Levenspiel ch1 solution

1. This is Example 1.1.
hydrogenUsed = 108 ê 18; H∗kmolês∗L
oxygenUsed = 108 ê 18 ê 2;
diam = 0.6; H∗m∗L
length = 0.75;
vol = Pi ê 4 diam ^ 2 length H∗m^3∗L;
rateH = 1 ê vol hydrogenUsed H∗kmolêm^3ês∗L
rateO = 1 ê vol oxygenUsed
28.2942
14.1471
This is Example 1.2. Note that in order to obtain the rate of oxygen consumption, we must multiply by 6, rather than
divide by 6; see p. 7, Levenspiel. This has been corrected in the newly printed version of the 3rd edition. (The "Fluid
cracking crackers" on p. 9 has not (yet) been corrected to "Fluid catalytic crackers.")
rhoHuman = 1000; H∗kgêm^3∗L
massHuman = 75; H∗kg∗L
volHuman = massHuman ê rhoHuman;
energyConsumed = 6000; H∗kJêday∗L
deltaH = 2816; H∗kJêmol∗L
oxygenConsumed = energyConsumed ê deltaH ∗ 6H∗mol O2êday∗L êê N
metabolicRate = 1 ê volHuman HoxygenConsumed ê 86 400L êê N
12.7841
0.00197285
This is Problem 2 in Chapter 1. Note that the result is in reasonable agreement with the very useful scale in Figure
1.3. Note also that 240 tons of coal per hour is the total consumption in the power plant; this can be verified by
assuming a typical plant efficiency of 35% and a heating value of coal of some 12000 BTU/lb.
oxygenConsumed = 0.1 H240 ∗ 0.5 ∗ 0.9 H32 ê 12L + 240 ∗ 0.5 ∗ 0.1 H16 ê 2LL; H∗tons of O2êhêreactor∗L
reactorVolume = 20 ∗ 4 ∗ 1; H∗m^3∗L
reactionRate = oxygenConsumed ∗ 1000 ê 0.032 ê reactorVolume ê 3600 H∗molêm^3ês∗L
4.16667