Question

Problem 3- A reducing section is attached to the end ofa straight pipe of 4 inch ID (1 inch 0.0254 m). The ID of the exit of the reducing section is 2 inch. Water (incompressible) is the flowing fluid, and the pressure difference before and after t bya mercury manometer, as sho wn in the graphon the right. The section is the atmo spheric pressure zero gage. Mercury's specific gravity is 13.6, and that of the flowing water is 1.0. Gravity- 9.81 m/s2 Pi : he reduction is measured pressure P2 downstream of the reducing When the flow at cross-section is at the velocity of 0.75 m/s, the monometer registers a height difference of1.5 inch. Please answer the follo wing questions a) What is the velocity of flow at cross-section 2 b) What is the gage pressure at cross-section I taken density of water into account? c Based on momentum balance, wt is the force acting on the reducing section from the fluid?

Answer 1

Given that Is attached to the end of straight pipe of 4 inch 1D (1 inch = 0.0254m). the reducing section is 2 inch. The specific gravity = 13.6 and that of the flowing water is 1.0 Gravity = 9.81 m/s^2 velocity = 0.75 m/s height difference = 1.5 inch From continuity Esuation A_1V_1 = A_2V_2 pi/4 (d_1)^2 v_1 = pi/4 (d_2)^2 r_2 v_2 = v_1(d_1/d)^2 v_2 = 0.75 (4/2)^2 v-2 = 0.75 (2)^2 v_2 = 0.75 (4) v_2 = 3m/sec From Bernoullis Equation p_1/rho g + v_1^2/2g + z_1 = p_2/pg + v^2_2/2g + z_2 + h_c \r\n �h_L' is unknown [neglect it] p_1/rho g + v^2_1/2g = p_2/rho g + v^2_1/2g Also rho_1 - rho_2/rho g = h = l [s_gn/rho l - 1] rho_1 - 0/rho g = 1.5 times 0.025[13.6/1 - 1] rho_1 = 4.709 kPa F_x = (p_1 A_1 - p_2 A_2) = 4709.38 times pi/4 (4 times 0.025)^2 - (0 times A_2)