Question

please solve and wire clearly

Problem 3 (4 pts.). A bathyscaphe a small free-diving self-propelled deep-sea submersible), is a sphere of radius 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 2R (0) H = 873.5m R=1.5m Figure 3: Schematic graph for Problem 3.

Answer 1

a) The pressure is giving by: P = PG + PgH = 1.01 x 10 + 1000 * 9.8 * 873.5 P = 8.66 X 10Pa The buoyant force: B = Pgv = pg (TRF) = (1000)(93((1.5)) B = 1.385 x 105N b) Consider a point in the surface and a point in the hole, using the Bernoulli equation: m c) P+ po? + PgH = P2 + pin + ng(0) At the top level V2 = 0 Then: PA + PgH = P2 + 5 pguž Using the giving data: 1.01 x 105 + 1000 * 9.8 + 873.5 = 1.3 * 1.01 x 105 +* 1000 * vž Solving for v2 v2 = 130.6 The flow rate is: v R3 41(1.5)3_001178 m Q = = = = =3*1200 = 0.01178 Then, the speed is: 0.01178 “A 1.07 x 10-4 = 110- Then, using the equation of the previous part: Pa + PgH = P2 +5pguil Using the giving data: 1.01 x 105 + 1000 * 9.8 * 873.5 = P, +=* 1000 * 1102 Solving we get: P2 = 2.61 X 10Pa