热力学课件6
- 格式：pdf
- 大小：367.66 KB
- 文档页数：34
3.4.2.利用立方型状态方程求MR
普遍化的立方型状态方程
RT a(T ) p= (V b ) (V + εb )(V + σb )
(2/5-3)
各项除以ρRT,并代入V=1/ρ,经过一些代数运算化简后
a(T ) q≡ bRT
ρb 1 q Z= (1 + ερb )(1 + σρb ) 1 ρb
(3/4-2)
HR
p B dB = RT R T dT
GR
Bp = RT RT
SR R
S
R
=
HR RT
GR RT
(3/3-8)
p dB = R R dT
(3/4-3)
例3/4-1: 利用下表的数据,试估算水蒸汽在573K,0.5 MPa下的剩余 焓和剩余熵.
解3/4-1: 根据(3/4-2)式
(3/4-16)
SR R
= ln(Z β ) +
d ln α (Tr ) qI d ln Tr
(3/4-17)
第一种情况: ε ≠ σ
Z + σβ 1 I= ln Z + εβ σ ε
(3/4-14)
第二种情况: ε = σ
ρb β I= = 1 + ερb Z + εβ
在第二种情况的讨论中 仅考虑van der Waals方程
(S )
R
R 0
= 0.8281
(S )
R
R 1
= 0.4428
SR = 0.8281 + (0.193)( 0.4428) = 0.9136 R S R = ( 0.9136)(8.3145) = 7.5958 J mol1 K 1
dB H R = B T p dT
HR
p B dT
T=573K, p=0.5MPa时 B=-119×10-6m3 mol-1.
而微分项可由表的数据近似计算
dB ΔB [( 113) ( 125)] × 10 ≈ = dT ΔT 583 563
(H )
RTc
R 0
(H )
RTc
R 1
SR R
= ∫
pr
0
Z 0 Tr Tr
dp pr Z 1 0 1 dp r r + Z 1 ω ∫ Tr T + Z p 0 pr r pr r pr
H H H = +ω RTc RTc RTc S S S = +ω R R R
例3/4-1: 利用下表的数据,试估算水蒸汽在573K,0.5 MPa 下的剩余焓和剩余熵. 解: 由附录查得水的临界性质
Tc = 647.3K pc = 22.05 MPa
ω = 0.344
573 Tr = = 0.885 647.3
0
0.5 pr = = 0.0227 22.05
0.422 B = 0.083 1.6 = 0.4301 Tr
3.4.以状态方程计算剩余性质
3.4.1.利用virial方程求MR 由Virial方程
Bp Z = 1+ RT
p
(2/4-7)
GR
dp = (Z 1) RT p o
∫
(等温情况下) (3/3-9)
GR
Bp = RT RT
(3/4-1)
HR
G R / RT = T RT T
(
)
p 1 dB B 2 = T R T dT T p
重写(2/5-12)式的RK方程式, ε 及σ 值由表2/5-1中得到:
Z β Z = 1 + β qβ Z (Z + β ) Z = 1 + 0.09695 (3.8689)(0.09695) Z 0.09695 Z (Z + 0.09695)
利用Z=1的起始值进行迭代,直到求得收敛结果Z=0.6852.
Redlich-Kwong (RK)方程:
1 ln α (Tr ) = ln Tr 2
σ=1,ε=0
d ln α (Tr ) 1 = d ln Tr 2
α(Tr)=Tr-1/2
Z + σβ 1 Z +β = ln I= ln σ ε Z + εβ Z
HR 3 Z +β = (Z 1) q ln 2 RT Z
由附录B3和B4 Lee-Kesler方程剩余性质分项参数表
(H )
RTc
R 0
= 1.3171
(H )
RTc
R 1
= 0.4366
HR = 1.3171 + (0.193)( 0.4366) = 1.4014 RTc
H R = ( 1.4014)(8.3145)(425.2) = 4954.28 J mol1
Z +β 0.6852 + 0.09695 I = ln = 0.13234 = ln Z 0.6852
HR
Z +β 3 = (Z 1) q ln RT Z 2 = 0.6852 1 + ( 0.5 1)(3.8689)(0.13234) = 1.0828
SR
1 Z +β = ln(Z β ) q ln R 2 Z = ln(0.6852 0.9695) (0.5)(3.8689)(0.13234) = 0.78661
dB RTc dB 0 dB1 = dT + ω dT dTr pc r r
dB0 B 0 dB1 B1 HR = Pr dT T + ω dT T RT r r r r
dB 0 SR dB 1 = Pr dT + ω dT R r r
本次课内容:
第三章 流体的热力学性质:焓与熵 3.1.纯流体的热力学关系 3.2.热容,蒸发焓与蒸发熵 3.3.剩余性质 3.4.以状态方程计算剩余性质 3.5.纯流体的焓变与熵变的计算 3.6.热力学性质图和表 3.7.多组分流体的热力学关系 3.8.偏摩尔性质及其与流体性质关系 3.9.混合性质与多组分流体性质 3.10.多组分流体焓变与熵变的计算
表2/5-1 状态方程 Vdw RK SRK PR α(Tr) 1 Tr-1/2 αSRK(Tr,ω)+ αPR(Tr,ω)++
状态方程中的参数值 σ 0 1 1 ε 0 0 0 Ω 1/8 0.08664 0.08664 0.07779 Ψ 27/64 0.42748 0.42748 0.45724 ZC 3/8 1/3 1/3 0.30740
H R = (8.314 )(500 )( 1.0828) = 4501.2 J mol 1 S R = (8.314 )( 0.78661) = 6.540 J mol 1 K 1
3.4.3.利用Lee-Kesler方程求MR
HR RT
HR RTC
= T
∫
∫
o
p
o
Z dp T p p
) (
)](
)
(
)(
)
J mol-1 K-1
B = B(T )
B
dB dT
图2/4-1
部分物质第二virial系数与温度的关系
如果无法得到实验数据,也可以用普遍化的方法.
HR
p = pc pr T = TcTr p B dB = RT R T dT dT = Tc dTr
HR
pc p r = RT RTc
0.172 B = 0.139 4.2 = 0.1483 Tr
1
dB 0 . 675 = = 0 . 9274 2 .6 dT r Tr
0
dB1 0.722 = 5.2 = 1.3628 dTr Tr
dB0 B 0 dB1 B1 HR = Pr dT T + ω dT T RT r r r r 0.4301 0.1483 = 0.0227 0.9274 + + 0.3441.3628 + = 0.044 0.885 0.885
Z T r dp r p pr r
SR R
= T
∫
p
o
Z dp T p p
(Z 1) dp ∫
p 0
p
= T
2 r
pr
T = TcTr p = pc pr
dT = Tc dTr dp = pc dpr
SR R
= Tr ∫
pr
o
Z T r
pr dp r dp ∫ (Z 1) r p 0 pr pr r
HR RT
SR RT
书第56页 至57页
= T ∫
ρ
0
Z dρ + Z 1 T ρ ρ
(3/4-10)
dρ + ln Z (3/4-11)
= T
∫
ρ
0
Z dρ T ρ ρ
∫ (Z 1)
p 0
ρ
GR R
HR
= Z 1 ln(Z β ) qI
(3/4-15b)
d ln α (Tr ) = (Z 1) + 1 qI RT d ln Tr
6
3 -1 -1 = 6 × 10 7 m mol K
H R = (573 ) 6 × 10 7 119 × 10 6 0 .5 × 10 6 = 2 .314 × 10 2 J mol-1
SR = p dB ≈ 0 . 5 × 10 6 6 × 10 7 = 0 . 3 dT
[
(
dB 0 SR dB 1 = Pr dT + ω dT = 0 .0317 R r r
H R = ( 0 .044 )(8 .3145 )(573 ) = 209 .63 J mol 1 S R = ( 0 . 0317 )(8 .3145 ) = 0 . 264 J mol 1 K 1
