Question

Water flows in the pipeline. Calculate maximum velocity umax in the pipe axis and discharge Q. Determine whether the flow is laminar or turbulent (T 12°C). The mercury differential manometer (E 13600 kgm 3) shows the difference between levels in Pitot tube Hn 0.02 m. Diameter of the pipe is D-0.15 m. Mean velocity is.considered to be v-0.4 ua Hm На

hydraulics

Answer 1

given data differential head of u - tube manometer delta p/v_f = h = x[s_m/s_f - 1] h rightarrow differential head, m x rightarrow differential level of fluid, H_m = 0.02 m s_m rightarrow specific gravity of mercury s_f rightarrow specific gravity of water therefore delta p/r_f = h = 0.02 [13.6/1 - 1] h = 0.252 m therefore delta p = r_f h = 9.81 times 0.252 p_1 - p_2 = 2.472 KN/m^2 from Bernoulli's equation p_1/r + v^2_1/2 g + i = p_2/r + v^2_2/2 g + 2 [p_1 - p_2/r] = [v^2_2 - v^2_1]/2g \r\n therefore 2.472/9.81 = v^2_2 - v^2_1/(2 times 9.81) v^2_2 - v^2_1 = 4.944 given pipe has uniform diameter therefore v_1 = v_2 = v therefore v_1 = 0 therefore v^2_2 = v^2 = 4.944 v = 2.22 m/s mean velocity v = 0.4 u_max therefore 0.4 u_max = 2.22 u_max = 5.56 m/s discharge Q = A v_avg = pi/4 times 0.15^2 times 2.22 Q = 0.039 m^3/s

Answer 2

sorry i dont have an answer bye and thanks 

Answer 3

sorry i dont have an answer bye and thanks 

Answer 4

sorry i dont have an answer bye and thanks