1. The corrected ultimate analysis shows 5.6% hydrogen on the as-received basis. Basis: 100 kg of coal as received The water formed on combustion is 5.6 kg Ha 1 kg mol H; | 1 kg mol H2O 2.02 kg Hz 18 kg H,O 1 kg mol H2 | 1 kg mol H,O 49.9 kg H20 The energy required to evaporate the water is 49.9 kg H,O | 2370 kJ _ 1183 kJ 100 kg coal | kg H20 kg coal The net heating value is 29,770 – 1183 = 28,587 kJ/kg The valuc 2370 kJ/kg is not the latent heat of vaporization of water at 25°C (2440 kJ/kg) but includes the effect of a change from a heating value at constant volume to one at constant pressure (-70 kJ/kg) as described in a later section of this chapter. Use the Dulong formulate to calculate the HHV of this coal. Do you get 30,000 kJ/kg?
1. The corrected ultimate analysis shows 5.6% hydrogen on the as-received basis. Basis: 100 kg of coal as received The water formed on combustion is 5.6 kg Ha 1 kg mol H; | 1 kg mol H2O 2.02 kg Hz 18 kg H,O 1 kg mol H2 | 1 kg mol H,O 49.9 kg H20 The energy required to evaporate the water is 49.9 kg H,O | 2370 kJ _ 1183 kJ 100 kg coal | kg H20 kg coal The net heating value is 29,770 – 1183 = 28,587 kJ/kg The valuc 2370 kJ/kg is not the latent heat of vaporization of water at 25°C (2440 kJ/kg) but includes the effect of a change from a heating value at constant volume to one at constant pressure (-70 kJ/kg) as described in a later section of this chapter. Use the Dulong formulate to calculate the HHV of this coal. Do you get 30,000 kJ/kg?
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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