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EPS 106A Problem Set 3

# EPS 106A Problem Set 3 - Nina Fitch SID#17849531 Z 50 75...

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Nina Fitch SID#17849531 1) A) Vaisala frequency: N 2 = (g / ρ )(d ρ / dz) ρ = sigma T + 1000 d ρ /dz = d(sigma t)/dz d ρ /dz max = d(sigma t)/dz max, where the slope of sigma t is the least negative sigma t is the least negative between the values of 125 meters and 175 meters where the slope -45 from the graph ρ = sigma t + 1000 27 Z dp/dz 50 -720 75 -720 100 -720 125 -90 150 -45 175 -90 200 -880 225 -880 250 -880 275 -880 300 -880 325 -880 350 -880 375 -880 400 -880 425 -880 450 -880 475 -880 500 -880 525 -880 550 -880 575 -880 600 -880

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N 2 = (g / ρ )(d ρ / dz) = (-9.81m/s / 27) x -45 4 B) The buoyancy period, T N = 2 π /N= 2 π /4 = 1.6 2) I 0 = 9 N 0 = 20 A= I 0 /3 = 3 B= N 0 /10 = 2 P 0 = 3 Z I(z) N(z) Pg 0 9 0 0 1 8.143536762 0.65567799 0.54128558 2 7.368576778 1.289860299 0.835893483 3 6.667363986 1.903251639 1.008874188 4 6.032880414 2.496533619 1.112446886 5 5.458775937 3.070365502 1.172356816 6 4.939304725 3.625384938 1.202836183 7 4.469267734 4.162208673 1.212458964 8 4.043960677 4.681433233 1.206758751 9 3.659126938 5.183635586 1.189519611 10 3.310914971 5.669373789 1.163461831 11 2.995839753 6.139187598 1.130627808 12 2.710747907 6.593599079 1.092609179 …0 …20 …0 The depth where Pg is maximum is 8 (1.21)

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b) as a decreases the depth where Pg is maximum is shallower, and when a increases the maximum depth is deeper this is because I is dependant on a, and when a is smaller the light intensity is greatest at a shallower depth so the phytoplankton growth (Pg) is greastest at a shallower depth a Max depth 5 4 10 7 20 11

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EPS 106A Problem Set 3 - Nina Fitch SID#17849531 Z 50 75...

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