Page 229 - Bird R.B. Transport phenomena

§7.6 Use of the Macroscopic Balances for Steady-State Problems 213

The x-component of the momentum balance then gives
F x = (viWi + piSJ - {v w 2 + p S ) cos в (7.6-22)
2
2 2
where F is the x-component of Fy_». Introducing the specific expressions for w and w , we
x s x 2
get
F = Viipv^) - v (pv S ) cos в + p S - p S cos в
x 2 2 2 x x 2 2
= pvlS (f3 - cos 6) + (pi - p )S + f? (Si - S cos в) (7.6-23)
2 2 l 2 2
Substituting into this the expression for p - p from Eq. 7.6-21 gives
x 2
2
F x = pv S (p - cos 0) + pv S p-\l Q - i(3 )
2
2
2
2
2
2
p
+ pg(h 2 - /z )S /3" 1 + S ((3-' - cos 61)
2 2
1
2
2
= w (pS y\j5p- 1 - cos в + lp)
2
1
+ pg№ - /z )S )S- + pzSzCiS" 1 - cos в) (7.6-24)
2 1 2
The y-component of the momentum balance is
p
F y = -{v w 2 + S ) sin 9 - m g (7.6-25)
2
tot
2 2
or
2
2
F y = -w (pS Y l sin в - p S 2 sin в - irR Lpg (7.6-26)
2
2
in which R and L are the radius and length of a roughly equivalent cylinder.
We now have the components of the reaction force in terms of known quantities. The nu-
merical values needed are
р = 60 lb /ft 3 s 2 = 6 4 ^ = 0.049 ft 2
w
2
2
w = (2.0X60) = 120 Ibjs p = Si/Si ~- = 3 /4 = 0.562
cos в = \ R~ 1ft
sin в = \Vb L « Ift
p = 16.2 lb/in. 2 /z - ft
2 2
With these values we then get
(120) 2 / 7 1 1 l 0.562\ f 1
x 2(0.049)(32.2) V10 0.562 :> 2 ; + (60)(i)(0.049)( .0.562
/ 1 _ 1
\0.562 2
= (152X1.24 - 0.50 + 0.28) + 2.6 + (144X1.78 - 0.50)
= 155 + 2.6 + 146 = 304 Vo = 1352N (7.6-27)
f
Ш э ( И - «6.2)(0.049)(144)(lV3) -
= - 132 - 99 - 2.5 = 234 \h f - 1041 N (7.6-28)
Hence the magnitude of the force is
| F | = У/ЩТЦ = V304 2 + 234 2 = 384 Ц = 1708 N (7.6-29)
The angle that this force makes with the vertical is
a = arctan(F /F ) = arctan 1.30 = 52° (7.6-30)
x ]/
In looking back over the calculation, we see that all the effects we have included are impor-
tant, with the possible exception of the gravity terms of 2.6 lfy in F and 2.5 lfy in F . y
x