Cv Ball=15170 Gate=8622 Globe=1310 Check=5857 A system using 10-inch (10) intern

Question

Cv Ball=15170 Gate=8622 Globe=1310 Check=5857 A system using 10-inch (10) internal diameter pipe requires a selection of flanged valves. Consider 0"ball valves, gate valves, globe valves, and check valves in the fully-open position; and buttertly valves in a variety of open positions. FNW has been chosen as the equipment supplier, the relevant data sheets can be found on Desire2Learn (see below for information on valve part numbers and cor responding files). The data sheets give valve coeficients (C): Bow rates in GPM corresponding to a pressure drop of I psi across the valve (for 60'F water). The flow through the system is normally wa er at 25C with a flow rate of 610 /min, through smooth-walled stainless steel tubing (a) Consider the four valves listed below (specification file names in parentheses). Determine the los oefficient (K) and equivalent pipe length (a)for each valve 10" Class 150 ball valve (file: FNW flanged StSt ball valve 600.pdf 10" Class 150 gate valve (ile: FNW Cv.pdt 10" Class 150 globe valve (6le: FNW Cvpd 10" Class 150 check valve (file FNW Cvpdf)

Explanation / Answer

Loss coefficient, K = 890*d^4/Cv^2

d = diameter of pipe in inches

Reynold Number,Re = density*flowrate*Diameter/Cross-sectional area/ kinematic viscosity

For water,

Density = 1000 kg/m3

given Diameter = 10 inch = 0.25m

Cross sectional area = 3.14/4*d^2 = 3.14/4*.25^2=0.049 m2

viscosity of water = 8.9*10^-4 Pa.s

Flowrate of water = 610 l/min = .61/60 = 1.01 m3/s

Re=1000*1.01*.25/(8.9*10^-4) = 283708

Because, flowrate is more than 3000, therefore flow is turbulent

For Turbulent flowrate, Darcy Friction factor,f = 0.316/(Re^.25) = 0.316/(283708^.25) = .013

Equivalent length, L/D = K/f

Valve Type Cv K L/D Ball 15120 .035 2.69 Gate 8622 0.12 9.2 Globe 1310 5.18 398.4 Check 5837 0.26 20

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