Thermo3 I need your help solving this problem A storage tank

Thermo.3/ I need your help solving this problem.

A storage tank for 150 lb_m n-octane has a volume of 20 ft^3. If the tank is rated for a maximum pressure of 400 psia, what is the maximum temperature that may be tolerated? Use the following methods for calculations: Redlich-Kwong Equation of State; Comment: If the solution requires iterative techniques (trial and error), be sure to explain your method.

Solution

Redlich Kwong Equation is:

P = RT/(V-b) - a / [T^0.5V(V+b)]

where a = 0.42748*R²T_c^2.5 / P_c

and b = 0.08664*RT_c / P_c

For n-octane, we have T_c = 569.4 K and P_c = 24.97 bar

Using R = 8.314 J/mol-K, we get

a = 91.55 m⁶ Pa K^0.5 / mol²

b = 0.1643*10^-3 m³ / mol

For n-octane (C₈H₁₈), molar mass M = 12*8 + 1*18 = 114 g/mol = 0.114 kg/mol

We have density rho = 150/20 = 7.5 lb_m / ft³ = 120.138 kg/m³

Molar volume V = 0.114 / 120.138 = 9.489*10^-4 m³ / mol

P = 400 psia = 27.58 bar

Using these in RK eqn,

P = RT/(V-b) - a / [T^0.5V(V+b)]

27.58*10⁵ = 8.314*T / (9.489*10^-4 - 0.1643*10^-3) - 91.55 / [T^0.5*9.489*10^-4(9.489*10^-4 + 0.1643*10^-3)]

27.58*10⁵ = 10596.48*T - 86669183.33 / T^0.5

This equation needs to be solved iteratively.

Assuming T = 550 K we get Right Hand side of eqn = 21.3*10⁵ which is lower than 27.58*10⁶

Assuming T = 600 K we get Right Hand side of eqn = 28.2*10⁵ which is higher than 27.58*10⁶

Hence, T shall be between 550 K and 600 K.

By solving it we\'ll get T = 595.4 K = 612 deg F