5. The gage pressure of the air in the tank shown in figure below is measured to be 80 kPa. Determine the differential height (h) of the mercury column. (Assumptions: The air pressure in the tank is uniform; the density of water to be 1,000 kg/m³). Oil 80 kPa SG=0.72 Air ↑ 30 cm 75 cm Water D Mercury SG= 13.6

Problem 2: A U-tube manometer is connected to pressurized tanks (Tanks A and B), as shown in Fig. 2. The pressure in Tank A is 500 kPa. The densities of water and mercury are 1000 and 13600 kg/m³, respectevely. Neglect the density of air and determine the pressure in Tank B. PB = Pa + Pm ghm + 2-P+ 1600000 + 13:0019.81)1,5) Pg=500000 + 13,600 (9.81.2.0m) 1000 (9.5Px1.0m) 500000 Pa + 13, 6000 (1791), (2.0) + 1000 Yw g h w PB = 776x6 kPa 690 kPa P. 500,000 113.600 (36819:42 000 + 1(1000) Xh+m + H) - Ym 13,600((500-X) Air- 13,000 2 Im B ܢܢܢܢܢ Mercury 4 ܠܠܠܠ Water water) = 0 6801499 Fig. 2 U-tube manometer +4 13,600 2+1 -1.0=0 Air 1.5 m

Determine the magnitude of the force on the inclined gate 1.5m by 0.5m shown in the figure. The tank of water is completely closed and the pressure gage at the bottom of the tank reads 90,000 N/m2. Use 9800 N/cubic meter for water? Ans: 48022.5N

For the manometer shown in the image, all fluids are at 20°C. If PB- PA = 90 kPa, determine the height H in centimeters. The specific weight of water is 9790 N/m³ and the specific weight of mercury is 133100 N/m³. Meriam red oil, SG = 0.827 Water 18 cm H 35 cm Numeric Response B Mercury