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Problem 3 (4 pts.). A bathyscaphe (a small free-diving self-propelled deep-sea submersible), is a sphere of radius R- 1.5 m located 873.5 m below the sea level. a) what is the pressure exerted on its surface (here, you can neglect the size of the bathyscaphe compared to the depth of its immersion) and the buoyant force acting on it in the water (b) If a small leak is opened at the top of the bathyscaphe, with what speed will the water ence u The pressure inside the bathyscaphe is p = 1.3 Datum where Patm = 1.01 10° Pa is the atmospheric pressure. (c) We need to make sure that our bathyscaphe will not be completely filled with the water for at least 20 minutes. What air pressure should be created inside the bathyscaphe if the area of the opening is 1.07 cm"? You may assume that the pressure inside the bathyscaphe does not change while it is being filled with water. 2R H = 873.5m R = 1.5m Figure 3: Schematic graph for Problem 3.

Answer 1

(a) Pressure on its surface is equal to the air pressure plus the water pressure. (density of sea-water = 1029 Kg/m³ and acceleration due to gravity = 9.81 m/s²)

$P=P_{0}+H\rho g=1.01\times 10^{5}\ Pa+873.5\times 1029\times 9.81\ Pa=89.19\times 10^{5}\ Pa$

Volume of the bathyscaphe is

$V=\frac{4}{3}\pi R^{3}$

Therefore, the weight of sea-water replaced by the bathyscaphe will be equal to buoyant force (F_b) acting on the bathyscaphe.

$F_{b}=V\rho g=\frac{4}{3}\pi R^{3}\rho g=\frac{4}{3}\pi\times 1.5^{3}\times 1029\times 9.81=1.43\times 10^{5}\ N$

(b) Bernoulli's equation says: The openings is horizontal, so both points are at the same height. Bernoulli's equation can be simplified in this case to

$P+0=p_{b}+\frac{1}{2}\rho v^{2}$

$\therefore v=\sqrt{\frac{2\left ( P-p_{b} \right )}{\rho }}=\sqrt{\frac{2\left (89.19\times 10^{5} -1.3\times 1.01\times 10^{5} \right )}{1029 }}=130.69\ m/s$

(c) Volume of water enters in time of 20 min, V = (130.69 x 1.07 x 10^-4) x 20 x 60 m³ = 16.78 m³

The volume of the sphere is V_b = (4/3)$\inline \pi$R³ = (4/3)$\inline \pi$ x 1.5³ = 14.14 m³

Hence, the velocity should be less than the velocity what we have calculated above. Let us consider the velocity will be u so that in 20 min the water volume enter inside will be 14.14 m³. Therefore, the velocity will be

(u x 1.07 x 10^-4) x 20 x 60 m³ = 14.14 m³

u = 110.13 m/s

Now, from Bernoulli equation,

$P+0=p_{b}+\frac{1}{2}\rho v^{2}$

$\therefore p_{b}=P-\frac{1}{2}\rho v^{2}=89.19\times 10^{5}-\frac{1}{2}\times 1029\times 110.13^{2}=26.79\times 10^5 \ Pa$

$\therefore p_{b}=\frac{26.79\times 10^5 \ Pa}{1.01\times 10^5 \ Pa}=26.52p_{atm}$