12_Control_Valves_Chap9S10_4

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online