homework 6 pe5623

7/28/2019 Homework 6 Pe5623

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5 MMscfd of NG of 0.7 SG is to be dehydrated in a solid bed dehydration plant. It is water

saturated at 500 psig and 100 F and must be dried to -120F dew point. Use 4A molecular sieve

of 1/8` beads (i.e., 4X8 mesh) The regeneration gas is taken from dehydrated gas which is

available at approximately 500 psia and 100 F. The bed must be heated to 500F for

regeneration.

Given Data :

Gas Specific Gravity 0.7 = MWgas /MWair = MWgas / 28.97

MWgas = 20.279 lb/lbmole

Z factor = 0.915

Pressure : 500 psig = 514.7 psia

Temperature : 100F = 560R

Calculation of Supervelocity and the Bed diameter

Z factor ( from GPSA 23-6 ) Z factors for SG 0.7 at 100 F ; Z = 0.915

`Density of gas = P x MW / Z x R x T

514.7 x 20.279 / 0.915 x 10.73 x 560 = 1.898 lb/ft3

Density of gas = 1.898 lb / ft3

7/28/2019 Homework 6 Pe5623


Viscosity of the gas from figure 23-23 GPSA = 0.0125 centipoise, below

The maximum supervelocity can be calculated by ( 20-16 GPSA ) equation.

Vmax = * ( ∆P / L )max / ( C p) ]½

- [ ( B / C ) ( µ / density ) / 2 ]

Let ∆P / L = 0.33 psi/ft

7/28/2019 Homework 6 Pe5623


Where :

∆P = pressure drop in psia

L =length of packed bed , ft

Cp = Heat Capacity Btu/(lb F)

C = constant in equation 20-11 --- 0.0000889 for 1/8` beads ( 4X8 mesh)

B = constant in equation 20-11 --- 0.0560 for 1/8` beads ( 4X8 mesh)

Vmax ={ ( 0.33 psi/ft ) / ( 0.000089 x 1.898 lb/ft3 ) } 1/2 - { ( 0.056 / 0.000089 ) x ( 0.0125 /1.898 lb/ft3 ) / 2 }

Vmax = 42.127 ft/min

m = ( 75 000 000 scf/day ) / ( 24 hrs/day ) ( 379.43 scf/lbmole )) x ( 20.279 lb/lb mole )

m = 167018.62 lb/hr of wet gas

q = ( 167018.62 lb/hr ) / (( 60 min/hr ) x ( 1.898 lb/ft3

)) --- [eq 20-13]

q = 1466.61 ft3

/ min

D min = [ ( 4 x ( 1466.61 ft3

/ min ) / ( 3.14 x 42.127 ft/min ) ]1/2

---- ( Eq 20-12 GPSA )

D min = 6.6595 ft

We round it off to 7 ft diameter and adjust V and ∆P/L values ;

V adjusted = ( 42.127 ) x { ( 6.6595 / 7 )2

} = 38.128 ft/min

( ΔP / L ) adjusted = 0.33 x { (38.128 / 42.127 )2

}= 0.2703 psi / ft

7/28/2019 Homework 6 Pe5623


Estimation of the amount of water to be removed from the

feed per cycle for each bed

Calculations are based on 24 hrs cycle consisting of 12 hours of adsorbing and 12 hours of

regenerating ( heating, cooling, standby and valve switching)

From the figure the water content of 514.7 psia gas at 100F is 103.5 lb/MMscf

The water content at dewpoint -120F is practically zero so the water removal should be

calculated as follows ;

w = ( 103.5 lb/MMscf ) x ( 75 MMscf/day ) / ( 24 hr/day ) = 323.43 lb/hr

7/28/2019 Homework 6 Pe5623


Wremoved = 323.43 lb/hr x 12 = 3881.16 lb of

water needs to be removed in 12 hrs drying cycle or 24 hours cycle per bed

Determination of Sieve Required and the Bed Height

We assume the bulk density of sieve 45lb/ft3, the feed gas is saturated with water, the relative

humidity is 100% so ;

Css ( Saturation correction factor for sieve) 1.0 from fig 20.84

from fig 20-85 the CT is 0.93 for 100F

Calculation of the amount of Molecular Sieve Required in

Saturated Zone

7/28/2019 Homework 6 Pe5623


water removed per cycle

Ss = amount of molecular seive req in sat zone lb = ------------------------------------------------------------( 0.13 ) x (Css is 1.0 ) x (CT is 0.93 for 100F)

Ss = 32102.23 lb of Sieve for each bed

We use two beds so we need 64204.46 lbs of Sieve

Calculation of the Length of Packed Bed Saturation Zone

( 4 ) x Ss amount of molecular sieve req in sat zone, lb

Ls = lenght of packed bed saturation zone, ft = -----------------------------------------------------------------

( 3.14 ) x ( D2

) x ( bulk density )

Molecular sieve bulk density is 42 to 46 lb/ft3 for spherical particles and 40 to 44 lb/ft3 for

extruded cyclinders. We assume the bulk density 44 lb/ft3

We assume that the bed will contain all the water to be removed .

Ls = lenght of packed bed saturation zone, ft = 18.9677 ft

Calculation of the Length of Mass Transfer Zone

L MTZ = ( V adjusted / 35 )0.3

x ( Z )

Z = 1.7 ft for 1/8 inch sieve

7/28/2019 Homework 6 Pe5623


L MTZ = (38.128 ft/min / 35 )0.3

x ( 1.7)

L MTZ = 1.7442 ft for mass transfer zone

Ls ( lenght of packed bed saturation zone ) + L MTZ = 20.711 ft of sieve for each bed

The total sieve = (20.711 / 18.9677 ) x ( 32102.23 ) = 35054.27 lb for each bed

Bed Design and the Pressure Drop Confirmation

Pressure drop in the bed can be calculated as ;

( ΔP / L ) adjusted ( 0.27 psi / ft ) x total bed height ( 20.711) = 5.592 psi

which meets the criteron of not exceeding 5-8 psi / ft

Calculation of Total Heat Required to Desorb Water

P design = 500 x 1.1 = 550 psia

7/28/2019 Homework 6 Pe5623


t ( thickness of vessel wall , inches )= 12 x 7 x 550 / 37600 - ( 1.2 x 550 ) = 1.25 inch

Approximate weight of steel 43207.53 lbs

Q st ( Heat required for steel ) = 43207.53 x ( 0.12 x (500 – 100) ) =2073961.7 Btu

Q w = 1800 x 3881.16 = 6986088 Btu

Q si = 35054.27x ( 0.24 ) x 400 = 3365210 Btu

Q heat loss = ( 6986088 + 3365210 + 2073962 ) x ( 0.10 ) = 1242526 Btu

7/28/2019 Homework 6 Pe5623


Q total req heat = ( 2.5 ) x ( 6986088 + 3365210 + 2073962 + 1242526) = 34169465

Total heat required is about 34.2 million Btu

Calculation of the Flow Rate For the Regeneration Gas

We assume the heating time be 60% of the total regeneration time

60% x 12 hrs = 7.2 hrs of heating

Cp

(Btu/lb/F) = 0.63 Btu/lb/F

mrg = { 34169465 / [ 0.63 ( 550 – 100 ) ( 7.2 ) ] }

mrg = 16739.89 lb/hr

Check that the ∆P/L > 0.01 psi/ft at 550F

Density of regeneration gas = 20.279 x 500 / 10.73 x 0.99 x 1010 = 0.945 lb/ft3

q = 16739.89 / 60 x 0.945 = 295.22 ft3/min of hot regenerative gas

V = 4q / π D2

= ( 4 x 295.22 ) / ( 3.414 ) x ( 72

) = 7.059 ft/min

µ = 0.023 cP from Fig 23-23 given above

7/28/2019 Homework 6 Pe5623


7/28/2019 Homework 6 Pe5623


q = 295.22 ft3/min

V = 7.059 ft/min

µ = 0.023 cP

Density of regeneration gas = 0.945 lb/ft3

∆P/L = 0.01328psi/ft which is > 0.01 , so the channeling will not occur

7/28/2019 Homework 6 Pe5623


Design Results

Number of Vessels : Two

Vessel Dimensions : 84 inches, 7 feet

Weight of Molecular Sieve : 2 x 35054.27 = 70108.5 lb

Total heat required is 34.2 million Btu

Approximate Heat Loss in the system : 1242526 Btu

Regeneration Gas Rate : 16739.89 lb/hr

Regeneration Gas Temperature : 550 F

Cycle time : 24 hours

12 hours of adsorbing

12 hours of regenerating ( heating, cooling, standby and valve switching)

Heating time : 7.2 hours

Density of regeneration gas : 0.945 lb/ft 3

Cp of regenerative gas (Btu/lb/F) : 0.63 Btu/lb/F

Approximate weight of steel for two towers : 86415.06 lbs

Thickness of vessel wall , inches : 1.25 inches

Water to be removed : 3881.16 lb

Super velocity of gas : 42.127 ft/min

Weight of wet gas per hour : 167018.62 lb/hr

homework 6 pe5623

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