CHAPTER 18
Transient Pressure Measurement
The difference between the pressure at the mouth of the
Pitot tube and that in the manometer decreases exponentially
with time.
H.L.Weissberg(1953)
18.1 GENERAL REMARKS
Most pressure-sensing systems involve small pressure holes
that open to tubing of small diameter.
Those are connected by other tubing (usually flexible and of
a larger diameter) to any of the mechanical or electrical
pressure transducers.
Often these orifices and the associated tubing are subjected
to time-varying pressures.
There may be a nonsteady flow, or the probe may be
moving, perhaps rotating, through a pressure gradient field.
Thus although there is usually no fluid flow in steady-state
pressure-sensing systems, when transients occur, fluid flow
generally accompanies them.
At the least this causes some flow resistance in the orifices
and tubing.
Also ,there are usually compression effects in the
transducer. As a result, the pressure signal at the transducer
tends to lag the impressed pressure at the orifice.
The purpose of this chapter is to indicate simple
expressions that can serve to predict the response behavior of a
pressure-sensing system when it is subjected to time-varying
pressure signals in both gases and liquids.
18.2 MATHEMATICAL DEVELOPMENT FOR A GAS
A single time constant can describe the response behavior of
only the simplest of systems. Fortunately, however, it is
usually sufficient to consider only two regions in a pressuresensing system.
FIGURE 18.1 Schematic representation of transient pressure
measure measurements.
Pe—environmental pressure.
All the fluid resistance in the system is assigned (lumped) to
one of these regions, called the capillary portion. If the system
has more than one capillary in series, the equivalent length of
the capillary portion is [1]
4
 d1 
 d1 
Le  L1  L2    ...  Ln  
 d2 
 dn 
4
(18.1)
where d1 is taken as the capillary reference diameter. All the
fluid mass in the system can be thought of as stored in a
second region called the transducer portion.
The volume of the transducer portion should include the
volume of the equivalent capillary portion.).
This system is illustrated in Figure 18.1, in which a
manometric-type transducer is included to represent the
general case of time-varying system volume.
The system can be characterized [2] by writing a flow
balance at the capillary-transducer interface (i.e.,at x = Le)
The mass flow rate into the transducer is equal to the rate of
increase of mass in the transducer,

r0
0
d t
dV
tVdA  t
V
dt
dt
(18.2)
where the subscript t indicates time in this section, ρt is the
fluid density at the interface, V is the system volume, r0 is the
capillary radius from center to wall, and A is the interface area.
The flow balance is localized in this manner to simplify the
analysis, since at this interface the density term has consistent
meaning.
Appropriate expressions for the various terms in equation
(18.2) are indicated next.
The fluid density at the capillary-transducer interface is
defined as
P1
t 
 f t 
(18.3)
RT
where R is the specific gas constant and T is the absolute
temperature of the fluid.
Under isothermal conditions, the rate of change of density
in the transducer is
d t  1   dpt 
  

dt  Rt  dt 
(18.4)
The volume of the entire system, when considered as lumped
in the transducer, is
V  V 
P

w
Ag
f
 Pt 
(18.5)
where V∞ is the steady-state volume from the orifice to the
surface of the manometric fluid; Ae is equivalent area that
depends on the geometry of the manometer (Figure 18.2);
Pf is the final environment pressure Pe at time t = 0+, that is,
after a step change takes place in pressure;
Pt is the desired transducer pressure such that at t = 0, Pt =Pi
and at t =∞, Pt = Pf ; and w is the specific weight of the
manometer fluid. The rate of change of volume is
dV  At  dpt 
  

dt  w  dt 
If we assume laminar flow in the capillary, the instantaneous
axial velocity at any radius r in the capillary is
1  p  2 2
V 
   r0  r 
4  x 
If we assume laminar flow in the capillary, the instantaneous
axial velocity at any radius r in the capillary is
1  p  2 2
V 
   r0  r 
4  x 
where μ is the dynamic viscosity of the fluid, r is the capillary
radius, and x is the distance from the orifice.
With each of the terms of equation (18.2) now defined,
the balance becomes
 pt   1   p  2 2
 
r0  r  2 rdr






0 RT

  4   x 
Ac  dp
1  Ac

pt
 V   p f  pt  

RT  w
w  dt
t
(18.8)
FIGURE 18.2 The definition of manometer equivalent area
Ae for various types of manometer.
(a) U-tube manometer, unequal diameters.
(b) Single leg on capillary side.
(c) Single leg on reference side.
(d) U-tube manometer, equal diameters.
Equation (18.8) is a first-order first-degree nonlinear
differential equation [3], and therefore cannot be used to
define a conventional time constant.
In order to get a usable time constant, we must find a
suitable linearized form of equation (18.8).
As a first step, we impose the limitation that
Pt  Pf  Pi
(18.9)
so that the pressure difference is always much smaller than
the initial pressure level Pi .
This implies that Pt , which is bounded by Pi and Pf , can be
well approximated by its average value
pt


t
0
Pdt
t
(18.10)
t
However, dPt ∕dt cannot be replaced by d P∕dt
(since the latter, of course, equals zero),
since system lag derives from the very fact that
Pt=f(t) .Making use of the approximation Pt ≈ Pt , and
separating the variables in equation (18.8), we obtain
  d14 Pt  pt
 At
 dpt

2 pt  Pf  V 
 p f dp  
w
 dt
 128 



l1
0
dx
(18.11)
On integration we get the first-order linear expression [4]-[6]
dpt 1
(18.12)
  Pf  Pt 
dt
K
The constants of the system are lumped in K. The general
solution for equation (18.12) is
Pf  Ce
t K
1 1 K t
 e  Pf e1 K dt
0
K
(18.13)
For the case of a pressure-sensing system subjected to a step
change in environment pressure, the initial condition is that Pt
= Pi at t = 0. This C = Pi , and at any time Pf is a constant. The
general solution then reduces to
Pf  Pt   Pf  Pi  e  t K
(18.14)
Thus K indicates the time required after the step change for
the reduction of environment-transducer pressure difference to
1/e times the initial difference.
In other words, K is the time required for the pressuresensing system to indicate a pressure corresponding to
approximately 63.2% of the initial pressure difference.
Since this is exactly the meaning of the time constant of a
first-order linear system, K is the time constant we seek.
For higher order variously damped systems, the time
constant has no such significance, that is, if the damping factor
of a second-order system is very small, there is a decaying
oscillatory response, and the 63% recovery has little meaning.
However, if the damping factor is ≥ 1, the second-order
response appears much the same as a fist- order response, and
the 63% recovery is significant (Figure 18.3).
For Reynolds numbers below 2000 a capillary acts like a
highly damped system and therefore behaves like a first-order
linear system.
The time required to reduce the pressure difference to any
acceptable value can be predicted from the expression
at 
Pf  Pt
Pf  Pi
 e t K
(18.15)
[Compare with equation (13.42) for temperature response.]
Equation (18.15) tabulates as
at
T
0.5 0.368 0.2
0.7K
K
1.6K
0.1
2.3K
0.05
3K
0.01
4.6K
FIGURE 18.3 Response of ideal second-order system to step
input of unit amplitude. (Source: From ANSI B88.1-1972
[12] .)
The average pressure level P t can now be evaluated, and
explicit expressions can be obtained for both the time constant
K and the recovery time t. Inserting Pt from equation (18.14)
into equation (18.10) results in
Pt t   pt dt    p f   p f  pi  et K dt
0
0
t
t
The latter term in parentheses is of the form
1 bx
bx
 e dx  b e
where b =-1/K. Hence equation (18.16’) becomes
P t t  p f t   p f  pi  1  e  t K  K
(18.16)
Now t is divided through, and its value from equation
(18.15), namely, t/K = ln(1/αt), is inserted in equation (18.16)
to yield
 1  ai 
p t  p f   p f  pi  

(18.16)
ln
1
a


t


Since Pt asymptotically approaches but can never equal Pf in
any practical time, an arbitrary cutoff point must be chosen for
αt to allow a realistic evaluation of Pt (Figure 18.4).
For simplicity, an αt corresponding to 98% recovery is
chosen, in which case
p  pf 
p f  pi
4
(18.17)
FIGURE 18.4 1 Approximation to Pt for a step change in environmental
pressure.
Pt ≈
t95%

0
Pt dt/t98%
FIGURE 18.5 Experimental confirmation of linearized theory at various
pressure levels. Experimental uncertainty envelope (±6%);
where P t represents the average value of Pt from zero to a
time corresponding to the 98% recovery point. The time
constant for a pressure-sensing system is then

128 Lc  Ac  Pf  Pi 
K
 
  V 
4
 d1 p t  w  2 

(18.18)
and the general expression for recovery time is
 Pf  Pi 
t  K ln 
 P  P 
t 
 f
(18.19)
These linear equation [equation (18.18) and (18.19)] reliably
predict response time at various pressure levels for pressure
differences on the order of 10% of the initial pressure level, as
indicated by the excellent agreement between experimental
points and the linear equation (Figure 18.5).
18.2.1 Nonlinear Solution
There are many situations where the pressure response
solution to equation (18.8) cannot be limited by the
approximation (18.9), wherein the step change in pressure was
stated to be much smaller than the pressure level.
This means that linearized differential equation (18.12), the
linearized solution (18.14), the linearized time constant
equation (18.18), and the linearized solution for time of
equation (18.19) may not be adequate.
Very briefly, for the nonlinear case, the counterparts of the
preceding linearized equation (for zero volume displacement
systems such as typified by diaphragm-type transducer which
do not allow system volume to vary with time) are, for the
nonlinear differential equation,
dpt
  2 B   p 2f  pt2 
dt
128 LeV
B
 d14
(18.12)
and, for the nonlinear solution,
  t

p f  pi   p f  pi  exp   
 ln bt  


  B p f

bt 
(18.17)
p f  pi
p f  pt
For the nonlinear (NL) time constant,
K NL
 B 
 1  ln bt  
 p 
 f 
(18.18)
and, for the nonlinear solution in terms of recovery time,
K NL
 B   1 

ln
 p   a b 
 f   t t
(18.19)
Example 1.
An air-filled pressure-sensing system is subjected to a small
step change in pressure of 1% at a pressure level of 15 psia.
The temperature is such that the viscosity is
3.85×10-7 lbf-s/ft2. If L1 =1 in., L2 =6 ft., d1 = 0.02 in.,
d2 =0.10 in., and V∞ = 8 in.3, find the time for the pressure to
reach within 1% of the final pressure by the linear and
nonlinear solutions.
Solution.
Linear.
4
By equation (18.1), L  1  72  0.02   1.1152in
e


 0.10 
By equation (18.18),
KL 
128  3.85 107  1.1152  8
 16 108  15 144 
B

pt
874.6687lbf  s ft 2

 0.405s
2
15 144 lbf ft
Since it takes about 4.6KL to be within 1% of the final pressure,
t99%  4.6  0.405  1.86s
Nonlinear.
αt = (Pf －Pt)/( Pf －Pi ) = 0.01 for 99% recovery. But Pf ,
for a 1% step change, id 15.15 psia, and Pt = Pf －( Pf －Pi ) αt
is 15.1485 psia. This,
15.15  15.00
bt 
 0.99509877
15.15  15.1485
By equation (18.18’),
K NL
 B 
 874.6687 
 1  ln bt  

1

0.00491

  0.403s
 p  
 15.15 144 
 f 
By equation (18.19’),
 B   1 
K NL  
ln
 p   a b 
 f   t t
1
 874.6687  


ln
 

 15.15 144   0.01 0.99509877 
 1.85s
Note how close tL and tNL are for the small pressure step of
1%.
Example 2.
For the same pressure system of Example 1, find the linear
and nonlinear solutions for 99% pressure recovery when the
step change in pressure is 50% of the initial 15 psia. αt = 0.01
for 99% recovery.
pi  15 psiaandp f  1.5 p f  22.5 psia
pt  p f   p f  pi  at  22.5  7.5  0.01  22.425 psia
 p f  pi 
pt  p f  
  22.5  7.5 4  2.625 psia
 4 
B 874.6687

 0.27 s
p f 22.5  144
22.5  15
 0.83472
22.5  22.425
B
KL 
 0.2945s
equation (18.18),
Pt
bt 
By
By equation (18.19),
 B  1
tL  
 ln    0.2945  4.605  1.356s
 pt   at 
By equation (18.18’),
K NL
 B 
 1  ln bt  
 1.18065  0.27  0.3188s
 p 
 f 
By equation (18.19’),
K NL
 B   1

ln
 p   a b
 f   t t

  0.27  4.78538  1.292 s

It must be emphasized that this approach, particularized by
equations (18.18) and (18.19) for linear solution and by
equations (18.18’) and (18.19’) for nonlinear solutions, is valid
only when there are two distinct regions in the pressuresensing system, namely,
a capillary portion (representing the inlet orifice, the probe,
and the pressure tubing) and a relatively large volume portion
(representing the transducer).