Question

A tank pressurised at 150 bar and 180 °C is used to supply CO2 in a urea plant. The pipe (3 cm in diameter and made of riveted steel with 0.1 mm roughness) that connects the CO2 tank with the reactor vessel has a length of 2 m. The adiabatic index of CO2 is 1.30 and the molecular weight is 44 kg/mol (Assume adiabatic conditions).

a) What is the physical state of the CO2 within the tank, i.e. liquid or gas?

b) If a puncture in the pipe occurs 1 m from the tank, what is the physical state of the CO2 escaping from the pipe (Assume 1.01 bar as the atmospheric pressure)?

c) Calculate the Ma number of the CO2 upstream.

d) What is the velocity of the gas upstream if the speed of sound is 340 m/s.

Answer 1

Initial temperature and pressure

T = 180°C

P = 150 bar

Adiabatic index = 1.30 = k

a) From Perry handbook

Critical temperature of CO2 (Pc)= 304 K (31°C)

Critical pressure of CO2 (Tc) 73.8 bar

It can be seen that initial temperature and pressure of carbon dioxide is greater than both Tc and Pc

​​​​Therefore the carbon dioxide is in the liquid state in the tank

B) puncture occurs 1 m from the tank

Diameter of pipe = 3 cm

Volume of pipe = A×L

L = 2 m

A = (3.142) (0.03) 2/4 = 7.068×10-4 m2

V = 7.068(2) ×10-4 = 1.413 ×10-3 m3

By using ideal gas equation

n = PV/RT

n = 150(1.013) (0.001413) (10) 5/((8.314×453) =

n = 5.7 mol/s

mass = 5.7(44) = 250.8 g/s= 0.25 kg/s

Density of gas = PM/RT

P= 150 bar , T = 180°C= 453 K

R = 8314 J/kmol K

Density of gas=

150(1.013×105)(44)/(8314×453)

Density of gas = 177.519 Kg/m3

Q = 0.2508/177.519 = 1.412×10-3 m3​​​​/s

Velocity = Q/A = 0.001412/(7.068×10-4) =

1.998 m/s

Nre = $\frac{Dv\rho}{ \mu }$

From Perry handbook

Viscosity of carbon dioxide at 180°C=

2.21×10-5 Pa s

Nre = 0.03(1.998) (177.519) /(2.21×10-5) = 481470.08

The flow is turbulent

Swamme jain equation is used to calculate darcy friction factor

(0.25 log ( 5 + ਸੈ))2

Roughness = 0.1 mm

Substituting all values we get

0.25 (log ( 30760,03) + 481470.080.9

f =0.02732

Gas is leaked to atmosphere at 1 m

So P2 = 1.01 bar

Using adiabatic relationship we get

$\frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}}) ^{\frac{k-1}{k}}$

P2 = 1.01bar

P1 = 150 bar

T1 = 180°C

K - 1.3

Substituting We get T2 = 56.76°C

The temperature is still above Tc but P<Pc

Hence the carbon dioxide will be in gaseous state escaping from pipe

C) The process is adiabatic

Applying bernaullu equation for adiabatic flow we get

k Pl. EP， པ + 0 = # - 1 + k: – 1 ༧12 པ

At stagnation point velocity = 0

Hence velocity at leak point is negligible

k = 1.30

v1= 1.998 m/s

v2 = 0

P1 = 150 bar

$\rho _{1}$ = 177.519 kg/m3

Substituting all values in above equation we get

Ps = 85487.90

P2= (85487.90) (177.219) = 15150081.6 Pa = 151.150 bar

At stagnation pressure

: +1) a = 4

Ps = 151.150 bar

P1 = 150 bar

k = 1.30

Substit all values in above equation we get

m = Mach no = 0.123

D )

Mach number is given by

m = v/c

c - speed of sound

V = mc

v = 0.123(340) = 41.82 m/s

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