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Mechanical and Materials Engineering Courses
MECH 230

Specific volume, =1/ [m /kg] U QW n Heat Exchangers If the specifPc hVats c and c are taken as constLiquids liquid are incompres1ib2e, so v =v = v if is uniform over the volume then M= V x =y m h h u*= satT*) Mdu Q pdV n = ln y/ln x 1 4 3 s s c ln T2 Rln v2 WCV Q12 m3 h1 h 2 2 1 V T v v(P 2 P )1 Absolute presabse, P , measured relative to a perMdu Q p(Mdv) m m(u2u 1W 12 1 1 m int Gauge pressgre, P , measured relative to thatmtmospheric pressure, P Q Q rev P* g P* abs atm= gL Q M(du Pdv) i (h h ) o (h4 h 3 | T2 P2 Gases where Pv = const m1 1 2 m3 s2 s 1 c lP Rln W 12 P(V2- 1 ) n = 0, Constant P process R K(T)R cP T1 P1 1 2 W = P V ln(V /V ) n = 1, Isothermal cV(T) K(T)1 cP(T) k(T) c cP Isentropic Pro1e2ses (s = s ) W CV n dP W 1=[(P V P V )/1-nn 1 if Ideal Gas then n = k = K(T)1 V T f TR kT R (const) 1 c /c2 2 2 1 1 cV o o P 2 m int 1 PV = constant n = constant 2 P V PV V Tanks s2s 1s 2 R1n 0 rev P n W pdV 2 2 1 1 M P1 For the special case of an ideal gas where Pv = RT The First Law of Thermodynamics states: 1 1n v dECV E Q W m (h ) m (h ) m (h ) E2 1 = Q W and u = q-w dt 1 1 2 2 3 3 s s o P W nRT T E = U + KE + PE 2 Recall,foranidealgasPV MRT Enthalpy 2 1 ln 2 CV 1 2 1 n1 , Adiabatic: Q=0 W pdV 2 ZRT R P1 m int n 1 T1 1 pdV MR(T -2 1 needT v P h u Pv h(P,T) = u(T) + Pv rev Constant Volume: W = PdV=0 1n 2 Enthalpy for Compressed Liquids(Never use compressed liq. aoles)o o Constant Pressure: W = PdV=PV 1 h* h1 v(P*P 1 P2 s2s 1 exp(s2/R) for a Polytropic process Isotherma1: 2 =T U=0, but not if the state is changing h(P*,T*) h (T*) v (T*) (P*P (T*)) exp o n1 Idea Gases Polytropic process: fsat fsat sat P1 R exp(s1/R) WCV nRT1 P2 n n 1 Work by or Heat to the system is positive n1 n1 Ideal Gas 1 Work on or Heat from the system is negative T2 P2 n T2 v1 W MR(T 2 1) o m int n 1 P1 1 P1 1 v2 1n Define relative Pr exp s /R rev W Fv If the process is reversible and adiabatic (isentropic) Pv = const Where W is power and v is velocity P2 Pr2 Substitute n= k in above equations to get work per unit mass for At any point between L and V, say C, we define tU M(u u ) Mx as u 2u 1 isentropic procesk const f (T) X= mass of vapor/(mass of vapov v liLuid) = M /(M + M ) 2 1 P1 sconst Pr1 For the case 1 1n=2 2P1v 2 P v T =T (isothermal) at point L x = 0 For ideal gas at point V x = 1 Vg Mg vap gxM) dM VA W CV Enthalpy h = u + P v CV m m where m VA Turbine/ Compressor/pump v2 T 2 P1 T2 Pr1 T 2 P r2 RT ln P 2 P1 n=1 2 2 2 dt i e v m int 1 Pa = 1 kg/ms = 1 N/m J = 1 kgm /s Mass flux defined as mass flow rate per unit area (h h )(V 2 2V 2 2) v1 T 1 P2 T1 Pr2 T1/ Pr1 rev 1 atm = 101,325 Pa = 760 mm Hg 1 2 1 2 1 bar = 100,000 Pa = 100 kPa = 0.1 MPa m V V units kg/m s h V 2 2V 2 2 w h h v v Carnot Cycle 1 ton = 711 KJ/mi 0C = 273K A v 1 2 1 2 1 2 Define T /P , so 2 r2 Process 1-2 Reversible Adiabatic Compression: Volumetric flow rate defined as volume per unit time r r v1 vr - Work done on the system: x mass of vapor M v Poweris workoutput perunit time sconst 1 U = Q (-W) U = +W mass of vapor +liquid M M m 3 Isentropic Process for ideal gav wiPh constant c- Gas temperature incrCasHs from T to T v L VA units (m /s) W mw m(h h 1 2 k M vap Throttling Device P2 v1 k k - Work done by the systemothermal Expansion M M vap xM dE (t) Applying First Law (steady-state, neglect heat tansfr and PE) or P1 1 P v 2 2 U =HQ (+W) HQ = +W specific internal eu= L + V (L - u ) CV QW m(u V /2 2 2 2 P1 sconst v2 - Heat traHsfer Q to the gas, teHperature stays at T specific volume, v= L + V (L - v ) dt i i i h1V 1 / 2 2 V 2 2 cpconst v(T,P)Lv (T) If the state 2 is taken far downstream from the blockage the change in Process 3-4 Reversible Adiabatic Expansion: u(T,P)Lu (T) gZi)m (e e V
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