Ch. 11: Introduction to Compressible Flow

• Focus on 1-dimensional, compressible, inviscid flows

• Liquids, = constant for us (1% increase in for every 1.6 km deep)

• Air, 1% change for every 26 m deep

• M = 0.3 ~ 5% = /; M = 0.3 ~ 100 m/s or 230 mph

• Significant density changes imply significant compression or expansion work on the gas, which can change T, e, s, …

•Compressibility: fluid acceleration because of friction, fluid deceleration in a converging duct, fluid temperature decrease

with heating• Ideal Gas: p = RT (simple, good approximations for our engineering

applications, captures trends)

-Smits/A Physical Introduction To Fluid Mechanics

Density gradients will affect how light is transmitted though medium (by affecting index of refraction). By applying the Gladstone-Dale formula it becomes evident that the shadowgraph is sensitive to changes in the 2nd derivative of the gas density.Strength of shock can be related to width of dark band.- Methods of Experimental Physics – Vol 18, Martin

Deflection of light caused by shock compressed gas ahead of a sphere flying at supersonic speed.

Thoughts on the increased complexity of incompressible flow

GOVERNING EQUATIONS FOR NEWTONIAN FLUIDSINCOMPRESSIBLE

/t + /xk(uk) = 0 becomes uk/xk= 0

uj/t + ukuk/xk = -p/xj+/xj( uk/xk)+/xi[(ui/xj+uj/xi)]+fj

becomes

uj/t + ukuj/xk = -p/xj + (2ui/xjxj) + fj

4 Equations: continuity and three momentum 4 Unknowns: p, u, v, wKnow: , , fj

GOVERNING EQUATIONS FOR NEWTONIAN FLUIDSCOMPRESSIBLE

/t + /xk(uk) = 0

uj/t + uk/xk = -p/xj +/xj( uk/xk) + /xk[(ui/xk + (uj/xi)] + fj

p = p(,T) Thermal ~ p = RT

e = e(,T) Caloric ~ e = CvT

e/t + uke/xk = -puk/xj +/xj(k T/xj) + (uk/xk)2 + (ui/xk + uj/xi)(uk/xk)7 Equations: continuity, momentum(3), energy, thermal, state 7 Unknowns: p, u, v, w, e, T, Know: , fj, , k

e is the internal energy

Thoughts on the speed of sound

COMPRESSIBLE FLOW

front

c2 = (p/)s

COMPRESSIBLE FLOW

If fluid incompressible, gas would behave like solid body and moveeverywhere at piston speed. If pressure disturbance is small relative to p1 then “front” propagates at speed of sound. If large shock waves occur where speed, temperature, density and pressure change significantly across shock. (Speed of shock is between the speed of sound in the compressed and undisturbed gas.)

front

~ SPEED OF SOUND ~

Sound waves are pressure disturbances << ambient pressure.

For loud noise: p ~ 1Pa whereas ambient pressure is 105 Pa

Speed of sound: c2 = (p/)s

Assumptions: ideal gas & isentropic p/k = const, or differentiating

dp/k – pk -k-1 d = 0 dp/(p/cont) – (const k) k -k-1 d = 0

dp/p – kd/ = 0

~ SPEED OF SOUND ~

dp/p – kd/ = 0

dp/d = kp/

p = RT for ideal gas then c2 = kRT

For 20oC and 1 atmospherec = 343 m/s = 1126 ft/s = 768 mph

M = 1

Dynamic Pressure Static Pressure

M = V/c

M2 = V2/c2 = V2/kRT (ideal gas)

p = RT (ideal gas)

M2 = 2(1/2V2/kRT) = 2(1/2 V2/(kp/))

M2 = 2[1/2 V2/(kp)] ~ 1/2 V2/p

M2 ~ dynamic pressure/ static pressure

Thoughts on the speed of sound

As related to the speed of the source

Regimes of flow:

(1) Acoustics – fluid velocities << c, speed of sound; fractional changes in p, T and are important.

(2) Incompressible flow – fluid velocities < c, speed of sound;fractional changes in are not significant; fractional changes in p and T are very important

(3) Compressible flow (gas dynamics) – fluid velocities ~ c, speed of sound; fractional changes in p, T and are all important.

SUB SONIC SUPER

Propagation of Sound Waves from a Moving Source

Some Assumptions

It is assumed that the system is always in equilibrium.

It has been found by experiment that as long as the temperatures and pressures are not too extreme, the flow attains aninstantaneous equilibrium. This continuesto hold even inside shock waves. For all the flows examined here, all systems willbe assumed to be in equilibrium at all times.

p1,1,T1,

s1, h1

p2,2,T2,

s2, h2

It is assumed that all gases obey ideal gas law:

p = RT

Not gauge pressure

Kelvin (or Rankine)

R = Ru/Mm = 287.03 m2/(s2-K) = (N-m)/(kg-K) = J/(kg-K)R = 1716.4 ft2/(s2-R)

Conservation of Energy

(note that u is now the internal energy)

FIRST LAW OF THERMODYNAMICS

Q + W = E = (KE + PE + U)

Q/m + W/m = E/m; q + w = u

W = - pdVU, internal energy, is energy stored in molecular bonding forces and random molecular motion. (KE and PE we will ignore)

Ideal gas is composed of point particles which exhibit perfect elastic collisions. Thus internal energy is a function of temperature only. U = f(T)

Enthalpy, h, defined as: h = u + pv ; h = f(T) since h(T) = u(T) + RT

Specific Heat for Ideal Gas

dQ = mCv,pdT

Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of substance by1oK. Different

for constant volume or pressure.

Definition of heat capacity at constant volume:

mCvdT = dQ or CvdT = dq

dq + dw = du

if Vol constant, w = -pdv = 0, then dq = du,

Cv = du/dT “It can be shown that du = Cv dT even if volume not held constant!!”

- pg 41, Thermal-Fluid Engineering, Warhaft

Definition of heat capacity at constant pressure:

mCpdT = dQ or Cp = dq/dT

dq + dw = dq – pdv = du h = u + pv

dh = du +pdv +vdp

if pressure constant, dh = du + pdv = dq

Cp = dh/dT~ again can be shown to be true even if pressure is not constant!!

Cv = du/dT* Cp = dh/dT*

h = u + pv = u + RT*dh = du + RdT

dh/dT = du/dT + RCp – Cv = R*

* IDEAL GAS

cp/cv = kcp – cv = R

cp/cp –cv/cp = R/cp

1 – 1/k = R/cp

(1– 1/k)/R = 1/cp

cp = R/(1-1/k) = kR/(k-1)

cp/cv = kcp – cv = R

cp/cv –cv/cv = R/cv

k – 1 = R/cv

(k – 1)/R = 1/cv

cv = R/(k – 1)

Another Assumption

It is assumed that cp/cv

is not a function of T

calorically perfect

For a perfect gas cp/cv = 1.4

cp/cv = k is not a function of temperature

k

cp/cv = k = 1.4 for perfect gas

The Second Law

“The second law of thermodynamics can be stated in several ways, none of which is easy to understand.”

– Smits, A Physical Introduction to Fluid Mechanics

dq = du + pdvTds = du + pdv = dh –vdp always true!

DEFINITIONS = rev Q/T or dS = (Q/T)rev

Change in entropy intimately connected with the concept of reversibility – for a reversible, adiabaticprocess entropy remains constant.

For any other process the entropy increases.

DEFINITIONS = rev Q/T or dS = (Q/T)rev

What we can do with:

Tds = du + pdv = dh –vdp+ ideal gas and constant specific heats

Cv = du/dT Ideal Gas Cp = dh/dTp = RT = (1/v)RT

Tds = du + pdv = dh –vdpds = du/T + RTdv/T

ds = CvdT/T + (R/v)dv

s2 – s1 = Cvln(T2/T1) + Rln(v2/v1)

s2 – s1 = Cvln(T2/T1) - Rln(2/1)

s2 – s1 = Cvln(T2/T1) - Rln(2/1)

If isentropic s2 – s1 = 0 ln(T2/T1)Cv = ln(2/1)R

Cp – Cv = R; R/Cv = k – 1

2/1 = (T2/T1)Cv/R = (T2/T1)1/(k-1)

assumptions

ISENROPIC & IDEAL GAS& constant cp, cv

Cv = du/dT Ideal Gas Cp = dh/dTp = RT = (1/v)RT

Tds = du + pdv = dh –vdpds = dh/T - vdp

ds = CpdT/T - (RT/[pT])dp

s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)

s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)

s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)If isentropic s2 – s1 = 0ln(T2/T1)Cp = ln(p2/p1)R

Cp – Cv = R; R/Cp = 1- 1/k

p2/p1 = (T2/T1)Cp/R = (T2/T1)k/(k-1)

assumptionsISENROPIC & IDEAL GAS

& constant cp, cv

Stagnation Reference (V=0)

BE: 1-D, energy equation for adiabatic and no shaft or viscous work.

(p2/2) + u2 + ½ V22 + gz2 = (p1/1) + u1 + ½ V1

2 + gz1

Definition: h = u + pv = u + p/; assume z2 = z1

h2 + ½ V22 = h1 + ½ V1

2 = ho + 0

Cp = dh/dT (ideal gas)

ho – h1 = cp (To – T) = ½ V12

T0 = ½ V12/cp + T = T (1 + V2/[2cpT])

T0 = ½ V12/cp + T = T (1 + V2/[2cpT])

cp = kR/(k-1)

T0 = T (1 + V2/[2(kR/(k-1)]T])

T0 = T (1 + (k-1)V2/[2kRT])

c2 = kRT

T0 = T (1 + (k-1)V2/[2c2])

V2/ c2

T0 = T (1 + [(k-1)/2] M2)

To/T = 1 + {(k-1)/2} M2

STEADY, 1-D, ENERGY EQUATION FOR ADIABATIC FLOW OF A PERFECT GAS

/o = (T/To)1/(k-1)

To/T = 1 + {(k-1)/2} M2

/o = (1 + {(k-1)/2} M2 )1/(k-1)

Ideal gas and isentropic

and constant cp, cv

(isentropic = adiabatic + reversible)

p/p0 = (T/To)k/(k-1)

To/T = 1 + {(k-1)/2} M2

p/p0 = (1 + {(k-1)/2} M2)k/(k-1)

Ideal gas and isentropic and constant cp, cv

(isentropic = adiabatic + reversible)

QUIZWhen a fixed mass of air

is heated from 20oC to 100oC –(a)What is the change in enthalpy?(b)For a constant volume process,

what is the change in entropy?(c)For a constant pressure process,

what is the change in entropy?(d)For an isentropic process what are the changes in p and ?(a)Compare speed of sound

for isentropic and isothermal conditions.

(a) h2 – h1 = Cp(T2- T1)(b) s2 – s1 = Cvln(T2/T1)(c) s2 – s1 = Cpln(T2/T1)(d) 100/ 20 = (T100/T20)2.5

2.5 = 1/(k-1) k = 1.4 for ideal gas p100 / p20 = (T100/T20)3.5

3.5 = k/(k-1) k = 1.4 for ideal gas

(e) c2 = {dp/d}But c2 = (p/)|T does not equal c2 = (p/|S)

If isentropic p/k = constant (ideal gas)Then c = {(p/)|S}1/2 = (kRT)1/2

= (1.4 * 287.03 * (20 + 273.15))1/2

= 343.2 m/s

If isothermal p = RT (ideal gas)Then c = {(p/)|T}1/2 = (RT)1/2

= (287.03 X (20 + 273.15)1/2 = 290.07 m/s 18% too low