Question

4. A two-liquid pressure gauge is connected to a tank filled with water. In the environment where the warehouse is located, Patm = 755 mmHgS. The height of the liquids is h1 = 0.2m h2 = 0.3m h3 = 0.45m and the density of the water is 1000kg/m', the density of the oil is 0.85 and the density of the mercury is 13.6. Calculate the pressure of the air inside the tank. hava yag hi h3 su h2 civa

Answer 1

0:1 Air Patu+ 755 mmHg water TV Hg Data h = 0,2m Pwater = 1000 ug/m² Yoil hz=0,3m Soil = 0,85 water h₂= 0,45m S Hg = 13,6 Pwater First step, calculate Specific weight of water. Using the next equation Pewater Soil- Poil SHg = Pug = YHg water Ywater = Pwater.g up s? p=Yug hag 3 Ywater- 1000 ug 9,81 m Ywater 9810 Nhu Second step, convert atmospheric pressure from muity to Pascal Patus - Sug. Ywater. hug Patus = 13,6,9810 N. 755X103 Patus= 100 729,08 Pa third step. Start with the pressure at point 3, advancing along the tube and add or subtract the pressure terme until reaching the res result Patur + SHO Ywater hz - Soil Ywater hz - Ywater his Pair Pair = 100729,08 Pa +13,6.9810.0,45m) -0,85.9810 N. 0,3 - 9810 N. 0,2 m Pair = 156302, 73 Pa the air Pressure is us 3 156302,73 Pa