Question

Saturated process gas containing CO( 5 wt%), CO2(15 wt%) and H2( 80 wt%) on dry basis enters compressor at 40oC and 8 atm . The process gas is compressed to 30 atm. For 100 NM3 (on dry basis) of process gas

(a) Estimate the weight of water condensed if gas is compressed isothermally

(b) Estimate outlet temperature with 15 degrees of superheat and amount of water condensed.

Answer 1

Let us repersent inlet as point (1) and outler as point (2)

At inlet : CO-5 wt% , CO₂ -15 wt% , H₂ -80 wt% ( on dry basis)

Let us first convert this weight percent to mole percent

Compound	Weight %	Molar mass	Moles	Mole fraction
CO	5	28	0.1786	0.0044
CO2	15	44	0.3409	0.0084
H2	80	2	40	0.9872
total	100		40.5195	1

                                  Table: mole fraction on dry basis.

Now Volume of process gas = 100Nm³

Here 'N' means normal temperature and pressure that is T=20 ^oC=293 K and P=1 atm

But our inlet is at 40 ^oC and 8 atm

Volume at T=40 ^oC=313 K and P=8 atm can be calculated as

(PV/T) [at Normal T &P)] =   (PV/T) [at 40 oC and 8atm]     

therefore V at inlet conditions = (T2/T1)*(P1/P2)Vo= (313/293)*(1/8)*100=13.3532 M3

Now at the inlet we know P, V,T. Hence we will use ideal gas law to calculate the total nummber of moles

At inlet , P=8 atm

              V= 13.3532 m3 =13.3532 *10³ litre

              T= 40 ^oC =313 K

              R =0.0821 Litre atm mole^-1 K^-1

by ideal gas law ,

P V = N R T

8 * 13.3532*10³=N *0.0821*313

N =4157.0748 moles at inlet

Cas(a) Isothermal process

     Outlet Temperature = 40 ^oC =313 K

     Outlet Pressure = 30 atm

Now we know V = 100 Nm³ at normal temperature and pressure,

Using this we will calculate volume at outlet condition

V (outlet)=(T2/T1)*(P1/P2)*Vo=(313/293)*(1/30)*100 = 3.56 m³

Hence outlet Volume = 3.56 m³ =3.56* 10³ litre

R= 0.0821 Litre atm K^-1 Mole^-1

Substiuting all the known values in ideal gas law to calculate the number of moles at outlet ,

30*3.56*10³= N *0.0821* 313

N ( at outlet) = 4156.0786 moles

Now the components present in the stream are CO,CO₂, H₂ , H₂0 only H₂0 is condensable

Hence the moles Which have been reduced in the oulet , will correspond to the mole sof H₂0 condensed

moles of H₂0 condensed =4157.0748-4156.0786=0.999 moles

therefore weight of H20 CONDENSED = 0.999*18 = 17.99 = 18 grams of water

case(b)

Degree of superheat =15 ^oC

Hence Outlet temperature =inlet temperature + degree of superheat

                                      = 40+15 =55 ^oC =55+273.15=328.15 K

Outlet temperature is 55 ^oC

Oultet Pressure= 30 atm

Here again, we will convert V = 100 Nm3 to V at 55 oC and 30 atm

V (at T=55 oC, P=30 atm) =(T2/T1)*(P1/P2)*Vo =(328.15/293.15)*(1/30)*100=3.73153 m³

V= 3.7315 m³ = 3.7315 *10³ Litre

R=0.0821 Litre atm Mole^-1 K^-1

Substituting all the known values in the ideal gas equation to obtain the number of moles at outlet

30*3.7315 *10³ = N *0.0821*328.15

N(at outlet) =4155.1732 moles

Since except H2O all the non-condensables

H2O condensed = Number of moles at inlet - number of moles at outlet

                        = 4157.0748-4155.1732

                        = 1.90156 moles

Therefore amount of water condensed = 1.90156*18=34.228 grams