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1Chemical Engineering Study Program – Faculty of Industrial Technology

ENERGY REQUIRMENT AND FRICTION IN

FLUID FLOW

Team Teaching:Dr. Yogi Wibisono Budhi, Dr. Ardiyan Harimawan, Dr. Dendy Adityawarman, Dr. Anggit Raksadjati, Dr. Haryo Pandu Winoto

TK2107 Fluid and Particle Mechanics (3 Credits)

Course Schedule:Tuesday, 13.00 – 14.00 -

Thursday, 10.00 – 12.00 -

Tutorial Schedule:Tuesday, 00.00 – 00.00 -

Wednesday, 00.00 – 00.00 -

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o Energy balance analysis using Bernoulli equation → doesn’t apply for real flows

Pressure drops along the flowInvolves the friction heat

Shaft work

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Real Fluid Flow–Incompressible and Steady State

Generated by the shear betweenthe moving fluid and stationary

walls

2 Pumps or blower Supplies energy in order to keep the fluid flowing

LiquidThe change of density caused by

pressure can be neglectedIncompressible fluid3

→ →

→→

→→

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o Gas → pressure drop due to friction can be neglected → gas density can be assumed to be constant → the flow can be assumed as incompressible fluid flow

Real Fluid Flow–Incompressible and Steady State

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Relation between Fluid Pressure Drop and Shear Stress

o friction force leads to the decrease of the pressure along the pipe line

o Straight pipe with constant diameter → pressure at inlet > outlet

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Centrifugal Pump

Gas Blower

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Fluid Flow Friction

o For constant x and position, the change of axial velocity along radial dimension r:

o Due to contact between fluid and stationary wall → the wall surface will give Fw force in the opposite direction with the fluid velocity

o Friction force Fs exists due to position difference between r and r+ r

o Shear Stress

o fluid flow towards x direction, cylindrical coordinate (x-r-)

o The axial velocity at radial position r and r+r dan lim

Δr→0

𝑢𝑥|𝑟+Δr − 𝑢𝑥|

Δr=𝜕𝑢𝑥𝜕𝑟

𝑢𝑥|𝑟+Δr𝑢𝑥|

Shear Stress

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Newton’s Lawo Correlate shear stress and

velocity gradiento For Newtonian flow → linear

correlation:

The friction area on fluid layer at layer r and length L is defined as:

𝑨𝒔 = 𝟐𝝅𝒓𝑳 Shear stress is defined as friction force Fs per friction area. For flow that goes to x direction with velocity Ux , Ux velocity changers along radial dimension r, so the shear stress τ rx

is defined as:

𝝉𝒓𝒙 =𝑭𝒔

𝟐𝝅𝒓𝑳

o 𝝉𝒓𝒙 = −𝝁𝛛𝒖𝒙

𝛛𝒓

Fluid Flow Friction

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At the pipe center line, r = 0, so the velocity gradient = 0

At the pipe wall,r = R, velocity gradient is maximum, shear stress at the wall surface is defined as:

𝛛𝒖𝒙

𝛛𝒓= 𝟎, 𝝉𝒓𝒙 = 𝟎

𝝉𝒘 = −𝝁𝛛𝒖𝒙𝛛𝒓

|𝒓=𝑹

Fluid Flow Friction

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Force Balance

𝐹𝑥|𝑥 = 𝑝|𝑥𝐴|𝑥𝐹𝑥|𝑥+Δx = 𝑝|𝑥+Δx𝐴|𝑥 = 𝑝|𝑥+Δx𝜋𝑟

2

𝐹 = 0

𝐹𝑥|𝑥 − 𝐹𝑥|𝑥+Δx − 𝐹𝑓 = 0

𝐹𝑓 = 𝐹𝑥|𝑥 − 𝐹𝑥|𝑥+Δx𝐹𝑓 = 𝑝|𝑥𝐴|𝑥 − 𝑝|𝑥+Δx𝐴|𝑥+Δx𝐹𝑓 = (𝑝|𝑥 − 𝑝|𝑥+Δx) 𝐴|𝑥

Δp = 𝑝|𝑥+Δx − 𝑝|𝑥𝐹𝑓|𝑥 = 𝜏𝑟𝑥2𝜋𝑟Δx

𝐴𝑥 = 𝜋𝑟2

𝜏𝑟𝑥2𝜋𝑟Δx = 𝑝|𝑥 − 𝑝|𝑥+Δx 𝜋𝑟2

𝜏𝑟𝑥 =𝑟

2−𝜕𝑝𝑓

𝜕𝑥

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Friction Heat

o The friction force in the fluid flow is transformed into friction heat

o The friction heat(Ef) is defined as friction force (Ff) multiplied by the fluid mileage (Δx):

o Expressed as friction heat rate:

o Friction heat rate per mass flow rate is defined as:

o Friction heat at pipes wall:

o Energy balance (including shaft work and friction):

𝑬𝒇 = 𝑭𝒇𝚫𝐱

ሶ𝑬𝒇 = 𝑭𝒇𝜟𝒙

𝜟𝒕ሶ𝑬𝒇 = 𝑭𝒇𝒖𝒙

𝒆𝒇 =ሶ𝑬𝒇

ሶ𝐦𝒆𝒇 =

−𝚫𝐩𝐟

𝝆

𝒆𝒇 = 𝟒𝑳

𝑫

𝝉𝒘𝝆

𝒈𝚫𝒛 +𝚫𝒑

𝝆+ 𝚫

𝒖𝟐

𝟐= 𝒘𝒔 − 𝒆𝒇

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Moody/ Darcy-WiesbachFriction Factor

Fanning Friction Factor (fF)

Friction Factor

Friction heat per mass flow unit

𝒇𝑭 =𝝉𝒘

𝟏𝟐𝝆𝒖𝟐

𝒆𝒇 = 𝟒𝑳

𝑫

𝝉𝒘𝟏𝟐𝝆𝒖𝟐

𝟏

𝟐𝒖𝟐

𝒆𝒇 = 𝟒𝑳

𝑫𝒇𝑭

𝟏

𝟐𝒖𝟐

𝒆𝒇 = 𝒇𝑴𝑳

𝑫

𝟏

𝟐𝒖𝟐

𝒇𝑴 = 𝟒𝒇𝑭

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Fanning Friction Factor

Average velocity

Friction Factor Correlation with Laminar Flow

Moody Friction Factor

𝒖 =𝟏

𝟑𝟐𝝁

−𝚫𝒑𝒇

𝑳𝑫𝟐

𝒇𝑭 =𝟏𝟔

𝑵𝑹𝒆

𝒇𝑴 =𝟔𝟒

𝑵𝑹𝒆

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Experiment by Nikurdase (1933) →friction factor is affected by Reynolds number and relative pipe surface roughness:

𝒇 = 𝒇 𝑵𝑹𝒆

𝜺

𝑫

Friction Factor

Friction Factor Correlation with Turbulent Flow

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Correlation Equation for Friction Factor

Prandtl Equation

For smooth pipe:

Coolbrook and White Equation

Nikurdase Equation

For very high NRe:

𝒇𝑴 = −𝟐 𝐥𝐨𝐠𝟐. 𝟓𝟏

𝑵𝑹𝒆 𝒇𝑭

−𝟐

𝒇𝑭 = −𝟒 𝐥𝐨𝐠𝟏. 𝟐𝟓𝟓

𝑵𝑹𝒆 𝒇𝑭

−𝟐

𝒇𝑭 = −𝟒 𝐥𝐨𝐠𝜺

𝟑. 𝟕𝑫

−𝟐

𝒇𝑭 = −𝟒 𝐥𝐨𝐠𝟏. 𝟐𝟓𝟓

𝑵𝑹𝒆 𝒇𝑭+

𝜺

𝟑. 𝟕𝑫

−𝟐

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Value of 𝜺

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Commercial Pipe Sizes

(i) Nominal size D (usually in inch with the symbol “)

(ii) Outside diameter Do (usually in inch)(iii) Schedule number Sch (dimensionless)(iv) Wall thickness, 𝓁𝑡ℎ (in inch)(v) Inside diameter Di (in inch)

Commercially, pipe sizes are expressed as parameters below:

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Commercial Pipe Sizes

(cont.)

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Fluid Flow in Straight Pipe Line and Constant Diameter

oVolumetric flowrate

oVariables need to be considered:

ሶ∨ =𝝅

𝟒𝑫𝟐𝒖

(i) Fluid volume flow rate, ሶ∨(ii) Internal pipe diameter diameter, D(iii) Pressure drop: −Δp = 𝑝𝑖𝑛𝑙𝑒𝑡 −

𝑝𝑜𝑢𝑡𝑙𝑒𝑡

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oPressure drop calculation based on energy balance equation

𝑃1𝜌+1

2𝑢12 + 𝑔𝑧1 =

𝑃2𝜌+1

2𝑢22 + 𝑔𝑧2 + 𝑒𝑓

𝑢1 = 𝑢2 = 𝑢

𝑒𝑓 = 4𝑓𝐹𝐿

𝐷

1

2𝑢2

ቇ𝑝1 − 𝑝2

𝜌= 4𝑓𝐹

𝐿

𝐷

1

2𝑢2 + 𝑔(𝑧2 − 𝑧1

ቇ−Δp = 𝑝1 − 𝑝2 = 4𝑓𝐹𝐿

𝐷

1

2𝜌𝑢2 + 𝑔𝜌(Δz

ቇ4𝑓𝐹𝐿

𝐷

1

2𝜌𝑢2 = (𝑝1 − 𝑝2) + 𝑔𝜌(Δz

𝑢 = 𝐷

2𝑓𝐹𝐿𝜌ሾ(𝑝

1− 𝑝2) + 𝑔(Δz)

ሶ∨ =𝜋

4𝐷2𝑢

ሶ∨ =𝜋

4𝐷2

𝐷

2𝑓𝐹𝐿𝜌ሾ(𝑝

1− 𝑝2) + 𝑔(Δz)

4𝑓𝐿

𝐷

1

2𝜌

ሶ∨𝜋4𝐷

2

2

= (𝑝1 − 𝑝2) + 𝜌𝑔Δz

𝐷 =32𝑓𝜌 ሶ∨2 𝐿

)𝜋2(𝑝1 − 𝑝2 − 𝜌𝑔Δz

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Exercise

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Solution

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Solution

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Solution

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Friction by Pipe Accessories (Fitting)

Valve

Enlargement

Contraction

Flow Velocity

…etc.

Flow Direction

…etc.

FITTINGS FUNCTION

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Fitting

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Strainer

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Gate Valve

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Plug Valve

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Globe Valve

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Angle Valve

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Butterfly Valve

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Ball Valve

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(Lift) Check Valve

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(Swing) Check Valve

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o Installation of pipe accessories (fitting) will cause pressure drop of the pipe line

oPressure difference caused by fitting → equivalent with a straight pipe at diameter D that has length at the value of Le → equivalent length

oPressure drop of a fitting is defined as:

Friction Loss by Equivalent Length

−𝚫𝒑𝒇

𝝆= 𝟒𝒇𝑭

𝑳𝒆𝑫

𝒖𝟐

𝟐𝝆𝒖𝟐

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oPressure drop due to fitting installation can be related to the loss of kinetic energy when fluids flow through the fitting

oFitting constant (KL) is expressed as:

oCorrelation:

Friction Loss Due to Kinetic Energy Loss

−𝚫𝒑𝒇

𝝆= 𝑲𝑳

𝒖𝟐

𝟐

𝑲𝑳 = 𝟒𝒇𝑭𝑳𝒆𝑫

𝑳𝒆 =𝑳𝒆𝟒𝒇𝑭

𝑫

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Friction Factor due to Geometry Changes

oD1<D2 → u1>u2 → p1<p2

oEnergy loss due to friction:

oThe mass flow rate is constant, thus:

𝑒𝑓 =𝑝1 − 𝑝2

𝜌+1

2ሾu1

2− u2

2

𝑒𝑓 = 1 −D14

D24

1

2u12 −

𝑝1 − 𝑝2𝜌

𝐶𝑝 =𝑝2 − 𝑝112𝜌u1

2

𝑒𝑓 = 1 −𝐷1𝐷2

4

− 𝐶𝑝1

2u12

𝐾𝐿𝑒 = 1 −𝐷1𝐷2

4

− 𝐶𝑝

𝑒𝑓 = 𝐾𝐿𝑒1

2u12

𝐴𝑟 =D14

D24 𝑎𝑛𝑑

)𝐾𝑠𝑒 = 𝑓(𝐴𝑟

Sudden Enlargement:

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Friction Factor for Sudden Enlargement

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Pressure Drop Evaluation for Fluid Flow in the Pipe Systems with Fittings

o A pipe line is constructed by N1 segments of straight pipes with diameter D1. Fitting like valves, elbows, etc are installed between segments. The pressure drop for this pipe line systems is determined using the following equation:

o The pipe line above is connected to second pipe line system with diameter D2 using a connector. The second pipe line system has N2 segment of straight pipe and M2 fittings. The pressure drop for the second pipe line system is determined using the following equation:

𝑒𝑓𝐷1 =−Δ𝑝𝑓

𝜌𝐷1

=

𝑗=1

𝑁1

4 𝑓𝐿𝑒𝑗

𝐷1

1

2𝑢𝐷12 +

𝑗=1

𝑁1

𝐾𝐿𝑖1

2𝑢𝐷12

𝑒𝑓𝐷2 =−Δ𝑝𝑓

𝜌𝐷2

=

𝑗=1

𝑁2

4 𝑓𝐿𝑒𝑗

𝐷2

1

2𝑢𝐷22 +

𝑗=1

𝑀2

𝐾𝐿𝑖1

2𝑢𝐷22

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oThe friction heat due to diameter change in the pipe line systems (from D1

to D2) is expressed as:

oTotal friction heat:

oThe pressure difference between inlet (point 1) and outlet (point 2) of the pipe line systems is determined using the following equation:

Pressure Drop Evaluation for Fluid Flow in the Pipe Systems with Fittings

−Δ𝑝𝑓

𝜌𝐷1→𝐷2

= 𝐾𝐿𝑢𝐷12

2

𝑒𝑓𝑡 =−Δ𝑝𝑓

𝜌𝐷1

+−Δ𝑝𝑓

𝜌𝐷1→𝐷2

+−Δ𝑝𝑓

𝜌𝐷2

𝑝1 − 𝑝2𝜌

=1

2u22 −

1

2u12 + 𝑔 𝑧2 − 𝑧1 + 𝑒𝑓𝑡

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oConservation of mass:

oFor constant density

oEnergy balance equation for a pipe segment I with diameter Di and length Li is expressed as:

Fluid Flow Calculation for Complex Piping System

𝑢𝑖 = 4𝑉𝑖

𝜋𝐷𝑖2

𝑒𝑓𝑡 = 4𝑓𝐹𝐿𝑖𝐷𝑖

𝑢𝑖2

2𝑔∆𝑧 +∆𝑝

𝜌+ 𝑒𝑓𝑡 = 0

ሶ𝑉4 = ሶ𝑉2 + ሶ𝑉3

ሶ𝑚4 = ሶ𝑚2 + ሶ𝑚3

where

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oMass balance:

oTotal pressure drop:

Fluid Flow Calculation for Complex Piping System

ሶ𝑚1 = ሶ𝑚2 = ሶ𝑚3 ሶ= 𝑚4 = ሶ𝑚8

𝑝1 − 𝑝0 = 𝑝1 − 𝑝𝐴 + 𝑝𝐶 − 𝑝𝐷 − 𝑝𝐷 − 𝑝𝐵 − 𝑝𝐵 − 𝑝0

Δ𝑝 𝑡 =

𝑖

𝑁

Δ𝑝 𝑖

or:

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o Pressure drop calculation:

o Total pressure drop calculation:

𝑝𝐴 − 𝑝𝐶 + 𝑝𝑐 − 𝑝𝐷 + 𝑝𝐷 − 𝑝𝐵= 𝑝𝐴 − 𝑝𝐸 + 𝑝𝐸 − 𝑝𝐹 + 𝑝𝐹 − 𝑝𝐵= 𝑝𝐴 − 𝑝𝐵

𝑝1 − 𝑝0 = 𝑝1 − 𝑝𝐴 + 𝑝𝐴 − 𝑝𝐵 − 𝑝𝐵 − 𝑝0

o Mass balance:

Fluid Flow Calculation for Complex Piping System

ሶ𝑚1 = ሶ𝑚2 + ሶ𝑚5

ሶ𝑚2 = ሶ𝑚3 + ሶ𝑚4

ሶ𝑚5 = ሶ𝑚6 + ሶ𝑚7

ሶ𝑚4 + ሶ𝑚7 = ሶ𝑚8

ሶ𝑚1 = ሶ𝑚8

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Example

?

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Solusi

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Solusi

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Friction in Non-Circular Pipes

oNon-circular →cross section area of the pipe is not a circle

oEquivalent diameter or hydraulic diameter is used

oThe Reynolds number is defined as:

o For pipes with rectangular cross section area:

𝐷𝑒 =4 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎

𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟=4𝐴

𝑙𝑝

𝑁𝑅𝑒 =𝜌𝑢𝐷𝑒𝜇

𝐷𝑒 =4𝐴

𝑤𝑝=4𝐷2

4 𝛼 −sin 2𝛼2

𝛼𝐷= 𝐷 1 −

sin 2𝛼

2𝛼

𝐷𝑒 =4𝑤ℎ

)2(𝑤 + ℎ

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oFriction factor for non-circular pipes

oDetermination of friction factor uses this following equation:

ofF is calculated using equivalent diameter

oknc (non circular factor) → function of the size and cross section geometry

Friction Factor for Non-Circular Pipes

𝑓𝑛𝑐𝐹 = 𝑘𝑛𝑐𝑓𝐹

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Friction Factor for Non-Circular Pipes

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Exercise

Natural gas will be transferred from a gas well located in Sengkang to the city of Pare-Pare, South Sulawesi. This well production flowrate is 50

MMSCFD. The pipe length required for this is 70 km long. Sengkang is 300 m higher than Pare-Pare. Natural gas has a molecular weight of 17. The gas temperature in the pipe is assumed to be fixed at 30 oC. The gas inlet pressure of this pipe is 60 kgf/cm2 gauge. Determine the gas outlet

pressure at various pipe diameters of 5, 10, and 15 inches. Graph the relationship between the pipe outlet pressure and the pipe diameter. The

allowable pipe roughness ε/D is 0.002.

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Thank You