70 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
A further assumed value ofdD 0 .05 m gives a value of the right hand side of
equation 4.55 of 25.9 and the procedure is repeated until this value is zero.
This occurs whendD 0 .08 m or 80 mm.
PROBLEM 4.10
In a vacuum system, air is flowing isothermally at 290 K through a 150 mm diameter
pipeline 30 m long. If the relative roughness of the pipewalle/dis 0.002 and the down-
stream pressure is 130 N/m^2 , what will the upstream pressure be if the flowrate of air is
0.025 kg/s? Assume that the ideal gas law applies and that the viscosity of air is constant
at 0.018 mN s/m^2.
What error would be introduced if the change in kinetic energy of the gas as a result
of expansion were neglected?
Solution
As the upstream and mean specific volumesv 1 andvmare required in equations 4.55 and
4.56 respectively, use is made of equation 4.57:
G/A^2 lnP 1 /P 2 CP^22 P^21 / 2 RT/MC 4 R/u^2 l/dG/A^2 D 0RD 8 .314 kJ/kmol K and hence:
2 RT/MD 2 ð 8. 314 ð 103 ð 290 / 29 D 1. 66 ð 105 J/kgThe second term has units ofN/m^2 ^2 /J/kgDkg^2 /s^2 m^4 which is consistent with the
other terms.
AD/ 4 0. 15 ^2 D 0 .0176 m^2∴ G/AD 0. 025 / 0. 0176 D 1. 414
and ReDdG/A/D 0. 15 ð 1. 414 / 0. 018 ð 10 ^3 D 1. 18 ð 104
For smooth pipes andReD 1. 18 ð 104 ,R/u^2 D 0 .0040 from Fig. 3.7. Substituting in
equation 4.57 gives:
1. 414 ^2 lnP 1 / 130 C 1302 P^21 / 1. 66 ð 105 C 4 ð 0. 0040 30 / 0. 15 1. 414 ^2 D 0Solving by trial and error, the upstream pressure,P 1 D 1 .36 kN/m^2
If the kinetic energy term is neglected, equation 4.57 becomes:P^22 P^21 / 2 RT/MC 4 R/u^2 l/dG/A^2 D0andP 1 D 1 .04 kN/m^2Thus a considerable error would be introduced by this simplifying assumption.