82-128. A boat having a mass of 80 Mg rests on the botto of the lake and displac

Question

82-128. A boat having a mass of 80 Mg rests on the botto of the lake and displaces 10.25 m* of water. Since the lifin capacity of the crane is only 600 kN, two balloons are atta to the sides of the boat and filled with air. Determine ing the air 3 smallest radius r of each spherical balloon that is needed lift the boat. What is the mass of air in each balloon if the and water temperature is 12°C? The balloons are at an average depth of 20 m. Neglect the mass of the air and the balloon. The volume of a sphere is3 600 kN Proh 2-128

Explanation / Answer

Weight of the boat = 80 * 106 * 10 = 800 KN (assuming g = 10m/s2)

Buoyant force acting on boat = unit weight of water * Volume of water displaced by boat = 10 KN/m3 * 10.25 = 102.5KN

Capacity of crane = 600KN

Therefore, force to be taken by the balloons = 800-600-102.5 = 99.75KN

Force taken by each balloon = 49.875KN

Therefore, ignoring mass of balloon and air, Buoyant force on each balloon = 49.875KN

49.875 = 10 KN/m3 * Vballoon

Vballoon = 4.9875 m3 = (4 * pi * r3)/3

Therefore radius of each balloon = 1.0598m

Gauge pressure at 20m water depth = 10 KN/m3 * 20 = 200 KN/m2 or 2 * 105 Pa

From the equation: PV = mRT

where T = 285K, R = 287.058 for air

m = 12.1926 kg

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