Chimney height and Draught head
(i) hw = 353[(1/Ta) - {(1/Tg)(ma + 1)/ma}] mm of water.
(ii) hw = P/(ρa - ρg)
Assuming that the same draught is produced by gas column of height H1
H1 = H[{(Tg/Ta) - 1}x {ma/(ma + 1)}]
Here
Ta = absolute temperature of atmosphere
Tg = Average absolute temperature of chimney gas
hw = equivalent mm of water head
ma = mass of air
H = height of chimney
Condition for Maximum Discharge through a Chimney
H1,max = H
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Example
A chimney is 28 m high and the temperature of hot gases inside the chimney is 320°C. The temperature of outside air is 23°C and furnace is supplied with 15 kg of coal burnt. Calculate the following: (i) Draught in millimetre of water (ii) Draught head in metres of hot gases
Solution
(i) hw = 353[(1/Ta) - {(1/Tg)(ma + 1)/ma}]
= 353 x 28 [(1/298) - {(1/593)(15 + 1)/15}]
= 15.61 mm of water
(ii) H1 = H[{(Tg/Ta) - 1}x {ma/(ma + 1)}]
= 28[{(593/296) - 1} x {15/(15 + 1)}]
= 26.34 m
Example
Determine the height of a chimney to produce a static draught of 22 mm of water if the mean flue gas temperature in a chimney is 290°C and ambient temperature in boiler house is 20°. The gas constant for air is 29.26 kgm/kgK and for chimney flue gas is 26.2 kgfm/kgK. Assume barometer reading as 760 mm of mercury.
Solution
Density of air at 290 K
ρa = p/RT = 1.033 x 104/29.26 x 290 = 1.217 kg/m3
Density of flue gas at 563 K
ρg = 1.033 x 104/26.2 x 563 = 0.7 kg/m3
Static draught
p = H(ρa - ρg)
22 = H(1.217 - 0.7)
H = 42.55 m will be the height of the chimney
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