BASIC COMPONENTS OF CONTROL SYSTEMSJUAN BLANCO CASTELLÓNDINELLY LANDÁZURY GALÉMAILIN MÁRQUEZ PÉREZZARA SANMARTÍN ÁLVAREZBAYRON VERBEL DE LEÓN

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5-2.3 Control Valve CharacteristicsIf the valve is closed If the valve is fully opened

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The particular function relating the coefficient to the valve position is known as the inherent valve characteristics.For linear characteristics

For equal percentage characteristics

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Valve Rangeability The valve rangeability is the ratio of the maximun controlable flow to the minimun controlable flow. Another definition uses 90 and 10% valve positions.

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Installed Valve Characteristics Assumptions: 1. The pressure drop in the line and equipment in series with the valve varies with the square of the flow. 2. There is a total pressure drop that is independent of flow

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Frictional pressure drop Pressure drop across the valve Where f= flow through the valve and line , gpm =constant friction coefficient for the line, fittings, equipment. Etc., = specific gravity of the liquid (water=1)

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Total pressure drop

Solving for the flow in (6)

We divide (7) by (8)

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EXAMPLE 5-2.4: Installed Flow Characteristics of a Liquid Valve For the valve of Example 5-2.3, find the maximun flow though the valve, the installed flow characteristics, and the rangeability of the valve. Assume both linear and equal percentage characteristics with a rangeability parameter of 50. Analyze the effect of varying the pressure drop across the valve at nominal flow.

8

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From the example 5-2.3. The valve is an 8-in valve with The line friction coefficient

The total pressure drop

The maximum flow

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To obtain the valve rangeability LINEAR CHARACTERISTICS EQUAL PERCENTAGE CHARACTERISTICS

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Results for pressure drops across the valve of 2 psi, 5 psi and 10 psiValve pressure drop (psi)2 5 10Total pressure drop (psi) 8 11 16

Calculated Cvmax 960 607 429Required valve size 10-in 8-in 8-in

Actual Cvmax 1000 640 640Maximum flow, gpm 779 870 1049Linear rangeability 5.4 7.9 7.9

Equal % rangeability 10.8 15.8 15.8 0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 100.20.40.60.8

1Equal percentage character-istics with α=50

2 psi5 psi10 psiInherent

% valve position

Flow, % o

f maximu

m

0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 100.20.40.60.8

1LINEAR INHERENT CHAR-ACTERISTICS

2 psi5 psi10 psiInherent

% valve position

flow, % o

f the max

imum

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5-2.4 Control Valve Gain and Transfer Function (10)

Dependence of vp on m If the valve fails closed (air-to-open) we use the plus sign (+) If the valve fails open (air-to-close) we use the minus sign (-)

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Dependence of Cv on vp For linear characteristics For equal percentage

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Dependence of f on Cv Constant valve pressure drop The gain of a valve with linear characteristics (replace 14, 12 and 11 in 10)

(15)

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The gain of a valve with equal percentage characteristics (replace 14, 13 and 11 in 10) (16)

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Variable pressure Drop across the Valve (17)

For the linear characteristics valve, the gain is obtanined by substituying equations (11, 12 and 17) (18)

For the equal percentage valve, the gain is obtanided by sustituying equations 11, 13 and 17 (19)

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EXAMPLE 5-2.5: Gain of steam valve with constant pressure drop Find the gain and the valve position at design conditions for the steam valve of Example 5-2.2. Assume that the 10-in. valve with =1000 is selected and that the pressures around the valve are independent of flow. Consider both a valve with linear characteristics and an equal percentage valve with rangeability parameter of 50. For the latter, find the gain at the nominal flow of 16,100 lb/h.Solution Assume valve fails closed to prevent overheating the reboiler.

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The valve position at design flow is found for:

The gain is obtained from:

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For the equal percentage valve with α=50, the valve position is calculated using:

The gain is obtained using:

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EXAMPLE 5-2.6: Valve gain with variable pressure drop.Calculate the gain of the valve in Example 5-2.4 at nominal flow. Consider both a linear valve and an equal percentage valve with rangeability parameter α=50.From Examples 5-2.3 and 5-2.4:

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SolutionAssume valve fails closedIn this case the gain is obtained by:

The gain of the equal percentage valve is obtained from:

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Valve transfer functionResulting the following transfer function

Where,

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Thanks for your atention!