Chapter 5 Two Phase systems Part 1

Example  CalculaIon:  SoluIon  Cont   Conversion  of  problem  parameters  to  SI  units   Temperature:  ... Example  CalculaIon:  SoluIon  Cont   Determine  gas  velocity  at  Lincoln, [r]

Two-phase Systems!

pA : partial pressure of species A

R : gas constant

P0, T0 P1, T1

Z

A

⇢CH4vvv · nnndA = 0

Z

A

⇢AvvvA · nnndA = 0

Z

A0

⇢CH4vvv · nnndA +

Z

A1

⇢CH4vvv · nnndA = 0

P0, T0 P1, T1

(⇢CH4)0 v0A0 = (⇢CH4)1 v1A1

v1 =



A0(⇢CH4)0

A1(⇢CH4)1 v0

˙

m0 = ˙m1 = (⇢CH4)0 v0A0

˙

M0 = ˙M1 = m˙ 0

M WCH4 =

(⇢CH4)0 v0A0

M WCH4

P0, T0 P1, T1

1 psia = 6895 Pa

1 F = 1.8 C + 32

1 K = 1 C + 273.16

(2900 psia) ⇥ 6895 Pa

psia ⇥ MPa

106 Pa = 20.00 MPa

(2100 psia) ⇥ 6895 Pa

psia ⇥ MPa

106 Pa = 14.48 MPa

(45 F 32 F) ⇥ 5 C

A0 = A1 = ⇡D

2

⇡(0.508 m)2

2

0 , P0

v1

n =

R T1

P1 =

8.314mol Km3Pa (280.4 K) 14.48 ⇥ 10 6 Pa = 1.53 ⇥ 10 4 m3/mol

v0

n =

R T0

P0 =

8.314mol Km3Pa (305.4 K) 19.98 ⇥ 10 6 Pa = 1.2 ⇥ 10 4 m3/mol

16.043 g/mol

3

16.043 g/mol

3

v1 =



A0(⇢CH4)0

A1(⇢CH4)1 v0

3

˙

˙

m0 = ˙m1 = (⇢CH4)0 v0A0

˙

M0 = ˙ M1 = 413 kg/s

16.043 g/mol = 25.74 ⇥ 19 3 mol/s

˙

M 0 = ˙ M 1 = m˙ 0

M W CH4

= (⇢CH4 )0 v0A0

M W CH4

pvap(T )

A, B, ✓

pM,vap = 102.10312 = 120.88 mmHg = 16.91 kPa

p B = x B P B,vap

pAV = nART

P V = n RT

P

yA = pA/P

yA = xA

✓

PA,vap P

◆

xA << 1