# Analyzing the Eco-Cooler, part 1

OK, that was ample time for you to have hit your fluid mech/heat transfer/thermo books, and to have it verified whether the 5 deg. C drop was believable or not… You must have made your notes, too, no?…  So, in this post, let’s cross-check our notes.

On my part, I will first present the simplest (and the most approximate) model, and also give you a simple Python script to play with, to see what kind of predictions this model makes. Then, we will go on considering more and more complicated but still approximate “engineering” models that hopefully become more and more realistic. We will cross-check their predictions too. We may eventually find that a full-fledged CFD analysis is called for. However, I will save that—I mean doing a full-fledged CFD analysis—for another day. (I in fact plan to write a paper on this problem, using CFD. (…Some day…).)

The reason we follow this method—from the simplest and crudest models to the more complicated and better ones—is because for problems related to fluid mechanics, it is this method which works best!

I mean to say, the full-fledged Navier-Stokes equations are too complicated to solve for, even when they are applied to the simplest of practically encountered geometries—e.g., the flow of air through the Eco-Cooler bottle. Since the NS equations cannot be solved exactly, the traditional engineering models (which are based on analytical or semi-analytical solutions) fall short, and then, a CFD analysis is called for.

But the fact of the matter also is, even CFD itself is only an approximate technique. CFD solutions sometimes even happen to carry more numerical artifacts than real physics. We therefore cannot approach CFD blindly. We ourselves have to have some good idea in the first place of what the desired solution should look like. We should have this idea right before we even think of setting up a CFD model/simulation. The traditional engineering models provide precisely these insights.

Yes, that’s right.

The traditional engineering models actually are more approximate than CFD. Yet, since they are also simpler than CFD, and since they explicitly carry conceptual connections with the major fluid mechanical phenomena in a more direct manner, they also make it easier to gain insights about both the nature of the problem, and the nature of the expected solution. No similar insights can be had by directly using CFD, for several reasons. The CFD theory itself is too complicated, and the CFD practice involves too many different analysis options. In the jungle of all those parameters, iterations, and convergence requirements, CFD happens to loose the directness of the conceptual connections with the basic analytical theory—with the fluid dynamical phenomena.

That’s why we first deal with the simplest engineering models, even though they are known to be approximate—and therefore, they are easily capable of giving us wrong results. But this way, we can build insights. Building insights is an art, and the process progresses slowly.

As I said, we will follow an iterative scheme of model building. In each phase or iteration of the model building activity, we will actually be applying the same set of principles: the conservation of mass, momentum, and energy. However, in going from a model-building phase to the next, we will aim to incorporate an increasing level of complexity or sophistication—and accuracy, hopefully.

Actually, in the fluid dynamics theory, all the three conservation equations come coupled to each other. You cannot solve for conservation of only the mass, or the momentum, or the energy, by neglecting any of the other two. However, for this particular problem (of the Eco-Cooler), for various reasons (which you will come to appreciate slowly), it so turns out that we can get away considering the mass, momentum and energy equations in a decoupled manner, and in the order stated: first mass, then momentum, and then energy. (It’s no accident that text-books spell out these three principles in this order. Many fluids-related phenomena with which we are well familiar through our direct experience of the world are such that for solving problems involving these phenomena, this order happens to be the best one to follow.)

So, with that big introduction, let’s now get going calculating—even though we will not shut up even while performing those calculations.

Model-Building Phase I: The Simplest Possible Model:

Geometry:

Consider the plastic-bottles used in the Eco-Cooler; they all lie horizontally. Consider one of these bottles. A tube has been obtained by cutting off the base of the bottle. Let the base-plane be identified by the subscript 1 and the neck-plane by 2. See the figure below:

Air flows from the base-plane (1) to the neck-plane (2).

Conservation of mass:

The simplest possible expression for mass conservation, applied to the bottle geometry, would be the continuity equation, given below:
$A_1 U_1 = A_2 U_2$
where $A_1$ and $A_2$ are cross-sectional areas of the bottle, and $U_1$ and $U_2$ are wind velocities, at the base and the neck, respectively. Rearranging for $U_2$, we get:
$U_2 = \dfrac{A_1}{A_2} U_1$             …(1)

We can use this simple an equation for mass conservation only if the flow is incompressible. To determine if our flow is incompressible or not, we have to calculate the Mach number for the flow. To do that, we have to first know the expected wind velocities.

Referring to the Wiki article on the Beaufort scale [^], we may make an assumption that the inlet speed can go up to about 60 kmph. Actually, the wind-speeds covered by the Beaufort scale go much higher (in excess of 118 kmph). However, practically speaking, the only times such high wind-speeds (gales etc.) occur in India is when rains also accompany them. The rains bring down the ambient temperature anyway, thereby obviating the need for any form of a cooler. Thus, we have to consider only the lower range of speeds.

Assuming the speed of sound in air to be about 340 m/s, we find that the Mach number (for the range of the winds we consider) goes up to about 0.65. Now, for $\text{Ma} < 0.33$, the flow is sub-sonic, and can be regarded as incompressible. For $0.33 < \text{Ma} < 1.0$, the flow is trans-sonic, meaning, the changes in pressures do not adjust “instantaneously” everywhere in the flow, and so, it is increasingly not possible to even idealize the flow as incompressible.

Therefore, in the interest of simplicity, for our first solution cut, we choose to consider only the wind-speeds up to 100 m/s, i.e. 28 kmph, so that the incompressibility assumption can be justified. Making this assumption about the highest possible wind-speed, we are then free to use the simplest form of mass conservation equation given above as Eq. (1).

For a typical one liter bottle, the base diameter is 7.5 cm, and the neck diameter is 2 cm (both referring to IDs i.e. inner diameters). (I measured them myself!) So, the area ratio $\frac{A_1}{A_2}$ turns out to be about 14.

Thus, the wind accelerates inside the bottle; the outlet velocity is about 14 times the inlet velocity!

This looks like a remarkable bit of acceleration to happen over just some 20 cm of length. (The bottle is cut somewhere in the middle.) More on it, later.

Conservation of momentum:

The simplest possible equation to use for momentum conservation is the steady-state Bernoulli’s equation:
$\dfrac{P_1}{\rho} + \dfrac{U_1^2}{2} = \dfrac{P_2}{\rho} + \dfrac{U_2^2}{2}$        …(2)
where we have ignored the potential head term ($gz$) because the tube is horizontal as well as symmetrical about its central horizontal axis (and because the air is so thin that its weight can be easily neglected here).

Carefully note the assumptions behind Eq (2). It holds only for a steady-state and laminar flow, and only after neglecting viscosity.

Since we are in a hurry, we will assume them all, and proceed!

Rearranging Eq (2) for $P_2$, we obtain:
$P_2 = P_1 + \dfrac{\rho}{2} \left( U_1^2 - U_2^2 \right)$

Conservation of energy:

We basically need the equation of energy conservation only in order to calculate the temperature at the neck ($T_2$), from a knowledge of: (i) the temperature at the base, $T_1$, and (ii) the pressures $P_1$ and $P_2$.

In the last line, I said “from.” This usage implies that there already is an assumption I made, viz., that the energy equation can be decoupled from the momentum equation. How reasonable is this assumption? It seems pretty OK. Think: can air flowing through a 7.5 cm or a 2.0 cm diameter tube at under 100 m/s get heated up to a significant fraction of 5 deg. C, over a length of just a foot or less? Not likely. Can heating up the neck region cause the air flow to come a halt, say because it helps build up a sufficient amount of pressure? Not even remotely likely. So, we may get away by decoupling the two.

The simplest equation to compute $T_2$ from the other three quantities would be: the ideal gas law, given as:
$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$.

We have already found that the flow can be considered incompressible (at least up to 28 kmph of wind-speeds). Hence, the volume of each fluid part remains constant, i.e., $V_1 = V_2$. (Note, in this equation, we have to consider the volume of a fluid element or a part, not the volume for a unit axial length at a given cross-section of the bottle.) For constant volume, the ideal gas law reduces to:
$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$,
from which we can conclude:
$T_2 = P_2 \dfrac{T_1}{P_1}$.

But we have to know whether the assumption of the ideal gas itself can be used in our problem—for the real air—or not. For doing that, we have to know the critical pressure and temperature of air. Cengel (“Thermodynamics: An Engineering Approach,” 8th SI units edition) lists them (Appendix A1) as 132.5 K, and 3.77 MPa. Using these values, the reduced temperature and the reduced pressure of air turn out to be $T_{1_R} = (30 + 273.15)/132.5 \approx 2.29$, and $P_{1_R} = (101.32\times10^3)/(3.77\times 10^6) \approx 0.027$. Further, the expected drops in the pressure and temperature would be just small fractions of the inlet values. Hence, the reduced quantities for the outlet also would not differ significantly from those for the inlet. Referring to the chart and the remarks on p. 139 in Cengel, it seems like we can get away using the ideal gas law.

Note, we are using the ideal gas law not as an approximation to the energy equation, but in place of it, simply because looking at the variables, we noticed that $T_2$ had to be determined from the other three variables, and this set of variables reminded us of the ideal gas law! Then, referring to the reduced pressure and temperature, the ideal gas approximation seemed to be OK. In short, we have not directly considered the energy conservation principle at all. We may subsequently have an occasion to revisit this issue.

Assumed data:

Now, let us make assumptions about the data to be used for our calculations.

Suppose that the ambient temperature is 30 deg C, and the Eco-Cooler is kept at the mean sea level (MSL), say by the sea-side (rather than somewhere on a slope going down into the Death Valley [^]). Now, seated at a sea-side, the evaporative cooler isn’t going to be feasible because of humidity, and that’s the reason why we are at all considering using the Eco-Cooler. Alright. So to wrap up this point, we have to use data values at the MSL.

Suppose that we can use the ambient MSL atmospheric pressure for the inlet of the bottle; it would then mean that $P_1 = 101330$ Pa. (Note, this is an assumption; like with many other assumptions, we may have occasion to revisit it later on.) The air density at MSL and at 30 deg. C may be taken to be $\rho = 1.169$ kg/m^3. (Minor changes to this value turn out to have minimal impact on the predictions, so it’s OK to use, even if we might have made a mistake in looking it up hurriedly.)

For wind-speeds, let’s assume that the speed at the inlet (i.e., at the base of the bottle, which is exposed to the outside) is the same as the ambient wind-speed. (Again, this is an assumption; we may have the occasion to revisit it later on.)

Since the wind-speeds vary, and since the pressure drop (and hence the temperature drop) obviously depends on the inlet wind-speed, we will have to repeat our calculations for each wind-speed, again and again.

Referring again to the Wiki article for the Beaufort scale [^], to have representative wind-speed values, we choose to take the averages of the lower and upper wind speeds which together define the range for each wind-grade on the Beaufort scale. Thus, our $U_1$ could be one of: 0.5, 3.0, 8.5, 15.5, 24.0, 33.5, 44.0, all in kmph.

We are now ready to do our calculations. To recap: First, we calculate $U_2$ from $U_1$ using the continuity (mass conservation) equation and the known area ratio (which is about 14). Then we substitute the data in the re-arranged Bernoulli’s equation (which brings in its own assumptions) and obtain $P_2$, the pressure at the outlet (i.e. the neck of the bottle). From the ideal gas law (being used in place of the energy equation proper), we then calculate $T_2$.

Python script:

We use a Python script only because the calculations for $T_2$ have to be repeated again and again for different wind-speeds. Anyway, here is the Python script:

'''
This Python script calculates the expected
outlet temperature for a bottle in the Eco-Cooler.
See the relevant blog posts by Ajit R. Jadhav.
All units are in SI.
'''

d1 = 7.5 # ID at the inlet (i.e. at the base), in cm
d2 = 2.0 # ID at the outlet (i.e. at the neck), in cm
AR = (d1*d1)/(d2*d2) # Area ratio for the bottle

T1 = 30.0 # Ambient temp., in deg. C
T1 = T1 + 273.15 # Conversion from deg. C to K
rho = 1.169 # Density of air, in kg/m^3
P1 = 101330 # Ambient pressure, in Pa

# The Beaufort Scale wind speeds in kmph, and their text descriptions
bsa = [0.5, 3.0, 8.5, 15.5, 24.0, 33.5, 44.0]
bsta = ["Calm","Light air", "Light breeze", "Gentle breeze", "Moderate breeze", "Fresh breeze", "Strong breeze"]

# Calculate the expected temperatures for various wind-speeds
for i in range(7):
s1 = bsa[i] # wind-speed, in kmph
sText = bsta[i] # wind-speed description

U1 = s1 / 3.6 # conversion from kmph to m/s
U2 = U1*AR

# Calculate P2 using Bernoulli's equation
P2 = P1 + rho*(U1*U1 - U2*U2)/2.0

# Calculate T2 using the ideal gas law
T2 = T1*P2/P1
TC = T2 - 273.15 # conversion from K to deg. C
print( "%20s %6.1f %6.2lf %10.0f %6.1f" % (sText, s1, U1, P2, TC) )



The program written using Python is so simple, that you can very easily modify it, say to report any additional calculations that were made along the way.

Output data:

Here is the output data I get. The columns are: wind description in words, wind speed in kmph, wind speed in m/s, outlet pressure, and outlet temperature:

    Wind Description        U_1        P2, Pa   T2, deg. C
kmph  m/s
Calm    0.5   0.14     101328   30.0
Light air    3.0   0.83     101250   29.8
Light breeze    8.5   2.36     100689   28.1
Gentle breeze   15.5   4.31      99198   23.6
Moderate breeze   24.0   6.67      96219   14.7
Fresh breeze   33.5   9.31      91372    0.2
Strong breeze   44.0  12.22      84151  -21.4



Interpretation of results:

Our model predicts that when the breeze is strong (but less strong than when the tube is held near the window of a car that is traveling on an expressway), we should get ice formation at the neck of the bottle—in fact, it should be a super-cooled ice!

From your knowledge of what happens on the Indian sea-side, do you expect a mere half-foot tube of variable diameter, to begin to have ice-formation at its neck?

Obviously, the model we dreamt up has gone wrong somewhere. …

… But precisely where? … We have made so many assumptions…. Which of these assumptions are likely to have impacted our analysis the most? In what all places should we bring in the corrections to our model?

I have already given a lot of explicit notes—not just hints—while writing down the analysis above. So, go through the entire post once again, now pausing especially at the assumptions, and think how we may go on to improve our model.

I will return to the business of improving our model, in my next post.

And as always, sure drop me a line if you think I am going wrong somewhere.

Alright, enjoy!

Ummm… Since I have given you Python scripts to play with, guess the usual section on a song I like has become redundant. So, let me mention the “other” song (re. my last post) when I finish this series of posts—which will take one, or at the most two more posts).

[I may come back and cross-check the latex entries in this post, grammar, etc., later on today itself, when I may perhaps edit this post just a bit, but not much. Done on 2016.10.02 itself.]

[E&OE]