ChE 441: Advanced Transport Phenomena
Problem Set #1: Solutions
1. It is very important to make a habit of checking equations for dimensional
consistency. By replacing the symbols in the formula by the dimensions corresponding
to the symbols in the table beginning on page 872 of the text, show that the following
are dimensionally consistent (omit any numerical factors that appear):
a. equation 1.4-14
Solution
b. equation 1.5-11
Solution
c. equation 1.7-2
(in part c. we can assume that the unit tensor
Solution
is a dimensionless quantity)
2. 1A.1 Estimation of dense-gas viscosity. Estimate the viscosity of nitrogen at 68oF
and 1000 psig by means of Figure 1.3-1, using the critical viscosity from Table E.1. Give
the result in units of lbm/ft-s. For the meaning of "psig," see Table F.3-2. (Answer:
1.4x10-5 lbm/ft-s)
Solution
T = 68oF = 20oC = 293K
P = 1000 psig
Table E.1 gives Tc = 126.2K, Pc = 33.5 atm, µc = 180x10-6 g/cm s
Figure 1.3-1, using this reduced state
Using Table F.3-4 for unit conversion, g/cm-s to lbm/ft-s
(Note - reading Figure 1.3-1 is critical to obtaining the result given. So your result may
vary depending on your reading of the figure.)
3. 1A.3 Computation of the viscosities of gases at low density. Predict the
viscosities of molecular oxygen, nitrogen and methane at 20oC and atmospheric
pressure. Express the results in mPa-s and compare your calculated results with
experimental data given in Chapter 1. (Answers: 0.0202, 0.0172, 0.0107 mPa-s)
Solution
Oxygen
Nitrogen
Methane
M
32.0
28.013
16.04
3.433
3.667
3.780
113
99.8
154
Oxygen at 20oC
Table E.2,
Equation 1.4-14:
Reported value in Table 1.1-3 is 2.04x10-2 mPa-s, so the results are very close.
Nitrogen at 20oC
Table E.2,
Equation 1.4-14:
Reported value in Table 1.1-3 is 1.75x10-2 mPa-s, so the results are very close.
Methane at 20oC
Table E.2,
Equation 1.4-14:
Reported value in Table 1.1-3 is 1.09x10-2 mPa-s, so the results are very close.
4. 1A.7 Molecular velocity and mean free path. Compute the mean molecular velocity
(cm/s) and the mean free path (cm) for oxygen at 1 atm and 273.2 K. A reasonable
value for d is 3 Å. What is the ratio of the mean free path to the molecular diameter
under these conditions? What would be the order of magnitude of the corresponding
ratio in the liquid state? (Answers: =4.25 x 104 cm/s, = 9.3 x 10-6 cm)
Solution
From Appendix F, page 867
Equation 1.4-1:
Equation 1.4-3:
Ratio of mean free path to molecular diameter is
under these conditions. At liquid states, the corresponding ratio would be on the order of
unity or less.