1. CHAPTER 18 Transient Pressure Measurement

2. 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)

3. 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.

4. 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.

5. 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 pressure- sensing system.

6. FIGURE 18.1 Schematic representation of transient pressure measure measurements. P e —environmental pressure.

7. 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] where d 1 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.). (18.1)

8. 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)

9. The mass flow rate into the transducer is equal to the rate of increase of mass in the transducer, where the subscript t indicates time in this section, ρ t is the fluid density at the interface, V is the system volume, r 0 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. (18.2)

10. The fluid density at the capillary-transducer interface is defined as 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 The volume of the entire system, when considered as lumped in the transducer, is (18.3) (18.4)

11. where V ∞ is the steady-state volume from the orifice to the surface of the manometric fluid; A e is equivalent area that depends on the geometry of the manometer (Figure 18.2); P f is the final environment pressure P e at time t = 0 +, that is, after a step change takes place in pressure; P t is the desired transducer pressure such that at t = 0, P t =P i and at t =∞, P t = P f ; and w is the specific weight of the manometer fluid. The rate of change of volume is (18.5)

12. If we assume laminar flow in the capillary, the instantaneous axial velocity at any radius r in the capillary is where μ is the dynamic viscosity of the fluid, r is the capillary radius, and x is the distance from the orifice.

13. With each of the terms of equation (18.2) now defined, the balance becomes (18.8)

15. FIGURE 18.2 The definition of manometer equivalent area A e 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 (18.9)

16. so that the pressure difference is always much smaller than the initial pressure level P i. This implies that P t, which is bounded by P i and P f, can be well approximated by its average value However, dP t ∕dt cannot be replaced by d ∕dt (since the latter, of course, equals zero), since system lag derives from the very fact that P t =f(t).Making use of the approximation P t ≈ t, and separating the variables in equation (18.8), we obtain (18.10)

17. On integration we get the first-order linear expression [4]-[6] The constants of the system are lumped in K. The general solution for equation (18.12) is For the case of a pressure-sensing system subjected to a step change in environment pressure, the initial condition is that P t = P i at t = 0. This C = P i, and at any time P f is a constant. The general solution then reduces to (18.11) (18.12) (18.13)

18. 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 pressure- sensing 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. (18.14)

19. 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.

20. The time required to reduce the pressure difference to any acceptable value can be predicted from the expression [Compare with equation (13.42) for temperature response.] Equation (18.15) tabulates as (18.15) a t 0.5 0.368 0.2 0.1 0.05 0.01 T 0.7K K 1.6K 2.3K 3K 4.6K

21. FIGURE 18.3 Response of ideal second-order system to step input of unit amplitude. (Source: From ANSI B88.1-1972 [12].)

22. t The average pressure level can now be evaluated, and explicit expressions can be obtained for both the time constant K and the recovery time t. Inserting P t from equation (18.14) into equation (18.10) results in The latter term in parentheses is of the form where b =-1/K. Hence equation (18.16 ’ ) becomes (18.16)

23. Now t is divided through, and its value from equation (18.15), namely, t/K = ln(1/α t ), is inserted in equation (18.16) to yield Since P t asymptotically approaches but can never equal P f in any practical time, an arbitrary cutoff point must be chosen for α t to allow a realistic evaluation of t (Figure 18.4). For simplicity, an α t corresponding to 98% recovery is chosen, in which case (18.16) (18.17)

24. FIGURE 18.4 1 Approximation to P t for a step change in environmental pressure. Pt ≈Pt ≈

25. FIGURE 18.5 Experimental confirmation of linearized theory at various pressure levels. Experimental uncertainty envelope (±6%);

26. where t represents the average value of P t from zero to a time corresponding to the 98% recovery point. The time constant for a pressure-sensing system is then and the general expression for recovery time is (18.18) (18.19)

27. 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.

28. 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, (18.12)

29. and, for the nonlinear solution, For the nonlinear (NL) time constant, and, for the nonlinear solution in terms of recovery time, (18.17) (18.18) (18.19)

30. 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/ft 2. If L 1 =1 in., L 2 =6 ft., d 1 = 0.02 in., d 2 =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. By equation (18.1), By equation (18.18),

31. Since it takes about 4.6K L to be within 1% of the final pressure, Nonlinear. α t = (P f － P t )/( P f － P i ) = 0.01 for 99% recovery. But P f, for a 1% step change, id 15.15 psia, and P t = P f － ( P f － P i ) α t is 15.1485 psia. This,

32. By equation (18.18 ’ ), By equation (18.19 ’ ), Note how close t L and t NL are for the small pressure step of 1%.

33. 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. α t = 0.01 for 99% recovery. By equation (18.18),

34. By equation (18.18 ’ ), By equation (18.19 ’ ), 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 pressure- sensing system, namely, By equation (18.19),

35. a capillary portion (representing the inlet orifice, the probe, and the pressure tubing) and a relatively large volume portion (representing the transducer).