A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

Solution 41E The efflux speed of the water from the bottom of the tank is given by, P 0P atm v = 2( ) + 2gh…..(1) Where, (P 0 P) is the gauge pressure, is seawater density, h is the height of water in the tank. 5 Given, gauge pressure P 0 P = 3.00 atm = 3.00 × 1.01 × 10 Pa Height h = 11.0 m Seawater density = 1025 kg/m 3 Acceleration due to gravity g = 9.80 m/s2 Substituting these value in equation (1), we get v = 2 × 3.00×1.01×10+ 2 × 9.80 × 11.0 1025 v = 805 m/s v = 28.4 m/s The water is moving at a speed of 28.4 m/s.