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Xelga [282]
3 years ago
10

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 p

ipe, 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.
Engineering
1 answer:
Viktor [21]3 years ago
6 0

Answer:

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}= \frac{nRT_{1} }{P}..................(2)

putting the value in eq(2)

V_{1}  =0.0629 m^{3}/s

to find destiny

ρ1=\frac{n}{V_{1} }...........................................(3)

putting value of n and V1 is eq 3

ρ1=31.76 kg/m^{3}

V_{2} =\frac{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)=\frac{P_{1} }{P_{crit} }=\frac{3.10^{6} }{7.39 . 10^{6}}=0.41

T(reduced 1)= \frac{T_{1} }{T_{crit} }=\frac{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} =\frac{nRT_{1} Z_{1} }{P} ....................................................(4)

Putting the values in eq(4)      

    =0.06171 m^{3}/s        

Now the density is

ρ=\frac{n}{V_{1} }

  =32.40966 kg/m^{3}

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

T(reduced 2)=\frac{T_{2} }{T_{crit} }=\frac{450}{304.25}=1.48

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

Z_{2}=0.97

V_{2} =\frac{nRT_{2} Z_{2} }{P}............................................(5)

putting value in eq (5)

    =0.05498 m^{3}/s

c) now find the error

Δn/n=modulus(ρ1V_{1}-n/n)*100%.....................(6)

putting the value in eq 6

      =0.004%

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