Sm1 34thru1 36

PROBLEM 1.34\np_atm = 1 bar\nInlet\nGas compressor\nExit\nReceiver\ntank at\n200 kPa\nFig. P1.34\n\nConverting the local atmospheric pressure to kPa, we get p_atm = 100 kPa. Since the pressure in the tank, 200 kPa, is greater than the atmospheric pressure, the Bourdon reading is a gage pressure. Using the following relationship, P_gauge = P_abs - P_atm the Bourdon reading is 100 kPa.\n\nPROBLEM 1.35\nSee Fig. P1.35. Applying the principles discussed in Sec. 16.1, the atmospheric\npressure is\n\np_atm = ρ m g L => L = \n\rho = \n\\rho\n2 g\n\nL = p_atm\n-----------\nρg\n\n= 100 \u00d7 10^3 N/m^2\n(13.57 kg/m3)(9.81 m/s2)\n= 0.75 m \n= 750 mm Hg\n\nConverting to inHg:\nL = 750 mm Hg \n1 cm\n----\n10 mm\n1 in.\n----\n2.54 cm\n= 29.53 in.Hg\n\nPROBLEM 1.36\np_atm = 1.147 lb/in2\n= 3200 lb/ft2\nWater\nρ = 0.01064 lb/ft3\nL = 10 in.\n\nFig. P1.36\n\nP2 = P_atm + ρ g L\nP0 = P_atm + ρ g (2 + L)\n\n:(P0-P2) = ρ g L\n\n= (0.01064 lb/ft3)(32.2 ft/s2)(\n2.0+0.5)(12 in)\n= 7.51(1.36 lb/in2)\n+ 0.36 lb/in2\n\nR___ increase