Answer:
Explanation:
Given that:
the temperature [tex]T_1[/tex] = 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:
[tex]\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}[/tex]
where; B = - [tex]152.5 \ cm^3 /mol[/tex] C = -5800 [tex]cm^6/mol^2[/tex]
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have
[tex]\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
[tex]4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
Multiplying through with V² ; we have
[tex]4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0[/tex]
[tex]4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0[/tex]
V = 2250.06 cm³ mol⁻¹
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*2250.06}{8.314*10^3*523.15}[/tex]
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :
[tex]T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345[/tex]
[tex]T__{\gamma}} = \frac{T}{T_c}[/tex]
[tex]T__{\gamma}} = \frac{523.15}{647.1}[/tex]
[tex]T__{\gamma}} = 0.808[/tex]
[tex]P__{\gamma}} = \frac{P}{P_c}[/tex]
[tex]P__{\gamma}} = \frac{1800}{22055}[/tex]
[tex]P__{\gamma}} = 0.0816[/tex]
[tex]B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}[/tex]
[tex]B_o = 0.083 - \frac{0.422}{0.808^{1.6}}[/tex]
[tex]B_o = 0.51[/tex]
[tex]B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}[/tex]
[tex]B_1 = -0.282[/tex]
The compressibility is calculated as:
[tex]Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}[/tex]
[tex]Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}[/tex]
Z = 0.9386
[tex]V= \frac{ZRT}{P}[/tex]
[tex]V= \frac{0.9386*8.314*10^3*523.15}{1800}[/tex]
V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At [tex]T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa[/tex]
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*10^3 *0.1249}{729.77*523.15}[/tex]
Z = 0.588
