Fluids are inherently unsteady, viscous and compressible

We limit our study to steady flows.

Under certain situations viscosity and compressibility is not important – inviscid, incompressible fluids are studied (Chap 6)– viscous, incompressible fluids are studied (Chap 8&9)– one-dimensional, inviscid, compressible fluids are studied(Chap 11&12)

Ch. 11: Introduction to Compressible Flow

• Introduction (focus on 1-dimensional, compressible, = 0 or = constant)

• Liquids, = constant for us (1% increase in for every 1.6 km deep)• Air, 1% change for every 85 ft deep; M = 0.3 ~ 5% = /• M > 0.3 , ~ 103 m/s or 230 mph• Significant density changes imply significant compression

or expansion work on the gas, which can change T, u, s, …• Compressibility “Paradoxes”: 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 engineeringapplications, captures trends)

Ch. 11: Introduction to Compressible Flow

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.

Compressible flows significantly more complicated

Incompressible ~ 4 equations and 4 unknowns

Compressible ~ 7 equations and 7 unknowns

GOVERNING EQUATIONS FOR NEWTONIAN FLUIDSINCOMPRESSIBLE

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

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

uj/t + ukuj/xk = -p/xj + (2ui/xjxj) + fj4 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

e/t + uke/xk = -puk/xj +/xj(k T/xj) + (uk/xk)2 + (ui/xk + uj/xi)(uk/xk)

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

e = e(,T) Caloric ~ e = CvT7 Equations: continuity, momentum(3), energy, thermal, state 7 Unknowns: p, u, v, w, e, T, Know: , fj, , k

IDEAL GAS

Good to 1% for air at 1 atm and temperatures > 140 K (-130 oC)or for room temperature and < 30 atm

Automobile engine the ratio of air to fuel is 15:1, and can be reasonably described by the ideal gas law.

pointmass

perfectelastic

collisions

p = RT [R=Runiv/mmole] (11.1)

du = cVdT (11.2)

u2- u1 = cv(T2 – T1) (11.7a)

dh = cpdT (11.3)

h2- h1 = cp(T2 – T1) (11.7b)

cp + cv = R (11.4)

k cp/cv ([k=] (11.5)

cp = kR/(k-1) (11.6a)

cv = R/(k-1) (11.6b)

IDEAL

GAS

EQUATION OF STATEFOR IDEAL GAS

p = RT (11.1)

= unique constant for each gas

[units of Kelvin]Static pressure

. .

..

.

.

(# of collisions/sec)1

p1, n1, m1, vx1, T1, L1

(# of collisions/sec)2

p2, n1 m1, vx1, T1, L2=2L1

L 2L

Daniel Bernoulli ~ PV = const

Hydrodynamics, 1738

.

.

.

.

.

.

If L doubled (system 2) but same v, then

(# of collisions/sec)1 = v x (1 sec)/L (# of collisions/sec)2 = v x (1 sec)/2L

(# of collisions/sec)2= ½ (# of collisions/sec)1

Daniel Bernoulli ~ Hydrodynamics, 1738

(system 1)

PV = const

Daniel Bernoulli PV = const

p = F/A

F {# collisions / sec}

p1 (# of collisions/sec)1/(L)2

p2 (# of collisions/sec)2/(2L)2

p2 ½ (# of collisions/sec)1/(2L)2

p2 = 1/8 p1

Vol2= 8Vol1

p2Vol2 = p1Vol1 QED

“ The elasticity of air is not only increased by compression but by heat supplied to it, and since it is admitted thatheat may be considered as an increasing internal motion of the particles, it follows that … this indicates a more intense motion of the particles of air.”

Daniel Bernoulli

Here was the recipe for quantifying the idea that heat is motion

– two generations before Count Rumford, but it came too early.

IDEAL GAS: p = RT (eq. 1.11)R = Runiv/mmole pV = N(# of moles)RunivT

1662: Boyle and Hooke experimentally showed that: PV = const for const T;

Boyle’s book laid the foundation for modern chemistry

Assume perfect elastic reflections so: - 2mvx is change of x-momentum per collision.

Initially assume vx is same for all particles.

What is Pressure ?

t = 2L/vx

= (mvx)/t =2mvx/(2L/vx) = mvx2/L

Time between collisions, t, of particle with samewall is equal to:

L

Force of one particle impact = Magnitude of momentum change per second due to one particle:

nmvx2/L

Magnitude of momentum change per second due to n molecules:

<vx2> = <vy

2> = <vz2>;

<vx2> + <vy

2> + <vz2> = <v2>

<vx2> = 1/3

1/3 nm<v2>/L

Pressure = F/A = [1/3nm<v2>/L]/L2

P = 1/3nm<v2>/L3

PV = 1/3nm<v2> = 2/3n (1/2 m<v2>)

average kinetic energy per particle

Empirically it is found that : PV = nkBT

n = #of particles; kB=1.38x10-23 J/K

We have reasoned:PV = 2/3n (1/2 m<v2>)

Empirically it is found that:

PV = nkBT

T(Ko) = [2/(3kB) ] [avg K.E.]

pV = (2/3) n <mv2/2>

Uinternal for monotonic gas

Uint = f(T) depending if p or V held constant

uint, v,… designate per unit mass

duint/dT = cv duint/dT =cp

(# of particles)

pV = nkBT

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

IDEAL GAS

p = RT pV = nRT

Same for all gases

Different for each gas

pV = nkBTn = [Nm][NAvag]

6.02x1023

nkBT = Nm x NAvag kBT = Nm x NAvag [Runiv/NAvag.] T

pV= NmRunivT

pV= NmRunivT

p=(1/V)Nmmmole{Runiv/mmole}T

p=(m/V){Runiv/mmole}T

p= {Runiv/mmole}T = RT (11.1)

WORK

The differential work dW done on the gas in compressing it by moving it –dx is –

Fdx.

dW on gas = F(-dx) = -pAdx = -pdVgoes into dT

FIRST LAW OF THERMODYNAMICS

Q + W = E = (KE + PE + U)Q/m + W/m = E/m; q + w = uU, internal energy, is energy stored in molecular bonding forces and random molecular motion. (KE and PE we will ignore)

W = - pdV

SPECIFIC HEATS

cV = dq/dT

du = cvdTcp = dq/dT

dh = cpdT

IDEAL GAS

h = u + pvq - w = du

The amount of heat per unit mass to raise the temperature of the system 1 degree (Kelvin) depends on if the process occurs at constant volume, or constant pressure.

For constant volume [definition]:

cvol = [ q / T]vol

but for constant volume, v = 0:

u = q + w = q - p v = q

So:

cvol = [q / T]vol = u / T

For perfect gas u(T)

cvol = du/dT or du = cvoldT (11.2)

for constant cvol: u2 – u1 = cvol (T2 – T1) (11.7a)

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.

mCvdT = dQ or CvdT = dq

q + w = u

if Volume 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

The amount of heat per unit mass to raise the temperature of the system 1 degree (Kelvin) depends on if the process occurs at constant volume, or constant pressure.

For constant pressure [definition]:

cpres = [ q / T]pres

but for constant pressure, p =0: h =(pv + u) = (p)v + p(v) + u =q +vp

h = qSo:

cpres = [q /T]pres = hpres / T

For perfect gas h(T) = pv + ucpres = dh/dT or dh = cpresdT (11.3)

for constant cpres, h2 – h1 = cpres (T2 – T1) (11.7b)

Definition of heat capacity at constant pressure:

mCpdT = dQ or Cp = dq/dT

q + w = u; dq = du + pdv; h = u + pv

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

IDEAL GASdu = w + q (CONSERVATION OF ENERGY)

q = cvdT + pdv pv = RT (R=Runiv/mmole)

pdv + vdp = RdTq = cvdT + RdT - vdpDivide by dT

[q]/dT = cv + R – vdp/dTIf isobaric, i.e. dp=0 then

{[q]/dT}p = cp = cv + R (11.4)

cp = cv + R; cp – cv = RDivide by cv, & let k = cp/cv;

k - 1 = R/cv, or

cv = R/(k-1) (11.6b)

Multiply by cp/cv = k cp = kR/(k-1) (11.6a)

(often k expressed as )

cp/cv = k = 1.4 for perfect gas

k

cp/cv = k = 1.4 for perfect gas

NOTE

Speed of propagation

Equilibrium

It is assumed that systems is always in equilibrium.

In compressible flows where high speeds are common and strong pressure and velocity gradients exist it has been foundby experiment that an instantaneous local equilibrium exist.

Assume all gases obey ideal gas law:

p = RT

Not gauge pressure Kelvin (or Rankine)

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

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

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

(a) Surrounding flow field is affected by pressure disturbances.

(b) Sound waves swept downstream, sound waves collect alonga front normal to the flow direction producing a normal shock.

(c) Sound waves swept downstream at a greater speed so waves confined to wedge shaped region – producing an oblique shock.

SUB SONIC SUPER

M = v/c

SUB SONIC SUPER

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.

SECOND LAW

Tds =Q/mreversibleprocess

Tds > Q/mirreversible

process

OF THERMODYNAMICS

“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

Tds = du + pdv = dh –vdp always truedq = du + pdv ds =q/T reversible

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

For any other process the entropy increases.

cv = du/dT cp = dh/dT

Tds = du + pdv = dh –vdpds = du/T + RTdv/Tds = CvdT/T + (R/v)dvs2 – s1 = Cvln(T2/T1) + Rln(v2/v1)s2 – s1 = Cvln(T2/T1) - Rln(2/1)

Ideal Gas

p = RT

Not Isentropic!!! (11.11a)

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)

T/k-1 = constant assumptions

ISENROPIC AND IDEAL GAS

(11.11a)

Cv = du/dT Cp = dh/dT

Tds = du + pdv = dh –vdpds = dh/T - vdpds = CpdT/T - (R/p)dps2 – s1 = Cpln(T2/T1) - Rln(p2/p1)s2 – s1 = Cvln(T2/T1) - Rln(p2/p1)

Ideal Gas

p = RT

Not Isentropic!!! (11.11b)

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)

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

Tp(1-k)/k = constantassumptions

ISENROPIC AND IDEAL GAS

(11.12b)

s2 – s1 = cpln(T2/T1) - Rln(p2/p1)T = pv/R cp-cv = R

s2 – s1 = cpln[(p2v2)/(p1v1)] – (cp-cv)ln(p2/p1)s2 – s1 = cpln[v2/v1] + cvln(p2/p1)

(11.11a)

s2 – s1 = 0 = cpln[v2/v1] + cvln(p2/p1)-(cp/cv) ln[v2/v1] = ln(p2/p1)

ln[v2/v1]-k =ln(p2/p1)p2v2

k = constant (11.12c)

To prove – (p2/2) + u2 + ½ V2

2 + gz2 = (p1/1) + u1 + ½ V12 + gz1

For steady flow ~dE/dt = dQ/dt + dW/dt

dm/dt[(u2 + ½ V22 + gz2) - (u1 + ½ V1

2 + gz1)] = dQ/dt + dW/dt

dW/dt = dW/dtpressure + dW/dtviscous + dW/dtshaft

Pressure work = pAds = pAVt = (p/)(dm/dt) t

dW/dtpressure = (p1/1)(dm/dt) - (p2/2)(dm/dt)

dW/dtpressure = (p1/1)(dm/dt) - (p2/2)(dm/dt)

dE/dt = dQ/dt + dW/dt

If no viscous or shaft work and no heat interaction:

dQ/dt = 0dW/dt = dWpressure/dt

dm/dt[(u2 + ½ V22 + gz2) - (u1 + ½ V1

2 + gz1)] = (p1/1)(dm/dt) - (p2/2)(dm/dt)

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

2 + gz1

the end

(for this part)

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

Cp = dh/dT (ideal gas)

CpdT2 + ½ V22 = CpdT1 + ½ V1

2

If pick stagnation conditions V = 0

CpdTo = CpdT1 + ½ V12

CpTo = CpT + ½ V2

Cp – Cv = R; R/Cp = 1- 1/kCp = R/(1-1/k) = kR/(k-1)

M2 = V2/c2 = V2/[kRT]

[kR/(k-1)]To/T = [kR/(k-1)] + ½ V2/TTo/T = 1 + (1/2 V2) / (T[kR/(k-1)])

To/T = 1 + {(k-1)/2} V2/(kRT)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)

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

(isentropic = adiabatic + reversible)

SECOND LAW

Tds =Q/mreversibleprocess

Tds > Q/mirreversible

process

OF THERMODYNAMICS

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

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

dp = (d)RT for T constant

dp/d = RT = c2

(/p)|T = c2 = RT

~ 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

Ideal gas: p/k = constant

or differentiating

dp/k – pk -k-1 d = 0 dp/p – kd/ = 0

~ SPEED OF SOUND ~

dp/p – kd/ = 0

dp/d = kp/

but remember this was for isentropic conditions

p/|s = c2 = kp/

p = RT for ideal gas c2 = kRT

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

If not concerned with sound propagation, low Mach number flows may be considered incompressible.

At what M does this occur?

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

M2 = V/cp = RT (ideal gas)

if isothermal (actually isentropic)1% change in density ~ 1% change in pressure

0.01p = 0.01*101325 N/m2 1/2 V2 = 1013.25 N/m2

V = 41 m/sM = 41/343 = 0.12

5% change in density ~ 5% change in pressure1/2 V2 = 0.05p = 0.05*101325 N/m2

V = 92 m/sM = 41/343 = 0.27 ~ 0.3

• The concept of absolute zero extends to a great many phenomena:– volume of a gas

(Charles law - 1800)

– electrical noise in a resistor

– wavelength of radiation emitted by a body

In the early 1800’s Lord Kelvin developed a universal thermodynamic scale based on the coefficient of expansion of an ideal gas.

Constant Pressure: Vol. vs Temp.

pV = 2/3 U for monotonic gaspV = (k - 1) U in general

k = cp/cv = 5/3 for monotonic gasU = pV/(k - 1)

dU = (pdV+Vdp)/(k - 1) – eq. of state

ASIDE: Want to derive important relation between p and V for adiabatic (i.e. Q = 0)

reversible (no friction) condition

Compression of gas under adiabatic conditions means all work goes into

increasing the internal energy of the molecules, so:

dU = W = -pdV for adiabatic (Q = 0)

Equation of statedU = (Vdp + pdV) / (k - 1)

Cons. of energy dU = W + Q

dU = W + QdU = (Vdp + pdV) / (k - 1)

-pdV = (Vdp + pdV) / (k - 1)

-(pdV)(k - 1) = Vdp + pdV

-(pdV)k + pdV = Vdp + pdV

-(pdV)k - Vdp = 0

0

-(pdV)k - Vdp = 0

(divide by -pV)(dV/V) + (dp/p) = 0

(integrate)kln(V) +ln(p) = ln(C)

ln(pVk) = ln(C)

pVk = C or pvk = c (11.12c)

pVk = C or pvk = p/k = c (11.12c)Ideal gas - p = RT

p/k = c p/[p/RT]k = p[1-k]Tk = cp[1-k]/k T = c (11.12b)

p/k = cRT/k = [1-k]RT = c{[k-1]/ [k-1]} [1-k]RT = cT/[k-1] = Tv[k-1] = c (11.12a)

ISENTROPIC & IDEAL GAS