Solution manual for chemical process equipment design by turton

Copyright c 2017 by Pearson Education, Inc.. Pressure head is a method for expressing pressure in terms of the equivalent height of a fluid, where the pressure head is the pressure at th

Solutions Manual for Chemical Process

Equipment Design

(2017)

by Richard Turton Joseph A Shaeiwitz

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

1

02



first term: enthalpy term

second term: kinetic energy, often written as (v)2)/2

third term: potential energy

fourth term: frictional losses

fifth term: shaft work

2

mass flowrate constant, so, if larger pipe is 1 and smaller pipe is 2

2 2 2 1 1

1A v A v

m    , where A is cross sectional area for flow, if density constant,

then

2 2 1

1v A v

A  , so if area goes down, velocity must go up in pipe 2

3

Pressure head is a method for expressing pressure in terms of the equivalent height

of a fluid, where the pressure head is the pressure at the bottom of a that height

of that fluid From the mechanical energy balance, the value is obtained by dividing

1-2

5

v Av

m  

mass in = mass out, so if single pipe, mass flowrate must remain constant

if density remains constant, volumetric flowrate remains constant

if density and area remain constant, velocity must remain constant

6

forcesviscous

forcesinertial

Re 

inertial forces keep fluid flowing, viscous forces resist fluid flow

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

all of the following have similar shape

f

laminar flow

flow in packed bed

1-4

pipe length: linear relationship, as pipe gets longer, friction increases proportionally

velocity in pipe (or flowrate): in turbulent flow, friction increases as square of velocity or flowrate, linear in laminar flow

pipe diameter: strong inverse relationship; in turbulent flow, friction goes with d-5;



for incompressible flow, density constant, so first term is simplified

02

u P



12

frictional drag due to “skin friction,” which is contact with solid object/surface

form drag is from energy loss due to flow around object (fluid changing direction takes energy)

13

in a packed bed,

void fraction = volume of bed not occupied by solid (void space)/total volume of bed

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

14

total volume is volume of bed if empty

solid volume is total volume of solids in bed

void volume is volume of bed not occupied by solids

total volume = solid volume + void volume

P manometer fluid  fluid flowing 

 (  ) , where h = is the difference in manometer fluid

height

So, for a large pressure drop, the height difference (and hence the height of the manometer required) decreases if the manometer fluid is dense, like mercury However, for very small pressure drops, accuracy is lost due to the small height difference in a mercury manometer, so a less dense manometer fluid is better

17

That is the flowrate at which the available net positive suction head equals the required net positive suction head For higher flowrates, the fluid will vaporize upon entering the pump, causing cavitation, which damages the pump However, it is physically possible to operate at higher flowrates

Energy in compression is minimized with isothermal compression, which is not possible, since compressing a gas causes the temperature to increase Isothermal operation could be approached with an infinite number of compression/intercooling stages with infinitesimal pressure and temperature increases, which is a nice limiting case, but impossible Staging compressors with intercooling is an attempt to approach the limiting case in a practical way The economics of a process determines the number of stages to use

There is also the problem that if the temperature in a compressor stage increases too much, the seals will get damaged, which is a good reason the keep the compression ratio low

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

For fully developed turbulent flow, it is assume that the friction factor has reached its asymptotic value The proportionalities are

5 2

D

v L

5

D

v  , so, for constant pressure drop, the flowrate increase significantly

d it is exactly a proportional increase

e v  L 0 5, so the flowrate decreases, but it is a one-half-power decrease, so the decrease in flowrate is less than the increase in length

f subscript 1 is for the original, long segment; subscript 2 is for the shorter parallel segments; take the ratio of pressure drops in both cases, noting that the diameters are constant, that the length of pipe 2 is half of pipe 1, and that the flowrates in each pipe are one-half of the original, so the pressure drop goes down by a factor of eight; note that minor losses due to the parallel piping are neglected

125.0)5.0(5

5 2

5 1 2 1

2 2 1 2 1

v L

L P

P





d it is exactly a proportional increase

e v  L 1, so the flowrate decreases in proportion to the increase in length

f subscript 1 is for the original, long segment; subscript 2 is for the shorter parallel segments; take the ratio of pressure drops in both cases, noting that the diameters are constant, that the length of pipe 2 is half of pipe 1, and that the flowrates in each pipe are one-half of the original, so the pressure drop goes down by a factor of four; note that minor losses due to the parallel piping are neglected

25.0)5.0(5.0

4 2

4 1 1 2 1 2 1

v L

L P

Av v

v Av m

/sm00706.0

/sm00706.0kg/m850

kg/s6

2 4 3

1

1 1

3 3

1 1

m v

1063.79

/sm0106.0

kg/s9.01/s)

m0106.0)(

kg/m850(

2 4 - 3

2

2 2

3 3

2 2

v m

m00529.0)(

kg/m850(

m/s00529.0)m10m/s)(13.13032

.4(

3 3

3 3

2 -4 3

3 3

A v v

4

4 3 2 1

/sm0124.0

/sm0124.0kg/m850

kg/s51.10

2 4 3

4

4 4

3 3

4 4

m v









m1013.13(kg/m1000

kg/s2.79m/s)

5)(

m10574.5(kg/m1000

2 4 3

2 2 2

2 4 3

1 1 1

v A m

4 3

4 3 2 1

m m m m

kg/s73.6

2 4 3





c

2 4 3

3

3

m/s)18.2(kg/m1000

kg/s73

2

i i

(0.02ft63

ft0.0491-

ft/sec)4(ft0.0218-

ft/sec)8(ft0.0873/sec

ft39.0

3

3 2 2

2 3

z g v

assume inlet and outlet at same pressure since no information provided

uniform pipe diameter, so kinetic energy term zero

pipe length not needed, since frictional loss given

gal/min487

)gal/ft487sec/min)(

60(lb/ft62.4

lb/sec7.67

lb/sec7.67

0sec))(hp/lb

ft hp)(5500.8(20

/lblb

-ft 80ft)50()seclb/(lb

ft 32.2

ft/sec2.32

3 3

f f

2 f 2

1-12

02

inlet and discharge both at atmospheric pressure, so pressure term is zero

inlet velocity is zero, since reservoir is like tank, so level is assumed constant

multiply 20 m of head by g to get frictional loss in correct units

M W84.7 W1084.7s

mkg1084.7

0)kg/m/s)(1000m

25(80.0)m/sm(9.8110

m)50(m/s81.92

0m/s)1

(

6 3

2 6

3 3

2 2

z g v

velocities both zero since tank levels

m6.7

0)kg/ms)(1000h/3600

/h(1m10

J/s)0.75(100-

J/kg3.5m) 10)(

m/s81.9

frictional loss assumed zero, since nothing stated

location 1 is reservoir, location 2 is discharge

hp3723

0lb/sec)65000(55.0

sec))/(hplb

ft 550()/seclb/lb

ft 32.2(2

sec/ft 13.26ft)

60(/seclb/lb

ft 32.2

ft/sec32.2

ft/sec26.134

ft)(10lb/ft

62.4

lb/sec65000

f 2

f

2 2 2

2 f 2

2 3

/sm10305.5kg/m

1000

N/m34500-3168

m/s10305.54

m)1.0(

mL)10/s)(m/60mL/min(min250

2 2 2 4 2

2 3

2

4 2

6 3 1



1-14

02

z g v

velocities both zero since tank levels

density of water in lb/gal = (62.4 lb/ft3)(ft3/7.48 gal) =8.3 lb/gal

a

hp37.1

0lb/gal)sec)(8.3in/60

gal/min)(m30

(

sec))(hp/lb

ft hp)(5500.8(

/lblb

-ft 200ft)928873()seclb/(lb

ft 32.2

ft/sec2

f 2

f 2

928-zft200

0/lblb

ft 200ft)()seclb/(lb

ft 32.2

ft/sec2.32

f 2

f 2

900

N/m)1000)(

6.1156.762(

kg/s0.7

2 4 - in





1-16

calculate velocity in 1-in pipe

2 2 2

1 2 2

2 4 - in

1

s/m62

pumpacross

m/s946.0)m10kg/s(82.19900

kg/s0.7

A

m v

J/s)1000(kg/m

1000

N/m)1000)(

5.346.551(

0

3 3

m

W P

z g v

1-18

02

z g v

0)lb/ft4.62(/secft5

sechp

lb

ft 55075

.02

secft 366.646.25)lb/ft0.88(62.4

lb/lb

ft 2381.2

ft/sec46.254

ft)5.0(

sec/ft5 ft/sec366.64

ft)1(

sec/ft5

lb/lb

ft 2381.2ft)

1012()seclb/(lb

ft 32.2

ft/sec2.32)

lb/ft4.62)(

88.06.13(

3 3

f 2

2 2 2

3 f

2 3 2

2 3 1

f 2

f

2 3

max height is z when pump is at 35.1 hp

ft8.79

0)lb/ft4.62(/secft5

sechp

lb

ft 550hp)1.35(75.0)seclb/(lb

ft 32.2

sec/ft2.32)

/ftin144()lb/ft0.88(62.4

ft/25)lb-(14.7

3 3

f

2 f

2 2 2

2 3

2 f

source tank and subtract 2 ft for destination tank, so answer is 77.8 ft

1-20

a

m/s27.1)m10(63.79kg/m

800

kg/s5.6

2 4 - 3





b

MEB from source tank level to pump inlet

first need velocity in suction line

m/s989.0)m10(82.19kg/m

800

kg/s5.6

2 4 - 3





kPa8.164

0J/kg30)m5(m/s81.92

m/s)989.0(kg/m

800

N/m000,150

0)

(2

3

2 2

3

2 3

1 3

2 3 1 3

0kg/s5.6

J/s)1500(7.0J/kg80)m(m/s81.92

m/s)27.1(kg/m

800

N/m000,150000,200

0)

(2

2 2

3 2

1 4

2 4 1 4

W e

z z g v P

so, height above ground = 6.86 m

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

2 2 1

2 1

1 2

liquidsinceconstant density

)(

v dt

dh A

A

A v dt

dh A

v A dt

h A d

h A h A V m

v A m

m

W e

z g v

f

0)(

02

1

2 1 2 2

1-22

integrate from initial height to any height from time zero to time t

2 5

0 0

0 5 0

5 0 2 1

2

2define

dt a h

dh

ah dt

dh

g A

A a

t h

5 0 0

2



Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

(2

02

4 5

2 4 4 5

2 3 2 4 3 4

4 3 2 1

e z z g v P P

W v

v P P

m m m m

first equation suggests mixing

second equation suggests pump with suction and discharge pipes having different diameters

third equation suggest pressure loss due to friction and height change, with deceleration upon entering tank, since zero velocity at point 5

1-24

1 6

2 5 5 6 5

6

2 4 2 5 4 5

1 6

2 2 1 2 1

2

5 4 2 3

02)(

02

0)

(

02)(

z z

v z z g P P

W v

v P P

W z

z g

v z z g P P

m m m m

s p

s p

first equation suggest mixing

second equation suggests height change with acceleration, probably flow out of tank, since no pump, assume downward

third equation suggests open-air tanks, since no KE term and no pressure terms

fourth equation suggests flow across pump with different pipe diameters

fifth equation suggest flow up to tank

inequality suggests destination tank level higher than source tank level

Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

02

1

2 1 2 2 1 2

m

W e

z g v

assume no friction, no height change, no pump (away from heart)

1 denotes normal artery, 2 denotes balloon region, where diameter is larger

so, if v 2 < v1, since velocity goes down in larger diameter region (from m  Av, at constant density and flowrate, of area goes up, velocity goes down), pressure must

go up, because if velocity term is negative, pressure term must be positive, so they can add to zero

pressure going up makes aneurysm worse, since it would keep expanding

2 2

2 2

Re

81.67

.3log41

equalmust power pump

andsection,pipe

on 02

02

toreduces

M EBdiameter,uniform

andhorizontalassumed

since

02

21

D f

A

v v

v m

P D

fLv P

m

W d

fLv

m

W D

fLv z

g v P

s p

s p

1064.7lb/ft/sec

1072.6

)lb/ft.4ft/sec)(6278

.4(ft12

067.2Re

lb/sec95.6gal48.7

ftsec60

gal50)lb/ft4.62(

ft/sec78.4ft

0.02330

gal48.7

ftsec60

gal50

4 4

3

3 2

2 3

1064.7

81.6in)

3.7(2.067

in0018.0log41

Re

81.67

.3log41

9 0 4 10

9 0 10



Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc.

Full file at https://.

ft 550

sechpsec/lb

ft 2004

seclb

ft 2004

0lb/sec95.6)sec/lb/lb

ft 2.32)(

ft12/067.2(

f f

f

2 f

.2

m/s51.2

00788.0

)kg/m1000(m)0525.0(

kg/m/s)001

.0(81.6m)

0525.0(7.3

m000045

0log41

0m)0525.0(

m)50(2m)10(m/s81.92

zeroassumed

in tank velocity

point,dischargeis

2location level,

tank is1location

uslysimultaneoequations

twosolvemust

81.67

.3log41

02

2

1

work termno

sopump,no

openis tank sourceand

catmospheri

at emergessince

droppressurezero

02

21

3 2

4 - 2

9 0 3 2

10

2 2 2

2 2

9 0 10

2 2

2 2

v f

v

dv D

f

D

fLv z

g v

m

W D

fLv z

g v

44.181

/hp/sec)lb

ft (5507

.0

)seclb/lb

ft ft)(32.212

/25.10(

ft/sec)ft)(5.8310

713.3)(

00776.0(2seclb/lb

ft 32.2

ft)100(ft/sec2.32)lb/ft4.62(87.0

ft/lb)144(25

ft106713.33.1676)5280(70so

ft3.1676)

12/25.10(1(50)0.55(50)

15(9)50(35)

exittank entrance

tank valves

elbowsmethod

length equivalent

00776.0

10057

81.6)25.10(7.3

0018.0log41

10057cP)

lb/ft/sec/

10cP(6.7240

)lb/ft87)(62.4ft/sec)(0

ft)(5.8312

/25.10(Re

lb/sec44.181)lb/ft62.4)(

87.0(gal48.7/ftsec60

gal1500

ft/sec83.5ft

12

25.10

4gal

48.7/ftsec60

gal1500

ninformatiopipe

20-schedulein,

10uplook must

-Re

81.67

.3log41

02

2

methodheads

velocity use

ifor,method,length

equivalentuse

if

02

zero termKEso tank,tank to

02

21

f

2 f

2 5

2 f

2 3

2 f

5

9 0 10

4 -

3

3 3

2 3

9 0 10

2 2

2

2 2

s p i

s p eq

s p

W

W L

m

W v

K D

fLv z

g P

m

W D

v fL z g P

m

W D

fLv z

g v P

the virtuallygive

methodsboth

that observe

hp3438

0lb/sec

44.181

/hp/sec)lb

ft (5507

.0ft/sec)83.5()seclb/lb

ft 32.2(2

)155.0)17.0(15)75.0(50(

)seclb/lb

ft ft)(32.212

/25.10(

ft/sec)ft)(5.835280

)(

70)(

00776.0(2seclb/lb

ft 32.2

ft)100(ft/sec2.32)

lb/ft4.62(87.0

ft/lb)144(25

f 2

2 f

2 f

2 2

f

2 3

2 f