Question

Water flows through the pipe in Figure 13-39 and exits to the atmosphere at C. The diameter of the pipe is 2.0 cm at A, 1.0 cm at B, and 0.8 cm at C. The gauge pressure in the pipe at A is 1.22 atm and the flow rate is 0.8 L/s. The vertical pipes are open to the air. Find the level of the liquid-air interfaces in the two ver- tical pipes. Assume lami- nar nonviscous flow 0) FIGURE 1 3-39 Problems 84,85

Answer 1

Flow rate = AV A = cross section area V = velocity At point A: 0.8 L/sec = 8 times 10^-4 m^3/sec = pi/4(2 times 10^-2)^2 V V = 2.546 m/sec we know, gauge pressure P_g = 1/2 rho v^2 + rho gh 1.22 times 1.01 times 10^5 = 1/2 times 1000(2.546)^2 + 1000 times 9.8 times h_A h_A = 123220 - 3241.06/9800 m h_A = 12.24 m \r\n flow rate = AV, A = cross section V = Velocity At point A: 0.8 times 10^-3 m^3/sec = pi/4 (2 times 10^-2)^2 times V_a V_A = 2.546 m/sec (1) At point B: 0.8 times 10^-3 m^3/sec = pi/4 (1 times 10^-2)^2 times V_B V_B = 10.186 m/sec (2) At point C: 0.8 times 10^-3 = pi/4 (0.8 times 10^-2)^2 times V_C V_C = 15.915 m/sec Now using bernoulli equation P_1 + 1/2 rho V_1^2 rho gh_1 = P_2 + 1/2 rho V_2^2 rho gh_2