Chimney 1
Chimney 1
Chimney 1
19 m
9.81
0. 23
m2
=H (1.2−0.79 ) (
1000 m3 )
hw =7.05 cm H 2 O
(
100
KPa ) H=57.18 m
1 cm H 2 O
3. The DENR is very strict in
hw =0.691 KP a terms of air pollution. The height
of the chimney’s is by then
m P regulated to be at 65m to 100m.
=
V RT Compute the mean draft pressure
if the gas density is determined to
101.325
¿ be at 0.91 kg/m3.
0.287(21+273)
hw =H ( ρa− ρg )
kg
¿ 1.2 3
m 9.81
hw =100 ( 1.2−0. 91 ) ( 1000 )
hw =0.284 KP a
m P
=
V RT hw =H ( ρa− ρg )
101.325 9.81
¿
8.314
(149+273)
hw =65 (1.2−0. 91 ) ( 1000 )
30
hw =0.185 KP a
kg h +h
¿ 0.866 3 h´w = w1 w 2
m 2
kg PV =mRT
mg=62550
hr 10 0 V =15 ( 0.287 ) ( 26+273 )
m3
V =12 . 87
s
Q= AV
m3 π 2 m
12.87 = d 10
s 4 ( )( )
s
d=1.28 m
101.325
¿
8.314
(250+273)
30
kg
¿ 0. 699
m3
m
ρ=
v
5(25)
0. 699=
v
m3
v =178.81
s
kg 0.00981 KN
)
V =7.67
Q= AV
m3 π 2
s 4
s
( )
7.67 = d ( 8 )
hw =37.12 2
m (
1 Kg ) d=1. 105 m
PV =mRT
40,000
80 V = ( 0.287 ) ( 21+ 273 )
3600
m3
V =11.72
s
Q= AV
m3 π 2
s ( )
11.72 = d ( 12.5 ( 0.4 ) )
4
d=1.728 m
kg 9.81 KN PV =mRT
(
hw = 37.12
m2 )( 1 kg ) 98 V =10.35 ( 8.314
10 )
( 250+273 )
hw =364.16 Pa m3
V =45.91
s
hw =ha +h f Q= AV
m3 π 2
364.16=200+ hf
( )
45.91 = d ( 7.5 )
s 4
1 KPa
hf =164.16 Pa ( 1000 Pa ) d=2 . 79 m
hf =0.164 KPa
220
CO2 =
O 2=
0.18
=
9
44 2200
0. 06
=
3
√
¿ 2(9.81) ( )
9.81
0 .814
m
32 16 00 V =23.24
s
0.76 19 m
N 2= = ρ=
28 700 v
¿ 0.033 7
0. 814=
v
m P
=
V RT
m3
v =8.59
98.2 s
¿
8.314 180+150
( 1
0.033 )( 2
+273 ) Q= AV
23.24
¿
0.4
¿ 0.814 m3 π
hw =H (ρa− ρg )
8.59 = d2 ¿
s 4 ( )
d=1.08 m
0.25=H ( 1.15−0.814 ) ( 19.00081 )
H=73.78 m
hw
V= 2g
√ ( )
ρg
√
3.2 1
m
V =30.46
s
Q= A V
17.52(2.8) π 2
0.673 4 ( )
= d ( 0. 4 ( 30.46 ) )
P −101.32
ρg =
RT 0.27 7 (25 0+273)
101.325 H=830.99 m
¿
8.314
(250+ 273)
30
23. What is the height of the
¿ 0.699
chimney if the driving pressure is
hw 500Pa and the gas and air
V= 2g
√ ( )
ρg densities are 2kg/m3 and 2.8kg/m3
respectively?
√
5 1
0 .699
m
500 ( 1001 )=H ( 2.8−2) ( 0.00981
1 )
V =23.69
s H=63.71 m
Q= A V
5 (25) π 2
0.699 4 ( )
= d ( 0.4 ( 23.69 ) )
d=4.9 m
1000
21.78
(100
=H ) (101.32
0.287 ( 26+273 ) )
995.68 ( 2.157
100 )=H
( 101.3 4
0.287 ( 31+273 ) )
−101. 34
−101.32
0.28 7( 260+ 273)
0.28 7(177 +273)
H=43.03 m
H=550 m
hw =H (ρa− ρg ) H=146.8 m
10 00 ( 2.89
100 ) 94
( 8.314 (130+273 ) )
=H ( 1.2 )−
30
H=89.6 m
PV =mRT
9.92(0.278)(57+273) π 2 37.33
97
= d
4 ( )(
3.281 ) 98 V =
10,000(19)
3600
( 0.287 )( 33+273 )
3
d=1.04 m m
V =47.296
s
32
( 3.381 9.81
)( 1000 )( 1000 ) 4 hw
d=1.15 m
V= 2g
√ ( )
ρg
√
8.62
for a furnace is 7.5 cm of water
¿ 2(9.81)
( 100 )
( (1000 )
)
and the frictional losses in the 0.892
stack is 13% of theoretical draft. m
V =23.69
Calculate the diameter of the s
hw =ha +h f d=1.25 m
hw =7.5+ 0.13 hw
hw =8.62 cm H 2 O
P
ρ=
RT
m
V =24.13
s
Q= A V
( 82800
3600 ) π
0.789 ( 4 )
2
= d ( 0.4 ( 24.13 ) )
d=1. 96 m
√
2.34
¿ 2(9.81)
((100 )
( 1000 )
) ρg =
P
RT
0 .789
101.325
¿
0.287(250+273)
¿ 0.675 65,000
97.53 V = ( 0.287 )( 27.78+273 )
3600(2.205)
h m3
V= 2g w
ρg√ ( ) V =7.02
Q= A V
s
√
5 1
Q= A V d=1.01 m
24.11
0.4( ¿)
12(25) π 2
0.675
= d ¿
4 ( ) 40. The theoretical velocity is
15m/s. Determine the required
d=4.94 m diameter if the mass flow of gas is
35,000kg per hour. Assume
39. The theoretical velocity of
atmospheric conditions.
gas entering in a chimney is
22m/s. The gas temperature is
Q= A V
82°F and pressure of 28.8 in Hg a
gas constant of 0.278KJ/kgK.
15
Determine the chimney diameter if 0.4(¿)
35,000 π
mas of flue gas is 65000 pounds
3600 (1.2) 4 ( )
= d2 ¿
per hour.
d=1.31 m
28.8∈Hg≈ 97.53 KP a
82 ℉=27.78 ℃
PV =mRT
98
¿
8.314
(250+273)
30
¿ 0.676
hw
V= 2g
√ ( )
ρg
hw
7.5
0.4 √
= 2(9.81)
0.676( )
kg
hw =12.12 2
m
ρ a=
P
RT
√
1
24.78 220
= H (1.14−0.795)
0.4(¿) 9.81
4 π
= d2 ¿
0.814 4 ( ) H=64.5 m
d=0.794 m
√
250 P
¿ 2(9.81)
( 9. 81 )
( ) ρg =
RT
0.779
101.325
¿
m 0.287(270+273)
V =25.33
s
¿ 0. 65
Q= A V
hw
25.33
0.4 (¿)
V= 2g
√ ( )
ρg
3 π
= d2 ¿ ( )
√
2.5
0.779 4
¿ 2(9.81)
(
( 100 )
(994.78)
)
0. 65
d=0.696 m
m
V =27.395
s
Q= A V
27.395
45. A steam boiler plant 0.4( ¿)
10000 (15) π 2
consumes 10000kg of coal per = d ¿
3600 (0.65) 4 ( )
hour and produces 15kg of dry
flue gas per kg of coal fired. d=2.73 m
46. 2.8kg of coal per second are
Outside air temperature is 33°C,
consumed by a steam boiler plant
average temperature of dry gas in
and produced 17.52kg of dry flue
the chimney is 270°C. The gage
gas per kg of coal fired. The air
fluid density is 994.78kg per cubic
temperature outside is 31°C, the
meter and the theoretical draft of
average temperature of the flue
2.5cm of water at the chimney
gases entering the chimney is What is the diameter of the
342°C and the average chimney if the driving pressure is
temperature of the flue gas in the 300Pa and the gas and air
chimney is 270°C. The gage fluid densities are 1kg/m3 and 1.5kg/m3,
specific volume is 1.005x10-3m3 per respectively?
kg and theoretical draft of 3.2 cm
hw
of water at the chimney base is
needed when the barometric
V= 2g
√ ( )
ρg
√
300
height of the chimney in meters. ¿ 2(9.81)
( 9.81 )
( ) 1
Assume molecular weight of 30.
m
V =24.49
P s
ρa= Q= A V
RT
101.325
¿
0.287(31+273) 24.49
0.4 (¿)
¿ 1.16 20,000
(17.25)
P 3600 π
ρg =
RT 1
= d2 ¿
4 ( )
101.325
¿
8.314 d=3.52 m
(270+ 273)
30
¿ 0.673
hw =H (ρa− ρg )
3.2 1
(
100 0.00 1 005 )
=H (1.16−0. 6 73)
H=65 .5 2 m
¿ 1.17
100
(
(1000)=H 1. 2−
8.314
30
( 130+270 ) )
H=86.5 m
hw =H (ρa− ρg )
2.35
(1000)=H (1.17−1.15)
100
H=1224.42 m
kg 1 hr
mg=5561.5 (
hr 3600 s )
kg
mg=1.54
s
P
ρg =
RT
98
¿
8.314
(130+273)
30
¿ 0. 877
hw
V= 2g
√ ( )
ρg
√
2.69
¿ 2(9.81)(( 100 )
(1000)
)
0 .877
m
V =24.5 3
s
Q= A V