Question 2.SP.3: At a depth of 8 km in the ocean the pressure is 81.8 MPa. As......

At a depth of 8 \mathrm{~km} in the ocean the pressure is 81.8 MPa. Assume that the specific weight of seawater at the surface is 10.05 \mathrm{kN} / \mathrm{m}^{3} and that the average volume modulus is 2.34 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} for that pressure range. (a) What will be the change in specific volume between that at the (c) What will be the specific weight at that depth?

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

(a) Eq. (2.2): \quad v_{1}=1 / p_{1}=g / \gamma_{1}=9.81 / 10050=0.000976 \mathrm{~m}^{3} / \mathrm{kg}

Eq. (2.3 a): \quad \Delta v=-0.000976\left(81.8 \times 10^{6}-0\right) /\left(2.34 \times 10^{9}\right)

=-34.1 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{kg} \quad

(b) Eq. (2.3b): \quad v_{2}=v_{1}+\Delta v=0.000942 \mathrm{~m}^{3} / \mathrm{kg} \quad

(c) \gamma_{2}=g / v_{2}=9.81 / 0.000942=10410 \mathrm{~N} / \mathrm{m}^{3} \quad

2.3

Related Answered Questions

Question: 2.SP.10

Verified Answer:

Table A.1 at 10^{\circ} \mathrm{C}: \quad \...
Question: 2.SP.11

Verified Answer:

From Appendix A, Table A.3, the pressure of the st...