Aside: M 2 = V 2 /(kRT) = directed K.E. / random K.E. Flow of air at standard sea level conditions and M=2, then internal energy of 1 kg of air = 2.07 x 10 5 J, whereas the directed KE is = 2.31 x 10 5 J; when flow decreases, part of this energy reappears as internal energy - hence increasing the temperature of the gas. For perfect gas, intermolecular forces can be ignored (on average gas molecules are more than 10 molecular diameters apart). Energy of molecule (e.g. O 2 ) = sum of translational, vibrational, rotational and electronic energies. For point particle only have translational energy. Sum of all the energies of the molecules = internal energy Really Interesting Fact ~

C d, (0 lift) vs M Northrup T-38 jet trainer

What causes fluid properties to change?

Ch.12 - COMPRESSIBLE FLOW WHAT CAUSES FLUID PROPERTIES TO CHANGE IN A 1-D FLOW?

Ch.12 - COMPRESSIBLE FLOW 1 st – assume flow can be affected by: areas change, friction, heat transfer, shock Use continuity, x-momentum, energy, entropy and eq. of state.

Fluid properties = T(x), p(x),  (x), A(x), v(x), s(x), h(x) What can affect fluid properties? Changing area, heating, cooling, friction, normal shock One-Dimensional Compressible Flow dQ/dt heat/cool RxRx Surface force from friction and pressure P1P1 P2P2 (+ s 1, h 1 ) (+ s 2, h 2 )

One-Dimensional Compressible Flow f(x 1 ) f(x 2 ) (+ S 1, h 1 ) (+ S 2, h 2 ) To make problem more tractable assume steady, fluid properties are uniform at each x-location, body forces are negligible, ideal gas, constant specific heats

V(s) V(x,y) (a) No viscous effects(b) Viscous effects

7 Variables = T(x), p(x),  (x), A(x), v(x), s(x), h(x) 7 Equations = mass, x-momentum, 1 st and 2 nd Laws of Thermodynamics, Equation of State (3 relationships) One-Dimensional Compressible Flow f(x 1 ) f(x 2 ) + S 1, h 1 + S 2, h 2

EQ 4.12 dA 1 dA 2 Velocity  (x),V(x),A(x) Quasi-1-Dimensional Flow =

Does the flow have to be inviscid? Does the flow have to incompressible? Does the flow have to one-dimensional? Does the flow have to steady?

Volume flow in over A 1 = A 1 V 1  t Volume flow in over A 2 = A 2 V 2  t mass in over A 1 =  1 A 1 V 1  t mass in over A 2 =  2 A 2 V 2  t  1 A 1 V 1  t =  2 A 2 V 2  t  1 A 1 V 1 =  2 A 2 V 2 If steady, what comes in must go out! (Cool way to derive 1-D continuity equation – steady) ? Vol 1 = Vol 2 ?? m 1 = m 2 ?

Glass #1 has volume V of red wine. Glass #2 has volume V of white wine. Take teaspoon of red wine from glass #1 and put into glass #2, then take same amount of wine from glass #2 and replace back in glass #1 so both glasses have volume V. Case 1: did not mix well Question – which glass has the purest wine? Case 2: mixed vigorously Question – which glass has the purest wine?

EQ 4.18a F = rate of change of momentum

EQ 4.18a (also steady; 1-D) R x is surface forces arising from friction and pressure on top* and bottom *

CONSERVATION of ENERGY

EQ Conservation of Energy rate of change of total energy of system = net rate of energy addition by heat transfer to fluid + net rate of energy addition by work done on fluid also steady, 1-D only pressure work, pg147

F shear  V on boundary since V = 0 ASIDE ~

W normal = F normal  ds work done on area element d W normal / dt = F normal  V = p dA  V rate of work done on area element - d W normal / dt = - p dA  V = - p(  v)dA  V + for going in + for going out ASIDE ~

dm/dt

EQ. 12.1c

dS/dt| system  (dQ/dt)/T

 h = c p  T ~ ideal gas p =  RT ~ ideal gas EQ EQ EQ b ~ ideal gas and constant c p EQUATIONS OF STATE

EQ. 12.1d EQ. 12.1c EQ. 12.1f EQ. 12.1e EQ. 12.1g EQ. 12.1b EQ. 12.1a Did NOT assume  S=0  A=0 No friction No shocks Most general case ~

Variables = T(x), p(x),  (x), A(x), v(x), s(x), h(x) Equations = mass, momentum, 1 st and 2 nd Laws of Thermodynamics, Equation of State (3 relationships) One-Dimensional Compressible Flow Cons. of mass Cons. of momentum Cons. of energy 2 nd Law of Thermodynamics equations of state – relations between intensive thermo. var. Ideal gas Ideal gas & constant c v, c p