Introduction to Compressible Flow Dρ ≠ 0 Dt
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Introduction to Compressible Flow Dρ ≠ 0 Dt The density of a gas changes significantly along a streamline Compressible Flow Definition of Compressibility: the fractional change in volume of the fluid element per unit change in pressure p p + dp p p p + dp − v v dv p + dp p p + dp Compressible Flow V local velocity 1. Mach Number: M = = c speed of sound 2. Compressibility becomes important for High Speed Flows where M > 0.3 • M < 0.3 – Subsonic & incompressible • 0.3 <M < 0.8 – Subsonic & compressible • 0.8 <M < 1.2 – transonic flow – shock waves appear mixed subsonic and sonic flow regime • 1.2 <M < 3.0 - Supersonic – shock waves are present but NO subsonic flow • M > 3.0 – Hypersonic Flow, shock waves and other flow changes are very strong 1 Compressible Flow 3. Significant changes in velocity and pressure result in density variations throughout a flow field 4. Large Temperature variations result in density variations. As a result we now have two new variables we must solve for: T & ρ We need 2 new equations. We will solve: mass, linear momentum, energy and an equation of state. Important Effects of Compressibility on Flow 1. Choked Flow – a flow rate in a duct is limited by the sonic condition 2. Sound Wave/Pressure Waves – rise and fall of pressure during the passage of an acoustic/sound wave. The magnitude of the pressure change is very small. 3. Shock Waves – nearly discontinuous property changes in supersonic flow. (Explosions, high speed flight, gun firing, nuclear explosion) 4. A pressure ratio of 2:1 will cause sonic flow Applications 1. Nozzles and Diffusers and converging diverging nozzles 2. Turbines, fans & pumps 3. Throttles – flow regulators, an obstruction in a duct that controls pressure drop. 4. One Dimensional Isentropic Flow – compressible pipe flow. 2 Approach • Control volume approach • Steady, One-dimension, Uniform Flow • Additional Thermodynamics Concepts are needed • Restrict our analysis to ideal gases Thermodynamics • Equation of State – Ideal Gas Law p = ρRT R Universal Gas Constant 8314 J/(kmol ⋅ K) R = u = = = 287 J/(kg ⋅ K) M m Molecular mass of air 28.97kg/kmol Temperature is absolute and the specific volume is (volume per unit mass): 1 v = ρ Thermodynamics – Internal Energy & Enthalpy • Internal Energy – individual particle kinetic energy. Summation of molecular vibrational and rotational energy. u~ = u~()v,T ∂u~ ∂u~ du~ = dT + dv ∂ ∂ T v v T • For an ideal gas u~ = u~()T ~ = du cv dT • Recall from our integral form of the Energy Equation for Enthalpy of an ideal gas: h = u~ + pv h = h(T ) = dh c p dT 3 Thermodynamics – Internal Energy & Enthalpy h = u~ + pv ~ p h = u + RT = RT ρ dh = du~ + RdT Substituting: = ~ = dh c p dT du cv dT dh = du~ + RdT = + c p dT cv dT RdT = + c p cv R − = = c p cv R const Thermodynamics – Internal Energy & Enthalpy c Define the ratio of specific heats: k ≡ p = const cv Then, kR c = p k −1 R c = v k −1 For Air: cp = 1004 J/kg-K k = 1.4 The 2nd Law of Thermodynamics & Isentropic Processes We define entropy by: δQ ds = T rev Combining the 1st and 2nd Laws gives us Gibb’s Equation dp Tds = dh − ρ dp Tds = c dT − dh = c dT p ρ p 2 2 2 dT dp 1 R ds = c − R = ∫ p ∫ ∫ ρT p 1 1 T 1 p 4 The 2nd Law of Thermodynamics & Isentropic Processes − = T2 − p2 s2 s1 c p ln R ln T1 p1 For an Isentropic process: adiabatic and reversible We get the following power law relationship k k p T k −1 ρ 2 = 2 = 2 ρ p1 T1 1 Control Volume Analysis of a Finite Strength Pressure Wave Moving Wave of Frontal Area A Stationary Wave c Reference frame moving with wave 12+ ∆ p p + ∆p p p p ρ ρ + ∆ρ ρ ρ + ∆ρ T + ∆T T T + ∆T T = = − ∆ V = 0 ∆V V c V c V Steady State Continuity Equation (Solve for the induced velocity ∆V): r 0 = ∫ ρ ()V • nˆ dA = −∫ ρcdA + ∫ ()()ρ + ∆ρ c − ∆V dA CS 1 2 ρcA = ()()ρ + ∆ρ c − ∆V A ρc = c()()ρ + ∆ρ − ∆V ρ + ∆ρ ∆ρ ∆V = c (A) ρ + ∆ρ The Speed of sound (c) is the rate of propagation of a pressure wave of infinitesimal strength through a still fluid. Control Volume Analysis of a Finite Strength Pressure Wave 12 p p + ∆p Steady State Momentum Equation: ρ ρ + ∆ρ (Find ∆p and c) T T + ∆T r V = c V = c − ∆V F = ρV ()V • nˆ dA = m()V − V ∑ x ∫ x & 2 1 CS pA − ()p + ∆p A = ρcA (c − ∆V − c ) ∆p = ρc∆V (B) Now combine A & B and solve for the speed of sound: ∆ ρ + ∆ρ ∆ ∆ρ 2 = p = p + c 1 ∆ρ ρ ∆ρ ρ ∂p c 2 = in the limit of ∆ρ → 0 ∂ρ p Small Amplitude moderate frequency waves are = const isentropic and ρ k 5 Control Volume Analysis of a Finite Strength Pressure Wave Calculating the Speed of Sound for an ideal gas: p = const ρ k ∂p p p = k c = k = kRT ∂ρ ρ ρ c = kRT Typical Speeds of Sound Fluid c (m/s) Gases: For Air: H2 1,294 c Air 340 k = p ≈ 1.4 cv Liquids: R = 287 J/(kg ⋅ K) Water 1,490 Ethyl Alcohol 1,200 Data From White 2003 Example 1: Speed of sound calculation Determine the speed of sound in Argon (Ar) at 120 oC. MW = 40 kg/kmol: c = kRT R 8314 J/(kmol ⋅ K) R = u = = 207.9J/(kg ⋅ K) M m 40kg/kmol c k = p ≈ 1.668 cv c = 1.668()()207.9J/kgK 393 K = 318 .8ms -1 Movement of a sound source and wave propagation Source moves to the right at a speed V Zone of silence Mach cone α 3 c∆t V < c V > c V = 0 V ∆t V ∆t V ∆t V 1 sinα = = c M 6 Example 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer. Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard M =3 200 m α x Example 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer. Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard c = kRT = 1.4()()287 300 = 347.2m/s V = Mc = 3()347.2 = 1041.6m/s − 1 − 1 α = sin 1 = sin 1 =19.5o M 3 200m tanα = x 200m x = = 565m tan19.5 Steady Isentropic Flow – Control Volume Analysis Applications where the assumptions of steady, uniform, isentropic flow are reasonable: 1. Exhaust gasses passing through the blades of a turbine. 2. Diffuser near the front of a jet engine 3. Nozzles on a rocket engine 4. A broken natural gas line 7 Steady Isentropic Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp dx Steady State Continuity Equation: r 0 = ρ ()V • nˆ dA = − ρ V A + ρ V A ∫ 1 1 1 2 2 2 CS ρVA = ()()()ρ + dρ V + dV A + dA ρVA = ρAV + ρVdA + VAd ρ + VdρdA + ρAdV + ρdAdV + AdρdV + dρdAdV Steady Isentropic Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp dx Steady State Continuity Equation: r 0 = ρ ()V • nˆ dA = − ρ V A + ρ V A ∫ 1 1 1 2 2 2 CS ρ = ()()()ρ + ρ + + VA d V dV A dA ~ 0 ~ 0 ~ 0 ~ 0 ρVA = ρAV + ρVdA + VAd ρ + VdρdA + ρAdV + ρdAdV + AdρdV + dρdAdV dA dρ dV 0 = + + A ρ V Only retain 1st order differential terms & divide By ρVA Steady Isentropic Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp dx Steady State Energy Equation with 1 inlet & 1 exit: − 2 − 2 Q& W& s = V2 V1 + ()()()− + ~ + − ~ + g z2 z1 u pv 2 u pv 1 m& 2 Neglecting potential energy and recalling: h = u~ + pv − 2 − 2 Q& W& s = V2 V1 + − h2 h1 m& 2 Assuming and ideal gas: & − 2 − 2 Q W& s = V2 V1 + ()− c p T2 T1 m& 2 8 Steady Isentropic Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp Steady State Energy Equation with 1 inlet dx & 1 exit, neglecting potential energy & assuming Isentropic duct flow: V 2 V 2 2 + h = 1 + h 2 2 2 1 Assuming and ideal gas: V 2 V 2 2 + c T = 1 + c T 2 p 2 2 p 1 V 2 k V 2 k 2 + RT = 1 + RT 2 k −1 2 2 k −1 1 Stagnation Conditions 2 Insolated walls 1 Assume the area A2is so big V2 ~ 0, then V 2 h = 1 + h = h Stagnation enthalpy 2 2 1 o Similarly, as we adiabatically bring a fluid parcel to zero velocity there is a corresponding increase in temperature V 2 V 2 2 + c T = 1 + c T 2 p 2 2 p 1 2 = V + To T Stagnation Temperature 2c p Stagnation Conditions – maximum velocity 2 = V + To T (+) 2c p If the temperature, T is taken taken down to absolute zero, then (+) can be solved for the maximum velocity: = Vmax 2c pTo No higher velocity is possible unless energy is added to the flow through heat transfer or shaft work.