ELEM PRIN CHEM PROCESS ETXT + WILEYPLUS
4th Edition
ISBN: 9781119807988
Author: FELDER
Publisher: WILEY
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Chapter 3, Problem 3.30P
Interpretation Introduction
Interpretation:
Calculate the molar flow rate of each element in the coal.
Concept introduction:
It is given that wet coal being used in a power plant at a rate of 8000 lbm / min has the following composition.
| Component | Weight % (dry basis) |
| Ash | 7.2 |
| Sulfur | 3.5 |
| Hydrogen | 5.0 |
| Carbon | 75.2 |
| Nitrogen | 1.6 |
| Oxygen | 7.5 |
In addition, the wet coal contains 4.58 lbm H2 O per lbm of dry coal.
Units can be converted as,
Molecular weight of each element is as follows,
Sulfur = 32 g/mol
Hydrogen = 1 g/mol
Carbon = 12 g/mol
Nitrogen = 14 g/mol
Oxygen = 16 g/mol
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Chapter 3 Solutions
ELEM PRIN CHEM PROCESS ETXT + WILEYPLUS
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