Chemical Engineering July 2021 - 37

Valve
Hr = 200 m
Pipe 1
L1 = 2,000 m
D1 = 1 m
a1 = 1,000 m/s
f1 = 0.026
Q1 = 1 m3/s
L4 = 1,000 m
D4 = 0.6 m
a4 = 1,000 m/s
f4 = 0.03
Q4 = 0.21 m3/s
Hr = 200 m
Pipe 1
Same as
Fig. 7
Pipe 2
Same as Fig. 7 except
Q2 = 0.79 m3/s
Pipe 3
Same as
Fig. 7
FIGURE 4. The diagrams accompany Example 4, depicting straight (top diagram)
and networked (bottom diagram) piping systems (from Ref. 1)
aware, when liquid pressure drops
to the local vapor pressure, the liquid
will flash and vapor will be generated.
This is called transient cavitation.
It is transient because in a
waterhammer situation, a reflecting
transient pressure wave will typically
and quickly repressurize the pipe and
collapse the vapor. It is usually not a
sustained, two-phase flow situation,
but a temporary two-phase situation
that quickly returns to single-phase
in most cases.
If enough vapor is generated and
flow conditions are right, the entire
cross section of fluid can vaporize
and the continuous column of liquid
will have a vapor gap. This is true
liquid column separation. In common
use, transient cavitation and
liquid column separation are synonymous
terms.
It turns out that predicting the waterhammer
pressure transients during
and after transient cavitation has
occurred is very difficult. The best
available models are not very accurate,
because the phenomenon is
quite complex. As a result, waterhammer
engineers treat predictions
from simulation software in cases
where transient cavitation occurs
with great caution [4].
Moreover, it has been shown experimentally
that when transient
vapor pockets collapse, the resulting
pressure spike can exceed the values
returned by using Equation (3). Example
2 discusses this.
Example 2: Transient cavitation
and vapor collapse. Figure 1 shows
Pipe 2
L2 = 1,000 m
D2 = 1 m
a2 = 1,000 m/s
f2 = 0.026
Q2 = 1 m3/s
Pipe 3
L3 = 1,000 m
D3 = 1 m
a3 = 1,000 m/s
f3 = 0.026
Q3 = 1 m3/s
Ho = 100 m
∆Hv = 91.4 m
Cv = 1,390
Pipe 4
Valve
Ho = 100 m
∆Hv = 92.2 m
Cv = 1384
the experimental
results [5] and
the commercial
software simulation
results [6]
for a 102-m (335
ft) coiled copper
tube experiencing
waterhammer.
Equation
(1) predicts a
maximum pressure
rise of 104
m (340 ft), resulting
in a peak
pressure (when
added to the
steady-state
pressure) of 171
m (560 ft) near 0.1 s. Thus, the initial
pressure rise agrees with Equation
(1). However, the reflecting wave
leads to a pressure decrease, and
transient cavitation begins at this
location at about 0.3 s. It lasts until
just after 0.4 s. This is the point at
which the system repressurizes and
collapses the vapor cavity. Both the
experiment and the simulation show
a peak pressure that exceeds the
result from Equation (1) at about 0.6
s. This peak is about 235 m (770
ft). Ref. 1 offers guidance on how to
quickly check for the possibility of
transient cavitation in your system.
Example 3: Line pack: recovery of
frictional pressure drop. The term
" line pack " is commonly used, but is
not very descriptive of the process.
So before trying to describe what the
term is, consider a simple conceptual
example. First, just think about
a steady-state situation. Figure 2
shows a 50-km horizontal pipeline
with conditions taken from Example
1. Additional details for the steadystate
input data for Example 3 are
shown here:
L = 50 km (31.1 mi)
D = 0.5 m (1.64 ft)
f = 0.018
Q = 0.4 m3/s (14.1 ft3/s), 1,440 m3/h
(6,340 gal/min)
V = 2.04 m/s (6.68 ft/s)
Pin 10,000 kPa (1,450 psi), fixed
Pvalve 6,638 kPa (962.8 psi), upstream
initial pressure
∆Ppipe 3,362 kPa (487.6 psi), initial
pipe pressure drop
ρ 900 kg/m3 (56.2 lbm/ft3)
CHEMICAL ENGINEERING WWW.CHEMENGONLINE.COM
JULY 2021
The steady-state pressure drop
is a simple calculation once the friction
factor is determined (the DarcyWeisbach
f = 0.018). The pressure
drop is thus 3,362 kPa (488 psid).
The question we ask here is: What
is the pressure at the valve after it
has closed, the flow has stopped,
and all transients have died out?
The answer is trivial. Since the entire
pipe is horizontal, the pipe will have a
pressure everywhere along its length
that is the same as the inlet pressure
of 10,000 kPa (1,450 psia).
Notice what happened here. The
pressure at the valve increased. By
how much? It increased by 3,362
kPa (488 psi). In fact, it increased
by the same amount as the frictional
pressure drop while it was flowing.
This pressure increase has nothing
to do with waterhammer. It is merely
the increase in pressure when frictional
pressure loss is no longer occurring.
We will call this pressure increase
the friction recovery pressure,
∆Pfr. In the present example, ∆Pfr =
3,362 kPa (or 488 psi).
During a pipeline transient, the
frictional behavior interacts with the
acoustic nature of the waterhammer
wave such that a passing wave does
not bring the fluid to a complete rest,
even after a valve is closed. Ref. 7
contains an excellent summary of
line pack:
" In pipeline transients, frictional resistance
to flow generates line packing,
which is a sustained pressure
increase in the pipeline behind the
waterhammer wave front after the
closure of a discharge valve. This
phenomenon is of interest to crosscountry
oil pipelines and long water
transmission mains because the
sustained pressure increase can be
very significant relative to the initial
sudden pressure increase by waterhammer
and can result in unacceptable
overpressures. "
Ref. 7 offers a powerful new
method to estimate the combined
pressure rise due to line pack and
waterhammer. The results will be
discussed in the next section.
Now let's see what happens to the
Figure 3 pipeline from a waterhammer
point of view when the valve is
closed. To determine this, we need
additional information as shown here
37
http://WWW.CHEMENGONLINE.COM
Chemical Engineering July 2021

Table of Contents for the Digital Edition of Chemical Engineering July 2021
