Answer:
a) 0.0629 ,ρ1=31.76
,0.05688
b) 0.06171 ,32.40966
0.05498
c) 0.004%
given data:
n=2 kg/s
R=188.92 J/kgK
=500 K
=450 K
solution:
a) as we know ideal gas relation
PV=nRT..........(1)
The volume flow rate of inlet is
..................(2)
putting the value in eq(2)
=0.0629
to find destiny
ρ1=...........................................(3)
putting value of n and V1 is eq 3
ρ1=31.76
............(A)
putting value in eq A
=0.05688
to find the rates from the compressibility factors we have to find the reduced pressure and temperature in both cases
P(reduced 1)==
=0.41
T(reduced 1)= =
b) After looking at the chart we obtain =0.98 now the volume flow rate at the inlet is:
....................................................(4)
Putting the values in eq(4)
=0.06171
Now the density is
ρ=
=32.40966
now the reduced pressure and compressibility factor in second case will be
T(reduced 2)==
=1.48
P(reduced 2)=P(reduced 1)=0.41
=0.97
............................................(5)
putting value in eq (5)
=0.05498
c) now find the error
Δn/n=modulus(ρ1-n/n)*100%.....................(6)
putting the value in eq 6
=0.004%