Question

Kerosene with a specific gravity of 0.823 is flowing downward through the venturi meter shown in Figure 4 a) 100 mm inside diameter 50 mm inside diameter Mercury (SG 13.6) Figure 4: Venturi meter ) By using the Bemoulli and the continuity equations, show that the manometer deflection, h, in the Figure 4 is given by 15吸 31.05g (8 Marks) manometer deflection h is 700 mm, compute the volume flow rate of the

Answer 1

P_A = P_B + Y_0 x + Y_m h - Y_0 [h + x + (z_A - z_B)] Y_0 rightarrow specific weight of Kerosene Y_m rightarrow for mercury (P_A - P_B) + Y_0 (z_A - z_B) = Y_m h - Y_0 h h = [(P_A - P_B)/Y_0 + (z_A - z_B)] [Y_0/Y_m - Y_0] Velocity ar A, V_A = A_B V_B/A_A A = pi D^2/4 D: diameter of flow A_B = pi D_B^2/4 = pi(0.05^2/4) = 1.96 times 10^-3 m^2 A_A = pi/4 (0.1^2) = 7.854 times 10^-3 m^2 V_A = 1.96 times 10^-3 times V_B/7.845 times 10^-3 = 0.25 V_B V_B = 4V_A P_A/Y_0 + z_A + V_A^2/2g = P_B/Y_0 + z_B + V_B^2/2g P_A - P_B/Y_0 + (z_A - Z_B) = V_B^2 - V_A^2/2g = 4^2 V_A^2 - V_A^2/2g = 15 V_A^2/2g Y_0 = (sg) Y_w = 0.823 Y_w = 8.07 KN/m^3 Y_m = (sg) Y_w = 13.6 Y_w = 133.3752 KN/m^3 h = 15 V^2_A/2g (8.07/133.375 - 8.07) h = 15 V_A^2/31.05g \r\n Volume flow rate Q = A_A A_B