Question

Propane (C3H8) burns completely with 150% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. Determine the mole fraction of water in the products, in lbmol(water)/lbmol(products).

Answer 1

Answer:

y=0.12 lbmol(water)/lbmol(products)

Explanation:

First we find the humidity of air. Using humidity tables and the temperature we find that is 0.01 g water/L air.

Now we set the equation assuming dry air:

$C_{3}H_{8}+7.5(O_{2}+3.76N_{2}) \longrightarrow 3CO_{2}+4H_{2}O+2.5O_{2}+28.2N_{2}$

With this equation we have almost all moles that exit the reactor, we are just missing the initial moles of water due to the humidity. So we proceed to calculate it with the ideal gas law:

PV=nRT

Vair=867.7L

With the volume and the fraction of water, we can calculate the mass of water:

0.01 * 867.7=8.677 g of water

Now we calculate the moles of water:

8.677 g / 18 g/mol = 0.48 moles of water

Now we can calculate the total moles of water in the exit of the reactor:

0.48 + 4 = 4.48 moles of water

And finally we just need to sum all moles at the exit of the reactor and divide:

3 moles of CO2 + 2.5 moles of O2+ 4.48 moles of H2O + 28.2 moles of N2

And we have 38.18 moles in total, then:

4.48/38.18=y=0.12 moles of water/moles of products

As the relation moles/moles is equal to lb-moles/lb-moles, we have our fina result:

y=0.12 lbmol(water)/lbmol(products)