Solving the pressure exerted by a diver in a given depth of water?

determine the pressure exerted of a diver at 30 m below the free surface of the sea. assume a barometric pressure of 101 kPa and a specific gravity of 1.03 fo sea water,this is my assignment and according to the book the answer is 404 kPa,plz help me with the solution

Answer:
10m fresh water = 1atm = 101kPa x 1.03(SG) = 104kPa
30m = 104kPa x 3 = 312kPa + 101kPa = 413kPa Absolute.

state 1, Pressure,p =101kPa, depth is zero, surface
state 2 , Pressure,p2 depth z2 = 30m
gg= 9800 N/m^3, sg =1.03
p2 - p1 = gg(z2 - z1) (1)
p2 = (gg*sg*d2) + p1
p2 = (9800*1.03*30) + 101
p2 = 404 kPa
equation 1 int dp/dz = gg

pressure at 30m below sea level is,

p = water head + atmospheric pressure.

water head = height x density of liquid
wh = 30 x 1.03 = 30.9 mH2O
convert, mH2O into kPa

wh = 30.9 mH2O x (100 kPa / 10.19 mH2O )
wh = 303.23 kPa

atmospheric pressure = 101 kPa

Therefore, p = 303.23 + 101 = 404.23 kPa
This is the pressure felt by the diver at 30m depth in sea water.

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Solving the pressure exerted by a diver in a given depth of water?

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