Question

A 1:4 scale model of a water pumping station pumping system is to be tested to determine overall head losses 5. Air at 30 °C, 1 atm, is available. For a prototype velocity of 600 mm/s in a 3-m-diameter section with water at 20°C, determine the air velocity and quantity needed. How may losses determined from the model be converted into prototype losses?

Answer 1

From scaling low & fluid flow in microscale rhov_md/mu = vd/upsilon theta = Kinematic Viscosity F or typical values upsilon = 10^-6 m/s(Approximately) d = diameter given d = 3m mu = mean velocity P or typical values mu = 10^-5cm^2s^-1 v = prototype velocity Given v = 0.6m/s (500 mm/s) v_m = volumetric flow (we need to find out) substitute all values in equation (1/4) times 3 times v_m/10^-5 = 0.6 times 3/10^-6 v_m = 0.6 times 3 times 10^-5 times 4/10^-6 times 3 = 2.4 times 10^1 v_m = 24 m/s (Reynolds No)_Prototype = (Reynolds Number)_Model Q_m = A_mV_m = pi/3[3(Y_u)]^2(24) = pi/3 times 13.5 Q_m = 14.1428 m^3/s losses are same when Expressed in velocity heads