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A water fountain is to be installed at a remote location by attaching a cast iron pipe directly to a water main through which water is flowing at 70F and 60psig. The entrance to the pipe is sharp-edged, and the 70-ft-long piping system involves three 90 miter bends without vanes, a fully open gate valve, and an angle valve with a loss coefficient of 5 when fully open. If the system is to provide water at a rate of 15gal/min and the elevation difference between the pipe and the fountain is negligible, determine the minimum diameter of the piping system. Answer: 0.713 in Properties The density and dynamic viscosity of water at 70F are =62.30lbm/ft3 and =2.360lbm/fth= 6.556104lbm/ft-s. The roughness of cast iron pipe is =0.00085ft. The minor loss coefficient is KL=0.5 for a sharpedged entrance, KL=1.1 for a 90 miter bend without vanes, KL=0.2 for a fully open gate valve, and KL=5 for an angle valve. Analysis We choose point 1 in the water main near the entrance where the pressure is 60 psig and the velocity in the pipe to be low. We also take point 1 as the reference level. We take point 2 at the exit of the water fountain where the pressure is the atmospheric pressure (P2=Pstm) and the velocity is the discharge velocity. The energy equation for a control volume between these two points is gP1+12gV12+z1+hpumb,u=gP2+22gV22+z2+hturtine,e+hLgP1eape=22gV22+hL where 2=1 and hL=hL,meal=hL,major+hL,minar=(fDL+KL)2gV22 since the diameter of the piping system is constant. Then the energy equation becomes (62.3lbm/ft3)(32.2ft/s2)60psi(1psi144lbf/ft2)(1lbf32.2lbmft/s2)=2(32.2ft/s2)V22+ The average velocity in the pipe and the Reynolds number are V2=AcV=D2/4VV2=D2/415/60gal/s(1gal0.1337ft3)Re=V2DRe=6.556104lbm/fts(62.3lbm/ft3)V2D The friction factor can be determined from the Colebrook equation, f1=2.0log(3.7/Dh+Ref2.51)f1=2.0log(3.70.00085/D+Ref2.51) The sum of the loss coefficients is Then the total head loss becomes hL=(fDL+KL)2gV2hL=(fD70ft+9)2(32.2ft/s2)V22