So, after a bunch of typhoons have struck the Philippines over the years, typhoon Maiyan being the latest and strongest, I've been hearing and seeing over some news media talks about a secret microwave weapon for generating tropical cyclones used by a rival neighboring country to wreak havoc on the Philippines. Well, it does seem logical. If microwaves can heat food, then it can be used to accelerate the heating of Pacific Ocean waters. Now let us not jump to that conclusion too fast. Before we make such a conclusion, let us first analyze the requirements and conditions for such a mechanism to work.

Microwave, as any electronics engineer would know, is an electromagnetic wave with a frequency ranging from 1 Ghz to around 110 Ghz (L, S, C, X, K, Ku, Ka, V, W bands) that is typically propagated directionally if it were to reach a significant distance (noting that the free space loss = (4*pi*f*D/c)^2 -- the square relationship making its value very high at high frequencies). The radiated power of

a typical directional parabolic antenna would be the input transmit power multiplied by the gain n(pi*D/lambda)^2 minus the free space loss and the fade margin.

Let us assume, under a worst case scenario, that a parabolic antenna with efficiency of 60% (typical value), diameter of 300 meters (extreme case) at 2.45 Ghz fed with 20kW (extreme case) tries to accelerate the heating of a part of the Pacific Ocean.

At the output of the parabolic antenna, the signal power will be .6*(pi*300/(299792458/2.45*10^9))=35.6MW.

Assuming a distance of 1 kilometer (extreme case), the FSL=50.11dB. The signal will be reduced to 347W.

Fade Margin = Multipath effect + Terrain Sensitivity - Reliability Objectives - Constant

FM = 30 log (1) + 10 log (6*4(over water)*0.5(hot humid areas-tropics)*2.45) - - 70

FM = -55 dB

But, since we are expecting the worst, let us forget about the fade margin.

So, 347 W of power is delivered on a pi*300^2 m^2 area of the Pacific ocean.

m=(1000kg/m^3)*(pi*300^2 m^2*0.5 m.) assuming the conduction of heat penetrates up to 0.5 meters deep. Thus, the mass affected is 141,371 kg.

Q=mc*(change in T)

Q=(141,371 kg.)*(4.186 J/(cal*g))*(change in T)

We need to a sample value for Q, say for 5 seconds.

P=Q/t, 347 W = Q/5 sec.

Q= 1735 Joules

Therefore, change in temperature is 2.93*10^(-6) degrees Celsius for a period of 5

seconds.

Wow, the factor 10^-6 is so insignificant its like comparing the length of a meterstick to the diameter of a strand of your hair. And we have done this calculation over an extreme case (not to mention neglecting the fade margin). And

the affected area is only 300 m. of the Pacific Ocean. And we treated the system as an isolated system (which in reality isn't).

Therefore, we conclude based on the immediate calculations that it is nearly impossible to accelerate the heating of a part of the Pacific Ocean. (If you also think about it intuitively, the law of energy and thermodynamics would be violated if this were to be possible.)

Of course, the calculations above were very rough, so the result is just a very rough estimate.

## Comments

## Post a Comment