fpp. finaldocx
DESCRIPTION
fluid particlesTRANSCRIPT
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