Kosky, Balmer, Keat and Wise: Exploring Engineering, Fourth Edition
Copyright ©2015, Elsevier, Inc
Each molecule of H2(g) requires two e–. Hence the flow of H2(g) = 1.02 x 1024/2.000 =
5.08 x 1023 molecules/s.
Divide by Avogadro’s number to obtain the kmols H2(g)/s = 5.08 x 1023/6.022 x 1026
[molecules/s][kmol/molecule] = 8.44 x 10-4 kmols H2(g)/s.
Use the standard molar volume 22.4 m3/kmol, to get the std3 volumetric flow of H2(g) =
22.4 x 8.44 x 10-4 [std m3/kmol] [kmols/s] = 1.89 x 10-2 std m3/s.
The O2(g) rate would have been 9.45 x 10-3 std m3/s and the air rate would have
been 4.76 x this or 4.50 x 10-2 std m3/s
3. Had we wanted actual flow rates we would have corrected by absolute temperature. Suppose the feed