Answer to Question #179512 in Mechanical Engineering for Johan Nieuwoudt

Question #179512

Given that oil, with density p=700kg/m3 and kinematic viscosity v = 0.00001m2/s flow at 0.3 m3/s through 500 m of 250mm diameter cast iron pipe and the roughness of cast iron pipe e = 0.26mm


Determine


  • The Reynolds number
  • The friction factor f
  • The head loss due to friction
1
Expert's answer
2021-04-12T02:32:16-0400
  • The Reynolds number

Re=udv=4×0.3π×0.25×0.00001=1527887.45Re= \frac{ud}{v}=\frac{4 \times0.3}{\pi \times0.25 \times0.00001}=1527887.45



  • The friction factor f

This a turbulent flow

f=1.325[ln(e3.7D+5.74Re0.9]2=1.325[ln(0.000263.7×0.25+5.741527887.450.9]2=0.021714033f=\frac{1.325}{[\ln(\frac{e}{3.7D}+\frac{5.74}{Re^{0.9}}]^2}=\frac{1.325}{[\ln(\frac{0.00026}{3.7 \times0.25}+\frac{5.74}{1527887.45^{0.9}}]^2}=0.021714033


  • The head loss due to friction

u=Q0.25πd2=0.30.25π×0.252=6.1m/su=\frac{Q}{0.25 \pi d^2}= \frac{0.3}{0.25 \pi \times0.25^2}=6.1 m/s

hf=fLdu22g=0.021714033×5000.256.122×9.81=82.4mh_f=f \frac{L}{d} \frac{u^2}{2g}=0.021714033 \times \frac{500}{0.25} \frac{6.1^2}{2 \times 9.81}=82.4 m


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