidal can be evaluated by using,

So, the total height of the tank is Step 4: Determine Design Pressure

Volume of Tank = 51.7 m

3

Volume of Liquid = 41.135 m3

Liquid Level = 3.8 m

Height of Tank = 4.908 m

a. Determine Hydrostatic Pressure

b. Determine Operating Pressure c. Determine Design Pressure

The design pressure of vessel must be 20% above the maximum pressure in the

column. (Walas, 1988):


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Step 5: Determine Thickness of Tank

a. Material Selection

SA 285 grade C is selected as the material for the vessel. The selection criteria are

based on the operating temperature and the maximum allowable stress of the

material. SA 285 grade C, can be operated in the range of -20 to 650oF with

maximum allowable stress is 18750 psi (ASME Code). For welding, we use Double

Welded Butt Joint, which have welded efficiency is 0.85 (Brownell, 1969).

Corrosion allowance is assumed to be 0.014 in per year (Perry, 1999).

b. Thickness of Shell (Body)

Double Welded Joint E=0.85

Using Material SA-285 Grade Callowable stress (S) = 18750 psi

Lifetime Equipment = 20 years

Corrosion Rate = 0.014 in per year

c. Thickness of Head


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Step 6: Determined Sparger and Agitator

Sparger

An orifice sparger was placed under the bottom impeller and its size was calculated to

be 75% the impellers diameter. The holed on this circular ring must not exceed 6 mm

(Doran, 1995) and therefore for design calculations their size was assumed to be 5 mm.

A magnetic drive was chosen as the driveshaft seal as this is more common in non-

viscous fermentation broths like Escherichia cell culture which is a very dilute culture

since the cell density is low. The location of sparger is 0.6 from the distance of impeller

from bottom of vessel (Walas,1988). Sparger clearance is 2.664 ft.

Agitator

a. Type : typically the impeller commonly used on the gas dispersion

system is radial turbine with six vertical blades (Walas, 1988)

b. Total Baffle : Baffle is essentially needed, commonly for gas dispersion 4

baffles are needed (Walas, 1988). The main purpose is to prevent dead zone on the

mixing process on the reactor or vessel.

c. Specification Design of agitator (Walas,1988):

Diameter Impeller

So, diameter impeller is 1.386m

Length and Width of blade on the turbine

The length of blades on turbine is 0.3465m

The width of blades on turbine is 0.2772m


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Width and offset of baffle

So the width of baffle is 0.3348 m while the offset of baffle is 0.048 m

Total Amount of Impeller:

Since

, we only need one single impeller.

Position of impeller:

The position of impeller is 0.933m from the bottom of vessel.

d. Power Needed

Data Needed:

- Viscosity = 3.8 cP = 0.00255 lb/ft.s

- Density = 1264.78 kg/m3= 78.96 lb/ft

3

- N = Assume to be 1 rev/s

- d = 4.54 ft

Determine NRe

Determine Power

Since the flow is turbulence we use the following equation (McCabe, 1999) to

determine power consumption of agitator:


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Step 7: Heater for Reactor Design

a. Determine Enthalpy on the Reactant Stream (T = 298K)

Table 3.1 Enthalpy on Reactant Precipitator Reactor Stream

Comp Mass (kg/h) g mol T(K) H Sensible H formation Total (kJ/h)

Lean Liquor 4377.91667 11415.69 298 0 0 0

Lignin 2626.875 1250.893 298 0 0 0

NaOH 437.8125 10945.31 298 0 -5134993.359 -5134993.359

Na2SO4 262.689583 1849.302 298 0 -2560357.964 -2560357.964

Na2CO3 1050.75833 9914.685 298 0 -11203594.23 -11203594.23

H2O 13134.4808 729693.4 298 0 -208575555.6 -208575555.6

H2SO4 0 0 298 0 0 0

NaHCO3 0 0 298 0 0 0

CO2 577.916667 13134.47 298 0 -5168545.17 -5168545.17

Total Enthalpy Reactant -232643046.4

b. Determine Enthalpy on the Product Stream (T=318K)

Table 3.2 Enthalpy on Product Precipitator Reactor Stream

Comp Mass (kg/h) g mol T(K) H Sensible H formation Total (kJ/h)

Lean Liquor 4377.91667 11415.69 318 113825.83 0 113825.8333

Lignin 2626.875 1250.893 318 84060 0 84060

NaOH 0 0 318 0 0 0

Na2SO4 0 0 318 0 0 0

Na2CO3 1210.88333 11425.58 318 27649.912 -12910909.29 -12883259.38

H2O 13095.0771 727504.3 318 1099986.5 -207949824.1 -206849837.6

H2SO4 181.4375 1851.403 318 5152.825 -1680166.792 -1675013.967

NaHCO3 976.391667 11623.71 318 20392.637 -10991380.48 -10970987.84

CO2 0 0 318 0 0 0

Total Enthalpy Product -232181213

c. Determined Transfer Heat around The Reactor Work done on the system, kinetic change, and potential energy can be neglected since

it doesnt give significant changes to the overall energy balance. So the following

equation can be simplified to:


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Based on these calculations, to maintain reactor operates at temperature 45

0

C, wehave to add heater with duty 318480.4 kJ/h to the reactor. Since it less endothermic,

so we can use external jacket as heater on the reactor (Sinnot, 2005).

d. Determined Steam Consumption

Specification of steam (Steam Table) :

Saturated Vapor Steam (1bar, 100

0C)

Enthalpy Vapor = 2693.6 kJ/kg

Enthalpy Liquid = 417.46 kJ/kg

Specific volume vapor = 1.694 m3/kg

Specific volume liquid = 1.043 x 10-3

m3/kg

Since there are 12 batch processes per day so, steam consumption is about 279.84

kg/batch.

Volumetric flow of steam on the inlet of Jacket:

Volumetric flow of liquid on the outlet of Jacket:

e. Specification of the Jacket

H = H of Liquid = 2.8 m = 9.184 ft


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Spacing Between Baffle and Vessel = 180 mm (Sinnot)

Spiral Baffle with pitch between spiral = 200 mm

The baffle forms a continuous spiral channel, section 75mm x 200mm

Number of Spiral = Height of jacket / Pitch = 2.8/0.2 = 14

Length of Channel = 14 x 3.14 x 4.018 = 176.63 m

Cross Sectional area of channel = (180 x 200) x 10-6

= 0.036 m2

Hydraulic mean diameter:

Calculating heat transfer coefficient on the jacket:


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B. Re-slurry Reactor

Operating Condition

- Pressure = 3 bar

- Temp = 318 K

- Flow Rate

o Black Liquor = 4928.4 kg/h

o H2SO4 = 445.4 kg/h

- Residence Time

Since we dont have enough information of kinetic reaction we cannot determine

total volume by using Fogler formula for batch reactor process. In this case, we get

reaction time information about 45 min (Per Tomani, 2001).

Step 1: Determine Batch Process per Day

Process Time (min)

Charging 30

Reaction time 45

Emptying 30

Cleaning Time 15

Total time needed for completing 1 batch process is 120 min or 2h. Based on this result,

total batch process per day can be calculated by using the following equation:

Step 2: Determine Volume Tank (basis: 1 batch process)

a. Volume Liquid - 1


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b. Volume Liquid - 2

c. Total Volume

After we evaluated the volume of the liquid-1 and liquid-2 on the feed of the reactor,

we are able to determine total working volume of the reactor. The total volume is

the sum of the volume of liquid-1 and liquid-2.

By applying 20% as vapor space on the reactor (Perry,1999), so the volume of tank

will be: Step 3: Determine Size of Tank

c.

Body (Shell)

Based on the rule of thumb on the Walas (1988), Brownell (1969), typical dimension for

reactor are about; So,

d. Head

Since this vessel are working on the high pressure around 11.5 bar, we choose

Ellipsoidal with major/minor = 2:1 , since this head are able to working on the pressure

3 bar (Brownell, 1969), so


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So total volume of the tank can be evaluated with the following equation:

Based on this equation, we can do a modification to get diameter of the tank as

following equation;

So, the height of the tank can be evaluated by using the following equation;

While for height of ellipsoidal can be evaluated by using,

So, the total height of the tank is

Step 4: Determine Design Pressure

Volume of Tank = 9.83 m

3

Volume of Liquid = 8.185 m

3

Liquid Level = 2.08 m


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Height of Tank = 2.83 m

a. Determine Hydrostatic Pressure

b. Determine Operating Pressure

c.

Determine Design Pressure

For the design pressure, we add safety factor around 20% (Walas, 1988) of the

operating pressure, so: Step 5: Determine Thickness of Tank

a.

Material SelectionSS 316 is selected as the material for the vessel. The selection criteria are based on

the operating temperature, the maximum allowable stress of the material and also to

accommodate the corrosion due to presence of acid (H2SO4) on the tank. SS 316,

can be operated in the range of -20 to 650oF with maximum allowable stress is

12650 psi (ASME Code). For welding, we use Double Welded Butt Joint, which

have welded efficiency is 0.85 (Brownell, 1969). Corrosion allowance is assumed to

be 0.014 in per year (Perry, 1999).

b. Thickness of Shell (Body)

Double Welded Joint E=0.85

Using Material SS 316 allowable stress (S) = 12650 psi

Lifetime Equipment = 20 years

Corrosion Rate = 0.014 in per year


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c. Thickness of Head

Step 6: Determine Agitator

a. Type : typically the impeller commonly used on the gas dispersion system is

radial turbine with six vertical blades (Walas, 1988)

b. Total Baffle : Baffle is essentially needed, commonly for gas dispersion 4

baffles are needed (Walas, 1988)

c. Specification Design of agitator (Walas,1988):

Diameter Impeller So, diameter impeller is 0.8 m

Length and Width of blade on the turbine

The length of blades on turbine is 0.2m


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The width of blades on turbine is 0.16m Width and offset of baffle

So the width of baffle is 0.1667 m while the offset of baffle is 0.027 m

Total Amount of Impeller:

Since , we need two impellers (Walas, 1988). Position of impeller:

o First Impeller

o Second Impeller

The position of first impeller is 0.3467 m and for the second impeller is 1.3867

m from the bottom of vessel.

d. Power Needed

Data Needed:

- Viscosity = 10.8 cP = 0.00725 lb/ft.s

- Density = 1290.78 kg/m3= 80.54 lb/ft

3


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- N = Assume to be 1 rev/s

- d = 2.624 ft

Determine NRe

Determine Power

Since the flow is turbulence we use the following equation (McCabe, 1999) to

determine power consumption of agitator:

To determine Power Number (Np), we can use the following graphic (Walas,1988) which depend on the type of impeller and Reynolds number.

By using this graphic Np is about 1.3, so Assume efficiency of motor is 80%, so:


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Since we use 2 impeller, so the total power consume to move an agitator is:


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Pump

A. Pump P-101

Operating Condition

Pressure = 1 atm.

Temperature = 250C

Mass Flow = 57570 kg/day 9595 kg/h (for 1 batch process which need

charging process 0.5m)

Density = 1000 kg/m

3= 62.4 lb/ft

3

Type = Centrifugal Flow

Viscosity = 1 cP = 0.00067 lb/ft.s

Volumetric Flow = 6.664 E-04 m3/s = 0.0235 ft3/s = 10.56 GPM

Calculation

1. Pump Design

From Appendix A3 (Noel De Nevers, 1991), we choose commercial steel pipe with

specification below:

Nominal Size = 1 1/2 in

Schedule Number = 40

Inside Diameter (ID) = 1.610 in = 0.1342 ft

Outside Diam. (OD) = 1.990 in = 0.1650 ft

Inside Sec. Area = 0.01414 ft2

2.

Check Reynold Number

a. Average Velocity


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b. Reynold Number

c. Fanningfactor

For commercial pipe, = 0.000046 (geankoplis, 1997), for Re = 14513.64 and = 0.00034 thefanning (f)factor is 0.0070.

3. Calculate Friction Loss

Assume there are accessories along the pipe which can be seen on the following table:

Item Amount L/D Total L/D

Elbow 90 4 30 120

Gate Valve 2 13 26

Elbow 45 1 16 16

Entrance 1 30 30

Exit 1 60 60

Total L/D 252

*(App. C-2a, Foust, 1980, Nevers, 1991)

So, the equivalent length; We assume, that length of the pipe needed to transport the fluid is around 20 m or 65.6

ft, so the total L is 99.318 ft. with this number, we are able to calculate total friction loss

along the pipeline with following equation:

4. Work Done by Pump


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Where for turbulence flow, , there is no velocity change after being pump sinceinlet pipe and outlet pipe have the same size. In this case we assume there is a height

difference where = 10m = 32.8 ft. and then . So,

To be noted negative sign means work done to the fluids.

5. Power of Pump

Efficiency pump, =70%

Mass flow through the pump is 6.131 lb/s, so the power of the pump:

So we choose the pump with power around 2 HP


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B. Pump P-201

Operating Condition

Pressure = 1 atm

Temperature = 25

0C

Mass Flow = 135000 kg/day = 45000 kg/h (for 1 batch process which need

charging time 0.5 h)

Density = 1226 kg/m3

= 76.53 lb/ft3


Viscosity = 3.3 cP = 0.00221 lb/ft.s

Volumetric Flow = 1.275 E-03 m3/s = 0.045 ft

3/s

Calculation

1. Pump Design From Appendix A3 (Noel De Nevers, 1991), we choose commercial steel pipe with


Nominal Size = 2 in


Inside Diameter (ID) = 2.067 in

Outside Diam. (OD) = 2.375 in


2. Check Reynolds Number

a. Average Velocity

b. Reynolds Number


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5. Power of Pump

Efficiency pump, =80%


So we choose the pump with power around 3.5 HP


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C. Pump P-202

Operating Condition

Pressure = 1 atm

Temperature = 25

0C

Mass Flow = 2291 kg/day

Density = 1181 kg/m3

= 73.69 lb/ft3


Viscosity = 1.12 cP = 0.00075 lb/ft.s

Volumetric Flow = 2.275 E-05 m3/s = 7.93 E-04 ft

3/s = 0.355 GPM

Calculation

1.

Pump Design From Appendix A3 (Noel De Nevers, 1991), we choose commercial steel pipe with


Nominal Size = 2 in


Inside Diameter (ID) = 2.067 in

Outside Diam. (OD) = 2.375 in


2. Check Reynold Number

a. Average Velocity

b. Reynold Number


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c. Fanningfactor

For commercial pipe, = 0.000046 (geankoplis, 1997), for Re = 14513.64 and = 0.00026 thefanning (f)factor is 0.0075.3. Calculate Friction Loss

Assume there are accessories along the pipe which can be seen on the following table:

Item Amount L/D Total L/D

Elbow 90 4 30 120

Gate Valve 2 13 26

Elbow 45 1 16 16Entrance 1 30 30

Exit 1 60 60

Total L/D 252

*(App. C-2a, Foust, 1980, Nevers, 1991)

So, the equivalent length; We assume, that length of the pipe needed to transport the fluid is around 20 m or 65.6

ft, so the total L is 109 ft. with this number, we are able to calculate total friction loss

along the pipeline with following equation:

4. Work Done by Pump


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5. Power of Pump

Efisiensi pompa, =80%


So we choose the pump with power around 0.3 HP


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D. P-301

PUMP

Identification: Item Centrifugal Pump

Item no. P-301

No. required 2 (1 active, 1 stand by)

Function: Pumping Water from Tank

Operation: Batch

Material handled: Hot Water

Composition (%):

Water 100

Design Data: Type Centrifugal Pump

Material of construction Carbon Steel

Mass flow (lb/s) 0.673450971Power (HP) 0.155650165

Suction pressure (bar) 1

Viscosity (cP) 1

Inlet temperature (oC) 110

E. P-302

SCREW PUMP


Item no. P-302



Operation: Batch

Material handled: Water

Composition (%):

Water 100



Mass flow (lb/s) 23.27

Power (HP) 4.17


Viscosity (cP) 1



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F. P-303

SCREW PUMP


Item no. P-303



Operation: Batch


Composition (%):

Water 100





Viscosity (cP) 1


G. P-304

SCREW PUMP


Item no. P-304



Operation: Batch


Composition (%):

Water 100



Mass flow (lb/s) 5.817754141

Power (HP) 0.471853844


Viscosity (cP) 1



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H. P-305

SCREW PUMP


Item no. P-305



Operation: Batch


Composition (%):

Water 100





Viscosity (cP) 1


I. P-306

SCREW PUMP


Item no. P-306



Operation: Batch


Composition (%):

Water 100



Mass flow (lb/s) 5.817754141

Power (HP) 0.471853844


Viscosity (cP) 1


Compressor, Pompa, Reactor

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