Final Compilation of Assing 3 - Vignes Shanmuganathan

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ASSIGNMENT 3

ASSIGNMENT 3

CHE 322 PROCESS PLANT ENGINEERING

5/8/2009

KUSHALA PRIYA RAMACHANDRAN- 7D6B4080VIGNES SHANMUGANATHAN- 7D6B4215KELVIN LIM FULIANG- 7D5B3402CHEN SOONG CHI- 7D7B5203

1 | P a g e

ASSIGNMENT 3

Question 1(i)

Crude oil is transported at a terminal to a refinery through a 0.3 m diameter pipeline at flowrate of 20,000 tonnes per day. As a result of frictional heating, the temperature of the oil is 20 K higher at the refinery end that at the terminal end of the pipe and the viscosity has fallen to one half its original value. What is ratio of the pressure gradient in the pipeline at the refinery end to that at the terminal end? Viscosity of oil at terminal = 0.090 kg/m.s. Density of oil (approximately constant) = 960 kg/m3.

Assumptions:1. Oil density taken to be constant and therefore the same at both ends of the terminal

and the refinery respectively.2. Pipe is assumed to be made of commercial steel and have a roughness value, of 4.6

x 10-5 m.

Flowrate = 20000 tonnes/day Terminal end Refinery end

Conditions and parameters and the terminal end Density, p1 of oil = 960 kg/ m3

Viscosity,µ1 of oil = 0.09 kg/ m.s Temperature = T1

Diameter of pipe, D1 = 0.3 m

Mass flowrate, m 1 = 20,000tonnes/day

=

20 , 000 tonnesday

⋅1000 kg1 ton

⋅1 day24 h

⋅ 1 h3600 s = 231.48kg/s

Conditions and parameters and the refinery endDensity, p2 of oil = 960 kg/ m3

Viscosity,µ2 of oil = 0.09 / 2 = 0.045 kg/ m.s Temperature = T1 + 20 KDiameter of pipe, D2 = 0.3 m

The pressure gradient is given by the equation [ dP

dL ]= 8 m¿

fg πρ D5

where m = mass flow rate in kg/s

f = friction factor

g = gravitational acceleration, 9.81 ms-2

p = density in kg/m3

D = diameter of pipe in m

2

ASSIGNMENT 3

Friction factor is unknown, we therefore use round’s equation friction factor correlation

which is given by

f =1 .6364 [ ln( 0 .135 εD

+ 6 . 5ReD,T

)]−2

It can be seen here that Reynolds number is needed in advance prior to using this equation

and therefore it has to be calculated first.

Re= ρuDμ

Where p = density in kg/m3

u = velocity in m/s

D = diameter of pipe in m

μ = viscousity of oil in kg/m.s

m¿

=ρ uA

Velocity of the flow u, = ❑pA

=

231.48

960 x π0.32

4¿

¿

= 3.441 m/s

Reynolds number for flow at terminal end,

Re1 = 960 x3.441 x0.3

0.090

= 10915.2 (turbulent flow as this value > 2100)

Using round’s equation, with roughness value, of 4.6 X 10-5 m.

Friction factor for flow at terminal end, f1 = 1 .6364 {ln [0 . 135 (0 .046×10−3 )

0 .3+ 6 .5

10915. 2 ]}−2

3

ASSIGNMENT 3

= 0.0183014

For the refinery end,

Flow velocity u2 = u1 = 3.441 m/s

Reynolds number, Re2 = 960 x3.441 x0.3

0.045

= 21830.4 (turbulent flow)

Again, using round’s equation

f2 = 1 .6364 {ln [ 0 . 135 (0 .046×10−3 )

0 .3+ 6 .5

21830 . 4 ]}−2

= 0.0154237

Pressure gradient ratio in the pipeline at the refirnery end to that at the terminal end

=

[ dPdL ]

Re finery

[ dPdL ]

Ter min al

=( 8m

¿f 2

g πρ D5 )( 8m

¿f 1

g πρ D5 )= f 2

f 1

=0.01542370.0183014

=0 .8428

Comments

1. The surface roughness, of 4.6 x 10-5 m is relatively a very small value and the pipe

in the above situation could be assumed to be smooth. If a smooth pipe were to be

assumed, the pressure drop obtained would be ❑❑ = 0.8359. This value does not differ

much if a surface roughness of commercial steel = 4.6 x 10-5 m were to be assumed

and therefore the assumption of smooth pipe is valid as well.

2. Instead of using round’s equation for friction factor, the friction factor could be

obtained from the moody diagram as well by relating Reynolds number and the x-

axis term which is equal to f/2.

Question 1(ii)

Oil of viscosity 0.01kg/m.s and specific gravity 0.90, flows through 60m of 100mm diameter

pipe and the pressure drop is 1.38kN/m2. What will be the pressure drop for a second oil of

4

ASSIGNMENT 3

viscosity 0.03kg/m.s and specific gravity of 0.95 flowing at the same rate through the pipe?

Assume the pipe wall to be smooth.

Answer:

Properties of the 1st oil:

Viscosity, μoil(1)0.01 kg/m.s

Specific gravity, S.G 0.90

Diameter, D 100 mm = 0.1 m

Length, L 60m

Pressure drop, ∆P1 1.38 kN/m2

Density of oil, ρoil900 kg/m3

Properties of the 2nd oil:

Viscosity, μoil(2)0.03 kg/m.s

Specific gravity, S.G 0.95 kg/m3

5

ASSIGNMENT 3

To find pressure drop, this formula should be used,

Since the pipe is smooth, the ε = 0; So,

To obtain the Reynolds number, Φ Re2 formula is used,

|Φ Re2=−Δ PD 3 ρ4 Lμ

|

=

1380×0 .13×9004×60×0 . 012

= 51750 ≈ 5.2 x 104

Form the moody chart, the Reynolds that is obtained ≈ 3000Equation (a) is substituted with the reynolds number to obtain the velocity,

NRe=9000 U3000=9000 UU=0 .3333 m /s

By using the velocity that is obtained, the pressure drop for the 2nd oil can be calculated,

ΔP2=32LU μ

D2

=32×60 m×0 .3333×(0 .03 kg/ms )(0 .1 m )2

= 1919.81 N /m2

∴ ΔP2 =1 .920 kN /m2

6

Δp1=12

U2 ρLD

f

NRe=ρ UDμ

=900×U×0. 10 . 01

=9000 U →(a )

ASSIGNMENT 3

Question 1 (iii)

Oil of viscosity 0.01kg/m.s. and density of 900kg/m3 is flowing through a 500mm diameter

pipe 10km long. The pressure difference between the two ends of the pipe is 106 N/m2. What will the pressure drop be at the same flow rate if it is necessary to replace the pipe by one, which is only 300mm in diameter? Assume the pipe surface to be smooth

Answer:

Oil Viscosity, µoil = 0.01kg/m.sOil Density, ρoil = 900kg/m3

Pressure different, ∆P = 1×106

Pipe data

Initial Pipe Replacing PipePipe Length = 10Km = 10,000m Pipe Length = 10km = 10,000mPipe Diameter = 500mm = 0.5m Pipe Diameter = 300mm = 0.3m

Assumptions1. Smooth pipe surface so = 0

7

ASSIGNMENT 3

2. The flow rate of initial pipe, Q1 = flow rate of the replacing pipe, Q2 as the pipe are to be at the same flow rate.

For this question, we will need to use the moody chart to solve it.

By using this formula,

φ Re2=−Δ Pd3 ρ4 lμ2

φ Re2=1×106 N /m2×(0 .5 m )3×900 kg /m3

4×10 , 000 m×0 .012

¿28125000¿2 .81×107

From the chart, we know that

Re≈1 .2×105=ρu1 d1

μWorking with the formula to found out the value of u1

8

ASSIGNMENT 3

u1=Re μd1 ρ

¿1 .2×105×0 .01 kg /m . s0 .5m×900kg /m3

¿2 .67 m/ s

Since we know that Q1=Q2, where Q = uA, where A is the cross-sectional area.u1 A1=u2 A2

u2=u1×A1

A2

¿2 .67 m/ s×π×(0 .5

2 )2

π×(0 .32 )

2

¿7 .416m /s

Reynolds number is determined for the replaced pipe

Re=ρ uDμ

=900 kg/m3×7 . 416 m /s×0. 3 m0.01 kg /m . s

¿200232¿2×105

9

ASSIGNMENT 3

From the chart we know that

R

ρu2=φ=0 . 0018

Using the equation 3.18 from the chart

−ΔP f=4 φld

ρu2

¿4×0. 0018×10000 m0 .3 m

×900 kg /m3×7 . 4162

¿11879364 N /m2

¿1 .19×107 N /m2

Question 2

Consider a situation wherein you are required to design a piping system for the pneumatic

transport of solids at the rate of 85 tonnes/day through a horizontal pipe of 0.4 km long.

Using typical solid to gas mass ratio of 5 and do not exceed the air flow rate of 30 m/s to

avoid excessive pressure drops, estimate the air flow requirements, the pipe diameters and the

pressure drops in the pipeline for transporting various solid materials given in Table 1.

Briefly comment on your results in terms of the energy requirements for handling various

types of solids.

Case Material Particle size (mm)

12345

CoalPerspex PolystyreneLeadBrass

0.751.50.360.300.40

10

ASSIGNMENT 3

678

AluminumRape SeedManganese dioxide

0.231.910.75

Table 1: Particle Sizes of Several Solid Particles.

The question above requires estimation of the air flow, the diameter of the pipe and

the pressure drops in the pipeline for transporting various solid materials stated in Table 1.

Brief comments must also be stated in terms of the energy requirements for handling various

types of solid.

Answer:

Values that can be obtained from the question above:

a) Mass flow rate of solid, Ms =

85 tonnesday

b) Ratio of mass flow rate of solid to gas,

M s

M g

=5

c) Pipe Length, L = 0.4 km

d) Maximum Air Flow Rate, ug=

30 msec

From the question above, we are required to calculate and determine:

a) Air flow requirements, Qair

b) Pipe diameter, D

c) Pressure drops in the pipeline, -∆Ptotal

Firstly, the mass flow rate of the solid, Ms which was given in the question is in tonnes per day. We have to convert it into kilograms per second (kg/sec).

Mass Flow Rate of Solid, M s =

85 tonnesday

×1000 kgtonne

× 1 day24 hours

×1 hour60min

× 1 min60sec

M s =

0 .983796 kgsec

≈0 . 9838 kgsec

Next, after converting the Mass Flow Rate of solid, ratio of solid to gas can be determined. As it is given,

11

ASSIGNMENT 3

M s

M g

=5,

Since Ms =

0 .9838 kgsec ,

∴ Mass Flow Rate of Gas, M g=

0 .98385

=0. 196759 kgsec

≈0 .1968 kgsec

Since the pipe length is given in kilometer, we can easily convert it into meter by:

1 km=1000 m

Pipe Length, L = 0 .4 km

L = 0 .4×1000 m=400 m

2.1 Minimum Pipe Diameter

In order to determine the diameter of the pipe, surface area of a pipe can be applied;

Surface Area of Pipe, A=π

4D2

Therefore, D=√ 4 A

π -------- (Equation 2.1.0)

D = Diameter of the Pipe

A = Area of the Pipe

From the equation, we have two unknowns which is the Diameter, D and also Area, A.

To get the value of Area, A, the volumetric flow rate of air, Qg must be calculated;

Volumetric Flow Rate of Air

Volumetric Flow Rate of Air, Qg=

M g

ρg

Mass Flow Rate of Gas, M g=0.1968 kg /s

Density of Air at standard ambient temperature, ρg=1. 168 kg /m3

∴The Volumetric Flow Rate of Air, Qg=

0 . 19681 .168

=0 .168458 m3 /s≈0 .1685 m3 /s

12

ASSIGNMENT 3

From here, Area, A can be determined as we have the value of Qg, Volumetric Flow Rate of

Air.

Volumetric Flow Rate of Air, Qg=ug A

Therefore, A =

Qg

ug

Volumetric Flow Rate, Qg= 0 .1685 m3 /s

Air Flow Rate, ug=30 m /s

Area, A=0 . 1685

30=0 . 0056164 m2≈0 . 0056 m2

Now with area, A known, substitute the value into equation 2.1.0 to obtain the minimum

diameter of the pipe.

Minimum Diameter, D=√ 4 A

π=√ 4×0.0056

π

D=0.08455 m

Due to only certain standard sizes of diameter of pipes that can be found, the diameter

obtained in this calculation is rounded to the closest values which is D = 0.1m.

2.2 Required Air Flow Estimation

By taking the new diameter D = 0.1 m, estimation of air flow required can be obtained

through:

Volumetric Flow Rate, Qg=ug A

Therefore the new Air Flow Rate, ug=

Qg

A

Volumetric Flow Rate, Qg=0 .1685 m3 /s

Area, A=π

4D2= π

4(0 . 12)=0 .07854 m2

13

ASSIGNMENT 3

∴ The estimated Air Flow Rate, ug=

0 .16850 . 07854

=21. 4488 m / s

2.3 Pressure Drop Calculations

From the following equation pressure drop can be obtained:

−Δpx

−Δp Air

us2

M s

=2805uo ------------- (Equation 2.3.0)

Pressure drop of solid = -Δp x

Pressure drop of air = -Δp Air

Velocity of solid = us

Particle free falling velocity = u0

Mass flow rate of solid = M s

In order to determine the pressure drop of air, the very first thing that needs to be done is to

calculate the Reynolds number.

Re=ρg Dug

μg

Density of air at standard atmosphere (ρg ) = 1.168 kg /m3

Air velocity (ug ) = 21.4488 m /s

Diameter of pipe (D) = 0.1 m

Viscosity of air at 25oC (μg ) = 1 .85 × 10−5 Ns/m2

Re=1 .168×21 .4488×0.1

1. 85×10−5

= 135417.12 which is > 2100 which indicates turbulent flow.

By using the Reynolds number which was calculated earlier, friction factor can be obtained

through:

14

ASSIGNMENT 3

f =1 .6364 [ ln ( 0 .135 εD

+ 6 . 5N Re

)]−2

Friction factor = f

Diameter of pipe (D) = 0.1 m

Reynold’s Number (NRe) = 135417.12

Pipe surface roughness (ε ) = 0

* We assume the pipe is smooth

f =1 .6364 [ ln ( 0 .135×(0)0. 1

+ 6 . 5135417 .12 )]

−2

= 0.016548

Density of air (ρg ) = 1.168 kg /m3

Friction factor (f ) = 0.016548

Pipe length (L) = 400 m

Diameter (D) = 0.1 m

Air Flow Rate (ug ) = 21.4488 m /s

Gravitational acceleration (g) = 9.81 m /s 2

−Δpg = f ( L

D)( ρu2

2)

−Δpg = (0 . 016548)×(400

0 . 1)×(21. 44882

2)(1.168)

=17783.79 N /m2

Hence, pressure drop of air −Δp g = 17783.79 N /m2

To determine the velocity of solid, the following formula is applicable:

15

ASSIGNMENT 3

ug−us=uo

0 . 468+7 .25√ uo

ρs

Velocity of gas = ug

Solid Velocity = us

Particle free falling velocity = u0

Solid density = s

The formula can be simplified to calculateus :

us=ug−uo

0 .468+7 . 25√ uo

ρ s ------------ (Equation 2.3.1)

Air Flow Rate, ug=21.4488 m /s

Free Falling Velocity = u0

Solid Density = ρ s

* Free falling velocity can be obtained from the table

Using (Equation 2.3.0), the pressure drop of solid can be calculated:

−Δp x=2805 M s

us2uo

(−ΔpAir ) ----------------- (Equation 2.3.2)

Mass Flow Rate of Solid, M s=0 . 1968 kg /s

Pressure Drop of Air, −ΔpAir = 17783.79 N /m2

Free Falling Velocity = u0

Solid Velocity = us

The total pressure drop:

16

ASSIGNMENT 3

−ΔpTotal=(−ΔpAir)+(−Δpx )=−ΔpAir(1+2805 M s

ux2uo

) ------------ (Equation 2.3.3)

Pressure Drop of Air, −ΔpAir = 17783.79 N /m2

Pressure Drop of Solid = −Δp x

Lastly, the total energy required for the solid can be determined through:

Power = QΔp ----------- (Equation 2.3.4)

Volumetric flow rate of air = Q

3.1 Case 1 (Coal)

Diameter, D = 0.00075 m

Density, ρ = 1400 kg /m3

17

ASSIGNMENT 3

Velocity of Air, ug = 21.4488 m /s

Free Falling Velocity, u0 = 2.8m /s

By substituting the value given into equation 2.3.1, the velocity of coal can be calculated.

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 2.8

0. 468+7 .25√ 2. 81400

us=17 .9145 m/ s

By replacing the value of us into 2.3.2, pressure drop of coal can be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

17 .91452×2 . 8(17783 .79 )

−Δp x=10924 .83 N /m2

Hence total pressure drop can be calculated by using eqn2.3.3−ΔpTotal=(−Δpg )+(−Δp x)

−ΔpTotal=17783 . 79+10924 . 83

−ΔpTotal=28708.62 N /m2

By applying 2.3.4, the total energy can be obtained

Power = QgΔp = 0.1685 ¿ 28708.62 = 4837.40 J / s=4 .837 kW

3.2 Case 2 (Perspex)



18

ASSIGNMENT 3



By substituting the values given into Equation 2.3.1, the Velocity of Perspex can be

calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 3 . 73

0. 468+7 .25√ 3 . 731185

us=17 .1848 m / s

By replacing the value of us= 17 .1848 m / s into Equation 2.3.2, pressure drop of Perspex

can be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

17 .18482×3 .73(17783. 79 )

−Δp x=8912. 19 N /m2

Hence, the Total Pressure Drop can be calculated by using Equation 2.3.3

−ΔpTotal=(−Δpg )+(−Δp x)

−ΔpTotal=17783 . 79+8912 .19

−ΔpTotal=26695.98N /m2

By applying Equation 2.3.4, the total energy required can be obtained

Power = QgΔp = 0.1685 ¿ 26695.98 = 4498.27 J / s=4 .498 kW

19

ASSIGNMENT 3

3.3 Case 3 (Polystyrene)





By substituting the values given into Equation 2.3.1, the Velocity of Lead can be calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 1. 62

0. 468+7 .25√ 1 .621080

us=19.2853 m /s

By replacing the value of us=17 . 9145 m / s into Equation 2.3.2, pressure drop of Lead can

be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

19 .28532×1. 62(17783 . 79)

−Δp x=16293 .50 N /m2



−ΔpTotal=17783 . 79+16293 .50

−ΔpTotal=34077 . 29 N /m2


Power = QgΔp = 0.1685 ¿ 34077 . 29= 5742 .02 J / s=5. 742 kW

20

ASSIGNMENT 3

3.4 Case 4 (Lead)





By substituting the values given into Equation 2.3.1, the Velocity of Lead can be calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 8 .17

0. 468+7 .25√ 8 . 1711080

us=9 .1607 m/ s

By replacing the value of us=9.1607 m / s into Equation 2.3.2 pressure drop of Lead can be

obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

9 . 16072×8 .17(17783 . 79)

−Δp x=14318 .68 N /m2



−ΔpTotal=17783 . 79+14318 .68

−ΔpTotal=32102 . 47 N /m2


Power = QgΔp = 0.1685 ¿ 32102 . 47= 5409 .27 J / s=5. 409 kW

21

ASSIGNMENT 3

3.5 Case 5 (Brass)





By substituting the values given into Equation 2.3.1, the Velocity of Brass can be

calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 4 . 08

0. 468+7 .25√ 4 . 088440

us=14 . 9458 m /s

By replacing the value of us=14 .9458 m /s into Equation 2.3.2 pressure drop of Brass can

be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

14 .94582×8. 17(17783 .79)

−Δp x=5 379 . 25 N /m2



−ΔpTotal=17783 . 79+5 379. 25

−ΔpTotal=23163 . 04 N /m2


Power = QgΔp = 0.1685 ¿ 23163 . 04 = 3902 .97 J / s=3. 902 kW

22

ASSIGNMENT 3

3.6 Case 6 (Aluminum)





By substituting the values given into Equation 2.3.1, the Velocity of Aluminum can be

calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 3. 02

0. 468+7 .25√ 3 .022835

us=17 .1628 m / s

By replacing the value of us=17 . 1628 m /s into Equation 2.3.2 pressure drop of aluminum

can be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

17 .16282×3 .02(17783. 79 )

−Δp x=1 1035 .68 N /m2



−ΔpTotal=17783 . 79+11035 . 68

−ΔpTotal=28819 . 47 N /m2


Power = QgΔp = 0.1685 ¿ 28819 . 47 = 4856 .08 J / s=4 .856 kW

23

ASSIGNMENT 3

3.7 Case 7 (Rape Seed)





By substituting the values given into Equation 2.3.1, the Velocity of Rape Seed can be

calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 5. 91

0. 468+7 .25√ 5 .911080

us=15.5642 m /s

By replacing the value of us=15 . 5642 m /s into Equation 2.3.2, pressure drop of Rape Seed

can be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

15 .56422×5 . 91(17783 . 79)

−Δp x=6857 . 11 N /m2



−ΔpTotal=17783 . 79+6857 .11

−ΔpTotal=24640 . 90 N /m2


Power = QgΔp = 0.1685 ¿ 24640 . 90= 4151 .99 J / s=4 .151 kW

24

ASSIGNMENT 3

3.8 Case 8 (Manganese Dioxide)





By substituting the values given into Equation 2.3.1, the Velocity of manganese oxide can be

calculated;

us=ug−uo

0 .468+7 . 25√ uo

ρ s

us=21.4488− 5. 27

0. 468+7 .25√ 5 .274000

us=14 .2410 m /s

By replacing the value of us=14 .2410 m / s into Equation 2.3.2, pressure drop of manganese

oxide can be obtained.

−Δp x=2805 M s

us2uo

(−ΔpAir )

−Δp x=2805×0 . 1968

14 .24102×5 . 27(17783 .79 )

−Δp x=9185. 24 N /m2



−ΔpTotal=17783 . 79+9185 .24−ΔpTotal=26969 . 03 N /m2


Power = QgΔp = 0.1685 ¿ 26969 . 03= 4544 .28 J / s=4 .544kW

25

ASSIGNMENT 3

Table 2: The free falling velocities and respective densities of the materials can be obtained from the table above.

CAS

E

MATERIALS PARTICLE SIZE

(mm)

PRESSURE DROP

(kW)

1 Coal 0.75 4.837

2 Perspex 1.5 4.498

3 Polystyrene 0.36 5.742

4 Lead 0.30 5.409

5 Brass 0.40 3.902

6 Aluminum 0.23 4.856

7 Rape Seed 1.91 4.151

8 Manganese Oxide 0.75 4.544

Table 3: Pressure drop value for each material

26

ASSIGNMENT 3

Comments

From the data obtained, it can be observed that the power losses or energy

requirement to transport is directly proportional on the air flow rate and the pressure drop

relative to the solid. Polystyrene have the largest energy requirement whereas rape seed have

the smallest. As the pressure drop is inversely proportional to the free falling velocity, it can

be said the lower the free falling velocity of a particular material particle, the higher the

energy consumption. The energy is also affected by density as it can be seen above that the

higher the density of a particular material, the lower the particle velocity which results in

higher pressure drop. The volumetric flow rate also must be manipulated in an efficient way

as a higher flow rate will result in higher energy consumption as on the other hand, if the

flow rate is too low, time consumption is higher which may incur an even higher cost in the

long term. So, it is always best to work out the optimum flow rate so as to incur the lowest

possible cost at an acceptable efficient speed of transport. It can be also concluded that for a

solid with low density and high free falling velocity, the energy consumption is lower.

27

ASSIGNMENT 3

Conclusion

It can be seen from the above calculations that calculations for single phase flow and

multiphase flows differ and that for a multiphase flow, the pressure drops for different phases

are calculated separately and in a different way. Pressure drop calculations for different phase

combinations in multiphase flow differ as well and use different correlations to estimate their

pressure drops. However, all the above calculations are estimations based on our assumptions

and may not be applicable in the real world such as the viscosity of oil in question 1 was

assumed to be constant throughout the flow at different temperature regions. Temperature is a

function of viscosity and in reality, the viscosity changes with temperature. Knowledge of

multiphase flow as well as single phase flow is highly essential for an engineer as this skill is

extensively applied in the real world.

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Final Compilation of Assing 3 - Vignes Shanmuganathan