6.14 Water at 20°C is to be siphonedthrough a tube 1 m long and 2 mm indiameter, as in Fig. P6.14. Is there anyheight H for which the flow might not belaminar? What is the flow rate if H = 50 cm?Neglect the tube curvature.(1)Water at 20° CFig. P6.14L=1m, d = 2 mm(2)H

Introduction In this problem, we will determine the height at which the flow may not be laminar by taking the Laminar Reynold number and the characteristics of water at the given temperature into account. The only type of head loss studied is significant head loss caused by friction. Given data: water temperatureT=20°Clength of the tubeL=1 mdiameter of tubed=2 mmheightH=50 cm Find out: The height (H) for which the flow might not be laminar. The flow rate if the height (H) is 50 cm.Calculation of the height (H) for which the flow might not be laminar. For the calculation of height (H), the following things we have to consider, Applying the Barnaulli equation between the free surface and the tube exit; P1ρg+V122g+z1=P2ρg+V222g+z1+hfP1=P2=Patmhf=head lossV1=0 velocity of free surface is zeroz1=height of tank from the datum levelz2=height of tube from the datum levelz1=V222g+z2+hfhf=z1-z2-V222gz1-z2=Hhf=H-V222g ...........equation(1) The head loss due to friction (hf) hf=f×L×V22g×df=friction factorf=64Re (for laminar flow)(Re)=ρ×V×dμhf=f×L×V22g×dhf=64×L×V2Re×2g×dhf=64×L×V2×μρ×V×d×2g×dhf=32×L×V×μρ×g×d2 Substituting the value of head loss (hf) in equation (1). hf=H-V222ghf=32×L×V×μρ×g×d232×L×V×μρ×g×d2=H-V222gV2=V32×L×V×μρ×g×d2=H-V22g .......... equation(2)The properties of water at 20oC are mentioned below; at T=20°Cρ=998 kg/m3μ=0.001 kg/ms Substituting the above values in equation (2). 32×L×V×μρ×g×d2=H-V22gL=1 md=2 mmd=0.002 mg=9.81 m/s2Re=2000 for laminar flowRe=ρ×V×dμ2000=998 kg/m3×0.002 m×V0.001 kg/msV=1.002 m/s32×L×V×μρ×g×d2=H-V22gH=32×L×V×μρ×g×d2+V22×gH=32×1 m×1.002 m/s×0.001 kg/ms998 kg/m3×9.81 m/s2×0.0022m2+1.002 m/s22×9.81 m/sH=0.87 m Hence the height is 0.87 m.Calculate of the flow rate if the height (H) is 50 cm. For the calculation of the flow rate we have to calculate the flow velocity by using the previous equation; H=32×L×V×μρ×g×d2+V22×gH=50 cmH=0.5 m0.5 m=32×1 m×V×0.001 kg/ms998 kg/m3×9.81 m/s2×0.0022m2+V22×9.81 m/s0.5=0.81713×V+0.05097×V20.05097×V2+0.81713×V-0.5=0V=0.59 m/s The flow rate (Q); Q=A×VQ=π4×0.0022 m2V=0.59 m/sQ=A×VQ=π4×0.0022 m2×0.59×3600(Q)=0.0067 m3/hr Hence the flow rate is 0.0067 m3/hr.Result: The height (H) for which the flow might not be laminar H=0.87 m The flow rate if the height (H) is 50 cm (Q)=0.0067 m3/hr