D | Question 33 20 pts The analysis of a coal is reported as follows (wt % Moist

Question

D | Question 33 20 pts The analysis of a coal is reported as follows (wt % Moisture 8.0%, Volatile matter 37%, Ash 3.1%, Carbon 78.5%, Hydrogen 4.5%, Nitrogen 1.6%, Sulfur 0.8% and Orygen 14.6%. What is the % Fixed carbon on a dry basis of this coal? D Question 34 20 pts A sample of coal has the following analysis (wt %). Moisture 60%, Fixed Carbon 42.0%, Volatile Matter 363%, Carbon 43.9% Hydrogen 5.0%, Nitrogen 1.0%, Sulfur 1.0%, Oxygen 21.7% and the rest as ash. What is the Lower Heating Value (in BTU/Lb) of this fuel Hint: Calculate HHV using Dulong Formulal

Explanation / Answer

Answer 33:

Fixed Carbon = 100 - (Volatile matter + Ash + Moisture content)

= 100 - [8+37+3.1]

= 100 - 48.1

= 51.9 %

Answer 34 :

Given : Moisture = 6.0%

Fixed Carbon = 42.0%

Volatile matter = 36.3%

Carbon = 43.9%

Hydrogen = 5.0%

Nitrogen = 1.0%

Sulfur = 1.0%

Oxygen = 21.7%

According to Dulong formula,

HCV = 1/100 [8080*C + 34500(H-O/8) + 2240*S]

= 1/100 [8080*43.9 + 34500(5-21.7/8) + 2240*1]

= 1/100 [ 354712 + 34500(5-2.7125) + 2240]

= 1/100 [ 354712 + 78918.75 + 2240]

= 1/100 [ 435870.75]

= 4358.7075 kcal/kg.

LCV = [HCV - 9/100*H*587]

= [4358.7075 - 9/100*5*587]

= [4358.7075 - 264.15]

= 4094.5575 kcal/kg.

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