Question

Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kg/s. CO2 is cooled at constant pressure as it flows in the pipe, and the temperature of the CO2 drops to 450 K at the exit. Determine the volume flow rate and the density of carbon dioxide at the inlet and the volume flow rate at the exit of the pipe using

(a) the ideal-gas equation and
(b) the generalized compressibility chart. Also, determine
(c) the error involved in the first case.

Answer 1

a) 0.0629
`m^(3)/s`,ρ1=31.76
`kg/m^(3)`,0.05688
`m^(3)/s`

b) 0.06171
`m^(3)/s`,32.40966
`kg/m^(3)`0.05498
`m^(3)/s`

c) 0.004%

given data:

n=2 kg/s

R=188.92 J/kgK

`T_(1)`=500 K

`T_(2)`=450 K

solution:

a) as we know ideal gas relation

PV=nRT..........(1)

The volume flow rate of inlet is

`V_(1)= (nRT_(1) )/(P)`..................(2)

putting the value in eq(2)

`V_(1)` =0.0629
`m^(3)/s`

to find destiny

ρ1=
`(n)/(V_(1) )`...........................................(3)

putting value of n and V1 is eq 3

ρ1=31.76
`kg/m^(3)`

`V_(2) =(nRT_(2) )/(P)`............(A)

putting value in eq A

=0.05688
`m^(3)/s`

to find the rates from the compressibility factors we have to find the reduced pressure and temperature in both cases

P(reduced 1)=
`(P_(1) )/(P_(crit) )`=
`(3.10^(6) )/(7.39 . 10^(6))`=0.41

T(reduced 1)=
`(T_(1) )/(T_(crit) )`=
`(500K)/(304.25K)=1.64`

b) After looking at the chart we obtain
`Z_(1)`=0.98 now the volume flow rate at the inlet is:

`V_(1) =(nRT_(1) Z_(1) )/(P)` ....................................................(4)

Putting the values in eq(4)

=0.06171
`m^(3)/s`

Now the density is

ρ=
`(n)/(V_(1) )`

=32.40966
`kg/m^(3)`

now the reduced pressure and compressibility factor in second case will be

T(reduced 2)=
`(T_(2) )/(T_(crit) )`=
`(450)/(304.25)`=1.48

P(reduced 2)=P(reduced 1)=0.41

`Z_(2)`=0.97

`V_(2) =(nRT_(2) Z_(2) )/(P)`............................................(5)

putting value in eq (5)

=0.05498
`m^(3)/s`

c) now find the error

Δn/n=modulus(ρ1
`V_(1)`-n/n)*100%.....................(6)

putting the value in eq 6

=0.004%

Answer 2

a=0.057m^3/s

b=0.061m^3/s, 32.40kg/m^3, 0.05498m^2/s

c=0.004%

Step-by-step explanation:

from the question, we have a set of data given

m= 2kg/s, p1 = 3mpa, T1= 500k, T2 = 450k

a)

`v_g=(mRT1)/(p)=(2*188.92*500)/(3*10^6)=0.06293m^3/s`

ρg= m/v1 =
`(2)/(0.0062973) =31.76kg/m^3`

v2=
`(mRT_2)/(p_2) =(2*188.92*450)/(3*10^6)=0.057m^3/s`

b)

to determine the rate of compressibility factor in the both case, we have to solve reduced pressure and temperature. The reduced pressure is calculated as 0.41 and reduced temperature is 1.64.

Using the chart, we found Z1 = 0.98 and let's find our volume flow rate

`v_1=(mRT_1Z_1)/(P_1)=0.98*0.06297=0.06171m^3/s\\p_1=(m)/(v_1) =(2)/(0.06171)=32.40kg/m^3 \\T_2=(450)/(304.24)=0.41\\v_2 =Z_2*V_2=0.97*0.05668=0.05498m^2/s\\`

The calculation for the reduced pressure and temperature is show above.

c)

The error involved in the first case can be calculated using

Δm/m=
`(pv_1-m)/(m) * 100 =0.004%`