Calculate the absolute pressure at an ocean depth of 1.0 x 103 m. Assume that the density of the water is 1.025 x 103 kg/m3 and that P0 = 1.01 x 105 Pa. Given: h = 1.0 x 103 m = 1.025 x 103 kg/m3 ρ Patm or Po = 1.01 x 105 Pa Unknown: Pabsolute Solution: Pabsolute = Patm + Pgauge Pabsolute = Patm + = 1.01 x 105 Pa + (1.025 x 103 kg/m3) (9.81 m/s2) (1.0 x 103 m) ρgh Pabsolute = 1.01 x 107 Pa A tank has a spigot near its bottom. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the spigot when the water level is 0.5 m above the spigot. Given: P1 = Patm = 1.01 x 105 Pa = P2 (both are open to atmosphere) v1 = 0 (negligible) h1 = 0.5 m h2 = 0 m Unknown: v2 Solution: P1 + ½ v12 + = P2 + ½ v22 + ρ ρgh1 ρ ρgh2 P1 + = P2 + ρgh1 ρgh2 v2 = sqrt(2gh1) v2 = sqrt(2 (9.81 m/s2) (0.5m)) v2 = 3.13 m/s
Calculate the absolute pressure at an ocean depth of 1.0 x 103 m. Assume that the density of the water is 1.025 x 103 kg/m3 and that P0 = 1.01 x 105 Pa. Given: h = 1.0 x 103 m ρ = 1.025 x 103 kg/m3 Patm or Po = 1.01 x 105 Pa Unknown: Pabsolute Solution: Pabsolute = Patm + Pgauge Pabsolute = Patm + ρgh = 1.01 x 105 Pa + (1.025 x 103 kg/m3) (9.81 m/s2) (1.0 x 103 m) Pabsolute = 1.01 x 107 Pa A tank has a spigot near its bottom. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the spigot when the water level is 0.5 m above the spigot. Given: P1 = Patm = 1.01 x 105 Pa = P2 (both are open to atmosphere) v1 = 0 (negligible) h1 = 0.5 m h2 = 0 m Unknown: v2 Solution: P1 + ½ ρ v12 + ρgh1 = P2 + ½ ρ v22 + ρgh2 P1 + ρgh1 = P2 + ρgh2 v2 = sqrt(2gh1) v2 = sqrt(2 (9.81 m/s2) (0.5m)) v2 = 3.13 m/s
uoh_fluidmechanics_lesson01_practiceproblemsas_v2_tedl_dwc.docx