Source: https://jmrplens.github.io/phonometry/devices/noise-control/valve-cavitation/ # Hydrodynamic valve noise (IEC 60534-8-4) Water is compressible enough to carry sound at 1 400 m/s and far too stiff to choke the way a gas does, so a control valve on a water line cannot build the shock cells that [the aerodynamic part](https://jmrplens.github.io/phonometry/devices/noise-control/control-valve-noise/) spends five regimes on. It makes noise two ways instead, and the whole of IEC 60534-8-4 is the sum of them: **turbulence** in the jet leaving the vena contracta, and **cavitation**, once the pressure there falls low enough for the liquid to flash to vapour and the bubbles to collapse again downstream. The difference between the two is not a matter of degree. Turbulent noise follows the jet velocity gently; cavitation arrives at a threshold and then climbs as a fifth power. Everything on this page turns on where the operating point sits with respect to that threshold. *The liquid case changes the source and not the geometry: the bubbles collapse where the jet is, and the level is quoted at the same metre.* ## 1. The threshold, and why the standard asks you to measure it Two ratios decide the regime. The first is the **differential pressure ratio**, Equation (1), which says how far towards flashing this operating point is: $$ x_F = \frac{p_1 - p_2}{p_1 - p_v} $$ The second is the **characteristic pressure ratio** $x_{Fz}$: the value of $x_F$ at which cavitation first becomes audible on this valve. It is a property of the valve, measured to IEC 60534-8-2, and 4.2 offers Equation (3a) only as an estimate for when no measurement exists. Equation (3c) then moves it from the 6 × 10⁵ Pa the estimate is drawn at to the working inlet pressure. ```python from phonometry import noise_control x_f = noise_control.differential_pressure_ratio( inlet_pressure_pa=1.0e6, outlet_pressure_pa=6.5e5, vapour_pressure_pa=2.32e3 ) print(round(x_f, 4)) # 0.3508 x_fz = noise_control.incipient_cavitation_ratio(90.0, 0.42, 0.92) print(round(x_fz, 4)) # 0.2543 at 6e5 Pa print(round(noise_control.corrected_incipient_ratio(x_fz, 1.0e6), 4)) # # 0.2386 at 10 bar ``` This valve is past its threshold, so it cavitates. A multihole trim of the same capacity would not be: Equation (3b) answers on the hole geometry rather than on the capacity, and a hundred and twenty holes of three millimetres push the threshold up by half. The comparison has to be made against the **corrected** value, which is the half of it a reader is most likely to skip. ```python holes = noise_control.multihole_incipient_cavitation_ratio(120, 0.003, 0.92) print(round(holes, 4)) # 0.3941 at 6e5 Pa print(round(noise_control.corrected_incipient_ratio(holes, 1.0e6), 4)) # # 0.3698 still above x_F ``` *The left panel is Annex A's own Figure A.1: the same valve at the same flow, with a characteristic pressure ratio 0,1 apart, and 14 dB between the two markers at $x_F = 0{,}35$.* That 14 dB is the reason 4.2 warns that a prediction built on the estimate "can create uncertainties". The estimate is not a small correction to the answer; on a cavitating valve it **is** the answer. ## 2. Two efficiencies, and one stream power to share The source half of the method is short. The differential of Equation (2) is capped at the choking point, Equation (5) turns it into a jet velocity and Equation (6) into the stream power the valve dissipates: ```python choked = noise_control.cavitation_differential( inlet_pressure_pa=1.0e6, outlet_pressure_pa=6.5e5, vapour_pressure_pa=2.32e3, pressure_recovery=0.92, ) speed = noise_control.vena_contracta_velocity(choked, 997.0, 0.92) power = noise_control.mechanical_stream_power(40.0, speed, 0.92) print(round(speed, 3), round(power, 1)) # 28.801 14042.1 ``` Then two efficiencies say what fraction of those fourteen kilowatts leaves as sound. Equation (8) makes the turbulent one linear in the jet Mach number, reaching $10^{-4}$ only when the jet reaches the speed of sound in the liquid, which for water is 1 400 m/s and never happens in a control valve: $$ \eta_{turb} = 10^{-4}\left(\frac{U_{vc}}{c_L}\right) $$ Equation (9) is the other one, and it is worth reading factor by factor: $$ \eta_{cav} = 0{,}32\, \eta_{turb} \sqrt{\frac{p_1 - p_2}{\Delta p_c}\cdot\frac{1}{x_{Fzp1}}}\; e^{5 x_{Fzp1}} \left(\frac{1 - x_{Fzp1}}{1 - x_F}\right)^{0,5} \left(\frac{x_F}{x_{Fzp1}}\right)^{5} \left(x_F - x_{Fzp1}\right)^{1,5} $$ Three of those factors do all the work. $(x_F - x_{Fzp1})^{1,5}$ starts the term at **exactly zero** on the threshold, so the two regimes meet without a step. $(x_F/x_{Fzp1})^5$ then makes it climb almost vertically. And $(1 - x_F)^{-0,5}$ sends it towards infinity as the valve approaches flashing, which is where the method stops. ```python turbulent = noise_control.turbulent_efficiency(speed, 1400.0) cavitating = noise_control.cavitation_efficiency( turbulent=turbulent, differential=3.5e5, choked_differential=choked, pressure_ratio=0.3508, corrected_ratio=0.2386, ) print(f"{turbulent:.3e} {cavitating:.3e}") # 2.057e-06 1.242e-06 ``` Just past the threshold the cavitation term is already comparable with the turbulent one, and Equation (7b) adds them before Table 2's acoustic power ratio $r_W$ takes the share that is radiated into the pipe rather than lost in the body: a quarter for every globe and rotary valve, a half for the butterflies, one for an expander. ## 3. The whole chain in one call ```python res = noise_control.valve_hydrodynamic_noise( noise_control.LiquidStream( mass_flow=40.0, # kg/s inlet_pressure_pa=1.0e6, # Pa, absolute outlet_pressure_pa=6.5e5, vapour_pressure_pa=2.32e3, density=997.0, # kg/m3 speed_of_sound=1400.0, # m/s ), noise_control.LiquidTrim( flow_coefficient=90.0, # C_v style_modifier=0.42, # F_d, from IEC 60534-8-3 pressure_recovery=0.92, # F_L incipient_ratio=x_fz, # x_Fz, at 6e5 Pa power_ratio=0.25, # r_W, Table 2 valve_diameter_m=0.1, # m seat_diameter_m=0.1, # m ), noise_control.LiquidPipe( internal_diameter_m=0.1071, # m wall_thickness=0.0036, # m density=7800.0, # kg/m3, steel ), ) print(res.regime) # cavitating print(round(res.sound_power, 5), "W") # 0.01158 W print(round(res.internal_level, 3), "dB") # 156.543 dB print(round(res.turbulent_peak, 2), round(res.cavitation_peak, 2)) # # 654.35 1088.94 print(round(res.external_level, 1), "dB at 1 m") # 81.0 dB at 1 m ``` Every one of those is a printed cell of Annex A's Table A.1. The same call on the first column, 30 kg/s into 8 bar, stays turbulent and answers 62,7 dB; the third column is the second one again with $x_{Fz}$ shifted by 0,1, and answers 66,9 dB. The two peak frequencies are Equations (11) and (13), and (13) is the surprising one: on the threshold the cavitating peak is **six times** the turbulent peak, and it falls from there as the valve is opened further into cavitation, because the bubbles grow larger and take longer to collapse. Just past onset, cavitation is a hiss; deep into it, a rumble. ## 4. The pipe, and a printed sign that cannot be right Inside the pipe this valve runs at 156 dB. What makes the outside habitable is the wall, and the standard anchors its transmission loss at the **ring frequency**, Equation (14), where one wavelength in the wall material wraps exactly once around the circumference: ```python print(round(res.pipe_ring_frequency, 1), "Hz") # 14860.4 Hz print(round(res.reference_transmission_loss, 2), "dB") # -44.71 dB print(round(res.transmission_loss, 2), "dB") # -62.86 dB ``` Both terms of Equation (15) are printed with a minus sign, so the transmission loss of this method is a **negative number that is added** all the way to the end. Away from the ring frequency Equations (16b) and (22b) only make it worse, and the peak frequencies of a control valve are more than a decade below the ring frequency of its pipe, which is where the other 18 dB of this valve's loss come from: Equation (16b) costs 27 dB at the turbulent peak, and Equation (17) hands 9 of them back because the valve cavitates. Band by band, 5.4 spreads the internal level with Equations (20a) and (20b) and gives the wall a frequency-dependent loss with Equation (22a): ```python import numpy as np band = int(np.argmin(np.abs(res.frequencies - 8000.0))) print(round(float(res.band_internal_level[band]), 1)) # 141.9 print(round(float(res.band_transmission_loss[band]), 2)) # -51.76 print(round(float(res.band_external_level[band]), 1)) # 77.4 ``` Those three are printed in Table A.1 as 141,9 dB, **51,76 dB** and 77,4 dB. The middle one is printed without its minus sign, and it cannot be: its own two inputs, $-44{,}71$ and $-7{,}053$ dB, sum to $-51{,}763$, and the row below only reproduces its printed 77,4 dB with the negative value. [The errata register](https://jmrplens.github.io/phonometry/reference/errata/) records it, along with three intermediates of the same table that its own equations do not reproduce. ## 5. One equation, printed two ways Equation (12), the Strouhal number that places the turbulent peak, appears twice in the document and not identically. Clause 5.1 prints a leading 0,02 and no valve style modifier; Table A.1 prints 0,036 and a factor $F_d^{0,75}$. Both are on the page, and they are not the same function of the valve. ```python form = dict( flow_coefficient=90.0, style_modifier=0.42, pressure_recovery=0.92, corrected_ratio=0.2386, valve_diameter_m=0.1, seat_diameter_m=0.1, inlet_pressure_pa=1.0e6, vapour_pressure_pa=2.32e3, ) print(round(noise_control.jet_strouhal_number(**form, form="annex"), 3)) # 0.399 print(round(noise_control.jet_strouhal_number(**form, form="clause"), 3)) # 0.425 ``` Only the annex form reproduces the annex's own printed $N_{Str} = 0{,}399$, so it is the default here and the clause form is one keyword away. For this valve the two differ by 6 %; for a single-port valve with $F_d = 1$ the annex form is 80 % above the clause one, which is five sixths of an octave in the peak frequency and a few decibels through the transmission loss. ## 6. Stage by stage Clause 6 is the same method with per-stage inputs. Each stage takes a share of the differential in inverse proportion to the square of its own flow coefficient, which is the series law $1/C^2 = \sum_i 1/C_i^2$: ```python stages = noise_control.stage_conditions( inlet_pressure_pa=1.0e6, outlet_pressure_pa=4.0e5, vapour_pressure_pa=2.32e3, stage_coefficients=[130.0, 160.0, 199.1], flow_coefficient=90.0, ) for stage in stages: print(round(stage.inlet_pressure_pa), round(stage.pressure_ratio, 3)) # 1000000 0.288 # 712426 0.267 # 522582 0.236 print(round(noise_control.combine_stage_levels(78.0, 74.0, 71.0), 1)) # 80.0 ``` Those three coefficients increase along the flow, which is the device of Figure 2: most of the pressure is taken in the first stages and the last one is left working at a differential small enough not to cavitate. 6.3.2 then calculates **only** that last stage, because the sound the earlier ones make is absorbed inside the trim before it reaches the pipe, and it caps the differential at the last stage's own cavitation threshold rather than at the choking point: ```python print(noise_control.last_stage_differential( inlet_pressure_pa=6.0e5, outlet_pressure_pa=4.0e5, vapour_pressure_pa=2.32e3, corrected_ratio=0.30, )) # 179304.0 print(noise_control.uniform_passage_style_modifier(16)) # 0.25 print(round(noise_control.last_stage_seat_diameter_mm(45.0), 1)) # 37.7 ``` The last of those carries its unit in its name for a reason. The formula it implements is the one display formula in the standard with no equation number, $d_o = 5{,}2\sqrt{N_{34} C_n}$, and Clause 3 declares $d_o$ in metres, which for any real last stage it cannot be: 37,7 of anything is millimetres. Divide by a thousand before handing it to Equation (12). ## See also - [Control Valve Noise (IEC 60534-8-3)](https://jmrplens.github.io/phonometry/devices/noise-control/control-valve-noise/): the same chain for a compressible fluid, where five flow regimes take the place of one threshold. - [Reactive Silencers](https://jmrplens.github.io/phonometry/devices/noise-control/silencers/): the four-pole chain that would be put downstream of a line this loud. - [Duct-Borne Noise](https://jmrplens.github.io/phonometry/devices/noise-control/duct-path/): the same question for a ventilation system, where the source is a fan rather than a jet. - [Errata in published sources](https://jmrplens.github.io/phonometry/reference/errata/): the six entries this page rests on, from a sign to an equation printed two ways. - API reference: [`noise_control.valves_hydrodynamic`](https://jmrplens.github.io/phonometry/reference/api/noise_control/valves-hydrodynamic/). ## Standards IEC 60534-8-4:2005, Clauses 4, 5 and 6. The preliminary calculations of Clause 4: the pressure ratio of Equation (1), the capped differential of (2), the characteristic pressure ratio of (3a) to (3c), the jet diameter of (4), the vena contracta velocity of (5) and the stream power of (6). The prediction of Clause 5: the two acoustical efficiencies of (8) and (9), the sound power of (7a) and (7b), the internal level of (10), the Strouhal number and both peak frequencies of (11) to (13), the transmission loss of (14) to (17) with the floor of the NOTE, the external level of (18a) and (18b), and the band route of 5.4. And Clause 6 for a multistage trim: the stage pressures and ratios of (23a) to (26), the energy sum of (27), and the last stage of 6.3.2. Validated against all three worked examples of Annex A in the [conformance report](https://jmrplens.github.io/phonometry/reference/conformance/); the six defects that annex and its clauses carry are in the [errata register](https://jmrplens.github.io/phonometry/reference/errata/). The laboratory measurement of x_Fz (IEC 60534-8-2) and the flashing regime past x_F = 1 are not implemented.