As discussed in a previous blog post, the frequency response of particles is dependent on the relaxation time:
If we replaced the particle with a helium-filled (or air-filled) soap bubble, how big can we make the bubble to have the particle have the same relaxation time? Well, let’s for this purpose assume that 
If the bubble is a hollow shell with outer radius 
Here I am seeking to find a bubble particle with the same relaxation time as the full-sphere droplet particle. Therefore, the relaxation times match:
This means that the ratio between the diameters of the drop and bubble particles are:
If we use the bubble shell thickness 
We can also express this as a function of the particle radius/thickness ratio:
Then:
I left this final formula as a ratio of the diameter of the bubble and drop squared, as the scattering cross-sections (and therefore scattering intensity) are a function of the scattering area instead of just the diameter.
Now let’s analyze the equation above. If one were to plot this function, it would look like this:
One thing that strikes immediately is the linearity, as well as the slope (1/3) at the latter part where the thickness ratio is large enough. Well, let’s physically interpret this for the Particle Image Velocimetry application: Say we have a bubble made of a water shell. I did some experiments at home and found that at ambient air conditions (20C, ~60% humidity) we get an evaporation rate of ~2-3mm/day; which is approximately ~23 to 34 nm/second. Of course, I don’t know if the evaporation rate scales down to the nanoscale that well, but it must be at least that rate otherwise I wouldn’t get the results I found by evaporating water on a dish. We also generally find that water evaporates pretty fast when we leave our wet kitchen counters for a little while; so I think this rate is not too far from reality. Further, given the spherical symmetry of the bubble and the rapid flow speeds we will look at, I can only expect the bubble to evaporate faster than what is measured in quiescent conditions.
Settling for 30nm/second of evaporation rate, if we make bubbles for use on a supersonic wind tunnel test we may afford maybe 5 to 10 seconds of lifetime between generation and the bubbles reaching the test section (by which time, we no longer need them so they can pop). This means that a bubble would have to be at least ~200-300nm thick so it can survive the run.
Let’s work with 200nm for now. If a bubble like that can be made on demand at scale with good repeatability, then the diameter ratio that such a bubble would have in comparison to a 1um oil particle would be about 1.2um (
That is not particularly great because we are producing bubbles such that we can have brighter particles, and 1.44 is not the kind of number we are after (we’re hoping for 1,000x brighter particles). Therefore, I have very little faith that bubbles can be a workable solution for tomographic PIV in high-speed flows (say, supersonic flows) that require particles to have very high frequency responses (of the order of 50kHz and more).
What do you think?
(i.e., the ISO number) is a function of the film exposure 
for a “specified contrast”. 
value of 
we should get 18% of the saturation value of the pixel.
is very confusing. So I will try to give a little summary of what the photometric units entail. First thing we have to understand is that the photometric units (candela, lumen, lux) differ from the radiometric units (W, W/m^2, etc.) because they consider a 
is plotted below for the photopic (black) and scotopic (green) curves, where the photopic corresponds to a brightly lit scene and the scotopic corresponds to a dimly lit place. Our eyes adapt, so apparently we are more sensitive to blue light in the dark.
(in lumens) is then defined for each wavelength as:
is the power of the light source in W, and 
. 
). The symbol for radiometric exposure is 
. We can, therefore, for assessing a point light source, break out of the ISO system as quickly as we can by performing the following calculations:
in lux-s, which will cause saturation:
, convert the exposure to radiometric units 
:
is in J/m2 and is the exposure required to saturate a pixel. Now that we’ve done that, we can work with the exposure in radiometric units and we can start talking about counting photons. If we know the pixel size 
, we can find the exposure required for saturation on each pixel in photons:
is the photon energy in Joule, and 
is the Planck constant and 
is the speed of light.
we need to saturate a pixel.
per photon 
can be found from the quantum efficiency 
:
. This could be found if such capacitance was known by using the electron charge 
Coulomb:
is sometimes also expressed as a sensitivity 
(in V/electron). The pixel voltage would be converted to an amplified voltage through an amplifier gain 
(in V/V), and then to counts given a saturation voltage 
:
and 
for a given sphere diameter for all scattering angles 
. We don’t care about the scattering efficiencies for PIV, so I just adapted Dr. Saviot’s Mie scattering code to output only the scattering constants. My implementation is in my 
(in W/m2) as a function of the intensity of the incoming light 
(also in W/m2), the wavelength 
and 
indicate scattered light in the perpendicular and parallel polarization directions, respectively. To find the polarization direction for your illumination, consider the following: (1) a polarizing filter made of an array of metallic wires will eliminate the electric field component along the wires; (2) imagine a plane in space constituted by the two vectors [a] the incident light direction and [b] the scattered light direction. If the light polarization is along this plane, then the 
and focal length 
to find the effective entrance pupil diameter 
:
is an energy spread ratio that depends on the spot size a point-like source produces on the sensor. If the particle has a Gaussian shape on the pixels, then this ratio is the peak of the 2D gaussian divided by its integral over the entire sensor.
at 532nm.
and is in a position of the flow where the flow velocity is 
:
:
![\displaystyle s V_{particle}(s)=\frac{1}{t_0} [V_{flow}(s)-V_{particle}(s)]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+s+V_%7Bparticle%7D%28s%29%3D%5Cfrac%7B1%7D%7Bt_0%7D+%5BV_%7Bflow%7D%28s%29-V_%7Bparticle%7D%28s%29%5D&bg=ffffff&fg=333333&s=0&c=20201002)
for a first order system. Thus, the 100 micron particle has a cut-off of 
, the 10 micron particle cuts off at 
and the 1 micron particle follows the flow with good fidelity until 
.
, we need to integrate:
and 
where the shock wave lies. Then, the integral becomes:
is taken as 
, we would get the “blurring distance” 
for three time constants; which would give us the settling distance to 99% of the velocity downstream of the shock. We then have:
shock wave in a flow with a stagnation temperature of 
, 
for a 1 micron particle, and 
for a 10 micron propyleneglycol particle. Thus, large particles (>1 micron) are typically not acceptable in supersonic flows.
is fixed, since the camera sensor position is not moving. 
, making a point light source (i.e., a small feature or edge in the model) look like a circle with the diameter 
, also known as the “circle of confusion”. The circle of confusion size is:
which, as expected, leads to 
. 
. Here’s a little diagram that may make it easier to go through the math:
) and red (
. Then we can plot 
for a given system. For example, say for a realistic system we have 
, 
, 
, 
:
, 
. Note, this is not specific for this system I calculated. It will always have to be the case. 
. And if the lens is to focus any closer than infinity, the lens distance from the sensor 
. Thus, 
. 
, and thus 
.
, thus:
). If we look at the energy conservation equation for a lumped element:
, a small capacitance 
). I will make this problem more 1-dimensional to consider the case where the stack of layers is much thinner than wide, because that’ll lead to interesting observations when we drop the “lumped element” part of this model. For now, let’s simply divide both left-hand and right-hand sides by the sheet area 
, to get everything in terms of heat fluxes:
then are what we will call the specific component’s capacitance (
), in the SI units of J/m2.K. One can then represent a layer of a stack like the following RC circuit:
, where 
. You can check the units: 
has units of m2.K.s/J, which times the units of 
, and we get a -20dB/decade low-pass filtering behavior.
, but to an elusive “ground” temperature. This is because, as long as you have a non-zero 
, you will have heat transfer to the thermal capacitance. Therefore, the ground level can be considered absolute zero temperature – for didactic purposes. The temperature of this ground level is not important, though: since capacitors always block DC, its absolute value will never play a role.
, which according to the energy conservation equation applied to this lumped element, can be split into the heat flux that stays in the element (
) and the heat flux that goes through the element (
). This heat flux then goes to the second layer and splits again into 
and 
. The energy conservation equations here are equivalent to the Kirchoff current laws applied to the nodes at 
and 
in the circuit representation (a):
– which is simply not the correct energy balance. 
time constant), the lumped element model would predict a decay rate of -20dB/decade, which is what is expected from a standard first-order system. 
(thermal resistance per unit length) and 
(thermal capacitance per unit length). This circuit immediately follows from the circuit presented in (a.1) in the previous figure; since one can consider a single layer can be decomposed into an infinite stack of infinitely thin layers. Not all mechanical engineers will recognize, however, that this corresponds to a (simple) version of the 1-dimensional transmission line model. Robertson and Gross (1958, “An Electrical-Analog Method for Transient Heat-Flow Analysis”, Journal of Research of the National Bureau of Standards) did recognize this in their paper, for example. Funnily enough, their figure shows they also got the RC network wrong, since the first resistor is shown between the first capacitor and T1 – an indicator of how tricky this business can be!
and conductance 
. If we come back to the thermal stack and consider there’s no conductance 
and no inductance 
, we recover the heat equation:
; as expected.
is the input impedance as seen by the node at 
and 
is the reflection coefficient due to the impedance mismatch of the aluminium slab and the convective boundary (also the load impedance) 
. 
in this case is the aluminium’s characteristic impedance, which in general has the form:
predicts a phase lag of 45º at high frequencies (not 90º!). Also referring to the Wikipedia article, the propagation constant 
is a material property:
, we get:
, we use the 
), and then iterate. Since we’re looking for the maximum temperature in the stack, 
, let’s calculate the overall input impedance of the stack 
:
. You can think about it like; its large capacitance combined with its (relatively) low conductivity prevents the thermal oscillations on one side of the material to immediately propagate to the opposite side – invalidating the lumped element assumption.
.
![20 \log_{10}[(T_p-T_h)/(T_\infty -T_h)]](https://s0.wp.com/latex.php?latex=20+%5Clog_%7B10%7D%5B%28T_p-T_h%29%2F%28T_%5Cinfty+-T_h%29%5D&bg=ffffff&fg=333333&s=0&c=20201002)
. This corresponds to a signal gain of 0.011 – or; for every 1 degree of temperature fluctuation in 
, we get one hundreth of a degree of fluctuation in the paint surface temperature; which is quite a small signal. Since the TSP gives a signal of about 1.4%/ºC, we are talking about an intensity signal of 
, or 0.6 counts/ºC of a 12-bit camera where 
in the previous sections – so let’s focus on the math. The thermal impedance of the slab in the -10dB/decade regime is simply its characteristic impedance (another interpretation is that in the -10dB/decade regime the slab is effectively infinite).
:
:
, but this is the same kind of transient phenomenon. If the Fourier number is larger than 1, the “time scales” are large (low frequencies) and the heat transfer of the slab can be modeled by a simple capacitor. If the Fourier number is smaller than 1, then the “time scales” are small (high frequencies) and the heat transfer needs the heat equation to be modeled, which we are simply doing by using the transmission line model. The analogy is quite powerful and enables simple, quick analysis of these systems.
equation because it is quite useful to determine whether our AC analysis needs to consider the transmission line model; knowing the frequencies of interest. Hope this provides the insight you needed for your analyses!
and the expected pressure fluctuation at 40kHz is 257 Pa, corresponding to a SPL of 142.2dB. Quite loud!
then becomes a density wave of the form:![\displaystyle \rho=\rho_a \bigg[1+\frac{P_0}{P_a} e^{i(\omega t - kx)} \bigg]^{1/\gamma}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Crho%3D%5Crho_a+%5Cbigg%5B1%2B%5Cfrac%7BP_0%7D%7BP_a%7D+e%5E%7Bi%28%5Comega+t+-+kx%29%7D+%5Cbigg%5D%5E%7B1%2F%5Cgamma%7D&bg=ffffff&fg=333333&s=0&c=20201002)
![\displaystyle \frac{\partial \rho}{\partial x} = -\frac{i\rho_a P_0/P_a e^{i\omega t} (k/\gamma) \big[1+(P_0/P_a) e^{i(\omega t-kx)}\big]}{P_o/P_a e^{i\omega t} + e^{i kx}}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cpartial+%5Crho%7D%7B%5Cpartial+x%7D+%3D+-%5Cfrac%7Bi%5Crho_a+P_0%2FP_a+e%5E%7Bi%5Comega+t%7D+%28k%2F%5Cgamma%29+%5Cbig%5B1%2B%28P_0%2FP_a%29+e%5E%7Bi%28%5Comega+t-kx%29%7D%5Cbig%5D%7D%7BP_o%2FP_a+e%5E%7Bi%5Comega+t%7D+%2B+e%5E%7Bi+kx%7D%7D&bg=ffffff&fg=333333&s=0&c=20201002)
![\displaystyle \frac{\partial \rho}{\partial x} \frac{\gamma}{\rho_a k} = -\frac{i P_0/P_a e^{i\omega t} \big[1+(P_0/P_a) e^{i(\omega t-kx)}\big]}{P_o/P_a e^{i\omega t} + e^{i kx}}=N(P_0)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cpartial+%5Crho%7D%7B%5Cpartial+x%7D+%5Cfrac%7B%5Cgamma%7D%7B%5Crho_a+k%7D+%3D+-%5Cfrac%7Bi+P_0%2FP_a+e%5E%7Bi%5Comega+t%7D+%5Cbig%5B1%2B%28P_0%2FP_a%29+e%5E%7Bi%28%5Comega+t-kx%29%7D%5Cbig%5D%7D%7BP_o%2FP_a+e%5E%7Bi%5Comega+t%7D+%2B+e%5E%7Bi+kx%7D%7D%3DN%28P_0%29&bg=ffffff&fg=333333&s=0&c=20201002)
represents the wave steepening as the amplitude becomes sufficiently high. Looking at the phasor in the denominator (
) we see the derivative operator split the time fluctuations from the spatial component of the fluctuation. For our detectability purposes, what really matters is the minimum and maximum value of ![\displaystyle \min |N(P_0)|=\frac{P_0/P_a \big[1-P_0/P_a \big]^{1/\gamma}}{1+P_0/P_a}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmin+%7CN%28P_0%29%7C%3D%5Cfrac%7BP_0%2FP_a+%5Cbig%5B1-P_0%2FP_a+%5Cbig%5D%5E%7B1%2F%5Cgamma%7D%7D%7B1%2BP_0%2FP_a%7D&bg=ffffff&fg=333333&s=0&c=20201002)
![\displaystyle \max |N(P_0)|=\frac{P_0/P_a \big[1+P_0/P_a \big]^{1/\gamma}}{1-P_0/P_a}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmax+%7CN%28P_0%29%7C%3D%5Cfrac%7BP_0%2FP_a+%5Cbig%5B1%2BP_0%2FP_a+%5Cbig%5D%5E%7B1%2F%5Cgamma%7D%7D%7B1-P_0%2FP_a%7D&bg=ffffff&fg=333333&s=0&c=20201002)
![\displaystyle N_{pk-pk}=\frac{P_0}{P_a} \frac{\big[1+P_0/P_a \big]^{1/\gamma} - \big[1-P_0/P_a \big]^{1/\gamma}}{1-(P_0/P_a)^2}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N_%7Bpk-pk%7D%3D%5Cfrac%7BP_0%7D%7BP_a%7D+%5Cfrac%7B%5Cbig%5B1%2BP_0%2FP_a+%5Cbig%5D%5E%7B1%2F%5Cgamma%7D+-+%5Cbig%5B1-P_0%2FP_a+%5Cbig%5D%5E%7B1%2F%5Cgamma%7D%7D%7B1-%28P_0%2FP_a%29%5E2%7D&bg=ffffff&fg=333333&s=0&c=20201002)
. However, we can do the opposite – isolate the frequency and plot the points with the axes flipped. To do so, let’s isolate the wavenumber:
![\displaystyle F=\frac{c}{\pi \rho_a } \frac{\gamma}{N_{pk-pk}} \bigg[\frac{1}{G(\lambda)} \frac{n_0}{L} \frac{a}{f_2} \frac{N_{lv}}{2^{n_{bits}-1}} \bigg]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%3D%5Cfrac%7Bc%7D%7B%5Cpi+%5Crho_a+%7D+%5Cfrac%7B%5Cgamma%7D%7BN_%7Bpk-pk%7D%7D+%5Cbigg%5B%5Cfrac%7B1%7D%7BG%28%5Clambda%29%7D+%5Cfrac%7Bn_0%7D%7BL%7D+%5Cfrac%7Ba%7D%7Bf_2%7D+%5Cfrac%7BN_%7Blv%7D%7D%7B2%5E%7Bn_%7Bbits%7D-1%7D%7D+%5Cbigg%5D&bg=ffffff&fg=333333&s=0&c=20201002)
(Schlieren)
m, 
mm, 
m and 
is the distance between the “screen” (or camera focal plane, in case of a focused shadowgraph) and the Schlieren object. Since the ray deflection occurs in the two directions perpedicular to the optical axis (
), we get a contrast in the 
) and another in the 
). The actual contrast seen in the image ends up being the sum of the two:
is dependent on the Schlieren object (as discussed in Part I), we can plug its definition to get:
derivative. We can plug in the Gladstone-Dale relation and the ideal gas law to isolate the pressure Laplacian:
![P=P_0 \exp(i[\omega t - \kappa_x x - \kappa_y y])](https://s0.wp.com/latex.php?latex=P%3DP_0+%5Cexp%28i%5B%5Comega+t+-+%5Ckappa_x+x+-+%5Ckappa_y+y%5D%29&bg=ffffff&fg=333333&s=0&c=20201002)
, now obviously considering the two-dimensional wavenumber in the directions perpendicular to the optical axis:![\displaystyle P_0 = -\frac{\big(\nabla_{xy}^2 P \big)\exp(-i[\omega t - \kappa_x x - \kappa_y y])}{\kappa_x^2 + \kappa_y^2} = -\frac{\big(\nabla_{xy}^2 P \big) P \exp(-i[\omega t - \kappa_x x - \kappa_y y])}{\kappa_{xy}^2}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+P_0+%3D+-%5Cfrac%7B%5Cbig%28%5Cnabla_%7Bxy%7D%5E2+P+%5Cbig%29%5Cexp%28-i%5B%5Comega+t+-+%5Ckappa_x+x+-+%5Ckappa_y+y%5D%29%7D%7B%5Ckappa_x%5E2+%2B+%5Ckappa_y%5E2%7D+%3D+-%5Cfrac%7B%5Cbig%28%5Cnabla_%7Bxy%7D%5E2+P+%5Cbig%29+P+%5Cexp%28-i%5B%5Comega+t+-+%5Ckappa_x+x+-+%5Ckappa_y+y%5D%29%7D%7B%5Ckappa_%7Bxy%7D%5E2%7D&bg=ffffff&fg=333333&s=0&c=20201002)
is the squared wavenumber vector magnitude, projected in the 
plane. Applying the same criterion of 
levels for minimum contrast, we get the following relation for the minimum detectable pressure wave:
based on the wave frequency 
:
[Shadowgraph] instead of 
[Schlieren]. This is obvious, since Schlieren detects first derivatives of pressure, whereas Shadowgraph detects second derivatives, popping out an extra 
m (i.e., the focal plane is 0.5 meters from the Schlieren object), we get the following curve for a 12-bit camera:
kHz. There might be applications where these ultrasound frequencies are actually of interest, but I believe in most fluid dynamics problems the Schlieren apparatus is the to-go tool. Remembering that, even though I show SPL values above 170dB in this chart, the calculations considered a constant-temperature ideal gas, which is not the case anymore at those non-linear sound pressure levels.
Zigunov Aero 