CBE 150A – Transport Spring Semester 2014 Compressible Flow.

CBE 150A – Transport Spring Semester 2014

Compressible Flow


Goals• Describe how compressible flow differs from

incompressible flow

• Define criteria for situations in which compressible flow can be treated as incompressible

• Provide example of situation in which compressibility cannot be neglected

• Write basic equations for compressible flow

• Describe a shape in which a compressible fluid can be accelerated to velocities above speed of sound (supersonic flow)


Basic EquationsFive changeable quantities are important in compressible flow:

1.Cross-sectional area, S

2.Velocity, u

3.Pressure, p

4.Density,

5.Temperature, T


Basic Equations

Restrict focus to those systems in which properties are only changing in flow direction.

Generally, cross-sectional area S is specified as a function of x. (S=S(x))

Need four equations to describe the other four variables.


Basic Equations1.1. Mass Balance relates Mass Balance relates , u, S, u, S

2.2. Mechanical Energy Balance relates Mechanical Energy Balance relates , , u, S, pu, S, p

3.3. Equation of State relates Equation of State relates T, p,T, p,

4.4. Total Energy Balance relates Total Energy Balance relates Q, TQ, T

What is different about compressible flow?

, u, p all change with position.

Need to use differential form of equations.


Mass Balance uSm constant

uSdSduudSuSd In differential form

Divide both sides by uS

0

d

u

du

S

dS

m

md


Mechanical Energy Balance

f

p

p

hdp

gZu

W

2

12

ˆ2

Differentiate and assume Ŵ = 0

02

2

fdh

dpgdz

ud


Viscous Dissipation

2

4 2u

D

Lfh f

2

4 2u

D

dLfdh f

For a short section of pipe: Assumes only wall shear (no fittings)

02

42

22

u

D

dLfdpgdz

ud


Equation of State

zRTpV zRT

PM

V

M

For simplicity it is assumed that z is either 1 (ideal) or a constant

0T

dT

V

dV

p

dp

0T

dTd

p

dp

Volume:

Density:


Total Energy Balance

For gases thermodynamics allows a better calculation of the heat transfer Q and changes in internal energy. These were terms that were previously included in the viscous dissipation term.

The temperature of a flowing gas depends on:

• Rate of heat transfer Q from environment.

• Rate of viscous dissipation (significant in compressors). Included in work term Ŵc

• Thermodynamic changes H.



cWm

QHgZ

u ˆ2

2

Q is the rate of heat addition along the entire length of the channel and Ŵc is the total rate of energy input into the system and includes efficiency to account for viscous dissipation.

For Ŵc to be in the correct units use:

fl bf tB T U 7 7 81


Compressible vs. Incompressible

When can simpler incompressible equations be used?

• Density change is not significant (<10%)• Fans, airflow through packed beds

Mach number is a measure of the importance of density changes for compressible fluids.

sound

fluidMa velocity

velocityN

Rule of Thumb: NMa < 0.3 assume incompressible


Isentropic Flow

Adiabatic (Q = 0) and ReversibleIsentropic (ΔS = 0)

Venturi meter, Rocket propulsion


Adiabatic Flow

Adiabatic (Q = 0), Frictional

Short Insulated Pipes

Mathematically more difficult


Isothermal Flow

Isothermal, Frictional

Long Uninsulated Pipes


Compressible Flow Through Pipes


Goals

• Describe equations useful for analyzing isothermal, compressible flow through a constant diameter pipe.

• Describe how Mach number and L are related for flow in a constant diameter pipe.

• Use equations for isothermal flow to compute the flow rate of compressible fluids in constant diameter pipes.


Isothermal FlowConstant Diameter Pipe

Goal is to analyze the friction section. Flow through pipes is irreversible so viscous dissipation is important.

P1, 1 P2, 2


Mass Balance

21 uSuS S is constant

21 uu

21 GG Mass velocity constant

Differential Balance

011

dx

du

udx

d


Mechanical Energy Balance

Wdx

dh

dx

dp

dx

dzg

dx

duu f ˆ1

turbulent horizontal no compressor

02

41 2

u

D

f

dx

dp

dx

duu



cp Wdx

dQ

dx

dTC

dx

dzg

dx

duum ˆ

turbulent horizontal isothermal

dx

dQ

mdx

duu

1

Note: This indicates that there must be heat transfer because dT = 0. This is the heat required to keep T constant.

no work


Equation of State

0111

dx

dT

Tdx

d

dx

dp

p

isothermal

011

dx

d

dx

dp

p


Isothermal Flow

02

0

2

121

12

1

Lp

p

p

pdx

D

fdpp

Gpp

dp

Combining Mass, MEB and EOS

Assume friction factor f is constant and integrate:

2

1

22

1

122

21 ln4

p

p

Gppp

D

Lf


Constant f ?

constantuG

constantT constant

constantuD

Re

constantf


Isothermal Flow

2

1

22

1

122

21 ln4

p

p

Gppp

D

Lf

P1, 1 P2, 2


Isothermal Flow

2

1

2

22

21

2

ln4

pp

DL

f

ppzRTM

G

For a fixed P1 this expression has a maximum at:

2max

11

112max

ln4

1Gp

DLf

pG


Maximum Flow

22max pG

2

2max

pu

M

zRT TSu ,

Thus for a constant cross-section pipe the maximum obtainable velocity is Mach one for any receiver pressure. This is said to be choked flow.

Ernst Mach (1838-1916)


“Choked” Flow

P1

PCritical

Vsonic

GMax

G

P1PCritical

Unattainable Flows

Attainable Flows0

P

GUG PipeofEndSonicMax

TR

MwtPUUG

Sonic Velocity


Example Problem (maximum flow)

An astronaut is receiving breathing oxygen at 10 C from his space capsule through a 7 meter long, 1.7 cm diameter, hose. The capsule supply pressure is 200 kPa and the suit pressure is 100 kPa. What is the flow rate of the oxygen to the suit ? If the hose breaks off at the suit, what is the flow rate of oxygen ? What is the pressure at the end of the hose ? The hose is “smooth”.


Calculation Approach (subsonic flow) Given P1, P2, and T

Assume subsonic flow at the end of

the pipe.

Assume GCalculate

NRE

Calculate f

Iterate

Calculate V at end of pipe

Calculate V sonic at

end of pipe

If V > V sonic - flow is unattainable - got to next page

Calculate G


Calculation Approach (sonic flow) Given P1, P2, and T

Assume sonic flow at the end of the

pipe.

Assume GMax

Assume FDTF

Calculate f

Iterate

Check FDTF assumption

If P2 (sonic) > P2 - flow is sonic at end of pipe and G = GMax

Calculate GMax

Calculate NRE

Calculate P2 (sonic)


10 Minute ProblemNitrogen (Nitrogen ( = 0.02 cP ) is fed from a high pressure cylinder through = 0.02 cP ) is fed from a high pressure cylinder through ¼ in. ID stainless steel tubing ( k = 0.00015 ft) to an experimental ¼ in. ID stainless steel tubing ( k = 0.00015 ft) to an experimental unit. The line ruptures at a point 10 ft. from the cylinder. If the unit. The line ruptures at a point 10 ft. from the cylinder. If the pressure in the nitrogen in the cylinder is 3000 psig and the pressure in the nitrogen in the cylinder is 3000 psig and the temperature is constant at 70 F, what is the mass flow rate of the temperature is constant at 70 F, what is the mass flow rate of the gas through the line and the pressure in the tubing at the point of gas through the line and the pressure in the tubing at the point of the break ?the break ?

P = 3014 psia

P = 1 atm

10 ft


Reversible Adiabatic Flow

0

0

T

p

r

r

T

p


Converging/Diverging Nozzle


Isentropic Flow of Inviscid Fluid

00 SQ

In this case The mass balance and MEB are the same as that for the isothermal case.

Now though the total energy balance will give a relation between the velocity and temperature



0

dx

dQ

dx

dTC

dx

dZg

dx

duum p

1 horizontal adiabatic

0dx

dTC

dx

duu p


Equation of State

0111

dx

dT

Tdx

d

dx

dp

p

Given the normal equation of state, the TEB, MEB, and the thermodynamic relation Cp – Cv = zR/M, isentropic flow gives the following useful values.


Useful RelationshipsGiven the normal equation of state, the TEB, MEB, and the thermodynamic relation Cp – Cv = zR/M, isentropic flow gives the following useful values.

1

00

0

0

00

T

T

p

p

pp

VppV

v

p

C

C


From Mechanical Energy Balance

01

dx

dp

dx

duu

0

1 dpudu

or

dpp

pdpudu

11

00

1

Integrating

1

00

020

2 11

2

p

ppuu u ↔ p


Isentropic Flow

0

020

2 11

2

T

T

M

zRTuu

u ↔ T

1

00

020

2 11

2

p

uu u ↔


Velocity, NMa, and Stagnation

For isentropic flow the definition of the speed of sound is:

M

RTp

d

dpu

S

SS

,

It is also convenient to express the relationships in terms of a reference state where u0 = 0. This is called the stagnation condition (u0 = 0) and P0 and T0 are the stagnation pressure and temperature.


Velocity – Mach Relationships

The previous relationships now become:

11

21

02

p

pNMa

1

1

2 02

T

TNMa

and


Cross-Sectional Areafor Sonic Flow

Application of the continuity (mass balance) equation gives:

121

2

* 1

121

Ma

Ma

N

NS

S

S* is a useful quantity. It is the cross-sectional area that would give sonic velocity (NMa = 1).


Summary of Equations for Isentropic Flow

121

2

* 1

121

Ma

Ma

N

NS

S

1

2

0 2

11 MaN

p

p

12

0 2

11

MaNT

T

1

1

2

0 2

11 MaN

These ratios are often tabulated versus NMa for air ( = 1.4). One must use the equations for gases with ≠ 1.4.

p0, T0, 0, are at the stagnant (reservoir) conditions.


Maximum Mass Flow Rate

Since the maximum velocity at the throat is NMa = 1, there is a maximum flow rate:

1

1

00*

max 1

2

pSm

Increase flow by making throat larger, increasing stagnation pressure, or decrease stagnation temperature. Receiver conditions do not affect mass flow rate.


Drug Injection via Converging / Diverging

Nozzle

Contour Shock TubePowdered drug cassette

Helium cylinderSupersonic jet


Shock Behavior


Shock Behavior

PoPt PR

PR = Pc

PR = Pe

PR = Pf

Isentropic Paths

Non-Isentropic Paths

Pe< PR < Pf

Sonic Flow at throat (maximum mass flowrate)


10 Minute Problem

Air flows from a large supply tank at 300 F and 20 atm (absolute) through a converging-diverging nozzle. The cross-sectional area of the throat is 1 ft2 and the velocity at the throat is sonic. A normal shock occurs at a point in the diverging section of the nozzle where the cross-sectional area is 1.18 ft2. The Mach number just after the shock is 0.70.

What would be the pressure (P1) at S = 1.18 ft2 if no shock occurred ?

What are the new conditions (T2 and P2 ) after the shock ?

What is the Mach number and pressure at a point in the diverging section of the nozzle where the cross-sectional area is 1.8 ft2 ?


CFD Simulation of Nozzle Behavior

CBE 150A – Transport Spring Semester 2014 Compressible Flow.

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Transcript of CBE 150A – Transport Spring Semester 2014 Compressible Flow.