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Mechanical and Materials Engineering Courses

MECH 230

Prof.

Fall

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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 i.eh 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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