Sm1 42thru1 43

PROBLEM 1.42 A = 0.01B m2 Spring Force varies linearly from 500 N when V = 0.003 m3 to zero when V2 = 0.002 m3 Pspring = 100 kPa Fig. P1.42 Air AL Initially, Fspring = 500 N. So P1 = 100 kPa &\left(\frac{900 N}{0.01m^2 \times 10^3 N/m^2}\right) = 150 kPa. At the initial and final state - state equilibrium. Using no effect of friction (between the piston and cylinder) can be assumed. Thus, focusing at the interface between the air and the piston (horizontal) we have: Pair A = Pspring = Padm A = Pairspring A Initially, Fspring = 500 N. So, P1 = 100kPa. Pf = 150kPa. Finally, Fspring = 0 N. So, P2 = 100kPa. = \frac{1atm}{101.3kPa} = 0.99 atm. PROBLEM 1.43 Patm = 14.7 lbf/in^2 Storage Tank h 50 ft Building \rho = 62.4 lb/ft^3 = 32.0 lbf/ft^3 The pressure at the bottom of the storage tank is p = Patm + \rho g L = 14.7 \left(\frac{lbf}{in^2}\right) + \frac{(62.4 \frac{lb}{ft^3})(32.0 \frac{ft}{sc^2})(50 ft)}{144 in^2} = 14.7 \left(\frac{lbf}{in^2}\right) + 8.6 \left(\frac{lbf}{in^2}\right) = 23.8 \left(\frac{lbf}{in^2}\right) Forward