Miguel is a chemical engineer working For a power plant industry and was asked to design a pollutant control equipment For the combustion products produced in the boiler. To be able to do this: a. Joel needs to determine the stack gas analysis (complete). The coal is assumed to contain 32% VCM, 53% FC and 18% ash. Aside From that, it also has 1.2% N and 6.2% S with a heating value of 23.78 MJ/kg. The air is supplied 125% in excess. She will assume that the C02 has a molar ratio of 5:1 with Co in the stack gas. b. How much stack gas (256°C, 7?6mmHg) in m3/kg of coal will Joel be able to estimate?

The answers are provided in the explanation section.

Step-by-step explanation

Basis:

100 kg of coal

 

Given:

Coal

32 kg VCM

53 kg FC

10 kg A

5 kg M (by difference)

1.2 kg N

6.2 kg S

CV = 23.78 MJ/kg

Air

125% excess

Stack Gas

CO2/CO = 5

250^oC = 523.15 K

770 mmHg = 102.66 kPa 

 

Solution:

 

Using Calderwood Equation,

C = 5.88 + 2.206(CV-0.094S) + 0.0053(80-100(VCM/FC))^1.55

C = 5.88 + 2.206(23.78-0.094(6.2)) + 0.0053(80-100(32/53))^1.55

C = 57.59 wt % (57.59 kg)

 

Using Dulong's Formula,

CV = 0.338C + 1.44NH + 0.094S 

23.78 = 0.338(57.59) + 1.44NH + 0.094(6.2)

NH = 2.59 wt % (2.59 kg)

 

VCM = VC + CW + NH + N + S

32 = (57.59 - 53) + CW + 2.59 + 1.2 + 6.2

CW = 17.42 kg

 

Ultimate Analysis

C = 57.59

H = 2.59 + 2/18 (5 + 17.42) = 5.08

O = 16/18 (5 + 17.42) = 19.93

N = 1.2

A = 10

S = 6.2

 

theo O2 = C + S + H/4 - O2

theo O2 = 57.59/12 +6.2/32 + 5.08/1 - 19.93/32

theo O2 = 9.45 kmol

 

excess O2 = 1.25 (9.45) = 11.81 kmol

 

O2 supplied = 9.45 + 11.81 = 21.26 kmol

N2 supplied = 21.26 (79/21) = 79.98 kmol

 

CO2 in SG = 5/6 (57.59/12) = 4 kmol

CO in SG = 1/6 (57.59/12) = 0.80 kmol

H2O in SG = (5.08/1) / 2 = 2.54 kmol

 

free O2 = excess O2 + CO/2 + H2/2

free O2 = 11.81 + 0.80/2 = 11.85 kmol

 

N2 in SG = N2 supplied = 79.98 kmol

 

Complete Analysis of Stack Gas

Component	Mole (kmol)	Mole %
CO2	4	4.03
CO	0.80	0.81
H2O	2.54	2.56
O2	11.85	11.95
N2	79.98	80.65
Total	99.17

By ideal gas law,

V = (99.17 kmol) (8.314 kPa-m³/kmol-K) (523.15 K) / (102.66 kPa) 

V = 4201.61 m³ 

 

m³ SG/ kg coal = 4201.61 m³ / 100 kg = 42.02