Reference ID: MET-661D | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
A second-order system model is the simplest representation that captures both inertia and restoring forces together with energy dissipation. In process plants the same mathematics describes:
Pressure surges inside rigid pipelines when a relief valve cracks open
Level oscillations in a tank with elastic gas cushioning
Flow/pressure pulsations in short, dead-ended pipe legs
Designers use the natural frequency \( \omega_n \) to check whether the period of oscillation coincides with mechanical eigen-frequencies of supports or pump speeds, and they use the damping ratio \( \zeta \) to predict the magnitude of the first pressure overshoot above the relief set-point. The calculation is therefore embedded in every hydraulic transient study, pulsation-dampener sizing exercise, and PSV stability check.
Methodology & Formulas
Geometric and thermodynamic constants
Pipe internal diameter \( D \), length \( L \), liquid density \( \rho \) and effective bulk modulus \( \beta_{\text{eff}} \) are treated as constants. Effective bulk modulus combines liquid compressibility and pipe-wall elasticity.
Equivalent mechanical system
The fluid column is modelled as a rigid mass
\[ m = \rho \left( \frac{\pi}{4} D^{2} L \right) \]
and the entrained fluid acts as a linear spring with stiffness
\[ k = \frac{\beta_{\text{eff}} A}{L} \quad \text{with} \quad A = \frac{\pi}{4} D^{2} \]
Natural frequency
The undamped natural frequency of the mass–spring system is
\[ \omega_n = \sqrt{\frac{k}{m}} \]
Damping coefficient
A target damping ratio \( \zeta \) (dimensionless) is specified by the user. The dimensional damping coefficient that must be supplied by friction, orifices, or dash-pot elements is
\[ c = 2 \zeta \sqrt{m k} \]
Damped oscillation frequency
Under-damped systems (\( \zeta < 1 \)) oscillate at
\[ \omega_d = \omega_n \sqrt{1 - \zeta^{2}} \]
First overshoot
If the pressure rises instantaneously by \( \Delta P = P_{\text{relief}} - P_{\text{line}} \) the first peak above the final steady value is
\[ \text{overshoot} = \Delta P \exp \left( \frac{-\zeta \pi}{\sqrt{1 - \zeta^{2}}} \right) \]
Flow regime check
Regime
Reynolds number
Friction model validity
Laminar
\( Re = \frac{\rho v D}{\mu} < 4000 \)
Linear viscous model required
Turbulent
\( Re \geq 4000 \)
Correlation used in code acceptable
Damping ratio limits
Range
Interpretation
\( 0.2 \leq \zeta \leq 5 \)
Correlation for overshoot factor valid
\( \zeta < 0.2 \)
Very lightly damped—higher overshoot possible
\( \zeta > 5 \)
Over-damped—no oscillation, overshoot formula not applicable
Linearity check
Condition
Interpretation
\( P_{\text{relief}} \leq 1.3 P_{\text{line}} \)
Small perturbation assumption holds
\( P_{\text{relief}} > 1.3 P_{\text{line}} \)
Large amplitude—non-linear stiffness & damping expected
Zeta is the dimensionless damping ratio. It tells you how energy is dissipated in the loop and therefore how the PV will behave after a set-point change or load upset.
ζ < 1 (under-damped): oscillations that decay with overshoot; faster but riskier for sensitive units.
ζ = 1 (critically damped): fastest response without overshoot; preferred for most temperature and pressure loops.
ζ > 1 (over-damped): sluggish, no oscillation; acceptable when overshoot could violate safety limits.
Choosing ζ ≈ 0.7–1.0 gives a good compromise between speed and stability in typical process skids.
Inject a small step ΔMV and record the PV until it settles.
Measure the peak time Tp between the first two peaks (or first peak and steady-state).
Estimate the damped period Td = 2 Tp, then ωd = 2π / Td.
Count the logarithmic decrement δ = ln(A1/A2) where A1 and A2 are successive peak heights.
Use the averaged result from two or three tests to reduce noise.
For an ideal PID interacting form Gc(s) = Kp(1 + 1/(Ti s) + Td s) controlling a second-order plant, the closed-loop characteristic equation becomes
s² + (1+Kp K Td)/(Ti τ) s + Kp K/(Ti τ) = 0.
Equate coefficients to the standard form s² + 2 ζ ωn s + ωn² = 0:
ωn = √(Kp K / (Ti τ))
ζ = (1 + Kp K Td) / (2 Ti τ ωn)
Solve these two expressions to obtain the effective ωn and ζ your tuning will impose on the loop.
Real processes deviate from the ideal model. Common causes include:
Valve stiction or backlash adding dead-band → limit cycling.
Measurement noise being amplified by high controller gain.
Un-modeled inverse response or dead-time making the effective ζ < 1.
Non-linear tank geometry changing gain with level.
Check the valve signature, filter the PV signal, and re-identify the model with a higher-order or dead-time term if necessary.
Worked Example – Sizing a Surge Vessel for a Rapid-Closure Scenario
A refinery cooling-water line (102 mm ID, 30 m long) feeds a heat exchanger at 1.5 m s-1. When the downstream valve shuts in 20 ms the trapped water acts like a spring-mass-damper system. We need to check whether the resulting pressure surge stays below the relief valve setting of 6 bar(g) and, if not, what volume of surge vessel is required to limit the overshoot to 35% of the static pressure rise.
Knowns
Gravity: 9.81 m s-2
Atmospheric pressure: 1.013 bar
Pipe inside diameter: 0.102 m
Pipe length: 30.000 m
Water temperature: 20 °C
Water density: 998 kg m-3
Dynamic viscosity: 0.001 Pa s
Mean velocity: 1.5 m s-1
Effective bulk modulus (pipe & fluid): 2.0 GPa
Line pressure (gauge): 5.000 bar
Relief pressure (gauge): 6.000 bar
Target damping ratio: 0.350
Convert pressures to absolute values:
Pline,abs = 5.000 + 1.013 = 6.013 bar = 601.325 kPa
Prelief,abs = 6.000 + 1.013 = 7.013 bar = 701.325 kPa
Compute the static pressure rise (ΔP) that would occur if the line were rigid:
ΔP = Prelief,abs − Pline,abs = 100.000 kPa
Estimate the acoustic velocity in the pipe:
\[ c = \sqrt{\frac{\beta_{\text{eff}}}{\rho}} = \sqrt{\frac{2.0 \times 10^9}{998}} = 1415 \text{ m s}^{-1} \]
(The code uses a slightly lower value to account for pipe-wall compliance.)
Calculate the characteristic impedance:
\[ Z = \rho \, c = 998 \times 1415 = 1.41 \times 10^6 \text{ kg m}^{-2} \text{s}^{-1} \]
Treat the 30 m water column as a lumped mass:
\[ m = \rho \, A \, L = 998 \times \frac{\pi}{4}(0.102)^2 \times 30 = 244.6 \text{ kg} \]
Model the entrained air pocket as a linear spring with stiffness:
\[ k = \frac{\beta_{\text{eff}} \, A}{L} = \frac{2.0 \times 10^9 \times 0.00817}{30} = 544.8 \text{ kN m}^{-1} \]
Natural frequency of the hydraulic spring-mass system:
\[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{544800}{244.6}} = 47.2 \text{ rad s}^{-1} \]
Damping coefficient required for ζ = 0.35:
\[ c = 2 \, \zeta \, \sqrt{k \, m} = 2 \times 0.35 \times \sqrt{544800 \times 244.6} = 8081 \text{ N s m}^{-1} \]
Maximum pressure overshoot for a step input:
\[ \text{Overshoot} = \Delta P \cdot \exp\left(-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}\right) = 100 \times \exp\left(-\frac{0.35 \pi}{0.937}\right) = 30.9 \text{ kPa} \]
Required surge volume to absorb the kinetic energy of the moving column:
\[ V = \frac{\frac{1}{2} m v^2}{\Delta P + \text{Overshoot}} = \frac{0.5 \times 244.6 \times 1.5^2}{130900} = 0.0021 \text{ m}^3 \]
Round up to the nearest standard size: 0.25 m3 (250 L) vessel.
Final Answer
Install a 250 L nitrogen-charged surge vessel on the line; it limits the transient pressure surge to 30.9 kPa above the relief setting, keeping the peak pressure below 7.01 bar(abs) and achieving the target damping ratio of 0.35.
"Un projet n'est jamais trop grand s'il est bien conçu."— André Citroën
"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise."— Charles de Gaulle
