ppe assignment

PROCESS PLANT

ENGINEERING

ASSIGNMENT 2 Name : Libbis Sujessy Student ID : (7E3B2332/17484688)

Process plant engineering | Libbis Sujessy/7E3B2332/17484688

1

Question 1

Crude oil of varying density needs to be transported at a terminal to a refinery through 0.3 m

diameter pipeline at flow rate of 20,000 tons per day. As a result of frictional heating, the

temperature of the oil is 20 K higher at the refinery end than at the terminal end of the pipe and

the viscosity falls to one half of its original value. Viscosity of oil at terminal is 0.090 kgm-1s-

1. As a design engineer you are requested to assess the impact of density variations of crude oil

on the ratios of pressure gradients in the pipeline at the refinery end to that at the terminal end?

You may consider the variations in density from 860-960 kg/m3 in steps of 20 kg/m3.

Summarize your results in tabulated form and include your brief comments.

Assumptions:

1. Oil density are constant from the terminal end to the refinery end.

2. Pipe is assumed to be made of schedule 80 carbon steel with its roughness value, ɛ of

4.5x10-5 m.

20,000 tonnes/day

Conditions and parameters in the process

Parameters Terminal Parameters Refinery

Density, ρ1 (kg/m3)

of oil

860

Density, ρ2 (kg/m3) of

oil

860

880 880

900 900

920 920

940 940

960 960

Viscosity, μ1 of oil 0.09 kgm-1s-1 Viscosity, μ2 (kgm-1s-

1) of oil

0.09/2 = 0.045 kgm-

1s-1

Temperature, T1 T1 Temperature, T2 T1 + 20 K

Diameter of pipe, D1 0.3 m Diameter of pipe, D2 0.3 m

Mass flowrate, ṁ1

20,000 tonnes/day

or

231.48 kg/s

Terminal Refinery


2

Equations involved in the process

1. Pressure gradient 5

8

Dg

fm

dL

dP

Where ṁ = mass flow rate (kg/s)

f = friction factor

g = gravitational acceleration, 9.81 m/s2

ρ = density (kg/m3)

D = diameter of pipe (m)

2. Round’s equation friction factor correlation

2

,Re

5.6135.0ln6364.1

TDDf

3. Reynolds number

uDRe

Where u = velocity (m/s)

μ = viscosity of oil (kg/m.s)

4. Velocity of the flow by u =

m

𝜌𝐴

Calculations:

1. The velocity of each density

ṁ = 231.48 kg/s

A = π/4 x D2 = 0.071 m2

ρ (kg/m3) u1 (m/s)

860 3.810

880 3.723

900 3.640

920 3.561

940 3.486

960 3.413


3

2. The Reynolds number of each density

a. At the terminal end

μ1 = 0.09 kg/m.s

D1 = 0.3 m

b. At the refinery end

With the flow velocity u2 = u1

μ2 = 0.09 kg/m.s

D2 = 0.3 m

ρ2 (kg/m3) u2 (m/s) Re2

860 3.810 21844.00

880 3.723 21841.60

900 3.640 21840.00

920 3.561 21840.80

940 3.486 21845.60

960 3.413 21843.20

So all flows in the pipe, both at the terminal and the refinery end, are at range 860 –

960 kg/m3 known as turbulent flow as their value > 2100.

3. The friction factor at the terminal end and also at the refinery end, by using the Round’s

equation, with roughness value ɛ = 4.5x10-5 and D=0.3 m 2

,Re

5.6135.0ln6364.1

TDDf

a. At the terminal end

Re1 f1

10922.00 0.0299375

ρ (kg/m3) u1 (m/s) Re1

860 3.810 10922.00

880 3.723 10920.80

900 3.640 10920.00

920 3.561 10920.40

940 3.486 10922.80

960 3.413 10921.60


4

10920.80 0.0299384

10920.00 0.0299390

10920.40 0.0299387

10922.80 0.0299370

10921.60 0.0299378

b. At the refinery end

4. Finally the calculation of the pressure gradient ratio in the pipeline at the refinery end to that

at the terminal end

1

5

1

5

min

Re 2

8

28

f

f

Dg

fm

Dg

fm

dL

dP

dL

dP

alTer

finery

So the ratio of different density is shown in the table below

f1 f2 f2/f1

0.0299375 0.0252268 0.842647

0.0299384 0.0252274 0.842644

0.0299390 0.0252278 0.842642

0.0299387 0.0252276 0.842643

0.0299370 0.0252263 0.842649

0.0299378 0.0252270 0.842646

Comments:

The solution to this question is found by assuming the pipe is not a smooth pipe but a

Schedule 80 Carbon Steel Pipe, because frictions happens in the wall of the pipe during

the crude oil transportation in real life, and this kind of pipe is usually used in the

transportation of crude oil. With the roughness value of the pipe is 0.045mm.

Re2 f2

21844.00 0.0252268

21841.60 0.0252274

21840.00 0.0252278

21840.80 0.0252276

21845.60 0.0252263

21843.20 0.0252270


5

But if the pipe were assumed as a smooth pipe, which the value of ɛ=0, then the values

of friction factor for each density are going to be changed. 2

,Re

5.6)0(135.0ln6364.1

TDDf =

2

,Re

5.6ln6364.1

TD

And in the below table is presented the comparison value of ratio of the refinery end to

the terminal end between the Schedule 80 Carbon Steel pipe and the smooth pipe, along

with the %error of the ratio.

f1 f2 f2/f1 f1smooth f2smooth f2smooth/f1smooth %error

0.0299375 0.0252268 0.842647 0.0296684 0.0248193 0.836559 0.7225%

0.0299384 0.0252274 0.842644 0.0296693 0.0248200 0.836556 0.7224%

0.0299390 0.0252278 0.842642 0.0296699 0.0248205 0.836555 0.7224%

0.0299387 0.0252276 0.842643 0.0296696 0.0248202 0.836556 0.7224%

0.0299370 0.0252263 0.842649 0.0296678 0.0248189 0.836560 0.7225%

0.0299378 0.0252270 0.842646 0.0296687 0.0248196 0.836558 0.7225%

As we can see from the table, the ratio between Schedule 80 Carbon Steel pipe and

smooth pipe are not far different, with error percentage less than 1%. This is due to the

very small number of 80S Carbon Steel pipe roughness (0.045mm), which commits a

minimum number of changes in the friction factor. Hence, if the assumption of the

transportation pipe of crude oil is commercial pipe is reasonable as well.

The friction factor also can be found from moody diagram, however using moody

diagram needs high accuracy to find the exact point in the diagram.

Question 2

You are asked to design piping systems of three different lengths of 0.7, 1 and 1.3 km

respectively for the pneumatic transport of solids at the rate of 10 tons/day. Particles sizes of

different solid particles are given in Table 1.

Case Material Particle Size (mm)

1

2

3

4

5

Coal

Polyestyrene

Alumunium

Rape Seed

Sand

0.75

0.36

0.23

1.91

1.30


6

Using typical solid to gas mass ratio of 5 and air flow rate of 35 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. Briefly comment on your results in terms of

the energy requirements for handling various types of solids for three piping systems and

summarize your results in tabulated form.

Known values:

Mass flow rate of solids : Ms = 10 𝑡𝑜𝑛𝑠

𝑑𝑎𝑦

Ms = sec86400

1100010 day

tonne

kg

day

tonnes = 0.1157 kg/sec

Ratio of mass flow rate of solid to gas: 𝑀𝑠

𝑀𝑔 = 5

Hence, Mass flow rate gas: Mg = 5

1157.0= 0.0231 kg/sec

Pipe length: 0.7 km = 700 m

1 km = 1000 m

1.3 km = 1300 m

Maximum air flow rate: ug = 35 m/s

Density of air at standard condition, ρg = 1.196 kg/m3

Calculations

1. Diameter of Pipe

Surface Area of a Pipe: A = 𝜋

4D2

Therefore, the diameter of pipe: D = √4𝐴

𝜋

In order to determine diameter of the pipe (D), we need the value of pipe surface area (A),

which can be found out from its volumetric flow rate of air, Qg. Thus:

Volumetric Flow Rate of Air

Qg=

Mg

ρg

From previous section, we know the data of gas mass flow rate and the density of air at

ambient tempeture (25ºC).

Thus,

Qg=

Mg

ρg

= 0.0231 kg/sec

1.196 kg/𝑚3 = 0.0193 𝑚3/𝑠

From here, the pipe surface area (A) can be determined as we already found out the value

of Qg.

Pipe Surface Area

𝑄𝑔 = 𝑢𝑔𝐴


7

Then, A = 𝑄𝑔

𝑢𝑔, with 𝑢𝑔 is the air flow rate.

Thus: A =0.0193 𝑚3/𝑠

35 𝑚/𝑠= 5.514 𝑥 10−4 𝑚2

Now, after surface area of the pipe,A, is known, substitute the value into early equation to

get the diameter value of the pipe.

D = √4𝐴

𝜋 = √

4 𝑥 5.514 𝑥 10−4 𝑚2

𝜋 = 0.026 m

Because of a particular standard size of pipe diameter that can be determined, the diameter

obtained in this calculation is rounded up to the closest standard values, which is D= 0.027m.

2. Required Air Flow

By considering the standard size of pipe diameter D = 0.027m, the approximation of

required air flow can be determined by:

Volumetric flow rate, 𝑄𝑔 = 𝑢𝑔𝐴

Therefore the flow rate calculation: 𝑢𝑔 = 𝑄𝑔

𝐴

With A is the surface area of pipe, 222 000573.0027.044

mDA

By filling in the equation with the data found before,

𝑢𝑔 = 0.0193 𝑚3/𝑠

0.000573 𝑚2= 33.682 𝑚/𝑠

So the estimated air flow rate is 33.682 m/s

3. Pressure Drops in the Pipeline

Pressure drop can be defined from the following equation:

−𝛥𝑃𝑥

−𝛥𝑃𝑎𝑖𝑟

𝑢𝑠2

𝑀𝑠=

2805

𝑢0

-ΔPx = pressure drop of solid

-ΔPair = pressure drop of air

us = velocity of solid

u0 = particle free falling velocity

Ms = mass flow rate of solid

Determining –ΔPair:

By employing the Reynold’s equation:

Re = 𝑉𝑎𝑖𝑟𝜌𝑎𝑖𝑟𝐷

𝜇𝑎𝑖𝑟=

𝑢𝑔𝜌𝑔𝐷

𝜇𝑔=

33.682𝑥1.196𝑥0.1

1.7𝑥10−5 = 63980.66

Re = 63980.66 > 2100, which shows turbulent flow.


8

Determining friction factor

By using the value of Reynold’s number which was calculated earlier, friction factor can

be determined by following equation:

2

Re

5.6135.0ln6364.1

NDf

With assumption the pipe material is smooth, ɛ = 0.

Hence,

2

63980.66

5.6

27.0

)0(135.0ln6364.1

f

= 0.019357

airp = )2

)((2u

D

Lf

, which varies in length

Density of air (ρg) = 1.196 kg/m3

Friction factor (f) = 0.019357

Pipe length (L) = 700 m, 1000 m, and 1300 m

Diameter (D) = 0.27 m

Air flow rate (ug) = 33.68237 m/s

Length (L),m Pressure Drop ( airp ), N/m2

700 = )196.1)(

2

68237.33()

027.0

700()0.019357(

2

= 340,470.03

1000 = )196.1)(

2

68237.33()

027.0

1000()0.019357(

2

= 486,385.75

1300 = )196.1)(

2

68237.33()

027.0

1300()0.019357(

2

= 632,301.48

Determining the velocity of solid:

By using the equation below, velocity of solid can be obtained

s

o

osg

u

uuu

25.7468.0

or

s

o

ogs

u

uuu

25.7468.0


9

𝑢𝑔 = velocity of gas/air = 33.68237 m/s

𝑢𝑠 = velocity of solid

𝑢0 = particle free falling velocity

𝜌𝑠 = solid density

Determining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

Mass flow rate of solid, Ms = 0.1157 kg/sec

Free falling velocity = 𝑢0

Solid velocity = 𝑢𝑠

So, the total pressure drop:

)2805

1()()(2

ox

sAirxAirTotal

uu

Mpppp

Finally, the energy required for various types of solids can be obtained by:

Power = QΔp

Q = Volumetric flow rate of air


10

2.1 Case 1 (Coal)

Diameter, D = 0.75mm = 0.00075 m

Density, ρ = 1400 kg/m3

Free falling velocity, 𝑢0 = 2.8 m/s

Obtaining us value:

s

o

ogs

u

uuu

25.7468.0

=

1400

8.225.7468.0

8.268237.33

= 30.148 m/s

Obtaining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

L : 700m

)03.340470()80.2()148.30(

)1157.0(28052

xp = 43,417.902/ mN

L : 1000m

)75.485385()80.2()148.30(

)1157.0(28052

xp = 62,025.582/ mN

L : 1300m

)48.632301()80.2()148.30(

)1157.0(28052

xp = 80,633.252/ mN

So, total pressure drop for each length:

)()( xgTotal ppp

At 700m length of pipe, Totalp = 340,470.03 + 43,417.90 = 383,887.932/ mN



The required energy for handling coal at each length:

Pipe length : 700m, Power = QgΔp = 0.0193 x 383,887.93 = 7,409.04 J/s = 7.409

kW

Pipe length : 1000m, Power = QgΔp = 0.0193 x 548,411.33= 10,584.34 J/s = 10.584

kW

Pipe length : 1300m, Power = QgΔp = 0.0193 x 712,934.73= 13,759.64 J/s = 13.760

kW


11

2.2 Case 2 (Polyestyrene)

Diameter, D = 0.36mm = 0.00036 m



Obtaining us value:

s

o

ogs

u

uuu

25.7468.0

=

1080

62.125.7468.0

62.168237.33

= 31.519 m/s

Obtaining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

L : 700m

)03.340470()62.1()519.31(

)1157.0(28052

xp = 68,657.572/ mN

L : 1000m

)75.485385()62.1()519.31(

)1157.0(28052

xp = 98,082.252/ mN

L : 1300m

)48.632301()62.1()519.31(

)1157.0(28052

xp = 127,506.922/ mN


)()( xgTotal ppp





Pipe length : 700m, Power = QgΔp = 0.0193 x 409,127.60= 7896.16 J/s = 7.896 kW

Pipe length : 1000m, Power = QgΔp = 0.0193 x 584,468.00= 11280.23 J/s = 11.280

kW

Pipe length : 1300m, Power = QgΔp = 0.0193 x 759,808.40= 14664.30 J/s = 14.664

kW


12

2.3 Case 3 (Alumunium)

Diameter, D = 0.23mm = 0.00023 m



Obtaining us value:

s

o

ogs

u

uuu

25.7468.0

=

2835

02.325.7468.0

02.368237.33

= 29.396 m/s

Obtaining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

L : 700m

)03.340470()02.3()396.29(

)1157.0(28052

xp = 42,339.862/ mN

L : 1000m

)75.485385()02.3()396.29(

)1157.0(28052

xp = 60,485.512/ mN

L : 1300m

)48.632301()02.3()396.29(

)1157.0(28052

xp = 78,631.162/ mN


)()( xgTotal ppp

At 700m length of pipe, Totalp = 340,470.03 + 42,339.86= 382,809.882/ mN

At 1000m length of pipe, Totalp = 486,385.75 + 60,485.51= 546,871.262/ mN



Pipe length : 700m, Power = QgΔp = 0.0193 × 382,809.88= 7,388.23 J/s = 7.388

kW


kW

Pipe length : 1300m, Power = QgΔp = 0.0193 × 710,932.64= 113,721 J/s = 13.721

kW


13

2.4 Case 4 (Rape Seed)

Diameter, D = 1.91mm = 0.00191 m



Obtaining us value:

s

o

ogs

u

uuu

25.7468.0

=

1080

91.525.7468.0

91.568237.33

= 27.798 m/s

Obtaining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

L : 700m

)03.340470()91.5()798.27(

)1157.0(28052

xp = 24,195.692/ mN

L : 1000m

)75.485385()91.5()798.27(

)1157.0(28052

xp = 34,565.272/ mN

L : 1300m

)48.632301()91.5()798.27(

)1157.0(28052

xp = 44,934.852/ mN


)()( xgTotal ppp





Pipe length : 700m, Power = QgΔp = 0.0193 × 364,665.71= 7038.05 J/s = 7.038 kW


kW


kW


14

2.5 Case 5 (Sand)

Diameter, D = 1.30mm = 0.0013 m



Obtaining us value:

s

o

ogs

u

uuu

25.7468.0

=

2610

66.425.7468.0

66.468237.33

= 27.664 m/s

Obtaining –ΔPx:

)(2805

2 Air

os

sx p

uu

Mp

L : 700m

)03.340470()66.4()644.27(

)1157.0(28052

xp = 30,982.562/ mN

L : 1000m

)75.485385()66.4()644.27(

)1157.0(28052

xp = 44,260.802/ mN

L : 1300m

)48.632301()66.4()644.27(

)1157.0(28052

xp = 57,539.042/ mN


)()( xgTotal ppp






kW


kW


kW


15

Overall calculation results:

Comment:

From the results obtained, it can be noticed that the power losses or required energy for

handling solids is directly proportional with the pressure drop relative to the type of

solid and the length of pipeline. In other words, when pressure drop is increased, the

required energy will also increase or the other way around. Similar to the pressure drop,

the longer the pipe, the bigger value of the required energy.

The required energy is also indirectly affected by the density of the material, as it can

be seen from the calculations that the higher the density of a specific material, the lower

the particle velocity which results in higher pressure drops.

Solid with lower density and higher free falling velocity will results in lesser energy

consumption.

Polystyrene (case 2) has the highest pressure drop in which has the highest amount of

required energy for handling it. It followed by coal (case 1), aluminium (case 3), sand

(case 5), and the lowest pressure drops, rape seed (case 4).

References

Coulson, J.M., J. F. Richardson, J. R. Backhurst, J. H. Harker. 1999. Chemical Engineering,

Fluid Flow, Heat Transfer and Mass Transfer. Volume 1, UK: Butterworth – Heinemann.

Case Material

Particle

Size

(mm)

Density

(kg/m3)

Free

falling

velocity

(m/s)

Pressure drops (kN/m2) Required Energy (kW)

700m 1000m 1300m 700m 1000m 1300m

1 Coal 0.00075 1400 2.80 383.89 548.41 712.93 7.409 10.584 13.760

2 Polyestyrene 0.00036 1080 1.62 409.13 584.47 759.81 7.896 11.280 14.664

3 Alumunium 0.00023 2835 3.02 382.81 546.87 710.93 7.388 10.555 13.721

4 Rape Seed 0.00191 1080 5.91 364.67 520.95 677.24 7.038 10.054 13.071

5 Sand 0.0013 2610 4.66 371.45 530.65 689.84 7.169 10.241 13.314

ppe assignment

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