Solutions
225Here Slat is the lateral surface of the inner tube,
Slat = 2xtrl, where r is the radius of the inner
tube, i.e. tr 2 = S, and r = VSla. The amount
of heat Q is spent for heating the liquid flowing in
tube 1. During the time t, the mass of the liquid
flowing in the outer tube 1 is at = !MS, and its
temperature increases from T^11 to Tn.. Conse-
quently,
Q = pvtSc (Tn —. T 11 ). (^) ( 3 )
Equating expressions (2) and (3), we obtain
2n - 17 — lk (7'1S 2 —Tn. ). pvSc (Tf 1 —T1 1 ).
Hence we can find Tr1, and using Eq: (1), Tr2 as
well. Therefore, we can write
Tri =-- Ti2+ Ti2) (
1+2 nIS lk )-I
pvc
Tf2 = Tii (Ti 2 - T11) (
- 1 2 173 - cIS lk )-t
mgasa
Pmax= s Psat,
where mgas is the mass of the substance in the
gaseous state contained in the vessel. Consequently,
for a < psatS/M, no condensation will take place,
while for a > PsatSIM, the mass of the gas will
pvc
2.19*. As a result of redistribution, the gas pressure
obviously has the maximum value at the rear
(relative to the direction of motion) wall of the
vessel since the acceleration a is imparted to the
gas just by the force of pressure exerted by this
wall. We denote this pressure by Amax. On the other
hand, Pmax < Psat• Considering that psat >> P
and hence neglecting the force of pressure exerted
by the front wall, on the basis of Newton's second
law, we can write
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