Par t One Consider 1 unit of incoming solar radiation that reaches the top of th

Question

Part One

Consider 1 unit of incoming solar radiation that reaches the top of the atmosphere. Calculate how much reaches into the ocean (just below the surface). Assume

1.Cloudy skies

2.Albedo of 5%

Part Two

Consider 1 unit of incoming solar radiation that reaches the top of the atmosphere. Calculate how much reaches into the ocean at depths of 2 m, 10 m and 100 m assuming:

1.Cloudy skies

2.An albedo of 5%

A coefficient of extinction (K) of 0.05 m-1

Part Three:

If there is 40 W m-2 of light at 10 m depth and directly below, 10 W m-2 of light at 50 m depth, calculate the coefficient of light extinction.

Explanation / Answer

Answer:1

Light intensity at depth below ocean surface can be calculated as follows;

K= (ln Io- ln ID) * D-1   --------------- equation 1

Where, K= Coefficient of extinction, Io= Light intensity at surface, ID= Light intensity at depth, D= Depth

Since Albedo is 5% means only 0.95 unit is strike at surface of ocean.

K for pure water is 0.035 and depth is about 1m

Put the values in equation 1, we get

0.035= (ln 0.95- ln ID)* 1-1

ln ID-1= -0.05- 0.035= -0.085

ID= 2.46 units

Answer:2

K= 0.05 m-1, Io= 0.95, D= 2m

Put the given values in equation 1, we get

0.05= ln 0.95- - ln ID)* 2-1

ln ID= -0.05-0.05= 0.1

ID= 2.30 units

For depth = 10m

0.05= ln 0.95- - ln ID)* 10-1

0.05= ln 0.95- - ln ID)* 9

0.05= (0.05- ln ID)* 9

0.005= 0.05- ln ID

ID= 3.10 units

For depth 100m

0.05= ln 0.95- - ln ID)* 100-1

5.05*10-4= 0.05- ln ID

ln ID= 0.049

ID= 3.01 units

Answer:3

Depth = 50-10= 40 m , Io= 40 W m-2, ID= 10 W m-2

Put these values in equation 1, we get

K= (ln Io- ln ID)* D-1

K= (3.68- 2.30)*39

K= 53.82

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