Basic Principles and Calculations in Chemical Engineering (Prentice Hall International Series in the Physical and Chemical Engineering Sciences) by David M. Himmelblau & James B. Riggs

Author:David M. Himmelblau & James B. Riggs [Himmelblau, David M.]
Language: eng
Format: azw3
Publisher: Pearson Education
Published: 2012-05-30T16:00:00+00:00

Moles of O2 in with air including excess:

(1 + 0.04)(0.5 mol O2) = 0.52 mol O2

Therefore, 0.52/0.21 = 2.48 mol air enters containing 1.96 mol nN2.

Specifications: 65% of the carbon burns to CO: (0.65)(2) = 1.30.

Basis: 1 mol of H2C2O4 enters

Element material balances:

The results:

nH2O = 1.0; nCO2 = 0.7; nCO = 1.3; nO2 = 1.82; nN2 = 1.96; total mol = 6.78

yH2O = 1 mol H2O/6.78 mol = 0.147

The partial pressure of the water in the product gas (at an assumed atmospheric pressure) determines the dew point of the stack gas; that is, the temperature of saturated steam that equals the partial pressure of the water is equal to the dew point of the product gas:

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