At 41.5 degrees north and 81.7 degrees west (lat-lon of Cleveland, Ohio), the te

Question

At 41.5 degrees north and 81.7 degrees west (lat-lon of Cleveland, Ohio), the temperature in Fahrenheit of the earth surface can be approximated as a function of time, t, and depth, y, with the following expression,

      T = 50 F + (40 F)[sin{Wt – (W/2a)1/2 y}][exp{- (W/2a)1/2 y}] ,

Where W = 2.0 x 10-7 radians/s and a = 10-7 m2/s,

At what depth will the temperature fluctuate no more than +/- 10 F?

At what depth will the temperature of the earth be warmest when the surface temperature is coldest and what is the value of that temperature?

Explanation / Answer

a)

T = 50 F + (40 F)[sin{Wt – (W/2a)1/2 y}][exp{- (W/2a)1/2 y}]

So, taking the derivative of T with respect to t,

dT/dt = 40*[cos({Wt – (W/2a)1/2 y})*W][exp{- (W/2a)1/2 y}]

So, dT/dt is a sinusoid with amplitude : 40*W*[exp{- (W/2a)1/2 y}]

So, for temperature to fluctuate no more than +- 10 F,

40*W*[exp{- (W/2a)^1/2 y}] = 10

So, 40*2*10^-7*exp(-y) = 10

So, y = -14.04 m <------ answer

b)

For Surface temperature , y =0

So, T = 50 + 40*sin(Wt)

So, for surface temperature to be lowest , sinWt = -1 . Thus Wt = 3*pi/2

So, T = 50 - 40 = 10 F,

At this time, temperature at a particular depth,

T = 50 F + 40*sin(3*pi/2 - y)*exp(-y)

Now, for warmest, dT/dy = 0 and dT^2/dy^2 < 0

So, 50 +40*[ cos(3*pi/2 - y)*(-1)*exp(-y) - exp(-y)*sin(3*pi/2 - y) ] = 0

So, y = -1.088 m, -3.909 m , 7.069 m

Now, for y = -1.088 m, T = 50 + 40*sin(3*pi/2 - (-1.088))*exp(1.088) = -5.12 F

for y = -3.909 m, T = 50 + 40*sin(3*pi/2 - (-3.909))*exp(3.909) = 1485.1 F

for y = -7.069 , T =  50 + 40*sin(3*pi/2 - (-7.069))*exp(7.069) = -33169 F

So, at depth, y = 3.909 m, its temperature is hghest and its temperature is 1485 F

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