Fluid mechanics Free Essay Example, 1500 words

Task Learning Outcome 3 Determine head losses in pipeline flow Q1: Calculate the flow of water under gravity in litres/sec between the two reservoirs in Figure 1 below. 5.6m Pipe à ¸ 11cm in steel (k=0.20mm) 15.78m Figure 1 Solution: The length of the pipe may be approximated to be ≈ ---ïÆ'   L = 16.7442 m. Assuming a completely turbulent flow (dependent on pipe roughness), the friction factor may be estimated by the von Karman equation: f = and, plugging in values gives f = ---ïÆ'   f = 0.022813 Then, by Darcy-Weisbach equation = f * * where = head loss due to friction = difference in levels of water (5.6 m) v = fluid’s average velocity (m/s); g = acceleration due to gravity (9.81 m/s2) Arranging the equation for v – v = = = 5.6249 m/s pipe’s area @ cross – section = * = 0.0095 m2 Thus, the volumetric flow rate (under gravity), V = 5.6249 m/s * (0.0095 m2) * = 53.44 L/sec Task 2 – Learning Outcome 3.2 Determine Reynolds’ number for a flow system and assess its significance Q2: As an engineer in an industry, you are required to pump oil (density 900 kg/m3, viscosity 0.12Ns/m2 and flow rate 0.2m3/s) in a 15 cm diameter pipe over a distance of 120m. (a) Calculate the critical velocity and the Reynolds’ number in the pipe. critical velocity, v = (volumetric flow rate per unit area of cross – section) cross – sectional area, A = * = 0.01767 m2 so that, v = = 11.32 m/s = = ---ïÆ'   = 12,735 (turbulent) (b) Calculate the power required (per metre) to pump the oil horizontally at a mass flow rate of 30kg/s. We will write a custom essay sample on Fluid mechanics or any topic specifically for you Only $17.96 $11.86/pageorder now For turbulent flow with < 100,000, Prandtl equation applies as— = 2.0 log whereupon substitution of the found: = 2.0 log yields f ≈ 0.0289927 (by trial & error) Since the flow is strictly horizontal (no turns/bends, elevation, nor expansion/contraction along pipe length) then, energy – balance with the Bernoulli equation reduces to: ÃŽ £F = (where pump work solely accounts for friction losses) And by Darcy-Weisbach equation, ÃŽ £F = f * * = ---ïÆ'   ≈ 1,486.08 Based on this, the required power would be the product of the work term and the mass flow rate, being Power = (1486.08 m*N/kg) (30 kg/s) = 44,582.4 watts For each meter of the pipe length, Power = ---ïÆ'   Power = 0.3715 kW/m (c) Calculate the power required (per metre) to pump the oil horizontally at a mass flow rate of 120kg/s.