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
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.
9 Stagnation Conditions – Mach number relations
V Recall, that the Mach number is defined as: M = c
2 = V + To T For Ideal gases: 2c p T V 2 kR 1 o = + 1 c T = T = kRT p − − T T 2c p k 1 k 1 2 2 c T k −1 V k −1 c p o = +1 = M 2 + 1 T 2 c 2 2
T k −1 o = M 2 +1 T 2
Stagnation Conditions – Isentropic pressure & density relationships
T k −1 o = M 2 +1 T 2 k k p T k −1 k −1 k −1 o = o = M 2 + 1 p T 2 1 1 ρ T k −1 k −1 k −1 o = o = M 2 + 1 ρ T 2
Critical Values: conditions when M = 1
* T = 2 + To k 1
k * − p = 2 k 1 + po k 1
1 * ρ k −1 = 2 ρ + o k 1 1 * c = 2 2 + co k 1
10 Critical Values: conditions when M = 1
For Air k = 1.4
T * 2 = = 0.8333 + To k 1
k p * 2 k −1 = = 0.5283 + po k 1
1 * ρ 2 k −1 = = 0.9129 ρ + o k 1 1 * c 2 2 = = 0.9129 + co k 1
In all isentropic flow, all critical values are constant.
Critical Values: conditions when M = 1
Critical Velocity: is the speed of sound c*
1 * c = 2 2 + co k 1 1 1 2 2 * = * = * = 2 = 2kRTo V c kRT co k + 1 k +1
Example 3: Stagnation Conditions Air flows adiabatically through a duct. At point 1 the velocity
is 240 m/s, with T1 = 320K and p1 = 170kPa. Compute (a) To (b) Po (c) ro (d) M
(e) Vmax (f) V*
11 Steady Isentropic Duct Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp
dx Recall, for Steady isentropic flow Continuity: dA dρ dV ρ (†) 0 = + + dA = − d − dV A ρ V A ρ V For compressible, isentropic flow the momentum equation is:
dp dV 2 dp (*) 0 = + = + VdV Bernoulli’s Equation! ρ 2 ρ neglecting gravity
Substitute (†) into (*)
dA dρ dp dp 1 dρ = − + = − 2 2 A ρ ρV ρ V dp
Steady Isentropic Duct Flow 12 h h + dh ρ ρ + dρ V V + dV T T + dT p p + dp dA dp 1 dρ = − 2 A ρ V dp ∂p c 2 = Recall that the speed of sound is: ∂ρ dA dp 1 1 dp V 2 = − = 1 − 2 2 2 2 A ρ V c ρV c V Substituting the Mach number: M = c dA dp Describes how the pressure = ()1 − M 2 behaves in nozzles and diffusers ρ 2 A V under various flow conditions
Nozzle Flow Characteristics dA dp = ()1 − M 2 A ρV 2
1. Subsonic Flow: M < 1 and dA < 0, then dP < 0: indicating a decrease in pressure in a converging P P channel. 2. Supersonic Flow: M > 1 and dA < 0, then dP > 0: indicating an increase in pressure in a converging P P channel. 3. Subsonic Flow: M < 1 and dA > 0, then dP > 0 : indicating an increase in pressure in a diverging P P channel. 4. Supersonic Flow: M > 1 dA > 0, then dP < 0 : indicating a decrease in pressure in a diverging P P channel.
12 Steady Isentropic Duct Flow – Nozzles Diffusers and Converging Diverging Nozzles
dA dp Describes how the pressure (††) = ()1− M 2 behaves in nozzles and diffusers ρ 2 A V under various flow conditions Recall, the momentum equation here is: dp dp 0 = + VdV = −VdV (**) ρ ρ Now substitute (**) into (††) : dA dV = ()M 2 −1 A V Or, dA A = ()M 2 −1 dV V
Nozzle Flow Characteristics
dA dV = ()M 2 −1 A V
1. Subsonic Flow: M < 1 and dA < 0, then dV > 0: indicating an accelerating flow in a converging channel. 2. Supersonic Flow: M > 1 and dA < 0, then dV < 0: indicating an decelerating flow in a converging channel. 3. Subsonic Flow: M < 1 and dA > 0, then dV < 0 : indicating an decelerating flow in a diverging channel. 4. Supersonic Flow: M > 1 dA > 0, then dV > 0 : indicating an accelerating flow in a diverging channel.
Converging-Diverging Nozzles
Amin Subsonic Supersonic M = 1
Amax M < 1 Subsonic Subsonic Supersonic M > 1 Supersonic
Flow can not be sonic
13 Choked Flow – The maximum possible mass flow through a duct occurs when it’s throat is at the sonic condition Consider a converging Nozzle: receiver pr po pe To ρ Ve o plenum
Mass Flow Rate (ideal gas): V V p M = = m& = ρVA = VA RT c kRT p k m& = M kRT A = p MA RT RT
k m& = p MA RT
Choked Flow Mass Flow Rate (ideal gas): k m& = p MA RT Recall, the stagnation pressure and Temperature ratio and substitute: k − − p k −1 k 1 To k 1 2 o = M 2 +1 = M +1 p 2 T 2 k +1 − 2()1−k = k + k 1 2 m& po MA1 M RTo 2 If the critical area (A*) is where M=1: k +1 + 2()1−k = * k k 1 m& po A RTo 2 The critical area Ratio is: k +1 A 1 2 + ()k −1 M 2 2()k −1 = * A M k + 1
14