Sm1 42thru1 43

PROBLEM 1.42 A = 0.0018 m² P₁atm = 100 kPa At the initial and final states equilibrium. using no effect of friction (between the piston and cylinder) can be assumed. Thus, looking at the interface between the air and the friction (shown dotted) we have: P₁/A = P₁atmA + Fspring Initially, Fspring = 500 N. So, P₁ = 100kPa + 500 N = 150kPa A = 150 kPa[ 1 m 0.0018 m² * 10⁵ N/m² ] = 1.48 atm. Finally, Fspring = 0 N. So, P₂ = 100 kPa. P₂ = 100 kPa [ 1 atm 101.325 kPa ] = 0.99 atm. PROBLEM 1.43 Patm = 14.7 lbf/in² Storage Tank φ = 62.4 lb/ft³ g = 32.0 ft/s² The pressure at the bottom of the storage tank is p = Patm + ρgL = 14.7 lbf/in² + ( 62.4 лб/ft³ ) [ 320 lbf (20 ft) ] ш ( ft )(144 in² ) = 14.7 lbf/in² + 8.6 lbf/in² = 23.3 lbf/in²