Chemical Engineering February 2014 - 39

0.01 0.05 0.1
Table 1b. Mach NuMber aT pipe iNleT (Mi1 )
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.01 - - - - - - - - - - -
0.05 15.56% - - - - - - - - - -
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
∆ptotal
∆p
=
1− f
2.41% - - - - - - - - - -
0.41% 12.97% - - - - - - - - -
0.13% 3.63% 20.26% - - - - - - - -
0.06% 1.49% 6.77% - - - - - - - -
0.03% 0.71% 3.02% 17.26% - - - - - - -
0.01% 0.36% 1.49% 7.14% 26.14% - - - - - -
0.01% 0.18% 0.75% 3.34% 9.36% 29.04% - - - - -
0.00% 0.09% 0.36% 1.52% 3.89% 8.66% 22.34% - - - -
0.00% 0.03% 0.13% 0.55% 1.35% 2.72% 5.19% 10.71% - - -
0.91 0.00% 0.03% 0.12% 0.48% 1.17% 2.35% 4.44% 8.81% - - -
0.92 0.00% 0.02% 0.10% 0.42% 1.01% 2.01% 3.75% 7.23% 21.16% - -
0.93 0.00% 0.02% 0.08% 0.35% 0.85% 1.70% 3.13% 5.88% 14.05% - -
0.94 0.00% 0.02% 0.07% 0.29% 0.71% 1.40% 2.56% 4.71% 10.16% - -
0.95 0.00% 0.01% 0.06% 0.24% 0.57% 1.13% 2.04% 3.68% 7.43% - -
0.96 0.00% 0.01% 0.04% 0.19% 0.44% 0.87% 1.56% 2.78% 5.33% - -
0.97 0.00% 0.01% 0.03% 0.14% 0.32% 0.63% 1.13% 1.97% 3.63% 9.12% -
0.98 0.00% 0.01% 0.02% 0.09% 0.21% 0.41% 0.72% 1.24% 2.23% 4.78% -
0.99 0.00% 0.00% 0.01% 0.04% 0.10% 0.20% 0.35% 0.59% 1.03% 2.03% -
0.992 0.00% 0.00% 0.01% 0.03% 0.08% 0.16% 0.27% 0.47% 0.81% 1.58% 5.37%
0.995 0.00% 0.00% 0.01% 0.02% 0.05% 0.10% 0.17% 0.29% 0.50% 0.95% 2.66%
0.999 0.00% 0.00% 0.00% 0.00% 0.01% 0.02% 0.03% 0.06% 0.10% 0.18% 0.44%
pp
p
22
2
() ()
(
isothermal −
incompressible
isothermam l )
(b) Calculated outlet pressure difference (at known inlet conditions) between using the incompressible and isothermal
equations (η >1% is highlighted green)
p2 using Equations (43) and (44).
By rearranging Equation (44), we
arrive at Equation (48), an expression
for p2/p1.


p
p
2
1


1
=
incomp
f
L
D
Mi +1
2
2
(48)
For given pipe characteristics
(length, diameter and roughness),
fluid flowrate, Mi2 and p2, fL/D
is first calculated - f can be read
from a Moody diagram or calculated
with various equations. Next,
we must calculate p2/p1 using
Equation (48). Afterwards, we determine
η from Table 1A based on
Mi2 and p2/p1(incomp.). If η is acceptable,
then p1 is reported as the
final result. If η is not acceptable,
the isothermal expression in Equation
(10) must be used to re-calculate
p2/p1(isothermal) and report a
new p1 value.
Modified incompressible flow
From Table 1 and Equations (43)
and (44), it can be seen that the isothermal
flow equation is more conservative
than the incompressible
equation because the latter does
not include the acceleration term.
Here, we present a modified incompressible
equation to account for the
missing acceleration term in Equations
(43) and (44). Equation (49)
defines fa, the acceleration factor,
where υm and pm are averaged velocity
and pressure.
fa =
Gυm
pm
(49)
The total pressure drop and pressure
drop due to friction can be linked as
in Equation (50) below [2].
×≥1
100%%
friction
a
(50)
Further expanding Equation (49)
gives Equations (51) and (52).
fa =
pp pp12 12
22G
++
G
fa =
4
=


1+
p
p
2
1


() ()
υ1
12
4G2
+
2
4
2 =


1+
Mi
p
p
2
1
2
1


2
(52)
The pressure drop due to friction
is expressed in Equation (39). By
substituting υm and pm, Equation
(39) becomes Equation (53).
∆pfriction = f
L
D
G
M
RT
2
= f
(p1 + p2
)
L
D
Mi p
p
2
1
2
1
1 + p2
(53)
Inserting Equations (51) and
(53) into Equation (50) yields
Equation (54).
pp f
−=
12
L
D
Mi p
pp Mi
p
p
2
1
12 1−
+


2
1
1
4
1+
2
1
2
1

 (54)
2
Moving fL/D to the lefthand
side brings us to the expressions in
Equations (55) and (56).

f
L
D
=
(p1 − p2 p1 + p2
2
)(
p1
1
=
Mi1
2 1−






p
p
2
1


2




− 4
1
1
)
1
Mi1
−
+
p
p
p
p
2 1−





2
1
2
1
(55)
With known outlet conditions, Equations
(55) becomes Equation (56).
f 14
DMi1
L
1
=−2






p
p
2
1


2




−
1
1
−
+
p
p
p
p
2
1
2
1

4


1+
Mi
p
p
2
1
2
1

2






 
2
pp
M
RT
RT
M
pp
p
12
2
+
1
RT
M
=
4
=
2
G
p
2
2
1
RT
M
()
12
4G2
+
pp
2
RT
M
(51)
(56)
ChemiCal engineering www.Che.Com february 2014 39
p2/p1(incompressible)
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Chemical Engineering February 2014

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