12_Control_Valves_Chap9S10_4

# 2 250 200 q galmin 150 100 50 0 0 chapter 9 figure

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Unformatted text preview: Thus for the following curve, pick “y-s = q values” and calculate “x-s = values”. qmin is at zero lift, qmax at lift = 1. 20 21 Installed Valve Characteristics, Example 9.2 250 q , gal/min 200 150 Chapter 9 Cv = 125; % units of gpm/(psi^0.5) q = 0:1:221; % q < qmax = 221.6 gpm delPv = 40 - 30*(q/200).^2 ; % psi lift = q./(Cv*sqrt(delPv)); % dimensionless, 0 to 1 figure, plot(lift,q,'b-') xlabel('lift'), ylabel('q , gal/min') title('Installed Valve Characteristics, Example 9.2') 100 50 0 0 Figure 9.12 Installed valve characteristics for Example 9.2. 0.2 0.4 0.6 0.8 1 lift 23 Installed Valve Characteristics, Example 9.2 250 q , gal/min Chapter 9 200 150 100 50 0 0 0.2 0.4 0.6 0.8 1 lift Which characteristics curve is more linear near the midrange of lift ? 24...
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## This note was uploaded on 04/17/2010 for the course CHE 461 taught by Professor Staff during the Winter '08 term at Oregon State.

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