Page 138 - Chemical Process Equipment - Selection and Design

110 FLOW OF FLUIDS
and the ideal gas relation
(6.67)
V = PIVJP and dV/V = -dP/P (6.73)
After these substitutions are made into Eq. (6.63), the results may so that Eq. (6.71) becomes
be solved for the mass rate of flow as
(6.74)

m/Az = (6.68)
This is integrated term-by-term between the inlet and outlet
conditions,
At specified mass flow rate and inlet conditions PI and V,, Eq.
fG2L
(6.68) predicts a relation between the area ratio AJA, and the Pg-P: G2 f-=O (6.75)
+-ln-
pressure ratio PJP, when isentropic flow prevails. It turns out that, 2P1V1 g, (2) 2gcD
as the pressure falls, the cross section at first narrows, reaches a
minimum at which the velocity becomes sonic; then the cross and may be rearranged into
section increases and the velocity becomes supersonic. In a duct of
constant cross section, the velocity remains sonic at and below a
critical pressure ratio given by
(6.76)
(6.69)
In terms of a density, pm, at the average pressure in the line,
The sonic velocity is gjven by f@L
P'' Pl -- . (6.77)
(6.70) 2gcDPm
The average density may be found with the aid of an approximate
where the last result applies to ideal gases and M, is the molecular evaluation of Pz based on the inlet density; a second trial is never
weight. justified. Eqs. (6.76) and (6.77) and the approximation of Eq.
(6.76) obtained by neglecting the logarithmic term are compared in
ISOTHERMAL FLOW IN UNIFORM DUCTS Example 6.12. The restriction to ideal gases is removed in Section
6.7.4.
When elevation head and work transfer are neglected, the
mechanical energy balance equation (6.13) with the friction term of
Q. (6.18) become
ADIABATIC FLOW
fi2
dL
VdP + (l/g,)U du + - = 0. (6.71)
2gcD The starting point for development of the integrated adiabatic flow
energy balance is Eq. (6.71), and again ideal gas behavior will be
Make the substitutions assumed. The equation of condition of a static adiabatic process,
u = G/p = GV (6.72) PVk = const, is not applicable to the flow process; the appropriate

EXAMPLE 6.U AP 4fpV2 4(O.O058)( 1693)( 1.22)'
Pressure Drop in Power-Law and Singham Flow -=-=
A limestone slurry of density 1.693 g/mL is pumped through a 4411. L 2gcD 2(0.152)
(152 mm) line at the rate of 4 ft/sec (1.22 m/sec). The pressure drop = 192.3 N/(m2)(m) [gc = kgm/sec2/N],
(psi/rnile) will be calculated. The slurry behavior is represented by + 192.3(14.7/101,250)1610 = 45.0 psi/mile.
a. The power-law with n = 0.165 and K = 34.3 dyn seco.165/cm2 Bingham:
(3.43 Pa se~O.*~).
b. Bingham model with zo = 53 dyn/cm2 (5.3 Pa) and pg = 22 CP
(0.022 Pa sec).
.-
Power law: He = toD'p/p% = 5.3(0.152)z(1693)/(0.022)2
= 428,000,
Re' = D"V2-"p/8"-'K critical Re, = 12,000 (Fig. 6.5),
= (0.152)0.'65(1.22)1.835(1693)(8)o~835/3.43 f- 0.007 [Fig. 6.6(b)],
= 2957, _-
"
'.Ow 45.0 = 54.3 psi/mile,
f= 0.0058 [Fig. 6.6(a)] L -0.oO58