Chemical Engineering July 2021 - 36

Pin = 10,000 kPa
(1,450 psi)
Pvalve = 6,638 kPa
(962.8 psi)
D = 0.5 m (1.64 ft)
L = 50 km (31.1 mi)
FIGURE 2. The diagram shows the system description
for the horizontal pipe in Example 3
in a form of the Joukowsky equation
that is more frequently used by
chemical and mechanical engineers
(Equation (3)).
∆PJ = - ρa∆V
(3)
Equations (1) and (3) are essentially
equivalent ways of presenting
the Joukowsky Equation, and
will be used interchangeably in this
article. Note that one limitation of
Equation (1) is that when the system
experiences zero gravity, Equation
(1) would have a divide-by-zero
problem. In such cases, Equation (3)
retains validity and is thus preferred.
During my time working in the aerospace
industry, I frequently worked
on zero-g systems, as well as those
with much higher and lower body
forces than 1 g (1 standard earth
gravitational acceleration). See Ref.
1 for more on this.
The Joukowsky equation traces its
history back to the 19th century. In
fact, Joukowsky is the lucky author
whose name is most often associated
with Equations (1) and (3). However,
a thorough review of the early
literature shows that others derived
Pvalve, initial +∆P1 + ∆Pfr
= 12367 kPa (1,794 psi)
Max Pressure Liou
= 11,949 kPa (1,733 psi)
Friction recovery Pressure
= 3,362 kPa (488 psi)
Pressure rise liou
= 5,133 kPa (770 psi)
12500
12000
11,500
11,000
10,500
10,000
9,500
9,000
8,500
8,000
7,500
7,000
6,500
this equation prior to Joukowsky -
for applications in blood flow [3].
Limiting assumptions
Before discussing the cases where
the Joukowsky equation does not
result in a conservative answer, it
should be realized that the equation
has numerous limiting assumptions.
From Ref. 1:
" In principle, Equation 1 only
claims validity the moment after the
velocity decrease (e.g., valve closure).
However, practicing engineers
often apply it as if it retains validity
both immediately after the velocity
decrease/valve closure, as well as
at all times thereafter, assuming that
no other independent transients
occur. Since Equation 1 is often applied
in this manner, the limitations
of this equation are discussed with
respect to its validity after the initial
transient occurs. These limitations
are as follows: "
* Straight, constant-diameter piping
of uniform material, wall thickness
and structural restraints
* Uniform pipe friction
* Minimal friction pressure drop in
piping (explained in a later section)
* Minimal fluid-structure interaction
with the piping and supports
* No cavitation or gas release
* No trapped, or entrained, gases in
the piping (that is, it is initially 100%
full of liquid)
* No external heat transfer that can
1,800
Max pressure at valve
= 11,967 kPa (1,736 psi)
Pressure rise from line pack
= 2,962 kPa (430 psi)
1,600
change any of the piping and fluid
physical properties or cause phase
changes
* Constant liquid density and constant
bulk modulus
* One-dimensional fluid flow
* Linearly elastic piping material
Example 1: Instant valve closure.
Let's take a look at a simple example
from Ref. 1. Consider a pipe of diameter
(D) 0.5 m (1.64 ft) conveying
oil with a specific gravity of 0.9 (the
density, ρ, is 900 kg/m3, or 56.2 lbm/
ft3). The volumetric flowrate, Q, is 0.4
m3/s (14.1 ft3/s) and the wavespeed,
a (the propagation speed of the fluid
transient), is 1,291 m/s (4,236 ft/s).
The pipe cross-sectional area, A, is
given by Equation (4):
A = πD2/4 = 0.196 m2 (or 2.11 ft2) (4)
Hence, the velocity change if a valve
is closed instantly can be calculated
with Equation (5).
∆V = ∆Q/A = -2.04 m/s
(or -6.68 ft/s)
(5)
Therefore, from Equation (1):
∆HJ =
-a∆V /g =
-(1,291 m/s)
(6)
( -2.04 m/s) / (9.81 m/s2) = 268 m
(or 880 ft)
And using Equation (3):
∆PJ = -ρa∆V = -(900 kg/m3)
(1,291 m/s)( -2.04 m/s) = 2,367 kPa
(or 343 psi)
(7)
Note that the value given by Equation
(7) is the pressure increase due
to waterhammer after an instant
valve closure. To find the maximum
pressure, the Equation (7) pressure
increase must be added to the fluid's
pre-existing, steady-state pressure.
1,400
Pressure after valve closes
= 9,005 kPa (1,306 psi)
1,200
Initial Pressure at Valve 6638 kPa (963 psi)
Pressure rise Eq. 3 = 2,367 kPa (343 psi)
50
100 150 200
Time, s
FIGURE 3. To illustrate Example 3, the graph shows the predicted pressure transient at the
valve from the system from Figure 2, with pressure rise estimation methods (from Ref. 1)
36
1,000
Cases where the equation does
not give a conservative result
Three situations where Equation (3)
may not be conservative are: transient
cavitation and liquid column
separation; line pack; and piping
system reflections (networks, components,
area changes and surgesuppression
devices).
These three cases will be explored
in three examples discussed below.
As chemical engineers are keenly
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