Page 1 SOLUTION MECH ENG 3101 & 7068 Applied Aerodynamics Incompressible Flow Assignment 2 SOLUTION Due: 5 pm, August 28, 2020 A chemical production process uses the device shown in Figure 1 to deliver a continuous flow of a chemical to a conveyor belt. The device consists of a chemical tank with an annular duct attached to the bottom. The annular duct is a circular pipe of internal diameter 20 mm with a concentric circular (solid) rod of outside diameter 19 mm. The loss coefficient of the duct entrance is 0.1. The internal surfaces of the annular duct have a roughness height of (e) = 0.001 mm. The depth of the chemical in the tank is held constant. Making suitable assumptions, calculate the velocity of the chemical leaving the annular duct. -Rod Tank ¢ 20 mm Kent=0.1 1 m 2 m ¢ 19 mm + V Conveyor belt Figure 1. Chemical tank and delivery tube arrangement. Notes: · The chemical has properties p = 1000 kg/m3 and p = 1.00 x 10-3 N.s/m2. . It is acceptable to assume that the kinetic energy correction factor o is 1.0 throughout the system. . The Moody Chart and laminar friction constant equations are given on the following pages. · Assessment will be based primarily on the method of approach, including the completeness of the analysis and the assumptions that are made. · You are expected to use the problem solving protocol.
Assignment 2, 2020. 1 Given: Chemical delivery tank and duct Find: mass flow rate of chemical Schematic diagram + data. Patur c=20mm 6=19mm p= 1000 kg/m3 1 2 = 1.00x10 m2/s 1m E = 0.001 mm 1 1z 2m 2) 14 Patm V2 Z =0 at exit Assume : Incompressible fluid Steady, fully-developed flow X = 1.0 (given) * Solution : Consider a streamline from 1 to 2 Mechanical energy equation gives PIL + 2, V,2 + Z1 = P2/ + 2, V2 + Z2 + h, 19 29 b 29 Put P1 = P2 = Patm V1 = 0 (surface where area is large) Z1= 3m, Z2 = 0 m X2 = 1 ( given in the problem statement) Mech energy equation reduces to 3 = V23 +hL * Note that the solution will find that the flow is laminar, so I should therefore be 2. We ignore this here for simplicity.
Losses : h1 =(Ik +I+41 12 2) V2 an 29 .2 = ( Kent + + 4/ dy ) 2/2g Kent = 0.1 (given) L= 2 m dh = d2-d1= 20-19=1mm using the So h1 = (0.1+2+) V2 3 1 x105 29 = (0.1+2000f) güven information. 2 29 Combine with Mech. Energy equation to give 3 = V2 + (0.1 + 2000 f) V2 29 29 12 = (1.1 +2000f 29 1 Assume initially that the flow is turbulent 0.001 mm fo 8/1 0.001 dn 1.0 mm For this relative roughness height, the Moody Chart gives for fully turbulent flow (Re -> 00), f = 0.0197 Substitute into