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x, fraction lift
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Effect of line loss on effective control valve characteristics from Example 20.2. I no pressure drop in supply line to valve, II pressure drop present in supply line to
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Example 20.2. Determine the flow versus lift relation for the linear control valve installed in the flow system of Fig. 20.4. The fluid is water at 5 C. The following data apply: pipe length inside pipe diameter density of water viscosity of water C, of control valve total pressure drop, pu - pr 1OOft 1.0 in 62.4 lb/ft3 1.5 cp 4.0 100 psi
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If there is no line loss as is the case for a large diameter line, the maximum flow can be calculated from Eq. (20.1): 4 =CV&=4.0~=40.0gpm
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To determine the flow/lift relation for the case of line loss, we arbitrarily start the calculation with a flow of 30 gpm. The pressure drop in the 100 ft pipe can be calculated from the well known expression from fluid mechanics: Ap =
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=fLPq*
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(20.11)
where Ap = pressure loss in line, psi q = flow through pipe, ft3/sec g, = 32.174 (lbm/lbf)(ft/sec2) L = pipe length, ft p = density of fluid, lb,/ft3 D = inside pipe diameter, ft f = fanning friction factor, dimensionless The fanning friction factor is a function of the Reynolds number and the pipe roughness. Equation (20.11) and a correlation for the fanning friction factor can be found in the literature (Perry and Chilton, 1973). We now calculate the Reynolds number (Re): Re = Duplp Replacing the velocity u with q/[(d4)D2] gives (20.12) q = 30/(60)(7.48)
Re =
= 0.0668 ft31sec or 240.6 ft3/hr
(W4WW.4) (7r)(1.50)(2.42)(1/12) = 63 224
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For this value of Reynolds number and for smooth pipe, the fanning friction factor f is 0.005. Equation (20.11) may now be used to calculate line loss: Ap = (32)(0.005>(l~)(62.4)(0.066~)2 (144)(&)(32.2)(1/12)5 therefore Ap across valve = 100 - 24.2 = 75.8 psi We next calculate the flow through the wide-open valve for a pressure drop of 75.8 psi: g,,=C,&=4.0~=34.8gpm Since the flow through the wide-open valve of 34.8 gpm at a pressure drop across the valve of 75.8 psi is greater than the selected value of 30 gpm, which was used to begin the calculation, we know the valve must be partially closed. Since the valve is linear, we calculate the lift x as follows: x = 30134.8 = 0.86 By means of similar calculations, several points on the effective characteristic curve of Fig. 20.5 can be found; the results are summarized in Table 20.1. The results shown in this table were used to obtain Curve II in Fig. 20.5. Example 20.3. A control valve is to be installed in the flow system of Fig. 20.4. The valve is supplied by water at 5 C through 200 ft of pipe having an inside diameter of 1.0 in. The total pressure drop, po - ~1, is 100 psi. When the valve is wide-open, the flow is to be 30 gpm. Determine C, for the valve. Plot the effective characteristic curve for the valve as flow versus lift. Do this problem for a linear valve and for an equal percentage valve. The equal percentage valve has an mu of 0.03. Linear Valve. To obtain the pressure drop in the line, use is made of Eqs. (20.11) and (20.12) as was done in Example 20.2. From Eq. (20.12), we obtain the Reynolds number as follows: q = 30/(60)(7.48) = 0.0668 ft3/sec or 240.6 ft3/hr = 24 2 psi *
TABLE 20.1
Effective characteristic for a linear valve with supply lie loss (Example 20.2). 4, am
0 20 30 33 x, fraction lift 0 0.53 0.86 1.0 Ap in line, psi 0 10.8 24.2 30.0
CONTROL VALVES
From a correlation for the fanning friction factor, we obtain f = 0.005. From Eq. (20.1 l) , the line loss is calculated to be: Ap = (32)(0.005)(200)(62.4)(0.0668)* (144)(~2)(32.2)(1/12)5 ApV = 100 - 48.5 = 51.5psi From knowledge of the maximum flow through the wide-open valve (30 gpm) and Ap,,, we calculate C, from Eq. (20.1) as follows: = 4.18 From CV, one can now calculate the stem position n needed for various flow rates m. For q = 20 gpm, one obtains from Eq. (20.11) Ap = 21.6 psi and Apy = 100 - 21.6 = 78.4 psi For a wide-open valve (x = l), across which the pressure drop is 78.4 psi, we obtain q,,=CV,@=4.18E=37.0gpm The fraction of lift needed to reduce the flow to 20 gpm is x = 20137 = 0.54 For other flow rates, one can repeat this calculation to obtain values of x . The results are shown in Table 20.2 and in Fig. 20.6. The latter also shows the inherent characteristic of the linear valve for comparison with the effective characteristic of the valve when line loss is present. TABLE 20.2 = 48Spsi
Eff ective characteristics for a linear valve and an equal percentage valve (Example 20.3).
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