Saturday, 29 August 2020

homework and exercises - How to measure percentage of nitrogen present in Nitro Coffee


Starbucks had launched nitro coffee. Unlike carbonated, this coffee contains nitrogen gas which makes that peculiar down draft waviness in the drink. In carbonated beverages, the gas effervescence and move upward. My question is, carbon dioxide percentage in carbonated beverages can be measured using a cuptester


How do we measure nitrogen percentage in nitro coffee.


I found this video of Professor Philip Moriarty on nitrogenated drink



Answer



The answer may depend on how accurate you need the result to be, and what fraction of nitrogen bubbles you expect to be present.


In principle, the presence of nitrogen will lower the density of the liquid, but so does temperature. You need to find a method to distinguish the two - recognizing that the nitrogen will slowly make its way to the surface and disappear.


Given that these are very tiny bubbles, I would probably look for the bulk modulus of the material. As you know, sound travels through a liquid with a speed given by the bulk modulus:


$$c = \sqrt{\frac{K}{\rho}}$$


Now if you add some bubbles, they will greatly affect the compressibility of the liquid. You may have noticed this phenomenon when you make a cup of hot chocolate from powder and boiled water. As you stir the cup, the sound of the stirring starts very low, and increases as the powder dissolves (and the tiny gas bubbles that are being created as the powder dissolves are moving to the surface and disappearing). As the bulk modulus increases, the speed of sound goes up and the resonant frequency of sound bouncing around in the cup increases.



You might be able to take advantage of this effect by setting up a liquid-proof transmit/receive device that you can immerse in the liquid. Measure the transit time of sound - it will relate to the nitrogen content. Effectively, the very small bubbles are "very compressible" compared to the liquid; so if we have a small fraction (by volume) $f$ of nitrogen ("air") in the liquid, we can consider the displacement at a given stress to give the effective bulk modulus. We find


$$\frac{1}{K_{eff}}=\frac{f}{K_a}+\frac{(1-f)}{K_w}$$


$$K_{eff} = \frac{K_{w} \times K_{a}}{f\cdot K_{w} + (1-f)\cdot K_{a}}$$


We can rearrange this, assuming that $f\ll 1$ and $K_a\ll K_w$, to


$$K_{eff} = K_w\left(1-\frac{K_w}{K_a}\cdot f\right)$$


Because the bulk modulus of air is so much lower than that of water, a small fraction of air has a large impact on sound propagation. So this is a very sensitive test.


We might want to take account of the change in density (but that is a much smaller effect):


$$\rho_{off} = (1-f)\rho_w$$


Putting this into the equation for the velocity of sound, we get


$$c=\sqrt{\frac{K}{\rho}}\sqrt{\frac{1-\frac{K_w}{K_a}\cdot f}{1-f}}$$



If we can assume that $K_{air}\ll K_{water}$, and that $f\ll 1$, then the reduction in sound speed for a given fraction $f$ will be roughly given by


$$c(f) = c(0) \left(1-\frac12\left(\frac{K_w}{K_a}-1\right)\cdot f\right)$$


So if you measure the drop in speed of sound, you can use this equation to get a good estimate of the volume fraction of bubbles. This assumes that the amplitude of the sound is small enough that the bubbles don't collapse / dissolve, and that you calculate $K_a$ correctly - it needs to be the adiabatic bulk modulus (since sound transmits through air adiabatically).


The ratio $\frac{K_w}{K_w}$ is roughly 15,000 so if you measure the resonant frequency of a cavity filled with your mixture (for example by tapping a spoon against the bottom of a cup filled with your coffee) the frequency shift can be calculated as follows:


$$\frac{\Delta \nu}{\nu}=\frac{\Delta c}{c}$$


Let's look for the change in fraction of gas volume required to cause a shift in the resonant frequency of a cavity filled with a gas/air mixture (assumptions as before) of one half a note (1/12th of an octave):


$$\frac{\Delta \nu}{\nu}=2^{1/12}\approx \frac{\log 2}{12}$$


Combining with the earlier expression for $c$ as a function of $f$ I get


$$f = 2\frac{K_w \log 2}{12 K_a} = 7.7\cdot 10^{-6}$$


So you can expect the tone to change by one half note (1/12th of an octave) for a change in fraction of air of 7.7 ppm (parts per million). That's a pretty sensitive test and it helps explain the "hot chocolate effect".



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