kg of Dry Blast Air

A P P E N D I X

O Conversion of Grams H2O(g)/Nm3 of Dry Blast Air to kg H2O(g)/kg of Dry Blast Air This appendix shows how to convert measured grams of H2O(g) in blast per Nm3 of dry air in blast to kg of H2O in blast per kg of dry air in blast.

so that n, kg mol of ideal gas, is related to volume, Nm3 of ideal gas, by; n 1 5 5 0:044 kg mol=Nm3 V 8:314 3 1022
273:15

(O.3)

from which we state that 1 Nm3 of ideal gas contains 0.044 kg mol of ideal gas.

O.1 kg mol OF IDEAL GAS PER Nm3 OF IDEAL GAS This section calculates kg mol of ideal gas per Nm3 of ideal gas. It uses the ideal gas law equation;

O.2 kg mol O2 AND N2 IN 0.044 kg mol OF AIR

P
V5n
R
T

(O.1)

Air is 21 vol.% (mol)% O2 and 79 vol.% (mol)% N2, Appendix B. So 0.044 kg mol of air contains;

n P 5 V R
T

(O.2)

0:044 kg mol of air


restated as;

where n 5 kg mol of ideal gas; P 5 pressure, bar; V 5 volume of ideal gas, m3; R 5 gas constant 5 8.314 3 1022 (m3 bar)/(K kg mol); and T 5 temperature, Kelvin. The conditions where m3 5 Nm3 are; 1. 1 bar pressure, and 2. 273.15K temperature (0 C)

21 mol% O2 in air 5 0:00924 kg mol of O2 100% (O.4)

and 0:044 kg mol of air


79 mol% N2 in air 5 0:0348 kg mol of N2 : 100% (O.5)

for a total of 0.044 kg mol of dry air.

715

716

APPENDIX O: CONVERSION OF GRAMS H2O(g)/Nm3 OF DRY BLAST AIR TO KG H2O(g)/KG OF DRY BLAST AIR

O.3 kg OF O2, N2, AND AIR IN 0.044 kg mol OF DRY AIR

"

0.00924 kg mol of O2 contains;

# mass through-tuyere

1 input H2 O g   mass dry air 5
C kg H2 O
g= in blast

kg of dry air


  C g H2 O g = mass dry air Nm3 of dry air 5
1:27
1000 in blast



0:00924 kg mol O2
32 kg of O2 =kg mol O2 5 0:296 kg O2 (O.6)

and 0.0348 kg mol of N2 contains;

which readily expands to; "

0:0348 kg mol N2
28 kg of N2 =kg mol N2 5 0:974 kg N2 (O.7) 3

for a total of 1.27 kg of dry air per Nm of dry air (32 and 28 are the molecular masses of O2 and N2).

O.4 kg H2O(g)/kg OF DRY AIR We now specify that the H2O(g) concentration of the blast furnace’s moist blast is;


  C g H2 O g = Nm3 of dry air
5 1:27
1000 in blast
   C g H2 O g = 3 mass N2 Nm of dry air (O.10)
1 1:27
1000 in blast    mass through-tuyere
 subtracting from
1 input H2 O g




 g H2 O g = Nm3 of dry air

"



or because we wish to work in kg of H2O(g)
 g H2 O g = Nm3 of dry air

C C kg H2 O
g= kg of dry air

5

1:27
1000

(O.8)

# mass through-tuyere

1 input H2 O g


  C g H2 O g = Nm3 of dry air 1
in blast 1:27
103
   C g H2 O g = mass N2 Nm3 of dry air
1 in blast 1:27
103



C

1:27 kg of dry air =Nm3 of dry air

mass O2

or both sides;

where C is concentration. Since, 1 Nm3 of dry air contains 1.27 kg of dry air; 5

# mass through-tuyere

1 input H2 O g

05 

Cg H2 OðgÞ=Nm3 of dry air

C g H2 O
g= kg of dry air

(O.9)

mass O2

where all masses are kg per 1000 kg of Fe in product molten iron. We use this equation wherever the ingoing air of a blast furnace is humid and wherever steam is injected into the blast. An example moisture value is 15 g H2O(g)/Nm3 of blast air for which C kg H O
g= 5 2

kg of dry air

O.5 MATRIX EQUATION The value in Eq. (O.8) may be applied to the blast furnace as

(O.11)

15 g of H2 OðgÞ=Nm3 of dry air 1:27 3 103

H2O(g)/kg of dry blast air.

5 0:0118

kg