Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."
In her Parade Magazine column of January 30, 2000, Marilyn assumed that the flow of water through a hole in the bottom of a tank does not change as the tank drains.
Wrong, Marilyn. You've done the simple task of applying a constant flow rate to a volume and coming up with the time required.
(5280 ^ 3 cu. ft.) / (0.1337 cu. ft. / sec) ~= 1,100,000,000,000 sec ~= 35,000 years.
But that's not what was asked.
The flow rate from such a tank will not remain constant; it varies as the square root of the height of the remaining water in the tank. So although the flow rate starts out at 1 gallon per second, it steadily decreases as the tank drains.
The velocity of water exiting the tank is given by SQRT(2gh) where g is acceleration and h is the height of the water.
SQRT(2gh) = SQRT((32 ft/sec^2) * 5280 ft) = SQRT (168,960 ft^2/sec^2) = 581 ft/sec.
The initial flow rate of 1 gallon per second (0.1337 cubic feet per second) allows us to calculate that the area of the hole is
(0.1337 ft^3/sec) / (581 ft/sec) = 0.00023 square feet = 0.0331 square inches.
If for simplicity's sake it's a square hole, it's 0.18 inches on a side. Not quite a garden hose as the question implies, although the stream of water is exiting it at half the speed of sound.
A numerical integration of the time required to drain the tank in one-foot increments shows that it takes 6.6 years to lower the water level one foot when the tank's full, 9.3 years when it's half full, 13.2 years when it's a quarter full. The last few feet each take several hundred years, with the flow rate continuing to drop asymptotically, but for the sake of argument let's say that the tank is essentially empty when 99.9% of the water is gone (5 feet left). It takes 67,700 years to get to this point, not 35,000 years.
Glenn Huebscher<firstname.lastname@example.org> sent the following:
Assuming the hole does not get bigger as the water drains out, the volume flow will decrease as the water drains out. This is due to the decreasing pressure from the water column above the hole (an orifice in engineering lingo). The relationship between flow through the orifice and the differential pressure across the orifice is:
Q = sqrt( deltaP / k ), for some constant k.
The pressure exerted at the bottom of a water column is:
P = R H, where R is the density of water and H is the height of the column.
The initial conditions allow k to be calculated:
Qinit = 1 gal/s = 0.13368 ft^3/s (initial volume flow rate) Hinit = 5280 ft (initial water column height) R = 62.4 lbf/ft^3 (density of water) deltaPinit = R Hinit = 62.4 (5280) = 329,472 lbf/ft^2 (initial differential pressure across orifice) (Note that the differential atmospheric pressure from the bottom to top of the cube should be included; however, this accounts for only a ~376 lbf/ft^2, which is negligable and results in about a 4 month increase in the result if included.) k = deltaPinit / Qinit^2 = 329,472 / 0.13368^2 = 18.43680 x 10^6
The time for the cube to drain out is generally the volume, V, divided by the flow rate, Q. However, the flow rate Q is a function of the deltaP across the orifice, which is in turn dependent upon the height, H, of water in the cube. This is solved with integral calculus (I use S for the integration 'squiggle'):
t = S (A / Q) dH
A is the area at any lateral plane in the cube (i.e. 5280' x 5280').
The function Q is listed above. Substituting the pressure relationship for a water column into this, we get (note that the change in differential pressure across the orfice due to changes in atmospheric pressure should be included here by adding a pressure function of H under the integral; ignoring this results in a 4 month error on the low side):
Q = sqrt( R H / k ) t = S (A / sqrt( R H / k )) dH
Rearranging and including the limits from the top to the bottom of the cube we get the definite integral:
5280 t = A sqrt( k / R ) S (sqrt H) dH 0
Integrating over the limits and substituting problem specific values:
t = A sqrt( k / R ) (2 sqrt(5280)) = 5280^2 sqrt( 18.43680 x 10^6 / 62.4 ) (2 sqrt(5280)) = 2.20224 x 10^12 seconds ~ 69,785 years
This result is double the 34,890 number published in the magazine.