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Chap 5 Quasi-One- Dimensional Flow
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5.1 Introduction Good approximation for practicing gas dynamicists eq. nozzle flow 、 flow through wind tunnel & rocket engines
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5.2 Governing Equations For a steady,quasi-1D flow The continuity equation :
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The momentum equation : Automatically balainced X-dir Y-dir
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The energy equation total enthalpy is constant along the flow Actually, the total enthalpy is constant along a streamline in any adiabatic steady flow
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PAuρPAuρ P +dP A +dA u +du ρ+dρ dx In differential forms Dropping 2nd order terms (1) (2)
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(1) - (2) = Euler’s equation
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5.3 Area-Velocity Relation ∵ adiabatic & inviscid ∴ no dissipation mechanism → isentropic
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Important information 1.M→0 incompressible flow Au=const consistent with the familiar continuity eq for incompressible flow 2.0 ≦ M < 1 subsonic flow an increase in velocity (du , +) is associated with a decrease in area (dA ,- ) and vice versa. 3.M>1 supersonic flow an increase in velocity is associated with an increase in area, and vice versa 4.M=1 sonic flow →dA/A=0 a minimum or maximum in the area
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A subsonic flow is to be accelerated isentropically from subsonic to supersonic Supersonic flow is to be decelercted isentropically from supersonic to subsonic
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Application of area-velocity relation 1.Rocket engines
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2.Ideal supersonic wind tunnel Diffuser is to slow down the flow in the convergent duct to sonic flow at the second throat, and then futher slowed to low subsonic speeds in the divergent duct. (finally being exhausted to the atmosphere for a blow-down wind tunnel) “chocking”“blocking” (When both nozzle with M=1) Handout – Film Note by Donald Coles
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5.4 Isentropic Flow of a Calorically Perfect Gas through Variable-Area Duct Stagnation density (constant throughout an isentropic flow)
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(1) (2) (3) Area – Mach Number Relation
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There are two values of M which correspond to a given A/A * >1, a subsonic & a supersonic value Boundary conditions will determine the solution is subsonic or supersonic
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1.For a complete shock-free isentropic supersonic flow, the exit pressure ratio Pe /P 0 must be precisely equal to Pb /P 0 2.Pe /P 0 、 Te /T 0 & P e /P 0 = f(A e /A*) and are continuously decreasing. 3.To start the nozzle flow, P b must be lower than P 0 4.For a supersonic wind tunnel, the test section conditions are determined by (A e /A*) 、 P 0 、 T 0 gas property & P b
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P b =P 0 at the beginning ∴ there is no flow exists in the nozzle Minutely reduce P b , this small pressure difference will cause a small wind to blow through the duct at low subsonic speeds Futher reduce P b , sonic conditions are reached (P b =P e3 ) P e /P 0 & A/A t are the controlling factors for the local flow properties at any given section
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for r=1.4
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Should use dash-line to indicate irreversible process What happens when Pb is further reduced below Pe 3 ?
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Note: quasi-1D consideration does not tell us much about how to design the contour of a nozzle – essentially for ensuring a shockfree supersonic nozzle Method of characteristics
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Wave reflection from a free boundary Waves incident on a solid (free) boundary reflect in like (opposite) manner, i.e, a compression wave as a compression (expansion wave ) and an expansion wave reflects as an expansion ( compression ) wave
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5.5 diffusers Assume that we want to design a supersonic wind tunnel with a test section M=3 A e /A * =4.23 P 0 /P e =36.7 3 alternatives (a) Exhaust the nozzle directly to the atmosphere
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(b) Exhaust the nozzle into a constent area duct which serves as the test section ∴ the resvervair pressure required to drive the wind tunnel has markedly dropped from 36.7 to 3.55 atm
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(c) Add a divergent duct behind the normal the normal shock to even slow down the already subsonic flow to a lower velocity For For
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∴ the reservoir pressure required to drive a supersonic wind tunnel (and hence the power required form the compressors) is considerably reduced by the creation of a normal shock and subsequent isentropic diffusion to M ~ 0 at the tunnel exit Note:
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Diffuser - the mechanism to slow the flow with as small a loss total pressure as possible Consider the ideal supersonic wind funnel again
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If shock-free →P 02 /P 01 =1 no lose in total pressure →a perpetual motion machine!!← something is wrong (1) in real life, it hard to prevent oblique shock wave from occuring inside the duct (2) even without shocks, friction will cause a lose of P 0 ∴ the design of a perfect isentropic diffuser is physically impossible Replace the normal shock diffuser with an oblique shock diffuser provide greater pressure recovery
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Diffuser efficiency (mostl common one) If η D =1→normal shock diffuser for low supersonic test section Me, η D >1 for hypersonic conditions η D <1 (normal shock recovery is about the best to be expected) Normal shock at M e
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Is very sensitive to A t2 >A t1 (due to the entropy increase in the diffuser) proof: assume the sonic flow exists at both throats always
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A t2 η D =max is slightly larger than (P 01 /P 02 )A t1 ∴ the fix- geometry diffuser will operate at an efficiency less than η D,m to start properly η D is low it is because A t2 is too large ∴ the flow pass though a series of oblique shock waves id still “very” supersonic ∴ a strong normal shock form before exit of the diffuser ∴ defeats the purpose of are oblique shock diffuser
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