8 compressible flow

HCMC University of Technology 08/12/200957:020 Fluid Mechanics 1

Compressible Flow

Supersonic Nozzle Design

Shock Wave Induced Condensation around a F18

Shock Tube

HCMC University of Technology 08/12/200957:020 Fluid Mechanics

Content

1. Basic Equations for 1D Compressible Flow

2. Isentropic Flow of an Ideal Gas – Area Variation

3. Flow in a Constant-Area Duct with Friction

4. Frictionless Flow in a Constant-Area Duct with Heat Exchange

5. Normal Shocks

6. Supersonic Channel Flow with Shocks

7. Oblique Shocks and Expansion Waves

8. Applications



• Continuity Equation

• Equation of State



• First Law of Thermodynamics

• Second Law of Thermodynamics



Basic Equations


2. Isentropic Flow of An Ideal Gas – Area Variation

• The flow is changed only by area variation

• No heat transfer (δQ/ δm = 0)

• No friction

• No shocks

Basic Equations



• Momentum Equation:

• Continuity Equation:



Reference Stagnation and Critical Conditions for Isentropic Flow of Ideal Gas



Relation between Mach number M and area A



Example 12.1: Isentropic Flow in a Converging Channel

GIVEN: Isentropic flow of air in a channel. M1 = 0.3, T1 = 62oC, p1 = 650kPa(abs), A1 = 0.001 m2, and M2 = 0.8

FIND: (a) The channel shape (b) Ts diagram(c) Properties at section 2

p02, T02, h02, ρ2, V2, A2


2. Isentropic Flow in a Converging Nozzle

(i) Valve is closed, there is no flow through the nozzle

(ii) The back pressure, pb, is reduced to slightly less than po, there will be flow

(iii) Continue decrease pb, exit-plane pressure pe will continue to decrease

(iv) Lower pb, the flow rate increase, is there a limit to flow rate through nozzle?



(iv) Maximum flow rate occurs when we have sonic conditions at the exit plane, when Me=1, and pe=pb=p* (choked flow)



(v) Continue to lower pb, the flow remains choked: the mass flow rate does not increase, pe = p* > pb. After exiting, flow adjusts down to pb (nonisentropic, series of expansion waves and shocks)



Example 12.2: Isentropic Flow in a Converging Nozzle

Flow is isentropic. At = 0.001m2

FIND: Me, mɺ

Example 12.3: Choked Flow in a Converging Nozzle

M1 = 0.52, T1 = 40oF, p1 = 60psia, A1= 0.013ft2

FIND: Mt, At, mɺ



Isentropic Flow in a Converging-Diverging Nozzle

(i) Opening the valve slightly (pb lightly less than p0), flow rate is low enough, the

flow will be subsonic everywhere (behave as a venturi)

(ii) Valve is opened farther, flow rate is increase, flow is subsonic everywhere




(iii) Valve is opened farther, at the section of minimum area M = 1, the nozzle is choked




(iv) If the exit pressure is set at piv, subsonic flow in convergence & supersonic flow in divergence nozzle

(v) Lowering the back pressure below piv, no effect on flow in the nozzle, additional expansion takes place outside the nozzle



Example 12.4: Isentropic Flow in a Converging-Diverging Nozzle

GIVEN: To = 350K, pp = 1.0MPa

pb = 954kPa, Mt = 0.68, Ae = 0.001m2

FIND: (a) Properties and area at nozzle throat (b) Me

Example 12.5: Isentropic Flow in a Converging-Diverging Nozzle: Choked Flow

GIVEN: To = 350K, po = 1.0MPa, pe(design) = 87.5kPa, pb = 50kPaAe = 0.001m2, At = 4.8x10-4 m2

FIND: (a) Me (b) mɺ



Basic Equations for Adiabatic Flow



Adiabatic Flow: The Fanno Line



Adiabatic Flow: The Fanno Line

Pressure drops rapidly Pressure gradient cancels the drag due to friction

Density decreases largely Velocity must be increasing!



Maximum Entropy Point – Choked Flow

• Entropy increases in the direction of flow, there is a maximumentropy point corresponding to M=1 for each Fanno line

• The maximum entropy point is reached by increasing the amount of friction, just enough to produce M=1 at the exit

• If adding duct beyond the critical duct length, inlet flow is subsonic, the additional length forces the sonic condition to move down.

• If inlet flow is supersonic, additional length causes a normal shock to appear somewhere in the duct, and the shock moves upstream as more duct is added



Example 12.6: Frictional Adiabatic Flow in a Constant-Area Channel

GIVEN: Air flow in insulated tube

FIND: (a) (b) Stagnation pressure at section 2(c) Force on duct wall

mɺ



Fanno-Line Flow Functions for 1D Flow of an Ideal Gas

Lmax = the maximum possible length of duct between a section where the Mach number is M and the section where sonic condition occur

p* is constant for all states in the same Fannoline



Example 12.7: Frictional Adiabatic Flow in a Constant-Area Channel

GIVEN: Air flow (with friction) in an insulated constant-area tube

FIND: L13, M2, L12, sketch Ts diagram

Gage pressures:p1=-18.9mmHg, p2=-412mmHg, M3=1

(Colebrook)



Basic Equations for Flow with Heat Exchange



The Rayleigh Line

Heating: entropy increases

Cooling: entropy decreases



The Rayleigh Line



Example 12.8: Frictionless Flow in a Constant-Area with Heat Addition

GIVEN: Frictionless flow of air in duct shown:T1=600oR, p1=20psia, p2=10psia, V1=360ft/sA1=A2=A=0.25ft2

FIND: (a) Properties at section 2(b) δQ/dm (c) s2-s1 (d) Ts diagram



Rayleigh Line Flow Functions for 1D Flow of an Ideal Gas

(Momentum Equation)



Example 12.9: Frictionless Flow in a Constant-Area Duct with Heat Addition

GIVEN: Frictionless flow of air shown:T1 = 333K M2=1.2 p1=135kPa(abs)V = 732m/s

FIND: (a) Properties at section 2 (b)δQ/dm(c) s2-s1 (d) Ts diagram


5. Normal Shocks

Knowledge of property changes across shocks and shock behavior is important in understanding the design of supersonic diffusers, e.g, for inlets on high performance aircraft, and supersonic wind tunnels.

Set of Basic Equations


5. Normal ShocksThe width of the control volume is infinitesimal (in reality about 0.2µm), so A2 ~ A1~A, the force due to the walls Rx~0, and the heat exchange with the walls δQ/dm~0.

Nature tells that in the absence of area change, friction, and heat transfer, flow properties will not change except in a very abrupt, irreversible manner, for which the entropy increases!


5. Normal Shocks

• The normal shock must satisfy all seven of basic equations. For a given state 1, the end state (state 2) of the normal shock must lie on both the Fanno line and Rayleighline passing through state 1.

• Flow through a normal shock involves a change from supersonic to subsonic speeds.

• Normal shocks can occur only in flow that is initially supersonic


5. Normal ShocksThe effects of normal shock on flow properties


5. Normal Shocks

Normal-Shock Flow Functions for 1D Flow of an Ideal Gas


5. Normal ShocksExample 12.10: Normal Shock in a Duct

GIVEN: Normal shock in a duct as shown:T1=5oC, p1=65 kPa, V1=668 m/s

FIND: (a) Properties at section 2(b) s2-s1 (c) Ts diagram


6. Supersonic Channel Flow with ShocksFlow in a Converging-Diverging Nozzle


5. Normal ShocksFind the position and strength of the shock

Me

M1

The area at which this shock occurs can then be found from the isentropic area relation for isentropic flow between the throat and state 1.


7. Oblique Shocks and Expansion WavesOblique Shocks



Continuity Equation:

Momentum Equation:



The oblique shock equations for an ideal gas

1st law of thermodynamics

2nd law of thermodynamics



Example 12.11: Comparison of Normal and Oblique Shocks

GIVEN: Air flow with p1=100kPa, T1=-2oC, V1=1650m/s

FIND: Downstream pressure, temperature, and speed if it experience (a) a normal shock and (b) an oblique shock at angle β = 30o. Also find the deflection angle Ө

Normal Shock

Oblique Shock


7. Oblique Shocks and Expansion WavesOblique Shock Deflection Angle



Example 12.12: Oblique Shocks on an Airfoil

GIVEN: Air flow over sharp leading edge p1=100kPa, δ =6o, T1=4oC, α = 1o, V1=600m/s

FIND: Pressure on upper and lower surfaces



Isentropic Expansion Waves

An isentropic wave differs from an oblique shock wave in two important ways:

• The wave angle is α = sin-1(1/M), instead of angle β for the oblique shock

• The changes in velocity and in density, pressure, etc., and the deflection angle, are all infinitesimals




Continuity Equation

Momentum Equation

1st law of thermodynamics




(Prandtl-Meyer Supersonic Expansion Function Value forRegion with Mach number M)

Total deflection caused by isentropic expansion from M1 to M2

2nd law of thermodynamics:



Basic Equations of Isentropic Expansion Waves



Example 12.13: Expansion Waves on an Airfoil

GIVEN: Air flow over sharp leading edge withp1=100kPa δ=6o α =6o T1=4oCV1=600m/s

FIND: Pressure on upper and lower surfaces

SOLUTION:

(a) Upper surface: isentropic expansion, deflection angle θu=3o

(b) Lower surface: oblique shock, deflection angle θt=9o



Example 12.14: Lift and Drag Coefficients of a Supersonic Airfoil

GIVEN: Air flow over symmetric section withp1=100kPa δ=6o α =6o T1=4oCV1=600m/s

FIND: Pressure distribution, lift and drag coef.

SOLUTION:

Prandtl-Meyer function value for the region 3u:



Example 12.14: Lift and Drag Coefficients of a Supersonic Airfoil

• We did not need to analyze the flow after the trailing edge expansion waves and oblique shock

• Unlike a subsonic flow, a supersonic flow can generate drag even in the absence of B.L and flow separation

• A supersonic flow can negotiate a sharp corner, even if we include the effect of a viscous B.L. This is because an expanding supersonic flow has a negative pressure gradient

• An actual airfoil is not likely to have planar surfaces, so more sophisticated techniques than we can cover here are needed


Applications

• Supersonic Nozzle Design

• Wings in Compressible Flow

• Jet Engine-Inlet Design


Applications• Supersonic Nozzle Design

• Wings in Compressible Flow

• Jet Engine-Inlet Design


Ramjet Engine

Applications

8 compressible flow

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