Recall the opening moments of the movie Water World, with Keven Costner?

http://www.imdb.com/title/tt0114898/

At the beginning is shown a graphic of the Earth with continents and the poles while some guy whines about mankind unleashing a great flood by melting the poles and flooding the lands. Hold on... wasn't that God's fault? He (She... it... what... existence proof not needed) already did that once already, right?

I digress...

Anyhow, the graphic shows that all the land disappears under water and humanity is forced to live on the world spanning ocean, tattooing maps on kids backs and drinking their own urine rather than filtering the sea water.

Now, in the Bible, God flooded the Earth cause he was pissed or something, and gave warning to Noah to build an ark and save his critters. (I guess God can flood the Earth, but can't be bothered to save the critters herself)

It rained for 40 days and 40 nights and covered the highest mountain to a depth of 2 fathoms. Wow, this sounds a bit like the depth of the supposed water world.

Fermi problem time!

---- Start Fermi Problem ----

Estimate the change in the level of the oceans if the poles were to melt. (Watch out here, there is a standard trick question involved... think of melting ice cubes)

Estimate how much water is needed to cover the Earth such that the highest mountain was under water.

Compare this volume of water to that of the current volume of the seas.

If it rained for 40 days and 40 nights, how much water per square meter per second would have to rain down?

---- End Fermi Problem ----

God, of course, can do anything... including violating common sense.

What is Hollywood's excuse?

Bismuth and Solar Cells

11 hours ago

well, the highest mountain (Mt everest) is about 8 km high, and the radius of the Earth is about 6400 km. Hence the volume of water needed to cover the highest Mtn would be

ReplyDelete4/3 * Pi* ((6400+8)^3 - 6400^3 }

about 3/4 th of Earth is covered with water is covered with seas. Seas and Oceans vary greatly in depth. lets confine ourselves to oceans (which ,lets say, occupy 2/3rd of earths surface) , and they are, on the average, probably a few kms deep (lets call it 8km which is an overestimatiob but might make up for not counting the seas ). Hence the volume of ocean waters would be

2/3 * 4/3 * Pi * {(6400^3 -(6400- 8)^3}

comparing this with the previous expression, we say that (without having to calculate the nos) its smaller compared to the total volume needed to cover the mtns ...

Now if it rains for 40 days and 40 nights, we would have to divide this no by (40*24*3600) an we would get how much water per square m per second it would have to rain. again, without even crunching the nos we can see that its ridiculously high compared to standard rainfall

Hi Samik,

ReplyDeleteWhere are your numbers? =)

If you divide your water volume by 40 days and 40 nights, you get volume per unit second, not water per square meter per second.

First you need to determine the surface area of the Earth, then you can use that to find out the rainfall per unit area per unit time.

To take it to the next level, note that if you sat in a boat during all this, the surface area would increase over those 40 days and nights, and thus the rain per square meter per second would decrease over those 40 days and nights. =)

But of course this is solving a differential equation.

Anyhow, do the calculations for rain per square meter per second. 8)

"Estimate the change in the level of the oceans if the poles were to melt."

ReplyDeleteMy quick calc:

Volume of the Earth:

V=4/3*PI*(R+h)^3

where R is Earth's average radius

h is an average depth of the oceans

The differential dV by h yields:

dV=4*PI*(R+h)^2*dh

Because h << R the formula simplifies:

dV=4*PI*R^2*dh

So dV represents total wold ice (grounding ice) to be melted and dh corresponding change in the level of the oceans.

Numerical answer:

dV is estimated as 29,960,000 km^3

(floating ice is negligible)

R=6371 km

dh=dV/(4*PI*R^2)=0.058 km or about 60 m.

P.S.

A related Fermi problem:

Consider an iceberg with average diameter of 1 km, freely floating in an ocean. Estimate how much time it will take the iceberg to melt completely.

The temperature of the surrounding water is t=4C

@Martin

ReplyDeleteInteresting problem!

I will think about it in more detail later, but quickly:

I will assume the snowball as a sphere of diameter z, who's diameter rate of change with respect to time is proportional its surface area:

dz/dt = k(4pi r^2)=k pi z^2

(since z = 2r)

Then... umm... I would take a spherical chunk of ice to a pond at 4C, and let it melt for an hour and measure the diameter change wrt time.

At this point I have an empirical time constant to work backwards to one km in diameter, and my answer to order of magnitude. =)

An a-priori way of determine this time constant is what I will think about when I have more time!

;)

Hi!

ReplyDelete@Michael

A remark to your approach:

A solution to the differential equation

dz/dt = -k(4pi r^2)=-k pi z^2

(z = 2r)

is t=const*(1/z-1/zo);z<=zo

where zo is an initial diameter of the iceberg

The only difficulty of this model is that

the limit of t as z approaches zero equals infinity.

Ok, my considerations:

The iceberg is exposed to a moving water having a temperature different from that of

the iceberg and energy is convected to the body by the water.

So it is reasonable to assume that the heat transfer is given by the so called Newton's law of cooling(heating):

W=k*A*(Ti-Tw)

where k is the convective heat transfer coefficient with units W/(m^2*K):K-Kelvin

A is body surface area (m^2)

Ti is the surface temperature of the ice

Tw is the upstream temperature of the water

The heat balance equation at the ice surface can be expressed as follows:

pi*z^2*k*(Ti-Tw)*dt=1/2*pi*z^2*d*L*dz =>

=> k*(Ti-Tw)*dt=1/2*d*L*dz

or

dz/dt=2*k*(Ti-Tw)/(d*L)

where d is the density of ice

L is the latent heat of melting of ice

Now for estimation purposes we assume that

dz/dt=const

After this assumption the integration of the equation is simple and the final result yields:

t=d*L*zo/[2*k*(Tw-Ti)]

Numerical answer:

Here the hardest part is to choose a reasonable value of the convective heat transfer coefficient k because it varies considerably depending on the on the type of media, gas or liquid, the flow properties such as velocity, viscosity and other flow and temperature dependent properties.

In general, k for water is within the ranges:

500 - 10,000 [W/(m^2*K)]

Lets choose k=500 W/(m^2*K)

This could provide the upper limit of melting time.

Latent heat of melting L= 334 kJ/kg

Density of ice d=920kg/m^3

Tw-Ti=4C (or 4K)

Answer:

t=d*L*zo/[2*k*(Tw-Ti)]=

=920*334*10^3*10^3/(2*500*4)=76820000 s=

=2.4 years

(By the way, i really enjoy your blog's content.)

Very cool! I sort of spaced out thinking about this problem more.

ReplyDeleteYes, the issue with my solution is the singularity issue of the differential. =)

And thank you on the blog!

Mike