Phase state equation of hydrocarbons Xuanqi zhang Petroleum engineering college
Phase state equation of hydrocarbons Xuanqi zhang Petroleum engineering college
油藏物理学一油气系统的溶解与分离 2.平衡常数equilibrium constant K=y:/X:与P、T及组成有关。 y一在压力P下，组分在气相中的浓度；（摩 尔分数) x:一在压力P下，组分在液相中的浓度； K一平衡常数，在压力P下，组分在气相中的 分配比例。 特点：在T一定时，随P的增加，平衡常数K逐 渐收敛于等于1的一点。该点即为收敛压力。 收敛压力：Convergence pressure
2. 平衡常数equilibrium constant Ki=yi/xi 与P、T及组成有关。 yi—在压力P下，i组分在气相中的浓度；（摩 尔分数） xi —在压力P下，i组分在液相中的浓度； Ki—平衡常数，在压力P下，i组分在气相中的 分配比例。 特点：在T一定时，随P的增加，平衡常数Ki逐 渐收敛于等于1的一点。该点即为收敛压力。 收敛压力: Convergence pressure 油藏物理学——油气系统的溶解与分离
「油藏物理学一油气系统的溶解与分离 set phase state equation 866666 a. P,T 1 moL Ng Yi 动平衡 NL Xi when P and T attain balance for 1mol of oil,balance equation of i component is Ng+N=1 (1) Nix+Ngyi=ni (2) Ki=yi/x (3)
set phase state equation 油藏物理学——油气系统的溶解与分离 a. when P and T attain balance for 1mol of oil, balance equation of i component is i i i L i g i i g L K y x N x N y n N N / 1 （1） （2） （3）
油藏物理学一油气系统的磙解与分离 Phase state equation for i component ni n Ni+K:Ng K;-(K,-I)N or Kin; Kini N1+K Ng 1+(K;-1)Ng
Phase state equation for i component 油藏物理学——油气系统的溶解与分离 L i g i i i L i g i i N K N K n y N K N n x i g i i i i i L i i K N K n y K K N n x 1 ( 1) ( 1) or
油藏物理李一油气系统的磙解与分离 total phase state equation ∑x,=1 So ∑K-(K-N n =1 ∑y,=1 Kini =1 1+(K,-)Ng
total phase state equation 油藏物理学——油气系统的溶解与分离 1 1 i i y x 1 1 ( 1) 1 ( 1) i g i i i i L i K N K n K K N n So
油藏物理李一油气系统的溶解与分离 application of equation 1.If we know ni,P and T,find Ne NLXi Y in the condition If there are m kinds of components for oil,we can got m equations n; x=K-(K:-1)N
application of equation 油藏物理学——油气系统的溶解与分离 1. If we know ni , P and T，find Ng、NL、Xi、 Yi in the condition. If there are m kinds of components for oil, we can got m equations i i L i i K K N n x ( 1)
油藏物理李一油气系统的磙解与分离 Because there are (m+1)unknown,we don't solve the equation.We use trial method We can choose NL n K,-(K,-1)N f ∑x=1 NL is true. ∑x,≠1 return
if 油藏物理学——油气系统的溶解与分离 xi 1 NL is true. xi 1 return Because there are (m+1) unknown, we don’t solve the equation. We use trial method. We can choose NL i i L i L i K K N n N x ( 1) Substitute
油藏物理学一油气系统的溶解与分离 2 find Pp和Pd a,Pp: based on saturated vapour pressure curve,N-0, NL=1,X-ni 工工晨 nKi =∑K,n,=1 By trial method,we can choose ，T,aem→Ke→∑Kn ∑Kn,=1 Pb is true. ∑Kn,≠1 return
2 find Pb和Pd 油藏物理学——油气系统的溶解与分离 a, Pb : based on saturated vapour pressure curve, Ng=0， NL=1，xi=ni 1 1 ( 1) i i i g i i i K n K N n K y By trial method, we can choose i i checkdiagram substitute Pb ,Tf Ki K n K 1 1 i i i i n K n Pb is true. return
油藏物理学一油气系统的溶解与分离 b.Pa based on dew point curve,NL-0,Ng=1,yini 2x-2张1 trial method like as above
b. Pd 油藏物理学——油气系统的溶解与分离 1 i i i K n x trial method like as above. based on dew point curve, NL=0，Ng=1，yi=ni
3.calculate phase state diagram If we can got corresponding P for T,we can got phase state diagram
If we can got corresponding P for T, we can got phase state diagram. 3. calculate phase state diagram