Pulse Dampener Size Analysis
Part I - Required Compliant Gas Volume Calculation
Part II - Additional Attenuation Using a Series Flow
Resistance
Craig E. Nelson - Consultant Engineer
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Goals
Part I:
It is desired to learn the required enclosed gas volume for a pressure pulse
dampener for use with small motor driven diaphragm pumps.
An equation is derived for calculating the required volume as a function of
pump stroke volume, allowable maximum pressure pulse amplitude and
absolute pump-up pressure.
The results for this expression are presented in graphic format for use by
pressure pulse dampener designers.
Part II:
Calculation of the frequency vs. pressure pulse amplitude roll-off rate as a
function of compliant volume and series flow resistance, is needed.
Suitable equations and useful display graphs are presented that can help the
pulse dampener designer produce an optimized resultant pulse dampening
device.
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Scenario for Analysis
Pump Piston or diaphragm
at end position
Pump Piston or
diaphragm at
start position
Ideal Gas filled
Compliant volume
Flow
Resistance
Variables
Stroke
Volume
(SV)
Pstart ; Vstart
Pend; Vend
Stroke Volume (SV)
Liquid height at end
of Pump Stroke
Liquid filled
volume
Rigid
Walls
Nelson Research, Inc.
Liquid height at start
of pump stroke
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Use Boyle’s Law to Solve for Required Initial Volume
Variables:
Pstart ; Vstart
Pend; Vend
Assume:
Pstart
Ppulse = pulse amplitude
Solve for:
Vstart = required pulse dampener volume
Boyle’s Law (Isothermal ideal gas law):
SV = Stroke Volume
Pstart * Vstart = Pend * Vend
Let:
Vend = Vstart – SV
Then:
Vstart = ( Pend/Pstart ) * Vend = { ( Pstart + Ppulse ) * ( Vstart - SV ) } / Pstart
After some algebra:
Pend = Pstart + Ppulse
Vstart =
Stroke Volume (SV)
=
Vstart – Vend
( 1 + Pstart / Pulse ) * SV
Example:
let
Pstart = 17 PSI absolute
then
Required Vstart (gas volume in dampener) = 8.55 cm^3
thus
Compliance = C = .05 cm^3 / .1 PSI = .5 cm^3 / PSI
Nelson Research, Inc.
Ppulse = .1 PSI
SV = .05 cm^3
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Required Pulse Dampener Volume
Damperner Size vs Stroke Volume for .1 PSI
Pressure Pulses
One
Atmosphere
Required Gas Volume (cm^3)
30
SV=.02 cm^3
25
SV=.05 cm^3
20
SV=.1 cm^3
15
10
5
0
0
5
10
15
20
25
Resting Pressure (PSI Absolute)
Region of
Interest
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Required Pulse Dampener Volume
Dampe ne r Vol. v s Stroke Volume for .05 cm^3 Stroke Vol.
Region of
Interest
Required Gas Volume (cm^3)
140
Ppulse = .01
120
Ppulse = .02
100
Ppulse = .05
Ppulse = .1
80
60
40
20
0
10
15
20
25
Re sting Pre ssure (PSI Absolute )
One
Atmosphere
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Conclusions – Part I
1.
It seems that at least 10 cm^3 gas volume will be needed to “snub” Model
XYZ Diaphragm pump pressure pulses to about .1 PSI
2.
It seems that 25 cm^3 gas volume would be the largest needed.
3.
Use of a fluidic resistor or small feed port restrictors ahead of the
dampener could possibly reduce these calculated volumes by a factor of 2
before average pressure drops in the resistor or port restrictors becomes a
problem.
4.
Reduction of pulse amplitudes by fluid resistances and compliance within
downstream system elements might allow further diminishment of
downstream pressure pulse
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Part II - Introduction
In the previous section, the compliant gas volume required to dampen pressure
pulses to a desired amplitude was calculated. This can be thought of the “DC” or
zero frequency of pressure excitation case. Now, using fluidic resistance and
compliance as variable parameters, I will calculate the pressure response of the
system as a function of frequency.
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Calculation Setup
Calculation of pressure pulse roll-off as a function of gas volume and series
flow resistance is made easier by converting the system to its electrical
analog … a series-shunt “RC” network.
Source
R is in Rsec = PSI/cm^3/sec
or Rmin = PSI/cm^3/min
(more on this later)
C
Patten
R
Pump
Pressure
Pulses
Attenuated
Pressure
Pulses
1 Ohm
S1
2
Fluidic_Resistor
1
1 V
1 Hz
0Deg
1 F
Compliance
0
Psource
Nelson Research, Inc.
Patten
0
Psource
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Complex Attenuation Amplitude
From basic network theory, Patten (having some magnitude and phase relative to
Psource) is given by:
Patten = 1 / ( Rseconds * C * omega * j + 1)
Complex Frequency Response
Where:
omega is angular frequency in radians/sec
j is the square root of -1
and
Rseconds is fluidic resistance in (Pressure / FlowRate) in PSI / (cm^3/sec)
C is compliance in (delta Volume / delta Pressure) in delta cm^3 / delta PSI
Pump
Pressure
Pulses
Attenuated
Pressure
Pulses
omega = 2*pi*frequency
Patten
1 Ohm
S1
2
Fluidic_Resistor
1
1 V
1 Hz
0Deg
1 F
Compliance
0
0
Psource
Nelson Research, Inc.
If Rsec = 1 then:
A flow of 1 cm^3/sec
through Rsec will cause a
pressure drop of 1 PSI
across Rsec
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Magnitude of Attenuation – Rminutes Units
It is now useful to convert from Rseconds and angular frequency to Rminutes and frequency:
The magnitude of
Patten(freq) = 1 / (2*pi*freq*Rmin*C*freq*j / 60 + 1)
is (by algebra):
|Patten(freq)| = 1 / sqrt( ( 2*pi*freq*Rmin*C*freq / 60 )^2 + 1)
Where:
frequency is in sec^-1
And
Rminutes is fluidic resistance (Pressure/Flow rate) in PSI / (cm^3/min)
Rseconds = Rmin / 60
C is compliance (delta Volume / delta Pressure) in delta cm^3 / delta PSI
Pump
Pressure
Pulses
j is the square root of -1
Attenuated
Pressure
Pulses
Patten
1 Ohm
S1
2
Fluidic_Resistor
1
1 V
1 Hz
0Deg
1 F
Compliance
0
0
Psource
Nelson Research, Inc.
If Rmin = 1 then:
an average flow of 1
cm^3/min through Rmin will
cause a pressure drop of 1 PSI
across Rmin
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Magnitude of Attenuation (Rmin =.01 C=.2,.5,1)
The roll-off would be the same if C was held at .01 and Rminutes varied from .2 to .5 to 1
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Log-Log (Bode) Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Semilog Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
1.0000
1.0000
Relative Attenuation (linear)
Relative Attenuation (linear)
0.9000
0.8000
0.7000
0.6000
0.5000
0.4000
0.3000
Rminutes =.01 C=.2
0.2000
Rminutes =.01 C=.5
0.1000
Rminutes =.01 C=1
0.0000
0.1000
0.0100
Rminutes =.01 C=.2
Rminutes =.01 C=.5
Rminutes =.01 C=1
0.0010
10
1,000
100
10,000
100,000
Pulse Frequency (Hz)
Corner frequency goes from
about 1000 to 5000 Hz
Nelson Research, Inc.
10
100
1,000
10,000
100,000
Pulse Frequency (Hz)
x 10 Roll-off per decade
of frequency increase
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Magnitude of Attenuation (Rmin =.1 C=.2,.5,1)
The roll-off would be the same if C was held at .1 and Rminutes varied from .2 to .5 to 1
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Log-Log (Bode) Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Semilog Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
1.0000
1.0000
Relative Attenuation (linear)
Relative Attenuation (linear)
0.9000
0.8000
0.7000
0.6000
0.5000
0.4000
0.3000
Rminutes =.1 C=.2
0.2000
Rminutes =.1 C=.5
Rminutes =.1 C=1
0.1000
0.0000
0.1000
Rminutes =.1 C=.2
Rminutes =.1 C=.5
Rminutes =.1 C=1
0.0100
1
100
10
1,000
10,000
100,000
1
10
Pulse Frequency (Hz)
Corner frequency goes
from about 50 to 500 Hz
Nelson Research, Inc.
100
1,000
10,000
100,000
Pulse Frequency (Hz)
x 10 Roll-off per decade
of frequency increase
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Magnitude of Attenuation (Rmin = 1 C=.2,.5,1)
The roll-off would be the same if C was held at 1 and Rminutes varied from .2 to .5 to 1
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Log-Log (Bode) Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
Pressure Pulse Attenuation Relative to the
"DC" Source Amplitude - Semilog Plot
( Rmin in PSI/cm^3/min - C in cm^3 / PSI )
1.0000
1.0000
Relative Attenuation (linear)
Relative Attenuation (linear)
0.9000
0.8000
0.7000
0.6000
0.5000
0.4000
0.3000
Rminues =1 C=.2
0.2000
Rminutes =1 C=.5
Rminutes =1 C=1
0.1000
0.0000
0.1000
Rminues =1 C=.2
Rminutes =1 C=.5
Rminutes =1 C=1
0.0100
0
10
1
100
1,000
10,000
0
1
Pulse Frequency (Hz)
Corner frequency goes
from about 5 to 50 Hz
Nelson Research, Inc.
10
100
1,000
10,000
Pulse Frequency (Hz)
x 10 Roll-off per decade
of frequency increase
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com
Conclusions – Part II
1.
With the high flow rates (50 ml/min), a series fluidic resistance large
enough to be helpful filtering out pressure pulses will induce too much
static pressure drop to be useful.
2.
Dividing the total flow into many parallel flow channels with separate
pulse dampeners for each channel would help the situation quite a bit.
3.
The best answer is to seek out or develop small pumps that do not
generate the relatively high pressure and low frequency pulses that the
Model XYZ diaphragm pumps do.
4.
In the mean time, two pressure pulse dampeners of approximately 9 cm^3
internal air volume, for each 50 ml/min flow loop, seem to be required for
adequate pressure pulse dampening.
Nelson Research, Inc.
2142 – N. 88th St. Seattle, WA. 98103 USA 206-498-9447 Craigmail @ aol.com