ass 2 and 3 solutions

8/8/2019 ASS 2 and 3 Solutions

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CHEMICAL ENGINEERING TECHNOLOGYCEM4M1C

ASSIGNMENT 2

Question 1

In the case of a runaway temperature in a reactor, cooling water (10oC) flows

from a tank (vented to the atmosphere, 100 kPa) through the reactor to a wastesystem at 50 kPa (a). The piping (L = 200 m) is 150 mm diameter smooth pipecontaining two gate valves, three standard 90o elbows and three couplings.

Assume the pressure loss through the reactor is equivalent to another 450 m ofthis pipe. The difference in elevation from the liquid level in the tank to thedischarge point into the waste system is 40 m. What will be the rate of water flowin an emergency (m3.hr-1)?

(10)

= 103 kg/m3

= 1.0010-3 kg/msgate valves L/D = 7standard 90o elbows L/D = 30couplings L/D = 0

ANSWER:(1) 100 kPa(a)

Tank

40 m

Reactor

(2)Waste 50 kPa(a)

ME balance

f2

ugz

pw

2

ugz

p 222

2s

2

11

1 +++=+++


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assume u1 is negligible and no mechanical work.

( )

( )( )

12

2

2

2

3

3

22

2121

Jkg450f2

u

f2

u1040

10

1050100

f2

ugzz

PP

=+

+=+

+=+

24 ud

Lf = (equation 3.19 from Coulson & Richardson, 6th edition)

1222 Jkg450u

d

L04

2

u.e.i =/+

( )( )

m6.665

m45003037215.0m200L

=

++++=

Assume u such that LHS = RHS

u 2 4

Re 3 x 105

6 x 105

0/ 0.00175 0.00155

dLu04

2u

22

/+ 126.2 448.2

U = 4m.s-1 is close enough

ud=Re

For a smooth pipes 25.0Re0396.0 = (equation 3.11 from Coulson & Richardson,6th edition)

flow rate through reactor = 4 m/s( )

4

15.02

= 0.0706m3.s-1= 254.5m3hr--1


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Question 2

A non-Newtonian Fluid is cooled in a jacketed pipe with cooling water. The non-Newtonian fluid flows through the annulus. Calculate the pressure drop of the

fluid if the jacketed pipe is 30 m long. The flowrate is 10 m3.hr-1. The inside pipeis a 40 mm outside diameter pipe, and the outside pipe is an 80 mm insidediameter. The physical properties of the fluid are:k = 0.208 Pa.sn (fluid consistency index)n = 0.28 (flow behaviour index (-))density = 1685 kg.m-3

(15)

ANSWER:

( )

( )( )

mmm

dDd

dD

dD

perimeterwetted

tionx

d

e

e

04.040

4080

4

4

sec4

22

==

=

=

+

=

=

k = 0.208 n = 0.28

app : apparent viscosity (N.s.m-2)

1

8

=

n

i

appd

uk (3.2.2.4 in Study Guide)

AuQ = A

Qu =

( )1

2

.21.204.0

4

3600

10

=

=

sm

u

Flow characteristic:( ) 1442

04.0

21.288 == sd

u

i


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( )

( )

)/(0026.01060.2

012.0208.0

442208.0

23

128.0

mNs

app

==

=

=

( )( )( )

5729

0026.0

21.204.01685Re

=

==app

n

du

Turbulent flow

Now f =bnRe

a(3.2.3.2.1 in Study Guide)

From Table (3.1 in Study Guide) a = 0.0677 and b = 0.33 for n = 0.28

( )0039.0

5729

0677.033.0== f

22 ud

LfP

i

f = (3.2.3.1 in Study Guide)

24

2u

d

LfP

i

f

=

( ) ( )

kPa

Pa

u

d

LfP

i

f

144.48

)(48144

2

21.21685

04.0

300039.04

24

22

=

=

==

Question 3

Calculate the required pressure at the delivery flange of a pump which feeds anorganic mixture to a distillation column (elevation = 16 m). The feed is pumpedfrom a feed tank that consists of 90 % carbon tetrachloride and a 10 % perchloroethylene. Physical properties given below.

The pressure in the feed tank is 100 kPa (g) while the column is pressurised at585 kPa (a). The proposed pipe outlay is:


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00115.004.0

000046.0==

d

6

3

10757.1

1008.0

212.204.01589Re

=

==

du

From chart = 0.0025

le= 45 + [ (1 ball valve) + (1 gate valve) + (12 90 elbows) ] di

= ( ) ( ) ( )[ ] 04.060127113145 +++ = 74.6 m

g

u

d

lh

i

efd

2

4=

( )( )

81.9

212.2

04.0

6.740025.04

2

=fdh

= 9.30 m

fdd

dd hg

PZh ++=

30.981.91589

1058516

3

+

+=dh

30.953.3716 ++=dh

= 62.83 m

( )aPa

hgP pd5108.9

83.6281.91589

=

==


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Question 4

Water flows at a rate of 10 m3/s in a channel shown in the Figure below. If the

canal is weedy (n=0.03) with a slope of 0.0014, determine the depth of flow.

[10]

40o 40o

ANSWER:

21

03

21SAR

nQ h=

A : area

P

ARh = ( P : wetted perimeter)

A = 2 ( ) +

( )

( )

( )

yyA

yy

A

yy

yA

ybhA

ybhA

66.319.1

66.340tan

66.340tan

66.3

65.32

12

2

2

+=

+=

+

=

+=

+

=

yP

yP

yP

11.366.3

6427.0

266.3

40sin266.3

+=

+=

+=

3.66 m

y


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y

yy

P

ARh

11.366.3

66.319.1 2

+

+==

21

03

21SAR

nQ h=

Q = 10 m3

.s-1

( ) ( ) 213

22

2 0014.011.366.3

66.319.166.319.1

03.0

110

+

++=

y

yyyy

( )( ) ( )( )

( ) 21

32

32

2

33

20014.0

11.366.3

66.319.166.319.133.3310

+

++=

y

yyyy

( )( ) 3

2

35

2

11.366.366.319.12471.110y

y++=

( ) ( ) 011.366.3018.866.319.1 3235

2 =++ yy

( ) ( ) 011.366.351566.319.1 252 =++ yy

by trial and error y = 1.5 m

Question 5

Water flows uniformly in a rectangular channel of width b and depth y.Determine the aspect ratio for the best hydraulic cross-section.

[15]

ANSWER:

Uniform flow

2

1

03

21SARnQ

h=

A = b yp = b + 2y

yb

by

P

ARh

2+==


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in term of A

( ) 22 222

2 yA

Ay

y

y

y

A

A

yy

A

A

yb

ARh

+=

+

=

+

=+

=

21

0

32

22

1S

yA

AyA

nQ

+

=

Rearrange and let

23

21

0

=

S

nQK

Then:

32

22

1

0 2

+

=

yA

AyA

S

nQ

( )

( ) 32

2

32

33

21

0 2yA

AyA

S

nQ

+=

( )

( ) 32

2

32

35

21

0 2yA

yA

S

nQ

+=

( )( ) 2

33

22

23

32

2

3

3

523

21

0 2

+=

yA

yA

S

nQ

( )( )2

252

3

21

02yA

yA

S

nQ

+=

( )( )AyKyA

yA

yAK +=

+= 22

5

2

25

22

best hydraulic section is when A is minimum for all y

0=dy

dA

+=+

dy

dAyKAy

dy

dAA 4

2

52

52

5


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For 0=dy

dA

y

AKKyA

44

25

25

==

( )22

5

2yA

yAK

+=

( )2

25

25

24 yA

yA

y

A

+=

( )

( ) AyyA

yA

yyAA

+=

+=

2

225

2

25

25

2

4

2

4

( ) 225

225

42 yAAyA =+

225

27

225

42 yAAyA =+

22 42 yAy =+

222 224 yyyA ==

21

2

=A

y

byA =

21

2

=

byy

bybyyby

y === 22

2

22

2=y

b

Gives smallest A & smallest wetted perimeter.


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ASSIGNMENT 3

Question 1Water is to be pumped from a dam at the rate of approximately 75 m3.hr-1through a pipeline 1 km in length and internal diameter 100 mm, and delivered atatmospheric pressure to a point 30 m higher than the surface of the dam.Homologous pumps are available for this duty with the impeller sizes of 100 mm,150 mm and 200 mm. Determine:

1.1. Which pump should be selected.1.2. The actual flowrate of water1.3. The maximum distance at which the pump can be situated from the

pipeline inlet without cavitation.(25)

Performance data for the 100 mm impeller pump are given below:

Q (m3.hr-1 ) Head (m) NPSH (m)

0 30 0.5

5 28 0.75

10 25 1.5

15 17.5 5.0

Suction line: 1 strainer, 1 check valve, 2 X 90o bendsPump is 3 m above the dam water level.

Discharge line: 2 globe valves, 5 X 90o bendsPipe roughness = 0.1 mmEquivalent lengths: strainer = 130 dCheck valve = 50 d

Assume all pumps run at same speedViscosity of water = 1 mPa.sVapour pressure of water = 3 kPa (abs)

Atmospheric pressure = 101.3 kPa.

ANSWER:

1.1.Use equations for centrifugal pump relationships:For 200 mm impeller:For N1 = N2

3

2

1

3

2

1

2

1

2

1

=

=

D

D

D

D

N

N

Q

Q


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( ) 125.05.0200

100 33

2

1 ==

=

Q

Q

125.0

12

QQ =

( ) 25.05.0200

100 222

2

1

2

1 ==

=

=

D

D

h

h

25.025.0 12

2

1 hhh

h ==

( ) 25.05.0200

100 222

2

1

2

11 ==

=

=

D

D

NPSH

NPSH

25.025.0 12

2

1 NPSHNPSHNPSH

NPSH==

For 200mmm impeller For 150mm impeller

Q

hr

m3 h(m)

NPSH(m) Q

hr

m3 h(m)

NPSH(m)

0 120 2 0 67.5 1.13

40 112 3 16.9 63 1.69

80 100 6 33.8 56.3 3.38

120 70 20 50.6 39.4 11.25

For Q = 75m3.hr

-1

( )1

2

s.m65.2

1.0

4.

3600

75u

=

=

( )( )

51065.2

001.0

65.21.01000duRe

=

==

001.01.0

0001.0==

d

e

0024.0=

g

u

dhf

2

4l

=


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le = 1000 + [ (2 globe valve) + ( 5 90 bends) + (strainer) +(check valve) ] di

= 1000 + [ (2 300) + (5 20) + 130 + 50 ] 0.1 = 1088 m

( )( )

m

g

u

d

hf

76.74

81.9

65.2

1.0

10880024.04

4

2

2

=

=

=l

( )

m

hfg

uuz

gPP

h

76.104

76.7430

22

212

12

=

+=

+

++

=

For Q = 80m3.hr-1

65 1085.275

801065.2Re ==

0024.0=

mhf 06.8575

8076.74

2

=

=

mh 06.11506.8530 =+=

Q = 70m3.hr-1 hf= 74.76

2

75

70

= 65.12m

mh 12.9512.6530 =+=

Only the 200mm impeller can provide sufficient head.


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1.2System head:

( )fsfd

sdsd

sd hhg

uuZZ

g

PPh ++

++

=

2

22

( )g

u

d

lZZh sd

2

4+=

( )gA

Qh

1

1.0

10880024.0430

2

+=

( ) 008.01.04

2==

A

( )81.9

1

008.01.0

10880024.0430

2

+=

Qh

231004.1330 Qh +=

From the pump curves and system curves the actual flowrate is 74m3.hr-1

( )

1

2

s.m62.2

1.0

4

3600

74u

=

=

1.3 Corresponding NPSH = 5.4m

fss

vpshZ

g

PPNPSH +

=

(4.1.2.1. in study guide)

( )fsh+

= 3

1081.9

1033.1014.5

3

3

g

u

dhr

h

s

fs

2

462.7

4.5302.10

l==

+=


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( )( )

( )( )

m

u

dgs

20.297

62.20024.04

81.91.062.7

4

62.7

2

2

=

=

=

l

Now fittings account for:

0.1 [ 1 strainer + 1 check valve + 2 90 bends ] =

0.1[130 + 50 + 2 20] = 22m

Max length at suction line = 297.20 - 22= 275.20 m

Assigment 3, Question 1, 2007

0

25

50

75

100

125

150

0 10 20 30 40 50 60 70 80 90 100 110 120 130

Q (m3/h)

Head(m)


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Question 2

A centrifugal pump is used to extract water from a cooling tower operating atatmospheric pressure (101.3 kPa).

At the rated discharge, the net positive suction head must be at least 4 m. If the

frictional head loss in the suction line is 5.5. m, what must be the least height ofthe liquid level in the cooling tower above the pump inlet? The vapour pressure ofthe water is 25 kPa at the operating temperature.

(10)

ANSWER:

NPSHR = 4mhfs = 5.5m

g

P

hNPSH

vp

sA = (4.1.2.2. in study guide)

fsss

s hZg

Ph +=

(4.1.1.1. in study guide)

( )

( )

s

s

s

fss

vps

vp

fsss

A

zm

z

z

hzg

PP

g

Phz

g

PNPSH

+=

+=

+

=

+

=

+=

27.2

5.577.7

5.5

1081.9

10253.1013

3

But NPSHA NPSHR

2.27 + zs 4m

zs 1.72m

Question 3

A vacuum system is required to handle 10 g.s-1 of vapour (MW = 56 kg.kmol-1) soas to maintain a pressure of 1.5 kPa in a vessel situated 30 m from the vacuum

pump. If the pump is able to maintain a pressure of 0.15 kPa at its suction point,what diameter pipe is required? The temperature is 290 K, and the isothermalconditions may be assumed in the pipe, whose surface can be taken as smooth.The ideal gas law is followed.

gas = 0.01 mPa.s(15)


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Isothermal Flow

)1(042

ln

2

11

2

1

2

2

2

1

2

Kl

=

+

+

A

G

dVP

PP

P

P

A

G (4.57) in Coulson &

Richardson, 6th Edition

1

1

11

kg.kJ05.43

kg.kJ56

290314.8

MM

RTVP

=

==

Since A, , d are unknown

Assume d = 0.1m

( ) 232 m1085.74

1.0A ==

==

A

Gd

d

G

4Re

( )

12740

1085.7

.1010

1001.0

1.0Re

3

13

3

=

=

=

skg

A

Gd

From friction chart

0035.0=

Substitute in equation (1)

( )( )

( )( )

01085.7

1010

1.0

300035.04

1005.432

105.115.0

15.0

5.1ln

1085.7

10102

3

3

3

6222

3

3

=

+

+

diameter too large

Assume d = 0.08

( ) 232 m10027.54

08.0A ==

( )

16000

10027.5

1010

1001.0

08.0Re

3

3

3

=

=

From chart = 0.00325 for smooth pipe.


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( )( )( )

( )

083.2

5.1987.253.22

10027.5

1010

08.0

3000325.04

1005.432

105.115.0

15.0

5.1ln

10027.5

1010

2

2

3

3

3

6222

3

3

=

+=

+

+

diameter of pipe is 0.08m = 80mm

Question 4

Determine the conditions (i.e. volumetric flowrate, density and pressure) at thesuction flange of a vacuum pump. The vacuum pump must maintain the pressure

in a stripping column at 5 kPa (a).

The 150 mm ID alloy pipeline between the vacuum pump and the column is 50 mlong and contains five 90o elbows and one ball valve. The absolute roughness (e)of the pipeline is 0.0003 m. The stripped gas is chlorine with a flowrate of 200kg.hr-1 and a temperature of 50oC. Assume isothermal flow in the pipeline. MWof Chlorine is 72 and Cp/Cv=1.3.

(20)

ANSWER:

( )

tablesfromcp015.0

m0177.04

15.0A

s.kg056.0hr.kg200m

2

2

11

=

==

==

31638

10015.0

15.0

0177.0

056.0d

A

GRe

3

=

=

=

002.015.0

0003.0de ==

From chart = 0.0034

( )

m95.96

95.4650

15.0131605m50L

=

+=

++=


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Assuming isothermal flow and kinetic term is negligible

0A

G

d4

vP2

PP2

11

2

1

2

2 =

+

l

MM

RTvPwhere

A

G

d4vP2PP

111

2

11

2

1

2

2

=

= l

( ) ( )( ) ( ) ( )( )

kPaPaP

P

62.51062.51056.31

1056.61025

0177.0

056.0

15.0

95.960034.04

72

32383142105

3

2

6

66

2

2232

2

===

+=

+=

From

( )

3

3

3

2

2

2

2

22

.68.150

323314.810

721062.5

1

=

=

=

=

=

=

mkg

RT

MMP

v

MMP

RTv

MM

RTvP

131

3

.1072.3

1068.150

056.0

=

==

sm

mflowrateVolumetric

&


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Question 4 cont

In a synthetic ammonia plant, the hydrogen is fed from a reservoir at 2 MNm-2pressure through a clean horizontal mild steel pipe 50 mm in diameter and 500 mlong to the synthesis converters. The hydrogen is required at a pressure of 2.5MNm-2and the pressure is raised to 3 MNm-2by a single stage compressor at

the upstream end of the pipe. The index of compression is 1.45 and thetemperature of the gas at the exit of the compressor is 320 K. If the flow of gas inthe pipeline is isothermal, determine:

4.1 the mass flowrate of the hydrogen

4.2. the indicated power of the hydrogen.

e/d = 0.001R = 8314 Jkmol-1K-1

(20)

ANSWER:

4.1 Flowrate unknown, need to assume and findA

G

2MPa

1 2

P3 = 2.5MpaT3

P1 P2 = 3MPaT1 T2 = 320K

From tables of H2 at 320K = 0.009mPa.s

Since flow is isothermal

0A

G

d4vP2

PP

P

P

lnA

G2

22

22

23

3

2

2

=

+

+

l

where P2v2 = RT2 / 2

Assume 0025.0=


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( )[ ]

2152

2

12222

57.10118.100

210168.5

0

22

1035.2

05.0

5000025.04

5.2

3

=

=

=

+

+

mkgsA

G

RTn

A

Gl

Check

001.0d

e

106.5d

A

GRe 5

=

==

From chart = 0024.0 assumed

Flowrate = 101.57kg.m-2.s-1

m& = 0.199kg.s-1

4.2

=

1P

P

M

RTm

1n

znip

zn

1n

1

21&

K16.282

2

3

320

P

P

TT

PPTT

45.1

45.0

n

1n

1

2

21

n

1n

1

2

1

2

=

=

=

=

( )( )( )

kW4.101

12

3

2

16.2828314199.0

45.0

45.1ip

45.1

45.0

=

=


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Question 5

A monolithic concrete inverted siphon on the Tugela River aqueduct is circular incross section and is 4.88 m in diameter. It is completely filled with water.

5.1. If n = 0.013, find the slope of the hydraulic grade line for a flow of 45 m3.s-

1.(5)

5.2. Solve the same problem using the friction factor charts, and assuming amean value of for concrete pipes. Compare your result with that of part5.1.

(10)

ANSWER:

5.1

( )

2

2

2

m7.18

88.44

D4A

=

=

=

m22.14

88.44

DRh ===

( )

km/m75.0or00075.0S

S22.1013.0

7.1845

SAR

n

1Q

o

2

1

o3

2

2

1

o3

2

h

=

=

=

5.2 Mean value of e for concrete

mm65.12

0.33.0=

+

000338.04880

65.1d

e ==

C15at10139.1

s.m41.27.18

45A

Qu

6

1

=

===


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CEM4M1C-102-0-2010

( )( )

7

6

10033.1

10139.1

41.288.4

DuRe

=

=

=

From chart f = 0.0153

( )

000924.0

81.92

41.2

88.4

10153.0

2

2

2

=

=

=

R

h

g

u

dfh

L

L

l

within 23% of value found in 5.1

ass 2 and 3 solutions

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