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 engineering applications, 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 2 nd 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 FLUIDS INCOMPRESSIBLE  /  t +  /  x k (  u k ) = 0  u k /  x k = 0  u j /  t +  u k  u k /  x k = -  p/  x j +  /  x j (  u k /  x k )+  /  x i [  (  u i /  x j +  u j /  x i )]+  f j  u j /  t +  u k  u j /  x k = -  p/  x j +  (  2 u i /  x j  x j ) +  f j 4 Equations: continuity and three momentum 4 Unknowns: p, u, v, w Know: , , f j

GOVERNING EQUATIONS FOR NEWTONIAN FLUIDS COMPRESSIBLE  /  t +  /  x k (  u k ) = 0  u j /  t +  u k /  x k = -  p/  x j +  /  x j (  u k /  x k ) +  /  x k [  (  u i /  x k + (  u j /  x i )] +  f j  e/  t +  u k  e/  x k = -p  u k /  x j +  /  x j (k  T/  x j ) + (  u k /  x k ) 2 +  (  u i /  x k +  u j /  x i )(  u k /  x k ) p = p( ,T) Thermal ~ p =  RT e = e( ,T) Caloric ~ e = C v T 7 Equations: continuity, momentum(3), energy, thermal, state 7 Unknowns: p, u, v, w, e, T,  Know: , f j,, k

IDEAL GAS Good to 1% for air at 1 atm and temperatures > 140 K (-130 o C) 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. point mass perfect elastic collisions

p =  RT [R=R univ /m mole ] (11.1) du = c V dT (11.2) u 2 - u 1 = c v (T 2 – T 1 ) (11.7a) dh = c p dT (11.3) h 2 - h 1 = c p (T 2 – T 1 ) (11.7b) c p + c v = R (11.4) k  c p /c v ([k=  ] (11.5) c p = kR/(k-1) (11.6a) c v = R/(k-1) (11.6b) IDEAL GASIDEAL GAS

EQUATION OF STATE FOR IDEAL GAS p =  RT (11.1) = unique constant for each gas [units of Kelvin] Static pressure

(# of collisions/sec) 1 p 1, n 1, m 1, v x1, T 1, L 1 (# of collisions/sec) 2 p 2, n 1 m 1, v x1, T 1, L 2 =2L 1 L 2L Daniel Bernoulli ~ PV = const Hydrodynamics,

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} p 1  (# of collisions/sec) 1 /(L) 2 p 2  (# of collisions/sec) 2 /(2L) 2 p 2  ½ (# of collisions/sec) 1 /(2L) 2 p 2 = 1 / 8 p 1 Vol 2 = 8Vol 1 p 2 Vol 2 = p 1 Vol 1 QED

“ The elasticity of air is not only increased by compression but by heat supplied to it, and since it is admitted that heat 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 = R univ /m mole pV = N (# of moles) R univ T 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: - 2mv x is change of x-momentum per collision. Initially assume v x is same for all particles. What is Pressure ?

 t = 2L/v x =  (mv x )/  t =2mv x /(2L/v x ) = mv x 2 /L Time between collisions,  t, of particle with same wall is equal to: L Force of one particle impact = Magnitude of momentum change per second due to one particle:

nmv x 2 /L Magnitude of momentum change per second due to n molecules: = = ; + + = = 1 / 3 1 / 3 nm /L

Pressure = F/A = [ 1 / 3 nm /L]/L 2 P = 1 / 3 nm /L 3 PV = 1 / 3 nm = 2 / 3 n ( 1 / 2 m ) average kinetic energy per particle Empirically it is found that : PV = nk B T n = #of particles; k B =1.38x J/K

We have reasoned: PV = 2 / 3 n ( 1 / 2 m ) Empirically it is found that: PV = nk B T T(K o ) = [2/(3k B ) ] [avg K.E.]

pV = ( 2 / 3 ) n U internal for monotonic gas U int = f(T) depending if p or V held constant u int, v,… designate per unit mass du int /dT = c v du int /dT =c p (# of particles ) pV = nk B T

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 = nk B T n = [N m ][N Avag ] 6.02x10 23 nk B T = N m x N Avag k B T = N m x N Avag [R univ /N Avag. ] T pV= N m R univ T

p=(1/V)N m m mole {R univ /m mole }T p=(m/V){R univ /m mole }T p=  {R univ /m mole }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 = -pdV goes into dT

FIRST LAW OF THERMODYNAMICS  Q +  W =  E =  (KE + PE + U)  Q/m +  W/m =  E/m;  q +  w =  u U, internal energy, is energy stored in molecular bonding forces and random molecular motion. (  KE and  PE we will ignore) W = -  pdV

SPECIFIC HEATS c V = dq/dT du = c v dT c p = dq/dT dh = c p dT IDEAL GAS h = u + pv  q -  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]: c vol = [  q /  T] vol but for constant volume,  v = 0:  u =  q +  w =  q - p  v =  q So: c vol = [  q /  T] vol =  u /  T For perfect gas u(T) c vol = du/dT or du = c vol dT (11.2) for constant c vol : u 2 – u 1 = c vol (T 2 – T 1 ) (11.7a)

Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of substance by1 o K. Different for constant volume or pressure. mC v dT = dQ or C v dT = dq  q +  w =  u if Volume constant, w = -pdv = 0, then dq = du, C v = du/dT “It can be shown that du = C v 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]: c pres = [  q /  T] pres but for constant pressure,  p =0:  h =  (pv + u) = (  p)v + p(  v) + u =  q +v  p  h =  q So: c pres = [  q /  T] pres =  h pres /  T For perfect gas h(T) = pv + u c pres = dh/dT or dh = c pres dT (11.3) for constant c pres, h 2 – h 1 = c pres (T 2 – T 1 ) (11.7b)

Definition of heat capacity at constant pressure: mC p dT = dQ or C p = dq/dT  q +  w =  u; dq = du + pdv; h = u + pv if pressure constant, dh = du + pdv = dq C p = dh/dT ~ again can be shown to be true even if pressure is not constant

C v = du/dT* C p = dh/dT* h = u + pv = u + RT* dh = du + RdT dh/dT = du/dT + R C p – C v = R * IDEAL GAS

IDEAL GAS du =  w +  q (CONSERVATION OF ENERGY)  q = c v dT + pdv pv = RT (R=R univ /m mole ) pdv + vdp = RdT  q = c v dT + RdT - vdp Divide by dT [  q]/dT = c v + R – vdp/dT If isobaric, i.e. dp=0 then {[  q]/dT} p = c p = c v + R (11.4)

c p = c v + R; c p – c v = R Divide by c v, & let k = c p /c v; k - 1 = R/c v, or c v = R/(k-1) (11.6b) Multiply by c p /c v = k c p = kR/(k-1) (11.6a) (often k expressed as  )

c p /c v = k = 1.4 for perfect gas k

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 found by 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 = m 2 /(s 2 -K) = (N-m)/(kg-K) = J/(kg-K) R = ft 2 /(s 2- R)

COMPRESSIBLE FLOW If fluid incompressible, gas would behave like solid body and move everywhere at piston speed. If pressure disturbance is small relative to p 1 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 an instantaneous equilibrium. This continues to hold even inside shock waves. For all the flows examined here, all systems will be assumed to be in equilibrium at all times. p 1,  1,T 1, s 1, h 1 p 2,  2, T 2, s 2, h 2

(a)Surrounding flow field is affected by pressure disturbances. (b) Sound waves swept downstream, sound waves collect along a 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/m reversible process Tds >  Q/m irreversible 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 true dq = du + pdv ds =  q/T reversible Change in entropy intimately connected with the concept of reversibility – for a reversible, adiabatic process entropy remains constant. For any other process the entropy increases.

c v = du/dT c p = dh/dT Tds = du + pdv = dh –vdp ds = du/T +  RTdv/T ds = C v dT/T + (R/v)dv s 2 – s 1 = C v ln(T 2 /T 1 ) + Rln(v 2 /v 1 ) s 2 – s 1 = C v ln(T 2 /T 1 ) - Rln(  2 /  1 ) Ideal Gas p =  RT Not Isentropic!!! (11.11a)

s 2 – s 1 = C v ln(T 2 /T 1 ) - Rln(  2 /  1 ) If isentropic s 2 – s 1 = 0 ln(T 2 /T 1 ) C v = ln(  2 /  1 ) R C p – C v = R; R/C v = k – 1  2 /  1 = (T 2 /T 1 ) C v /R = (T 2 /T 1 ) 1/(k-1) T/  k-1 = constant assumptions ISENROPIC AND IDEAL GAS (11.11a)

C v = du/dT C p = dh/dT Tds = du + pdv = dh –vdp ds = dh/T - vdp ds = C p dT/T - (R/p)dp s 2 – s 1 = C p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) s 2 – s 1 = C v ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) Ideal Gas p =  RT Not Isentropic!!! (11.11b)

s 2 – s 1 = C p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) If isentropic s 2 – s 1 = 0 ln(T 2 /T 1 ) C p = ln(p 2 /p 1 ) R C p – C v = R; R/C p = 1- 1/k p 2 /p 1 = (T 2 /T 1 ) C p/ R = (T 2 /T 1 ) k/(k-1) (p 2 /p 1 ) (1-k)/k = T 1 /T 2 Tp (1-k)/k = constant assumptions ISENROPIC AND IDEAL GAS (11.12b)

s 2 – s 1 = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) T = pv/Rc p -c v = R s 2 – s 1 = c p ln[(p 2 v 2 )/(p 1 v 1 )] – (c p -c v )ln(p 2 /p 1 ) s 2 – s 1 = c p ln[v 2 /v 1 ] + c v ln(p 2 /p 1 ) (11.11a) s 2 – s 1 = 0 = c p ln[v 2 /v 1 ] + c v ln(p 2 /p 1 ) -(c p /c v ) ln[v 2 /v 1 ] = ln(p 2 /p 1 ) ln[v 2 /v 1 ] -k =ln(p 2 /p 1 ) p 2 v 2 k = constant (11.12c)

To prove – (p 2 /  2 ) + u 2 + ½ V gz 2 = (p 1 /  1 ) + u 1 + ½ V gz 1 For steady flow ~  dE/dt = dQ/dt + dW/dt dm/dt[(u 2 + ½ V gz 2 ) - (u 1 + ½ V gz 1 )] = dQ/dt + dW/dt dW/dt = dW/dt pressure + dW/dt viscous + dW/dt shaft Pressure work = pAds = pAV  t = (p/  )(dm/dt)  t dW/dt pressure = (p 1 /  1 )(dm/dt) - (p 2 /  2 )(dm/dt)

 dE/dt = dQ/dt + dW/dt If no viscous or shaft work and no heat interaction: dQ/dt = 0 dW/dt = dW pressure /dt dm/dt[(u 2 + ½ V gz 2 ) - (u 1 + ½ V gz 1 )] = (p 1 /  1 )(dm/dt) - (p 2 /  2 )(dm/dt) (p 2 /  2 ) + u 2 + ½ V gz 2 = (p 1 /  1 ) + u 1 + ½ V gz 1

the end (for this part)

BE: 1-D, energy equation for adiabatic and no shaft or viscous work. (p 2 /  2 ) + u 2 + ½ V gz 2 = (p 1 /  1 ) + u 1 + ½ V gz 1 Definition: h = u + pv = u + p/  ; assume z 2 = z 1 h 2 + ½ V 2 2 = h 1 + ½ V 1 2 C p = dh/dT (ideal gas) C p dT 2 + ½ V 2 2 = C p dT 1 + ½ V 1 2 If pick stagnation conditions V = 0 C p dT o = C p dT 1 + ½ V 1 2

C p T o = C p T + ½ V 2 C p – C v = R; R/C p = 1- 1/k C p = R/(1-1/k) = kR/(k-1) M 2 = V 2 /c 2 = V 2 /[kRT] [kR/(k-1)]T o /T = [kR/(k-1)] + ½ V 2 /T T o /T = 1 + ( 1 / 2 V 2 ) / (T[kR/(k-1)]) T o /T = 1 + {(k-1)/2} V 2 /(kRT) T o /T = 1 + {(k-1)/2} M 2 STEADY, 1-D, ENERGY EQUATION FOR ADIABATIC FLOW OF A PERFECT GAS

 /  o = (T/T o ) 1/(k-1) T o /T = 1 + {(k-1)/2} M 2  /  o = (1 + {(k-1)/2} M 2 ) 1/(k-1) p/p 0 = (T/T o ) k/(k-1) T o /T = 1 + {(k-1)/2} M 2 p/p 0 = (1 + {(k-1)/2} M 2 ) k/(k-1) Ideal gas and isentropic (isentropic = adiabatic + reversible)

SECOND LAW Tds =  Q/m reversible process Tds >  Q/m irreversible process OF THERMODYNAMICS

QUIZ When a fixed mass of air is heated from 20 o C to 100 o C – (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) h 2 – h 1 = C p (T 2 - T 1 ) (b) s 2 – s 1 = C v ln(T 2 /T 1 ) (c) s 2 – s 1 = C p ln(T 2 /T 1 ) (d)  100 /  20 = (T 100 /T 20 ) = 1/(k-1)k = 1.4 for ideal gas p 100 / p 20 = (T 100 /T 20 ) = k/(k-1)k = 1.4 for ideal gas

(e) c 2 = {dp/d  } But c 2 = (  p/  )| T does not equal c 2 = (  p/  | S ) If isentropic p/  k = constant (ideal gas) Then c = {(  p/  )| S } 1/2 = (kRT) 1/2 = (1.4 * * ( )) 1/2 = m/s If isothermal p =  RT (ideal gas) Then c = {(  p/  )| T } 1/2 = (RT) 1/2 = ( X ( ) 1/2 = m/s 18% too low

If isothermal p =  RT (ideal gas) Then c = {(  p/  )| T } 1/2 = (RT) 1/2 = ( X ( ) 1/2 = m/s 18% too low dp = (d  )RT for T constant dp/d  = RT = c 2 (  /  p)| T = c 2 = RT

~ SPEED OF SOUND ~ Sound waves are pressure disturbances << ambient pressure. For loud noise: p ~ 1Pa whereas ambient pressure is 10 5 Pa Speed of sound: c 2 = (  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 = c 2 = kp/  p =  RT for ideal gas c 2 = kRT For 20 o C and 1 atmosphere c = 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 M 2 = V 2 /c 2 = V 2 /kRT (ideal gas) p =  RT (ideal gas) M 2 = 2( 1 / 2 V 2 /kRT) = 2( 1 / 2 V 2 /(kp/  )) M 2 = 2[ 1 / 2  V 2 /(kp)] ~ 1 / 2  V 2 /p M 2 ~ dynamic pressure/ static pressure

M 2 = V/c p =  RT (ideal gas) if isothermal (actually isentropic) 1% change in density ~ 1% change in pressure 0.01p = 0.01* N/m 2 1 / 2  V 2 = N/m 2 V = 41 m/s M = 41/343 = % change in density ~ 5% change in pressure 1 / 2  V 2 = 0.05p = 0.05* N/m 2 V = 92 m/s M = 41/343 = 0.27 ~ 0.3

The concept of absolute zero extends to a great many phenomena: –volume of a gas (Charles law ) –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 gas pV = (k - 1) U in general k = c p /c v = 5/3 for monotonic gas U = 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 state dU = (Vdp + pdV) / (k - 1) Cons. of energy dU =  W +  Q

dU =  W +  Q dU = (Vdp + pdV) / (k - 1) -pdV = (Vdp + pdV) / (k - 1) -(pdV)(k - 1) = Vdp + pdV -(pdV)k + pdV = Vdp + pdV -(pdV)k - Vdp = 0 0

(divide by -pV) (dV/V) + (dp/p) = 0 (integrate) kln(V) +ln(p) = ln(C) ln(pV k ) = ln(C) pV k = C or pv k = c (11.12c)

pV k = C or pv k = p/  k = c (11.12c) Ideal gas - p =  RT p/  k = c p/[p/RT] k = p [1-k] T k = c p [1-k]/k T = c (11.12b) p/  k = c  RT/  k =  [1-k] RT = c {  [k-1] /  [k-1] }  [1-k] RT = c T/  [k-1] = Tv [k-1] = c (11.12a) ISENTROPIC & IDEAL GAS