Report on Hydrodynamic Solution

Added on - 21 Apr 2020

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HYDRODYNAMIC SOLUTIONAnswer -1)(a)Sketch and fully labelyourdesign of the piping system based on the criteriadescribed above;(b)the Static Head;elevation between the two tanks is 6m.So suction head = 6m(c)the Reynold’s number;Re=SρVDμS =0.9ρ=1000Kgm3As continuity equationQ =A*V
Q =2L/sA=πr221000=π(0.01)24VSo V =0.398 m/sμ=0.005PasPut all values we get ,Re=0.910000.398800.051000Re=5731.2(d)the relative roughness of the pipe;RR=DεRR=800.045RR=1777.8(e)the friction factor (using the Moody formula);f=0.316Re14f=0.316(5731.2)14f=0.0363
(f)the total K factor for the fittings;As k factor is given for all different conditionSo total k factor = 0.5+1+0.2+6+ (15*0.03)+(4*0.6)= 10.55(g)the total head loss for pipe and fittings;hl=10.55v22ghl=10.550.398229.81h¿=0.851m(h)the Dynamic Head;hd=P2P1ρghd=5.81m(i)the System Head;hsys=hd+hshsys=6+5.81hsys=11.81m(j)the System Head Equation in terms of the volumetric flow rateh=Pρg+Q22A2g+C(k)the Fluid Power;
FP=SρghQFP=0.910009.8111.8121000FP=0.117KW(l)the Pump shaft Power, given that it has an efficiency of 72% at the given flowrate.eff=SPFP0.72=SP117SP=84.24WAnswer 2
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