Question

SU = WATER , YAĞ = OIL , CIVA = Mercury

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 hl = 0.2m h2 = 0.3m h3 = 0.45m and the density of the water is 1000kg / m³, 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 yağ hl h3 su h2 civa

Answer 1

solution Purm ازه air te Parm = 755 mmHg. -hy = 0.2 m - 0.3m hz = 0.45 m ha x I ha wwtoo Sw=1000 kg/m3 • density of oil Po= 850 kg m3 density of mercury Sm- 13600 kg Mercury m3 Foom Fig. we com write Pit Swghy + 809.ha Im ghz Putm Now Parm = 775 mm of Hg Smg. (0.775) so P, + (1000) (9.51) (0,0) + (850) (9.81) (0.3) - 13600 C 9.81) C0.45) 13600 (9.81) C0.775) P = 13600 C 9.81) (0.775) + 55573.65 P = 15897.09 Mime Pi= 0,1589 mpa. 0. Pressure of wo inside the temki