chapter 05 compressible fluid flow

7/23/2019 CHAPTER 05 Compressible Fluid Flow

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CHAPTER 5

Compressible Fluid Flow

Fluid Flow and Transport Processes (CDB 1033) May 2014

[email protected]

CLO4:Analyze problems relating to

incompressible an

d compressible fluid flow.


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8-9CHAPTER 5: COMPRESSIBLE FLUID FLOW

Speed of sound and Mach number

Processes of compressible flow: Isentropic Flow, Adiabatic Flow,Isothermal Flow

CLO 4

10 11

CHAPTER 6: FLOW PAST IMMERSED BODIES

Flow around submerged object.

Drag force, terminal velocity, Stokes law

Flow through porous media.

Blake-Kozeny / Carman Kozeny / Ergun equation

Fluidization

CLO 5

12 TEST 2 August 5, 2015 ( Time: TBA) CLO4 & CLO5

12-13

CHAPTER 7: TRANSPORTATION OF FLUID

Positive displacement pumps and compressors

Centrifugal pumps and compressors

Axial flow pumps and compressors

Compressor efficiencies

CLO 5

14

CHAPTER 8: FLUID MIXING

Types of mixing problems

Mixing in stirred tanks

Agitator, impellers, turbine

Power number, Blending and mixing, Suspension, dispersion

CLO 5

14 Group project presentation and report submissionCLO 3, CLO4 &

CLO5

Subject Planning (Week 8- 14)


Weeks Topics Outcomes


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Review on incompressible Fluid flow

DV

Re

Reynolds Number, Re

Laminar flow Transition flow Turbulent flow

Re < 2000 2000 < Re < 4000 Re > 4000

Fluid move in

orderly manner

A rapid, chaoticmotions in all

directions

Formation ofeddies / wakes in

the flow

= inertial forces / viscous forces


4/70

When the fluid is flowing in a pipe bound to shear

stress which are quantified by the term friction factor, f.

Laminar flow (Re < 2000):

VD

16

Re

16 f

Shear Stress Friction Factor


Review on incompressible Fluid flow (Contd)

Turbulent flow (Re > 4000):

For turbulent only, considering pipe roughness, , the

best approximation is given by:

3

1

6

Re

10

D200001001375.0f

Moodys Diagram can be used to Calculate friction factor, f


5/70

MZA@UTPChemEFluidMech

0.00475



Moodys Diagram


6/70

For a pipeline system consists of fittings such as elbows,

valves, enlargement and contraction, overall friction lossand head loss is determine through:

2

V

D

Lf4F

2

efittings

KKK c

Friction Loss (Head loss) for pipeline system

Contraction

Enlargement

Elbows

Valve




7/70



For a system that involves pump, overall energyequation is given as:

FW pump2

2

Vzg

P

For a system that involves turbine, overall energy equation

is given as:

FWtur2

Vzg

P2

ggg F//W2g

Vz

g

P/W

2

Vz

g

Pturb

2

22

2p

2

11

1

System Head

W /g is also known as system head.



8/70

For pump, actual work and actual power requirement,can be determined through:

where = efficiency

ideal

idealactual

wmWPower,

ww

Pump efficiency




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CHAPTER 5

Compressible Fluid Flow


[email protected]


10/70

The concept of compressible fluid Processes of compressible fluid flow:

Isentropic flow without friction

Adiabatic flow with friction Isothermal flow with friction

Mach number, Ma

Analysis of compressible flow


Chapter Outline


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At the end of this chapter, you should be able to:

Define compressible fluid flow

Determine Mach number of a systems

Characterize the compressible flow

Calculate the properties of gas flow throughvarious flow processes, such as pressure,

temperature, velocity, flow rate, etc.

Learning Outcome



12/70

Factors that should be considered in compressible flow

( but not considered in incompressible flow);

Fluid density

Changes in temperature (e.g., internal energy change)

Compressible flow

The density changes that result from pressure

changes, have a significant influence on the flow.

The changes in the flow that result from the density

changes are often termed compressibility effects.



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Importance of compressible flow

Design of high speed aircraft

Gas and stream turbines

the flow in the blading and nozzles is treated as compressible

Natural gas transmission linescompressibility effects important in calculating the flow

Reciprocating engines

the flow of the gases through the valves and intake and exhaust systems



14/70

Ideal gas thermodynamics: Quick review

Specif ic heats: Q = cmT

Q = heat added; c = specific heat;

m= mass andT = changes in temperature

The unit is J/kg Celsius

The specific heat of water is 1 calorie/gram Celsius. What does it mean?



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Ideal gas specific heat at constant volume is defined as

Ideal gas specific heat at constant pressure is defined as

From enthalpy definition, h = u + P/ = u + RT

Take derivative with respect to T of above Eqn. we have

cp = cv + R

Again the ratio of specific heat, k = cp/ cv (For air and other diatomic

gases k = 1.4 )

Combining above to relations, w can also obtain,

Ideal Gas thermodynamics: Quick review (Cont`d)

where, u = internal energy

where, u = enthalpy



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Ideal Gas thermodynamics: Quick review (Cont`d)

Internal energy (u): The energy of unit mass of fluid due to molecular

activity. Change of internal energy,

u2-u1 =cv (T2-T1)

Enthalpy (h) : Enthalpy is a measure of the total energy of a

thermodynamic system. It is represented as a sum of pressure per unit

mass (P/) and internal energy per unit mass (u).

h = u +p/

for ideal gas, h = cpT

Entropy (s): It is defined as a measure of the availability of energy for

conversion into mechanical work. The entropy changes ds for a perfect

gas is

Tds = dh - dp/



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Class example 1

Compute the change in internal energy and enthalpy of 101.94 kg of

CO2 if its temperature is increased from 15C to 65C. Take cp =

858 J/(kg.K) and cv = 670 J/(kg..K)

Solution:

Mass of CO2 = 101.94 kg; T1 = 288K and T2 = 338 K

(1) Change in internal energy per nit mass, u = cv (T1-T2)

= 33,500 J/kg

Total change in internal energy = m u = 3414 kJ

(2) Change in enthalpy per unit massh =cp(T2-T1)= 42900 J/kg

Total change in enthalpy = m h = 4373 kJ



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Self assessment exercises

A gas has a molecular weight of 16 has a cv = 1730 J/(kg.K).

Find the value of the specific heat ratio.


( )


19/70

The energies involve of a simple compressible

closed system :

Kinetic Energy

V2/2

Potential

Energy, gz

Internal

Energy, u

Energy an object has byvirtue of its motion

Energy an object has byvirtue of its position in a

field of force

Sum of all microscopicforms of energy, related to

molecular structureof a system and the

degree ofmolecular activity

Thermodynamics of a Compressible System


l id l d (C 033) 20


20/70

When the system is in a control volume (opensystem), i.e. mass and energy in/out; there will beflow energy involved.

Flow energy

P/Energy due to the

quantity of mass flowingin/out


Fl id Fl d T P (CDB 1033) M 2014


21/70

Total energy,

= flow energy(F.E)+ internal E + Kinetic E +Potential E

gz2

V

u

P

2

uRT

u

P

We know, Enthalpy, h

gz2

Vh

2

Total energy in a compressible open systems


Fl id Fl d T t P (CDB 1033) M 2014


22/70

Energy balance for an open system, with unit

mass flow in/out:

out

2

outout

in

2

inin

gz2Vhwq

gz2

Vhwq

outoutoutininin wqwq


Fl id Fl d T t P (CDB 1033) M 2014


23/70

Characteristic of Compressible flow: Mach number

In fluid mechanics, Mach number represents the ratio

of velocity of an object moving through a fluid and thelocal speed of sound.

c

VMa

V= velocity of the source relative to the medium

C = speed of sound in the medium

F/A-18 breaking the sound barrier

http://en.wikipedia.org


Fl id Flo and Transport Processes (CDB 1033) Ma 2014
http://localhost/var/www/apps/conversion/tmp/scratch_3//upload.wikimedia.org/wikipedia/commons/d/d0/FA-18_Hornet_breaking_sound_barrier_(7_July_1999).jpghttp://localhost/var/www/apps/conversion/tmp/scratch_3//upload.wikimedia.org/wikipedia/commons/d/d0/FA-18_Hornet_breaking_sound_barrier_(7_July_1999).jpghttp://localhost/var/www/apps/conversion/tmp/scratch_3//upload.wikimedia.org/wikipedia/commons/d/d0/FA-18_Hornet_breaking_sound_barrier_(7_July_1999).jpg


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A property of a material / compound.

Where:

c = sonic velocity

k = specific heat ratio, cp/cvP = absolute pressure of the fluid (kPa, psi or equivalent)

= density of the fluid (kg/m3 or equivalent)

R = specific gas constant (kJ/kgK or equivalent)

T = absolute temperature of the fluid (K or oR)

kRT

kP

Pc

21

21

S

Find the speed of sound in oxygen at a pressure of 100kPa(abs)

and 25C. Take R= 260 J/ kg. K and k = 1.40.

[329.4 m/s]

Speed of sound




25/70

Characteristic of compressible fluid flow

Ma < 0.3 : Incompressible flow

0.3 < Ma < 0.8 : Subsonic flow 0.8 < Ma < 1.2 : Transonic flow

1.2 3.0 : Hypersonic flow

Boeing 747, Ma = 0.85-0.95

(high speed, Transonic aircraft)

Concorde, Ma = 2

supersonic aircraft)


http://www.google.com.my/url?sa=i&source=images&cd=&cad=rja&docid=d6GHzCc-gFBWAM&tbnid=RGYoUgeEiMnieM:&ved=0CAgQjRwwAA&url=http://en.wikipedia.org/wiki/Concorde&ei=A605Uc31MMytrAftxoBY&psig=AFQjCNGn7jFEqBSeg3EIfO2Ds8VFSvJpAg&ust=1362820739838918http://www.google.com.my/url?sa=i&source=images&cd=&cad=rja&docid=d6GHzCc-gFBWAM&tbnid=RGYoUgeEiMnieM:&ved=0CAgQjRwwAA&url=http://en.wikipedia.org/wiki/Concorde&ei=A605Uc31MMytrAftxoBY&psig=AFQjCNGn7jFEqBSeg3EIfO2Ds8VFSvJpAg&ust=1362820739838918


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Dependency of Ma number on temperature




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Class example 2

An aeroplane is to move at Mach number of 1.5 at a pressure of 89.89 kPa.

If density of fluid is 1.112 kg/m3, calculate the speed of the plane in km/h.

Given, k = 1.4 (if the value of k is not specified, it is usual to assume 1.4.)

[1817 km/h]

Solution: Sonic velocity, = 336.4 m/s

Mach number, Ma1 = V1/C

or, V1 = C Ma1 = 504.6 m/s= 1817 km/h




28/70


1. A airplane is cruising at a speed of 800 km/h at altitude where the air

temperature is 0C. Calculate the Mach number of the flight.



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30/70

RESERVOIR RECEIVER

Conduit

Flow

Stagnation state

Stagnation state is a reference state used in compressible flow

calculations.

It is the state achieved if a fluid at any other state is brought to restisentropically.

For an isentropic flow there will be a unique stagnation state.

Fluid in this large reservoir is

almost stagnant. This reservoir

is said to be at stagnation state.




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The process is a steady flow.

Flow is one-dimensional.

Velocity gradients within a cross section are

neglected.

Friction is restricted to wall shear.

Shaft work is zero.

Gravitational effects are negligible.

Fluid is an ideal gas of constant specific heat.

Assumption for compressible flow analysis




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Convergent Divergent

FlowReservoir Receiver

Thermal insulation

Processes of compressible flow: Isentropic process

Fig. :Steady, frictionless reversible adiabatic flow

An adiabatic process is any process occurring without gain or loss of heat

within a system (e.g.,Q = 0)

thermodynamically isolated

no heat transfer with the surroundings


Throat



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Adiabatic flow with friction

Processes of compressible flow: Adiabatic process


Thermal insulation

Isentropic Friction section




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Flow

Friction section

Isothermal flow with friction

ReceiverReservoir

Processes of compressible flow: Isothermal process

An isothermal process is a change of a system, in which the

temperature remains constant: T= 0.




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As the process is isentropic, therefore it is:

Steady flow. Frictionless.

Adiabatic, q = 0

No work interaction, wf= 0. For gas flows potential energy change is

negligibly small compared to kinetic energychange, hence z = 0.

Isentropic Compressible Flow Analysis

Fluid Flow and Transport Processes (C 033) May 0 4



36/70

One dimensional isentropic flow

Applying the steady flow energy equation between 1 and 2 we have:

Applying conditions for Isentropic adiabatic flow, q= 0 and if no work is done then Wf=0,

we have,

(1)

Consider gas flowing in a duct which varies in size . The pressure and

temperature may change

1 2

(2)

p ( ) y

For horizontal flow, Z1=Z2

Bernoullis equation for gas



37/70

Stagnation enthalpy

The sum (+2/ 2) is known as stagnation enthalpy and it is constant

inside the duct.

It is called stagnation enthalpy because a stagnation point

has zero velocity and the enthalpy of the gas is equal to R at

such a point.

p ( ) y



38/70

Consider, gas in the reservoir is in stagnation state

1T

T

1k

2

RkT

V

1T

T

1k

2

RkTV

TT1k

2RkV

1

R

1

2

1

1

R1

2

1

1R

2

1

(3)

(4)

Isentropic Flow Analysis (Cont.)

or,

or,

or,

1RP TT2C

Flow

Reservoir Receiver

VR, TR. PR, hR, V1, T1, P1, h1,

From Eq. 2, we can write,

h

h2V 1R2

1

p ( ) y



39/70

Using the definition of speed of sound

Substitute the above Eq. into Eq. (4), and using Ma term,

1

2

1

11

kRTc

kRTc

1

T

T

1

2

V

1

R2

1

21

kc

1

2

1Ma

T

T 21

1

R

k


(5)Relat ion between stagnat ion

temperature and mach number

p y



40/70


For compressible fluid, pressure and density change

accordingly to the change in temperature:

The isentropic (frictionless, adiabatic) relation is givenby:

1k

k

1

R

1

R

T

T

P

P

T

T

1k1

1

R

1

R

(a)

(b)

Relat ion between stagnat ion

temperature and stagnat ion pressure

Relat ion between stagnat ion

temperature and stagnat ion density



41/70


Substituting Eq. (5) into (a) and (b), respectively:

1k

k2

1

1

R 12

1kMaP

P

1k

12

1

1

R 12

1kMa

(6)

(7)

Relat ion between s tagnat ion

pressure and mach num ber

Relat ion between s tagnat iondens i ty and m ach n umber



42/70

In any flow, mass is conserved.

From continuity equation:


RR

11

1

R

1R

V

V

A

AAVAV

(8)




43/70

Class example 3An air plane is moving in an atmosphere with pressure 44 kPa( abs) and

Density 0.63 kg/m3 . A pitot tube on the plane records the stagnation

pressure as 70 kPa(abs). Estimate the speed of the airplane and stagnationTemperature. (k = 1.4 and R = 287 J/kg. K)

1kk

21

1

R

12

1kMa

P

P

Solution: (1) stagnation pressure PR is given by

Ma1 = 0.8422

Sonic velocity, = 312.7 m/s

Mach number, Ma1 = V1/C

or, V1 = C Ma1 = 263.4 m/s



44/70

Class example 3 (cont.)An air plane is moving in an atmosphere with pressure 44 kPa( abs) and

Density 0.63 kg/m3 . A pitot tube on the plane records the stagnation

pressure as 70 kPa(abs). Estimate the speed of the airplane and stagnationTemperature. (k = 1.4 and R = 287 J/kg. K)

Solution: (2) stagnation temperature TR is given by

12 1MaTT2

1

1

R

k

Temperature of the atmosphere

T1 = p/R = 243.35 K

or, TR = 277.87 K = 4.87 C



45/70


1. A supersonic plane flies at 1900 km/h in air having a pressure of 28.5 kPa (abs)

and density 0.439 kg/m3. Calculate the (a) temperature, (b) pressure and (c)

density of air at the stagnationpoint on the nose of the plane.(a) 91.8 C; (b) 151.85 kPa (abs) and (c) 1.45 kg/m3

2. A conduit conveys air at a Mach number of 0.70. At a certain section the static

pressure is 30 kPa (abs) and the temperature is 25C. (a) calculate the stagnation

temperature and pressure (b) if the stagnation temperature is 90C, what would be

the Mach number of the flow.(a)TR = 59.7C; PR = 41.61 kPa (abs) (b) M1= 0.995

3. An aircraft cruises at 12 km altitude. A pitot-static tube on the nose of the aircraft

measures stagnation and static pressures of 2.6 kPa and 19.4 kPa. Calculate

a) the flight Mach number of the aircraft

b) the speed of the aircraft

c) the stagnation temperature that would be sensed by a probe on the aircraft.



46/70

Critical state is the special state where Ma = 1.

It is shown with an asterisk, like , , etc. Ratios derived previously can be written using the critical

state

Critical State

1kk

2

1

1

R 12

1kMa

P

P

12

1Ma

T

T 21

1

R

k

1kk

R 12

1k

*P

P

12

1

*T

TR

k

Ma = 1

Solve problems given in the handout



47/70

Relation between area perpendicular to the flow and area at criticalstate:

At subsonic to get the fluid go faster, one must reduce the cross

sectional area perpendicular to the flow.

At supersonic to get the fluid go faster, one must increase the crosssectional area perpendicular to the flow.

1k21k

21

1

1

12

1k

12

1kMa

Ma

1

A

A

Simple area change flow



48/70

Mass flow rate of fluid can be determined from

continuity equation with respect to the critical state:

For air with k = 1.4:

12

1

2

1

12

1

kk

R

R

k

RT

kP

A

m

RT

AP6847.0m

R

R

Simple area change flow (Cont`t)



49/70

Self assessment exercise

1. A converging duct is fed with air from a large reservoir where the

temperature and pressure are 350 K and 200 kPa. At the exit of theduct, cross-sectional area is 0.002 2 and Mach number is 0.5.

Assuming isentropic flow

a)Determine the pressure, temperature and velocity at the exit.

b)Find the mass flow rate

2.Air is flowing isentropically in a diverging duct. At the inlet of the duct,

pressure, temperature and velocity are 40 kPa, 220 K and 500 m/s,

respectively. Inlet and exit areas are 0.002 2 and 0.003 2.

a) Determine the Mach number, pressure and temperature at the exit.

b) Find the mass flow rate.



50/70

Friction loss are involved when a gas flows through a

length of pipe at high velocity,

If pipe is insulated or flow is fast, heat transfer isconsidered negligible adiabatic.


Thermal insulation


Adiabatic flow with friction



51/70

Effect of friction due to flow will cause the entropy offlowing gas to increase (entropy is not constant)

Therefore isentropic relation cannot be applied in theanalysis.


Thermal insulation


Adiabatic flow with friction (Cont`d)



52/70

Applying the momentum balance:

Mass flow rate x (Velocity out Velocity in) = Net

pressure force Force due to wall shear stress

PT

V

P + dPT + dT

V + dV

+ d

dFfriction

dx

frictionFPdPPAVdVVm d




53/70

Applying the continuity equation:

AV = constantV = constant (since A is constant)

V = ( + d)(V + dV)

P

T

V

P + dP

T + dT

V + dV

+ d

dFfriction

dx


Fluid Flow and Transport Processes (CCB 1033) January 2013


54/70

Applying the energy balance:

Equation of state: (P = RT)

P + dP = ( + d) R (T + dT)

P

T

V

P + dP

T + dT

V + dV

+ d

dFfriction

dx

2

dVVdTTC

2

VTC

2

P

2

P




55/70

Also from Mach number definition:

PT

V

P + dPT + dT

V + dV

+ d

dFfriction

dx

dTTkRdVV

dMaMa

kRT

VMa

c

VMa

22

22




56/70

The equations represents a set of equations with

unknown dP, dT, d, dV and dMa Have to be solved accordingly to obtain

appropriate expressions.

P

T

V

P + dP

T + dT

V + dV

+ d

dFfriction

dx



57/70

In momentum balance, there exist the term wallshear stress, twall.

In pipeline system, this is expressed asdimensionless value friction factor, f

Most compressible gas flows in duct involve

turbulent flow

P

T

V

P + dP

T + dT

V + dV

+ d

dFfriction

dx




58/70

Solving from the equation (for circular pipe):

The equation describe the change of Ma over agiven length.

When friction is involved, flows tend to reachsonic condition Ma 1.

2

1

2

2

22

2

1

22

21 Ma1k

2

11

Ma1k2

11

Ma

Ma

ln2k

1k

Ma

1

Ma

1

k

1

D

x4 f




59/70

When friction is involved, flows tend to reachsonic conditionMa 1 (Ma2 approaching 1)

By setting Ma2 = 1, the length of duct required togive the value of Ma1 is obtained as maximumlength, Lmax (or critical length, L*)

2

1

2

1

2

1

2

1

Ma1k2

112

Ma1kln

2k

1k

Mak

Ma1

D

L4 *f


Adi b i fl i h f i i (C `d)



60/70

Air flows in a 5 cm diameter pipe. The air enters at Ma = 2.5 and is to

leave at Ma = 1.5. Determine the length of pipe required. What would

be the maximum length of pipe?Assume f = 0.002 and adiabatic flow.

m1850L

2.51412

11

1.5141211

1.5

2.5ln

2(1.4)

141

1.5

1

2.5

1

1.4

1

050

L00204

Ma1k2

11

Ma1k2

11

Ma

Maln

2k

1k

Ma

1

Ma

1

k

1

D

x4

2

2

2

2

22

2

1

22

2

2

2

1

2

2

2

1

.

.

..

.

.

f

Flow is adiabatic




61/70

Maximum pipe length, L*:

m72L

2.51412

112

2.5141ln

1.42

11.4

2.541

2.51

050

L00204

Ma1k

2

112

Ma1kln

2k

1k

Mak

Ma1

D

L4

2

2

2

2

2

1

2

1

2

1

2

1

.*

.

.

..

*.

*f




62/70

Occurs in long, small, uninsulated pipe in contact

with environment transmit sufficient heat to keepthe flow isothermal.

E.g.: flow of natural gas through long distancepipelines.

Flow

Friction section

ReceiverReservoir

Isothermal flow with friction



63/70

From Bernoullis equation:

For long pipelines, V dV is negligible compared to the

other terms:

P

V

P + dP

V + dV

+ d

dFfriction

dx

dQ

D

dx

2

V-4fdPdVV

2

D

dx

2

V-4fdP

2

(I-1)

(I-2)

Isothermal flow with friction (Con`t)



64/70

From continuity equation:

P

V

P + dP

V + dV

+ d

dFfriction

dx

dQ

TR

PM

RT

P

AVm

u

w

(I-3)

(I-4)




65/70

Substitute (I-3) and (I-4) into (I-2):

(I-5)

D

dx

A

m

P

RT

2

4-dP

D

dx

A

m

2

4

-dP

D

dx

A

m

1

2

4-dP

D

dx

A

m

2

4-dP

2

2

2

2

2

f

f

f

f




66/70

Rearrange equation (I-5):

Integrate, and taking limit between two locationsin the pipeline:

(I-6)D

dx

A

m

2

RT4-PdP

2

f

1 2

dx




67/70

Rearrange equation (I-6)

Rearranging, by bringing to the LHS :

(I-7)D

L

A

m

2

RT4-

2

P2

2

f

2

122

222

RTL4

DAPm

RTL4

DAPm

f

f

m

(I-8)




68/70

If the pipe is circular:

21

252

2

2

1

2

122

4RTL4

DPPm

RTL4

DAPm

f

f

(I-9)




69/70

Friction factor can be approximated throughWeymouth equation:

Where, D is in inches.

f

dx

3

1

D0.0080f (I-10)


Ch k li t f thi h t



70/70

Check list for this chapter

Compressible fluid

Mach number and characterization of compressible flow

Steady isentropic adiabatic flow without friction loss.

Stagnation conditions

Critical state

Adiabatic flow with friction loss

Isothermal flow analysis

End of this Chapter

chapter 05 compressible fluid flow

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