Unformatted text preview: AAE 439 4.2 THERMODYNAMICS  Review Ch4 20 AAE 439 PERFECT GAS Definition: Particles posses three translational degrees of freedom. Intermolecular forces are negligible.
Equation of State p = RT p pv = = RT = T M = 8.314 M J mol K is universal gas constant: R is gas constant for a particular gas: R =
Characteristics: Pressure is a result of kinetic energy exchange with molecules in movement. As T rises, kinetic energy goes up, pressure goes up. Perfect gas approximation is good at low p and high T. More accurate "equations of state" (e.g., van der Waals equation) can be used.
Ch4 21 AAE 439 PERFECT GAS
General
u = u p,T h = h ,T Internal Energy Enthalpy ( ) Perfect Gas
u=u T h=h T ( ) h = u + pv h = u + RT = u + p ( ) ( ) Specific Heat Definition u cv T v h cp T p du = c v dT, dh = c p dT, u = c vT h = c pT Thermally Perfect Gas
Gas in thermodynamic equilibrium Not chemically reacting u, h and cp/v functions of temp ONLY cp = cv + R cv = R 1 cp cv cp = 1 Calorically Perfect Gas
All of above Specific heats are constant R Ch4 22 AAE 439 LAWS OF THERMODYNAMICS
e = u + e potential + ekinetic dw = dw shaft + dw flow de = dq  dw dh + v dv = dq  dw shaft dq rev = du + p dv = dh  v dp = dh + v dv ds dq T dq ds = T rev Total Energy: Total Work: 1st Law of TD (Energy Equation)
for e pot = const. de pot = 0 Expresses the universal law of conservation of energy. Nowork reversible interaction
2nd Law of TD (Entropy) Expresses the universal law of increasing entropy. No work reversible interaction
1st & 2nd Laws of TD Tds = dh  v dp v dp + v dv = dq rev  Tds
Spec. Volume Velocity !
Ch4 23 AAE 439 THERMODYNAMIC PROCESSES
Q=0 U = W Adiabatic Process No heat or other energy crosses system boundary. Adiabatic processes include isentropic and throttling processes.
Isobaric Process p = 0 T = 0 V = 0 S = 0 Q = H Q=W Q = U W =0 Q=0 Constant pressure process.
Isothermal Process Constant temperature process.
Isochoric/Isometric Process Constant volume process.
Isentropic Process An adiabatic process in which there is no change in system entropy. This is an reversible process.
Ch4 24 AAE 439 THERMODYNAMIC CYCLES : Reversible isothermal expansion of the gas at the "hot" temperature, TH (isothermal heat addition). During this step the expanding gas causes the piston to do work on the surroundings. The gas expansion is propelled by absorption of quantity Q1 of heat from the high temperature reservoir : Isentropic (Reversible adiabatic) expansion of the gas. For this step, we assume the piston and cylinder are thermally insulated, so that no heat is gained or lost. The gas continues to expand, doing work on the surroundings. The gas expansion causes it to cool to the "cold" temperature, TC. : Reversible isothermal compression of the gas at the "cold" temperature, TC (isothermal heat rejection). Now the surroundings do work on the gas, causing quantity Q2 of heat to flow out of the gas to the low temperature reservoir. : Isentropic compression of the gas. Once again we assume the piston and cylinder are thermally insulated. During this step, the surroundings do work on the gas, compressing it and causing the temperature to rise to TH. At this point the gas is in the same state as at the start of step 1) Carnot : Isentropic process; ambient air is drawn into the compressor, where it is pressurized. : Isobaric process; compressed air then runs through a combustion chamber, where fuel is burned, heating that aira constantpressure process, since the chamber is open to flow in and out. : Isentropic process; heated, pressurized air then gives up its energy, expanding through a turbine (or series of turbines). Some of the work extracted by the turbine is used to drive the compressor. : Isobaric process; heat Rejection (in the atmosphere) Brayton : Isentropic compression (blue), Work in (Win) is done by the piston compressing the working fluid : Reversible constant pressure heating (red), Heat in (Qin) is done by the combustion of the fuel : Isentropic expansion (yellow), Work out (Wout) is done by the working fluid expanding on to the piston, this produces usable torque : reversible constant volume cooling (green), Heat out (Qout) is done by venting the air Diesel : The working fluid is pumped from low to high pressure, as the fluid is a liquid at this stage the pump requires little input energy. :, The high pressure liquid enters a boiler where it is heated at constant pressure by an external heat source to become a dry saturated vapor. : The dry saturated vapor expands through a turbine, generating power. This decreases the temperature and pressure of the vapor, and some condensation may occur. , The wet vapor then enters a condenser where it is condensed at a constant pressure and temperature to become a saturated liquid. The pressure and temperature of the condenser is fixed by the temperature of the cooling coils as the fluid is undergoing a phasecha Rankine Stirling , Isothermal Expansion. The expansionspace is heated externally, and the gas undergoes nearisothermal expansion. , ConstantVolumw heatremoval. The gas is passed through the regenerator, thus cooling the gas, and transferring heat to the regenerator for use in the next cycle. , Isothermal Compression. The compression space is intercooled, so the gas undergoes nearisothermal compression. , ConstantVolume heataddition. The compressed air flows back through the regenerator and picksup heat on the way to the heated expansion space. Ch4 25 AAE 439 4.3 GENERAL DESCRIPTION OF FLUID IN MOTION Ch4 26 AAE 439 GOVERNING EQUATIONS Compressible Flow through Arbitrary Duct: SteadyState
in v A p T x y z h u p wall out v+dv A+dA p+dp T+dT +d x+dx y+dy z+dz h+dh u+du wall
m in m out Q W ( vA ) = ( vA ) ( + d ) ( v + dv ) ( A + dA ) = ( vA )
out in CONTINUITY m out = m in CONSERVATION OF ENERGY p Q  W  d A = u + 1 v2 + gz d A v v 2 Q  W = h out  h in + No Work Interaction
2 dq = h + dh  u + 1 v + dv  1 v2 + g z + dz  gz 2 2 Depending on case, can be neglected ( CS ) ( ( CS ( ) 1 2 v2  1 v2 + g zout  z in m out 2 in ) ( ) dA dv d + + =0 A v ( ) ) ( ) dq = dh + vdv Ch4 27 AAE 439 GOVERNING EQUATIONS
CONSERVATION OF MOMENTUM F = dA v v (p A) ( in + Si deWall pWall dA  p A ( ) CS ( ) out  Si deWall Wall c(x) dx = m out vout  m in v in Local Circumference p A  p + dp A + dA + p + 1 dp dA  Wall c(x) dx = vA v + dv  vA v 2
Average pressure in xdirection times projected area in xdirection )( ) ( ) ( )( ) 2nd LAW OF TD dq ds T Tds = dh  v dp = dh  dq = dh + vdv = Tds +
1st Law of TD A dp  Wall c(x) dx = vA dv dp + v dv =  Wall c(x)
A dx dp dp + vdv T ds = dq + Wall c(x)
A dx dp + vdv = dq  Tds Ch4 28 AAE 439 EXAMPLE Consider a (calorically) perfect gas flowing adiabatically and without friction in a constant area duct. Changes occur as a result of changes in elevation in a gravity field. Using the differential form of the conservation laws and 2nd law of TD, determine the change in velocity, pressure, temperature, (total pressure, and total temperature) for a positive change in z, for both subsonic and supersonic initial Mach numbers. Comment on the tendency of the flow to choke under these conditions. Ch4 29 AAE 439 EXAMPLE Derivation of Speed of Sound Ch4 30 ...
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 Fall '09
 Thermodynamics, perfect gas, calorically perfect gas

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