《化工热力学》课程授课教案（讲义）Chapter 6 Thermodynamic properties of fluids

f=C-P+2=1-1+2=2 That is the thermodynamic state function can be calculated by two variables only for homogeneous phase For example U=U(T.V) UU(P n UU(P (1 Enthalpy ad Entropyas Functions Second H-H(T.P) S=S(T.P) d=欲n+0n (.c 0=n+w dU =TdS-Pdv dH TdS+VdP m-c.mv-ror) dA=-PdV-SdT dG=VdP-SdT H-H(T.P) S=S(T.P) n (,dr(),dp dS-(dp 0=C, , 0-n+w w.civ-) (2)Internal energy as Function of P dU=TdS-Pdl U=UP.T) dH =TdS+Vdp dA=-PdV-Sdl When T=constant,U-U(P) dG=VdP-SdT the change rate of interal energy with pressure 02 U=H-PV -】 -) .r-n P=-P-7

f = C-P+2 = 1-1+2=2 That is : the thermodynamic state function can be calculated by two variables only for homogeneous phase For example U=U1 (T,V) U=U2 (P,T) U=U3 (P,V) etc. （1）Enthalpy and Entropyas Functions of T and P (The Second dS, dH Equation) （2）Internal energy as Function of P When T=constant, U=U(P) the change rate of internal energy with pressure: H=H(T,P) S=S(T,P) d ( ) ( ) P T H H H dT dP T P   = +   d ( ) ( ) P T S S S dT dP T P   = +   ( )P P H C T  =  ( ) ( ) T T H S T V P P   = +   dP T V dH C dT V T P P               = + − dU TdS PdV dH TdS VdP dA PdV SdT dG VdP SdT = − = + = − − = − H=H(T,P) S=S(T,P) d ( ) ( ) P T H H H dT dP T P   = +   d ( ) ( ) P T S S S dT dP T P   = +   ( )P P H C T  =  ( ) ( ) T T H S T V P P   = +   dP T V dH C dT V T P P               = + − P T V S T P           = −       P P C V dS dT dP T T    = −     U=U(P,T) dU TdS PdV dH TdS VdP dA PdV SdT dG VdP SdT = − = + = − − = − S P T V T S P V P S V S T P T V P SV P V S T T         = −         =                        = −              =             ( ) ( ) ( ) T T P H S V T V V T P P T    = + = −    ( ) ( ) ( ) T T P U V V P T P P T    = − −    ? U=H-PV ? ( ) ( ) ( ) T T T U H V P V P P P    = − −   

(3)Alternative Forms for Liquids (av) =BV Attention (1)Although general,these equations are -别一 usually applied to liquid (2)For liquid not near the critical point. B and x is small.thus at most 党 conditions.pressure has little effect on =-=-Tm=-m =-A0-0=Pr-mr=P-m 号r-,即号r-m dP=C,dT+v(1-BT)d Example 6.1 Solution: △H=H,+△H, dH=C dT+V(1-BT)dP AH,AS △H=∫cdr+jv(I-BT)dr 4h4 AH,=[C,dT+v(1-BT)dP=[C,dT=(C)(T-T) AH,-jc,d+jv(1-BT)adp=fv(-BT)r By data (1)C is a weak function -了y-(Bz)dP=w-z)水R-P (2)Both V and 6 are △H=(C)I,-T)+(W）1-(E)T）B-R) weak function ofp ()Enthalpy and Entropy,internal energy as Functions of T and V(The First dS.d.dH Equation) H=H(TV) S=S(T.V) U=U(T.V) s-c() =cr+r号pr m-crw

（3）Alternative Forms for Liquids Example 6.1 Solution： （4）Enthalpy and Entropy, internal energy as Functions of T and V (The First dS, dU,dH Equation) ( ) ( ) T P V P T V S    − =  = −  ( ) ( ) (1 ) T P H V V T V T V P T  T V   = − = − = −   ( ) ) ( ) ( ( ) T T P U V V P T PV TV P T V P P T        = − − = − = −    Attention : (1)Although general, these equations are usually applied to liquid (2)For liquid not near the critical point, β and κ is small, thus at most conditions, pressure has little effect on the properties of liquid 1 V V T p     =      V T p  =        V 1 T V V P     −      = T V V P          =- P P P C dT V C V dS dT dP T T dP T     = − =   −    P (1 ) P P V dH C dT V T dP T C dT V T dP       = + − =         + −  P1 T1 H1 S1 P2 T2 H2 S2 ΔH ΔS P1 T2 ΔH2ΔS2 ΔH1 ΔS1  =  +  H H H 1 2 dH = C dT T P P + − V (1  ) d  = + − H C dT V T dP P (1  )  ( ) 2 1 1 1 2 1 1 1 2 1 T P P T P T P T  = − H C dT V T d +  P = C dT = CP T T    （ - ） ( ) ( ) ( ) ( ) 2 1 2 2 2 2 1 1 2 2 2 2 2 2 1 1 1 1 1 T P P P P T P P P V T dP V H C T dP V T P P dT V T dP      = + = − = − − = −     （ - ）  = − + − − H C T T V T P P P ( ) 1 ) 2 1 2 1 (  （) By data (1) CP is a weak function of T (2)Both V and β are weak function of P H=H(T,V) U=U(T,V) S=S(T,V) V V dT P dS C dV T T    = +     ( ) ( ) ( ) V V V T P P P dH C V dT T V dV T T V        = + + +            ( ) V V P dU C dT T P dV T    = + −     