Question

Pure nitrogen at 160°C and 2 atm enters a cylinder whose inner walls are coated with naphthalene. The nitrogen flow is turbulent. The Dittus-Boelter relation for turbulent flow heat transfer in a pipe is

Nu = 0.023Re^0.8Pr0.4

The velocity through the 6 cm diameter tube is 10 m s-1.

(a) Derive an equation relating concentration of naphthalene to tube length;

(b) What tube length would be necessary to obtain a naphthalene mole fraction of 0.05 in the exit stream?

Additional data: Molecular weight of naphthalene: 128 g mol-1; Vapour pressure of naphthalene at 160°C: 200 mmHg; Diffusivity of N2-naphthalene at 298 K and 1 atm: 0.061 cm2 s-1; Viscosity of nitrogen at 300 K and 0.1 MPa: 1.8∙10-5 Pa s.

The flow is fully developed

Answer 1

density of nitrogen at given condition will be=1.557 kg/m3.

specific heat capacity=1048.9 J/Kg K.

viscosity=given viscosity*(T/T*)^(1.5)=3.12*10^-5 Pa s

Reynold number =(density*diameter*velocity/viscosity)

=29942.3 (turbulent condition)

schmidit number=0.2003/.061

=3.28

sherwood number=

Sh=.023*Re^(0.8)*Sc^(0.4)

Sh=Diameter/film thickness

Putting value , we get thickness=6/140.96

=.043cm

mass transfer cofficient=diffusivity/film thickness

=1.433cm/s

mass flux =mass transfer cofficient*(mole fraction difference)

=1.433*135.25(0.263-0)

=50.97mole/m2 s

moler flow rate=50.97*2*3.14*.03*length

(A) mole flow rate=9.6*length

(B)length=41.28/50.97 cm