Question 1

Given:

Upstream Pressure, P1 = 300 psia Downstream Pressure, P2 = 30 psia Flow rate, Q= 55000 ft3/hour Length, X= 4000m Temperature =150℃ Viscosity of Methane, μ = 0.02 cP

Unit conversion

Upstream Pressure , P1=300 psia × 6.893× 103 Pa1 psia

=2.0679 ×106 Pa

DownstreamPressure , P2=30 psia × 6.893× 103 Pa1 psia

=2.0679 ×105 Pa

VolumetricFlowrate ,Q=55000 ft3

hour× 0.02832m3

ft3 × 1hour3600 s

=0.433 m3

s

Length , X=4 km× 1000 m1km

=4000 m

Viscosity of Met h ane , µ=0.02 cP=0.02×10−3 Pa . s

Molecular weight (C H 4 )=16g/gmol

Temperature =150 + 273 = 423K

Assumption:

(a) Commercial steel pipe, ε=0.046 mm

(b) Constant properties of the pipe roughness

(c) Natural gas (CH4) is ideal gas

(d) Consistent pressure drop along the pipe

(e) No leakage

(f) Non-insulated steel pipeline

(g) Isothermal flow.

Analysis:

To calculate density of methane:

ρ= PMRT

=2.0679× 106× 16 ×10−3

8.314 × 423=9.408 kg/¿m3

To calculate pressure drop:

∆ P=P1−P2=2.0679 ×106 Pa−2.0679 ×105 Pa=1.86111× 106 Pa

6 inch

( i ) For D1=6∈× 0.02541∈¿=0.1524 m :¿

To calculate the mass flux in the pipe, G:

G = ρv

V =0.433

π (0.1524)2

4 = 23.74 m/s

G = 9.408×23.74 = 223.346 kg

m2 . s

To calculate Reynolds number,ℜ:

ℜ= ρVDµ

¿ 9.408×23.74 × 0.15240.02 ×10−3 ¿1.701895 ×106 (Thus flow is turbulent)

εD

= 0.046 mm( 0.1524 ×103 ) mm

¿0.0003

From Moody Chart, f = 0.016

To calculate the pressure drop:

Assuming isothermal flow and the kinetic energy is small and thus can be neglected. Rearranging gives:(P1

2−P22 )= f G2 X

DP1

ρ1

∆ P=√0.016× 223.3462 × 40000.1524

× 2.0679 ×106

9.408

∆ P=2145813 = 2.145813 ×106 Pa

∴However ∆ P ≠ ∆ Pactual . The pressure drop obtained for this particular diameter is too high compared to the actual pressure drop which is1.86111× 106 Pa , therefore the diameter 6 inch is not applicable in this case.10 inch( ii ) For D 2=10∈.× 0.0254

1∈¿=0.254 m :¿

G = 223.346 kg

m2. s

To calculate the pressure drop:

Assuming isothermal flow and the kinetic energy is small and thus can be neglected. Rearranging gives:(P1

2−P22 )= f G2 X

DP1

ρ1

∆ P=√0.016× 223.3462 × 40000.254

× 2.0679 ×106

9.408

∆ P=1662139 Pa

= 1.662139 Pa ×106 Pa

∴However ∆ P ≠ ∆ Pactual . The pressure drop obtained for this particular diameter is too low compared to the actual pressure drop which is1 .86111 ×106 Pa , therefore the diameter 10 inch is not applicable in this case.

Since both 6 inch and 10 inch diameter of pipe are not suitable, trial and error method is

used. The appropriate diameter should be in the range of 6 in. to 10 in.

8 inch

( iii ) For D 3=10∈.× 0.02541∈¿=0.2032m :¿

G = 223.346 kg

m2. s

To calculate the pressure drop:

Assuming isothermal flow and the kinetic energy is small and thus can be neglected.

Rearranging gives:

(P12−P2

2 )=f G2 XD

P1

ρ1

∆ P=√0.016× 223.3462 × 40000.2032

× 2.0679× 106

9.408

∆ P=1858328 Pa

= 1.858328 ×106 Pa

∴However ∆ P ≈ ∆ Pactual .The pressure drop obtained for this particular diameter is close to the

actual pressure drop which is1.86111× 106 Pa , therefore the diameter 8 inch is the most

appropriate pipe diameter for this case

Question 2

Given:

Standard Pressure,Pstd=1atm Volumetric flowrate,Q=600000m3/day Downstream Pressure,P2=660kPa Length, X=500m Diameter=300mm Molecular Weight (C3H8) = 44g/gmol Temperature Line ,T1 = 22˚C Standard Temperature, T=15oC Viscosity, µ = 9×10−6Pa.S

Unit conversion

Volumetric flowrate , Q=600000 m3

day× 1day

24 hours× 1 hour

3600 s=6.94 m3 /s

Temperature Line,T 1 =22oC+273 =295 K

Standard Temperature, T =15oC+273=288 K

P = 1 atm = 101.3 kPa

Length, X= 0.5 km

Diameter, D = 300 mm ×1m

1000 mm= 0.3 m

Assumption:

a) Standard condition properties: Pstd=1 atm=1.01× 105 Pa,T std=15℃=288 K

b) Wrought iron , ε=0.046 mm

c) Isothermal flow

d) Kinetic energy is negligible

e) Ideal gas

f) Constant pipe properties

Analysis:

(a) To calculate pressure 0.5 km upstream for 300 mm diameter pipeline with downstream pressure of 660 kPa.

To calculate volumetric flow rate at downstream, Q2:

P2V 2

RT 1=

P stand VR T stand

Velocity , V=QA

P2

Q2

ART 1

=Pstand

QA

R T stand

Q2=Pstand

T stand×

T1

P2×Q=1.01 ×105

288× 295

660 ×103 ×6.94=1.088 m3/s

To find downstream density:

ρ2=P2 MR T 1

=(660000 )(44×10−3)

8.314 ×295=11.84 kg

m3

To calculate the velocity of propane in pipeline:

v=Q2

A= 1.088

π× 0.32

4

=15.39 m /s

ℜ=ρ2 vD

μ=11.84 ×15.39 ×0.3

9×10−6

¿6.07 ×106 (turbulent)

εD

=0.046300

=0.00015

From Moody Chart, f =0.013

According to the assumption made, isothermal flow equation is applied to find pressure drop.

G= ρ2 v

¿11.84× 15.39

¿182.22 kg/m2 s

Assuming that the kinetic energy is small and thus can be neglected. Rearranging gives:

(P12−P2

2 )=f G2 XD

P1

ρ1

P1

ρ1=RT

M = 55741.6

(P12−(660000)2 )=0.013× 182.222× 500

0.3×

P1

ρ1

¿4.01×1010

P12=4.01×1010+(660000)2

P1=¿ 689.711×103 Pa

(b) Calculate the pipe diameter required if the upstream pressure is to be no less than 850 kPa.

P1 = 850 kPa

To find upstream density:

ρ1=P1 MR T 1

=(850000 )(44 ×10−3)

8.314 ×295=15.25 kg

m3

To find the diameter:

(P12−P2

2 )= f G2 XD

P1

ρ1

(850,0002−660,0002 )=0.013 ×182.222 × 500D

× 850 ×103

15.25

2.869 ×1011=1.835 × 1011

15.25D

D=¿0.0419m

Comment: For the isothermal flow equation, (P12−P2

2 ) is inversely proportional to pipe diameter, D. Therefore, therefore, D must be greater than 23.84m when downstream pressure, P1 is not less than 850 kPa.

(c)What downstream pressure is required to handle 600000m3/day in the 300 mm diameter pipeline if the upstream pressure is to be 850 kPa.

P1 = 850 kPa

(P12−P2

2 )= f G2 XD

P1

ρ1

(850,0002−P22 )=0.013× 182.222× 500

0.3× 850 ×103

15.25

P22= 8500002- 4.01×1010

¿6.824 ×1011 Pa

Question 3

Schematic Diagram:

Feed Information

CO2 (g)

T: 120ºC

P: 6 atm

V: 35 m/s Pipe Details

Material: Steel

Internal Diameter: 12 mm

Other given details:

For CO2

γ = 1.33

µ = 0.019 cP = 0.019×10-3 Pa.s

Molecular Weight = 44

Assumptions

From the question

Adiabatic Flow

Commercial Steel Pipe, ε=0.046 mm

Other Assumptions

Constant pressure drop along the pipe

Ideal gas law applied

Chocked flow

Find : a) Maximum possible length of pipe

b) Pressure and stagnation temperature of the gas at the end of the pipe at

maximum length

Calculations (Part a):

Relative Pipe Roughness:

εD

=0.046 mm12mm

=0.0038 33

Density of CO2,

ρ=PMRT

= 607950 × 0.0448.314 × 1000 ×393.15

=8.184 kg/m3

ℜ= ρvDμ

=8.184 × 35× 0.0120.019 ×10−3 =180909(Turbulent Flow)

From Moody Chart, ∴ f =0.028

Acoustic velocity for CO2,

a=√ γRTM

=√ 1.33× 8.314 ×393.150.044

=314.3m / s

Ma1=va= 35

314.3=0.1114

Hence, x=xmax

f xmax

D=1

γ ( 1Ma1

2−1)− γ +12 γ

ln [1+

1Ma1

2 −1

1+ γ−12

]¿ 1

1.33 ( 10.1114 2−1)− 1.33+1

2(1.33)ln [1+

10.1114

−1

1+1.33−12 ]

¿58.03

xmax=58.03 × D

f=58.03 ×0.012

0.028=24.87 m

Calculations (b):

P1=607950 Pa

T 1=393.15 K

Ma1=0.1114

Ma2=1

Pressure at maximum length:

P1

P2=

Ma2

Ma1 √ 1+γ−1

2 Ma22

1+ γ−12

Ma12

P1

P2=

10.1114 √1+

1.33−12

¿¿¿

607950P2

=9.679

P2=62811

Temperature at maximum length:T1

T2=

1+ γ−12

Ma22

1+ γ−12

Ma12

T1

T2=1+ 1.33−1

2¿¿

393.15T 2

=1.163

T 2=393.151.163

=338.04 K

Question 4

Given information:

Material Flow Properties Air Water

Q, Volumetric Flowrate (m3/hr) 5 1.5

Temperature (°C) 30

Pressure (kPa) 200

Pipe Properties

Diameter (m) 0.02

Length (m) 5

Assumptions

From the question

Isothermal

Commercial Steel Pipe, ε=0.046 mm

15% from the initial values density for both phases along the pipe

Assume Separated Flow model

Other Assumptions

Constant pressure drop along the pipe

Ideal gas law applied

Dynamic viscosity is independent from the pressure

Find : a) Mass flux for both phases and determine the flow pattern

b) Pressure gradient for liquid and gas phase

Analysis (a):

Initial values of the density for the air and water phase are:

ρg is calculated using the equation below

ρg=PMRT

=200000× 0.0298.314 ×303.15

=2.301 kg /m3

Water density at 30°C is 966 kg /m3

From the question, it is understood that the value of the density dropped 15% along the pipe

Hence, the new values for liquid and gas density are:

ρg=0.85 ×2.301=1.956 kg/m3

ρl=0.85× 966=846.6 kg /m3

Conversion of Volumetric Flow rate:

Air:

5 m3

h | 1 h3600 s

= 1720

m3 s−1

Water:

1.5 m3

h | 1h3600 s

= 12400

m3 s−1

Mass flow rate

mg=Qg × ρ= 1720

×1.956=2.717 ×10−3 kg /s

ml=Ql × ρ= 12400

×846.6=0.3528 kg /s

Mass Flux

Gg=mg

A=

mg

π D 2

4

=4 × 2.717× 10−3

π× 0.022 =8.648 kg s−1m−2

Gl=ml

A=

ml

π D 2

4

=4× 0.3528π × 0.022 =1123kg s−1m−2

The flow pattern of the flow is determined through Chart 1.

y−axis=Gg

2

ρg=8.6482

1.956=38.23

x−axis=G l

2

ρl= 11232

846.6=1490

Hence, we can deduce that the flow pattern of water and air flow is annular from the graph.

Chart 1

Analysis (b):

Velocity of the flow

vg=Qg

A=

Qg

π D2

4

=4 × 1

720π ×0.022 =4.421 m /s

vl=Ql

A=

Ql

π D2

4

=4 × 1

2400π× 0.022 =1.326 m / s

Using the values above, dynamic viscosity is determined to find the Reynolds number value for

both phases.

ℜg=Gg D

μg=8.648× 0.02

1.983 x 10−5 =8722(Turbulent )

ℜl=Gl D

μl= 1123× 0.02

0.798 x 10− 4 =28145(Turbulent )

Friction factor of the gas and liquid values were retrieved from the moody chart using the

relative roughness.

εD

=0.046 mm20mm

=0.0023

f g=0.035∧f l=0.030 from the moody chart

Pressure drop per length calculated for both phases using the following equation

∆ Pg

X=

f G2g

2 D ρg= 0.035 ×8.6482

2× 0.02×1.956=33.45 Pa/m

∆ Pl

X=

f G2l

2 D ρl= 0.03× 11232

2× 0.02× 846.6=1117 Pa /m

The above calculated values were used to calculate Lockhart-Martenelli parameter, X ;

X=√ ∆ Pl

∆ Pg=¿ 5.779 ≈ 6

Y g & Y L is determined from the Lockhart-Martnelli graph where it uses the turbulent-turbulent

curve values and X parameter value.

Hence, ϕ2g and ϕ2

l values were 13 and 2 respectively.

Pressure gradient for both phases is calculated using the equation below

( dpdx )

gl=( dp

dx )g=ϕ2

g( dpdx )

g=132 ×33.45=5653 Pa /m

( dpdx )

gl=( dp

dx )l=ϕ2

l(dpdx )

l=¿ 22× 1117.23=4460 Pa/m

Question 5Given:

D = 100 × 10-3 m

mliq=20 kg/ s

mgas=0.5 kg /s

T = 100˚C

P1 = 400 kPa

P2 = 130 kPa

ρliq=1220 kg /m3

μliq=0.5 ×10−3 Pa s

μgas=1.8 × 10−5 Pa s

Upstream:

ρgas=PMRT

¿ 400 ×103 × 44 ×10−3

8.314 × (100+273 )

¿5.6754 kg /m3

A=π D2

4

¿ π (100 ×10−3 )2

4

¿2.5 ×10−3 π m2

G=mgas

A

¿ 0.52.5× 10−3 π

¿63.662 kg /m2 s

L=mliq

A

¿ 202.5× 10−3 π

¿2546.48 kg /m2 s

ℜg=GDμgas

¿ 63.662×100 × 10−3

1.83× 10−5

¿347879.78

ℜl=LDμ liq

¿ 2546.48× 100× 10−3

0.5× 10−3

¿509295.82

Assumption: Commercial steel pipe is used,

ε = 0.046 × 10-3 m.

For gas:

f = 0.02

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

347879.78√0.02 ]f =0.0177

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

347879.78√0.0177 ]f =¿0.0178

For liquid:

f = 0.02

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

509295.82√0.02 ]f =0.01733

1√ f

=−2 log10[ 0.046× 10−3

3.7 × 100× 10−3 + 2.51509295.82√0.01733 ]

f =0.0174

∆ Pg

x= f G2

2 Dρ

¿ 0.0178× 63.6622

2× 100× 10−3 ×5.6754

¿63.55 Pam

∆ Pl

x= f L2

2 Dρ

¿ 0.0174 × 2546.482

2× 100× 10−3 ×1220

¿462.424 Pam

X=√ ∆ Pl /x∆ Pg/ x

¿√ 462.42463.

¿2.6975

Chisholm equation:

∅ l2=1+ 20

2.6975+ 1

2.69752

¿8.6015

∆ Pgl=∅ l2 ∆ Pl

¿8.6015 × 462.424

¿3977.5 Pam

Downstream:

ρgas=PMRT

¿ 130× 103× 44 × 10−3

8.314 × (100+273 )

¿1.8445 kg /m3

Assume mass flux G and L do not change, Re

and f remain constant

∆ Pg

x= 0.0178× 63.6622

2×100 ×10−3 ×1.8445

¿195.556 Pam

X=√ ∆ Pl/x∆ Pg/ x

¿√ 462.424195.556

¿1.538

Chisholm equation:

∅ l2=1+ 20

1.538+ 1

1.5382

¿14.427

∆ Pgl

x=∅ l

2 ∆ P l

¿14. .427× 462.424

¿6671.23 Pam

Average pressure gradient=6454.31+3832.762

¿5143.54 Pam

Distance=( 400−130 ) ×103

5143.54

¿49.87 m

∴Yes, it is consistent.

Question 6

Given:

D = 100 × 10-3 m

x = 50 m

mliq=25kg/ s

mgas=1.5 kg /s

T = 100˚C

P2 = 145 kPa

ρliq=1220 kg /m3

μliq=0.5 ×10−3 Pa s

Assume μgas=0.018× 10−3

G=mgas

A

¿ 1.52.5× 10−3 π

¿190.98 kg /m2 s

L=mliq

A

¿ 252.5× 10−3 π

¿3183.09 kg /m2 s

ℜg=GDμgas

¿ 190.98× 100× 10−3

0.018× 10−3

¿ 1061000

ℜl=LDμ liq

¿ 3183.09× 100× 10−3

0.5× 10−3

¿ 636618

Assumption: Commercial steel pipe is used, ε = 0.046 × 10-3 m.

For gas:

f = 0.02

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

1061000√0.02 ]f =0.01734

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

1061000√0.01734 ]f =0.0168

For liquid:

f = 0.02

1√ f

=−2 log10 [ 0.046× 10−3

3.7 × 100× 10−3 +2.51

636618√0.02 ]f =0.0166

1√ f

=−2 log10[ 0.046× 10−3

3.7 × 100× 10−3 + 2.51636618√0.0166 ]

f =0.0166

Downstream:

ρgas=PMRT

¿ 130× 103× 44 × 10−3

8.314 × (100+273 )

¿1.844 kg /m3

∆ Pg

x= f G2

2 Dρ

¿ 0.0168× 190.982

2× 100× 10−3 ×1.844

¿1661.48 Pam

∆ Pl

x= f L2

2 Dρ

¿ 0.0166 × 3183.092

2× 100× 10−3 ×1220

¿689.31 Pam

∆ Pg = 1661.48 × 50

= 83074 Pa

∆ Pl = 689.31 × 50

= 34465.5 Pa

X=√ ∆ Pl /x∆ Pg/ x

¿√ 34465.583074

¿0.6441

Chisholm equation:

∅ l2=1+ 20

0.6441+ 1

0.64412

¿34.46

∆ Pgl=∅ l2 ∆ Pl

¿34.46 ×34465.5

¿1187732 Pam

P1−P2=1187.732 kPa

P1=1187.732+130

¿1317.732 kPa

Trial 1:

Average Pressure=P1+P2

2

¿ 1317.732+1302

¿723.87 kPa

ρ=723.87 ×103× 44 ×10−3

8.314 × 373

¿10.26 kg /m3

∆ Pg=f G2 x2 Dρ

¿ 0.0168 ×190.982×502× 100× 10−3 ×10.26

¿14930.62 Pa

X=√ ∆ Pl /x∆ Pg/ x

¿√ 34465.514930.62

¿1.519

∅ l2=1+ 20

1.519+ 1

1.5192

¿14.599

∆ Pgl=14.599 ×34465.5=503.162 kPa

P1−P2=503.162

P1=503.162+130

¿633.162 kPa

Trial 2:

Average Pressure=P1+P2

2

¿ 633.162+1302

¿381.581 kPa

ρ=381.581 ×103 × 44 ×10−3

8.314 × 373

¿5.408 kg /m3

∆ Pg=f G2 x2 Dρ

¿ 0.0168 ×190.982×502× 100× 10−3 ×5.408

¿28326.20446 Pa

X=√ ∆ Pl /x∆ Pg/ x

¿√ 34465.528326.204

¿1.103

∅ l2=1+ 20

1.103+ 1

1.1032

¿19.95

∆ Pgl=19.95 ×34465.5=687.736 kPa

P1−P2=687.736

P1=687.736+130

¿817.736 kPa

Trial 3:

Average Pressure=P1+P2

2

¿ 817.736+1302

¿473.868 kPa

ρ=473.868× 103× 44 × 10−3

8.314 × 373

¿6.7159 kg /m3

∆ Pg=f G2 x2 Dρ

¿ 0.0168 ×190.982×502× 100× 10−3 ×6.7159

¿22809.768 Pa

X=√ ∆ Pl /x∆ Pg/ x

¿√ 34465.522809.768

¿1.2292

∅ l2=1+ 20

1.2292+ 1

1.22922

¿17.93

∆ Pgl=17.93 ×34465.5=618.029 kPa

P1−P2=618.029

P1=618.029+130

¿748.029 kPa

Question 7Given:

Water Flow Rate, Q = 700gpm× 0.003785 m3

1 gallon=2.6495 m3

min× 1min

60sec=¿0.044158m3/sec

Diameter of 6 in 80 sch commercial pipe (taking inner diameter)

= 5.76 in = 5.76∈× 0.0254 m1∈¿=0.1463 m ¿

Temperature = 90 F = ¿90−32

1.8=32.222+273.15=305.372 K

Pressure Drop = 2.23 psi =2.23 psia14.7 psia

× 101.3 Kpa1 = 15.367 Kpa

Length X1= 100ft = 100ft×0.3048 m

1 ft=¿30.48m

Assumptions:

Incompressible flow

Taking µ of water= 0.8 x10-3 Pa.s

Density of water = 1000kg/m3

Surface Roughness for commercial steel pipe Ɛ = 0.0457 mm

Surface Roughness for cast iron pipe Ɛ = 0.259 mm

i) Reynolds Number, Re¿ρvD

μ

Velocity,v = Water Flow

Cross sectional area= 0.044158

π (0.1463)2

4= 2.62682 m/s

ii) Pipe Wall Roughness = εD

=0.0457146.3

=3.12× 10−4

Thus, from the moody chart, we can find the friction factor.

f = 0.0041

We know that pressure drop,

∆ P=f ρ v2

2 DX

15.367 ×103=f 1000 (2.62682 )2

2 (0.1463 )30.48

15.367 ×103=f ×718788.747 1000 (2.62682 )2

2 (0.1463 )30.48

f =0.02138

Thus, i) Reynold’s number from moody chart= 6.5x 104

We know, Re ¿ρvD

μ

6.5 ×104=1000× 2.62682× 0.1463μ

μ=5.9124 ×10−3

iii) Surface roughness when pipe is made of cast iron, Ɛ = 0.259

Then, Pipe Wall Roughness ,

εD

=0.259146.3

=1.77 × 10−3

The friction factor from moody chart from same Reynolds number but new pipe wall roughness

f= 0.0251

Now to find the pressure drop for 100 m

∆ P=f ρ v2

2 DX

∆ P=0.0251 1000 (2.62682 )2

2 (0.1463 )100=59191.593 pa

iv) To find the pressure drop,

When pressure difference is the static head of water 175 ft above pipe

175 ft = 175ft× 0.3048 m1 ft = 53.34 m

Pressure head ¿p

ρg

53.34= p1000 × 9.8

∆ p=522732 pa

For 100 ft of pipe with internal diameter 0.1463, the pressure drop is too small compared to

522732 pa. The appropriate diameter can be found by iteration

Trial 1

Taking Diameter 3 in= 3∈× 0.0254 m1∈¿=0.0762 m¿

Velocity,v=Water Flow

Cross sectional area= 0.044158

π (0.0762)2

4

=9.68299 m /s

Re ¿ρvD

μ=1000 ×9.68299 × 0.0762

5.9124 ×10−3 =124795.9945

Thus flow is turbulent

Relative roughness

εD

=0.045776

=6.013× 10−4

Thus, from Moody chart,

Friction factor,f= 0.0208

∆ P=f ρ v2 X2 D

V2= ∆ P× 2 D

ρfX

V2= 522732× 2× 0.0762

1000× 0.0208 ×30.48=125.6567

V = 11.209m/s

Trial 2

Taking V = 11.209m/s

V=0.044158π (D)2

4

11.209=0.044158π ( D )2

4

D=0.07082m

Then, Reynolds number, Re¿ρvD

μ=1000 ×11.209× 0.07082

5.9124 ×10−3 =134270.1219

Flow is turbulent.

Relative roughness

εD

=0.045770.82

=6.453× 10−4

Thus, from Moody chart,

Friction factor,f= 0.0205

∆ P=f ρ v2 X2 D

V 2=∆ P ×2 DρfX

V 2= 522732 ×2×0.070821000 ×0.0205 ×30.48

=118.494

V = 10.885m/s

Trial 3

Taking V=10.885m/s

V=0.044158π (D)2

4

10.885=0.044158π ( D )2

4

D=0.07187 m

Then, Reynolds number, Re¿ρvD

μ=1000 ×10.885 ×0.07187

5.9124 ×10−3 =132315.3303

Flow is turbulent.

Relative roughness

εD

=0.045771.87

=6.359× 10−4

Thus, from Moody chart,

Friction factor, f = 0.02048

∆ P=f ρ v2 X2 D

V 2=∆ P ×2 DρfX

V 2= 522732× 2× 0.071871000 ×0.02048 ×30.48

=120.368

V = 10.97m/s

Trial 4

Taking V = 10.97m/s

V=0.044158π (D)2

4

10.97=0.044158π ( D )2

4

D=0.0716 m

Then, Reynolds number, Re¿ρvD

μ=1000 ×10.97 × 0.0716

5.9124 × 10−3 =132848.2511

Flow is turbulent.

Relative roughness

εD

=0.045771.6

=6.383× 10−4

Thus, from Moody chart,

Friction factor, f= 0.02049

∆ P=f ρ v2 X2 D

V 2=∆ P ×2 DρfX

V 2= 522732× 2× 0.07161000 ×0.0248 ×30.48

=119.916

V = 10.95m/s

Finding the diameter with V = 10.95m/s

V=0.044158π (D)2

4

10.95=0.044158π ( D )2

4

D=0.0716 m

Thus, after the fourth iteration, the diameter converges to 0.0716m which is the required

diameter.

Question 8

i) Stokes Flow- Stokes flow (named after George Gabriel Stokes), also named creeping flow or

creeping motion, is a type of fluid flow where advective inertial forces are small compared

with viscous forces. The Reynolds number is low. This is a typical situation in flows where

the fluid velocities are very slow, the viscosities are very large, or the length-scales of the

flow are very small

Terminal Velocity- When an object in a fluid body is subject to motion only with the

influence of gravity or in certain cases, centrifugal force, then the body accelerates to a

certain velocity when the friction and other forces are balanced. This is the terminal

velocity.

ii) a. To show terminal velocity u = d2g(ρs - ρf)/18μ

In general, a particle is acted upon by the forces of gravity, buoyancy and drag. When

terminal velocity is reached, all these forces balance out.

The drag force of a sphere Fd= 3πµvd

Fb= Mass of fluid displaced = 43

π r 3× ρf × g

Mass of particle= mg=43

πr 3× ρp× g

Then,

6 πµvd+( 43

π r3 × ρf × g)−¿

v=43

π ( d2 )

3

× g ( ρp−ρ f ) ÷ 6 πµd

v=4 g ( ρp−ρf ) d2

3×3 πµ×8=

g ( ρp−ρ f ) d2

18 πµ

So Terminal velocity u¿g ( ρp−ρ f ) d2

18 πµ Hence proven.

b. To show Cd= 24/Re

FdFb

mg

In 1851, Stokes solved the hydrodynamic equations of motion for a case of creep flow (slow velocity) and

derived

F= 3πµvd

If R’ is the force per unit projected area of the particle, and the drag coefficient is given by,

CD= R '1 /2 ρ v2 [1]

Now , R '=FA

=3 πμvd

π d2

4

=12 μvd

[2]

Then substituting 1 and 2 we get,

CD=

12 μvd

1 /2 ρ v2

CD=24 μρvd

[3]

But, we know, ℜ= ρvDμ

Then 3 becomes,

CD=24ℜ

Hence proved

iii) Given:

d = 1.5 mm

Ρg= 2500kg/m3

Ρw=1000kg/m3

µ= 1cP= 1x10-3Pa.s

When the terminal velocity is not given, the following equation is used to give the

relationship between Cd and Re. Also the Archimedes number is

Ar=(ρg−ρw) ρw gD3

μ2 =(2500−1000 ) ×1000 × 9.8×(1.5 ×10−3)3

(1 ×10−3)2

Ar=49612.5

Since 18< Ar<330000 ,

We use the Ar value chart to find the value of Reynolds number.

Re = 295

We know,

ℜ= ρvDμ

295=1000 × v× (1.5 ×10−3 )

1× 10−3

v=0.196667 m /s

Thus the terminal velocity is 0.196667 m/s