Example 1=1.75slug/ft3Q=1500 gpm xv=1.15 x 10-4ft2/sRe=DV0.000115=3.3422459893ft3/svD=0.667ft=0.667 ft x 9.58 ft/s1.15 x 10-4 ft2/sV=Q=0.00055564A=A=5.55 x 10-4=3.3422459893ft3/s=3.14x(0.667)20.349237865ft2=8.5 x 10-4ft4=9.5701134506ft/sD0.667ft=0.349237865ft2=0.0012743628=1.27 x 10-31=f1=12.785724035=f(1)2=(12.78572)2=12.785724035(f)2which yields1=163.4747390991f=0.00612ff=0.00611715310.00612The head loss is obtained by using:1 HorsePower=550(ftlb)/shf=2f (L/D)(V2/g)x HorsePower=15766.569(ftlb)/s=2 x 0.00612 (1,600 ft/0.667 ft)[(9.58 ft/s)2/(32.2 ft/s2)]x=15766.569/550=83.6855959101ft=28.666489090983.7ft28.7hpThe mass flow rate isthe efficiency of the pump () = 0.85=Q=1.75 slug/ft3 x 3.34 ft3/sBreak Horse Power = Power to fluid/=5.845slug/s=28.7/5.85slug/s=33.7647058824hpPower to fluid33.7hp=.hf.g=5.85 x 83.7 x 32.2=15766.569(ftlb)/s

D2 41 ft3/s 448.8 gpm

Example 2For water at 15C, ==999kg/m3D=0.25m=1.6 x 10-3PasL=450mso that the kinematic viscosity can be calculated asv=/=0.0000016016m2/sRe=DVA=D2/4v=3.14 x (0.25)2=0.25 m x V41.16 x 10-6 m2/s=0.0490625m2=2.16 x 10-5 V--> where V must be in m/shf=0.01990.14106735987.3=fV2=1.99 x 10-2 m2/s2--> where V must be in m/sfV2=1.99 x 10-2Vf=0.141f=0.141VWe can see that the product Ref can be calculated, even though we do not know the velocity V.Ref=2.16 x 10-5V x 0.141V=

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