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At a depth of 7 km in the ocean, the pressure is 71.6 MPa. Assume the specific weight of the sea water at the surface is 10.05 kN/m3 and its average Bulk Modulus is 2.43 GPa. Determine:

i. The change in specific volume between the surface and depth of 7 km
ii. The specific volume at 7 km depth
iii. The specific weight of sea water at 7 km depth

1 Answer

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Final answer:

At a depth of 7 km in the ocean, the change in specific volume is -0.307 m^3/kN, the specific volume is 9.743 kN/m^3, and the specific weight is 97.43 kN/m^3.

Step-by-step explanation:

To determine the change in specific volume, we can use the formula:

Δv = -v * ΔP / K,

where Δv is the change in specific volume, v is the specific volume at the surface, ΔP is the change in pressure, and K is the bulk modulus. In this case, the depth is 7 km and the pressure at that depth is 71.6 MPa.

With the given values, we can calculate the change in specific volume:

Δv = -10.05 kN/m^3 * (71.6 MPa - 0) / (2.43 GPa) = -0.307 m^3/kN.

To find the specific volume at 7 km depth, we can use the formula:

v = v0 + Δv,

where v0 is the specific volume at the surface. Using the given values, we get:

v = 10.05 kN/m^3 + (-0.307 m^3/kN) = 9.743 kN/m^3.

The specific weight at 7 km depth can be calculated using the formula:

w = γ * v,

where w is the specific weight, γ is the specific weight at the surface, and v is the specific volume at the given depth. Substituting the values, we get:

w = 10.05 kN/m^3 * 9.743 kN/m^3 = 97.43 kN/m^3.