Page 140 - Chemical Process Equipment - Selection and Design

112 FLOW OF FLUIDS

EXAMPLE 6.12
Adiabatic and Isothermal Elow of a Gas m a Pipeline
Steam at the rate of 7000 kg/hr with an inlet pressure of 23.2 barabs = 0.9997,
and temperature of 220°C flows in a line that is 77.7mm dia and
305 m long. Viscosity is Z8.5(10-6)N sec/m2 and specific heat ratio is 0.9997(23.2)(10’)(0.7715)
k = 1.31. For the pipe, E/D = 0.0006. The pressure drop will be
found in (a) isothermal flow; (b) adiabatic flow. Also, (c) the line = 17.89(1@) N/mz,
diameter for sonic flow will be found. AP=23.2- 17.89=5.31 bar.
Vl = 0.0862 m3/kg, (c) Line diameter for sonic flow. The critical pressure ratio is
G = 7000/(3600)(n/4)(0.0777)2 = 410.07 kg/mz sec,
= 0.5439, with k = 1.31,
f = 1.6364/[ln(O.l35)(0.0006) + 6.5/1.2(106)]2 = 0.0187. G=--- 7000/3600 2.4757
Inlet sonic velocity: (n/4)D2 - D2 ’
M --_
GV, - 2.4757(0.0862) - 3.909(W4)
us, = qm= q1(1.31)(8314)493.2/18.02 = 546 m/sec Us, 5460’ - D2 .
MI = ul/uSl = GV,/usl = 410.07(0.0862)/546 = 0.0647.
Equation (6.89) becomes
As a preliminary calculation, the pressure drop will be found
by neglecting any changes in density: o.5439(vz/v1) = 1 + 0.1183M:[1 - (V2/V1)”], (3)
fL/D = 0.0187(305)/0 = 5.7035/D (4)
=rhs of Eq. (6.88).
... P2 = 23.2 - 5.32 = 17.88 bar. Procedure
(a) Isothermal flow. Use Eq. (6.76): 1. Assume D.
2. Find Ml [Eq. (2)].
2P V G2 3. Find Vz/Vl from Eq. (6.89) [Eq. (3)].
=
11 2(23.2)(105)(0.0862)(410.07)2 = 6.726(1010),
gc 4. Find rhs of Eq. (6.88) [Eq. (l)].
5. Find D = 5.7035/[rhs of Eq. (6.88)] [Eq. (4)].
6. Continue until steps 1 and 5 agree.
= lo5 538.24 - 6.726 Some trials are:
Eq. (6.89) Eq. (6.88)
= 17.13(1@) N/mz, D nn, 414 rhS D
and 0.06 0.1086 0.5457 44.482 0.1282
AP = 23.2 - 17.13 = 5.07 bar. 0.07 0.0798 0.5449 83.344 0.06843
0.0697 0.08046 0.5449 81.908 0.06963
When the Iogarithmic term is neglected,
Pz = 17.07(10)5 N/m2. :. D = 0.0697 m.
(b) Adiabatic flow. Use Eq. (6.88): lG1
20
36
4r3
0.0187(305) -- 1 2 50
0.0777 - 2.62 (Om3l m) 60
-k
r9
7
89
73.4 = 182.47[ 1 - + 0.8817 In( gr, 98
160
119
:. 5 = 0.7715.
v,
Equation (6.89) for the pressure:

148 01515 Xi
150 RETURN