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36
APPLICATION OF THE LAWS OF HYDRAULICS
When the gas passes from a larger channel into a smaller
channel the velocity changes from 7mm to Fmax- If the velocity
Fmm corresponds to a velocity head Amm and Fmax to a velocity
head hmaiK} then to produce the increase in velocity from Fmto to
Fmax a velocity head will be required equal to /tmax~^min.
This quantity represents the resistance, or the velocity head
required by one tier of the checkerwork.
If there are N tiers, their total resistance or velocity head
required will be
5=
in which At is the weight of 1 cu m of the gas in motion at the
temperature t.
B. What will be the velocity head required to overcome the
resistance offered by the changes of velocity of the air in passing
through a checkwork 6 m in height (52 tiers)? The average
temperature of the air is
50+1050
= 550°.
The velocities are Fmm = 0 m 50 per second, Fmax=l rn 00 per
second.
The weight of 1 cu m of air at a temperature of 550° is
Fmm =0 m 50 per second corresponding to hmln = 0.01274,
Fmax=l m 00 per second corresponding to Amax = 0.05097,
from which
ftrnax -fcmta = Om 03823 ; ;
&-egen = 0.03823X52X0. 43 = 0 mm 85 of water.
And therefore the resistance due to changes of velocity in the
checkerwork is equal to 0 mm 85 of water column, while the
regenerator 6 m in height produces a hydrostatic pressure of
A = 6(1. 29-0. 43) =5 mm 16 of water.