I ' !
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
in which At is the weight of 1 cu m of the gas in motion at the
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
The velocities are Fmm = 0 m 50 per second, Fmax=l rn 00 per
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,
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.