12_Control_Valves_Chap9S10_4

12_Control_Valves_Chap9S10_4 - Chapter 9 Chapter 9 Dynamic...

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Unformatted text preview: Chapter 9 Chapter 9 Dynamic Measurement (vs. steady-state) A “lumped” energy balance with T and Tm of the same units for the thermowell with m = mass, C = heat capacity (per mass basis) dT mC m = UA (T − Tm ) (9-13) dt where U is the heat transfer coefficient and A is the heat transfer area. Rearranging gives ⎛ mC ⎞ dTm + Tm = T ⎜ ⎟ ⎝ UA ⎠ dt (9-14) Converting to deviation variables and taking the Laplace to deviation variables and taking the Laplace transform gives ′ Tm ( s ) 1 = , not simply equal to 1. T ′ ( s ) τs + 1 where τ Figure 9.16 Schematic diagram of a thermowell/thermocouple. (9-15) (mC / UA) has time units used in U definition. 1 2 Control Valves • However, a simple and widely used method of accomplishing this result with fluids is to use a control valve, also called an automatic control valve. Chapter 9 Chapter 9 • There are many different ways to manipulate the flows of material and energy into and out of a process; for example, the speed of a pump drive, screw conveyer, or blower can be adjusted. • The control valve components include the valve body, trim, seat, and actuator. Air-to-Open vs. Air-to-Close Control Valves Figure 9.7 A pneumatic control valve (air-to-open). • Normally, the choice of A-O (air-to-open) or A-C (air-toclose) valve is based on safety considerations. … a few more pictures 8 9 Valve Positioners Chapter 9 Chapter 9 Pneumatic control valves can be equipped with a valve positioner, a type of mechanical or digital feedback controller that senses the actual stem position, compares it to the desired position, and adjusts the air pressure to the valve accordingly. Specifying and Sizing Control Valves A design equation used for sizing control valves relates valve lift to the actual flow rate q by means of the valve coefficient Cv, the proportionality factor that depends predominantly on valve size or capacity: q = Cv f () ΔPv gs (9-2) 10 11 • Here q is the flow rate, f ( ) is the flow characteristic, ΔPv is the pressure drop across the valve, and gs is the specific gravity of the fluid. • Specification of the valve size is dependent on the so-called flow of the valve size is dependent on the so flow (or valve) characteristic f. • Three flow characteristics are mainly used for control valves. • For a fixed pressure drop across the valve, the flow “characteristic” f ( 0 ≤ f ≤ 1) is related to the lift ( 0 ≤ ≤ 1) , that is, fraction of valve opening, by one of the following that is, fraction of valve opening, by one of the following relations: Linear: f= Quick opening: Equal percentage: f= f =R (9-3) Chapter 9 Chapter 9 • This relation is valid for nonflashing fluids (single phase). Figure 9.8 Control valve characteristics. −1 f is the flow characteristic, not flow rate = q ! 12 13 where R is a valve design parameter that is usually in the range of 20 to 50. The rangeability of a control valve is defined as the ratio of maximum to minimum input signal level. For control valves, rangeability translates to the need to operate the valve within the range 0.05 ≤ f ≤ 0.95 or a rangeability of 0.95/0.05 = 19. Control valves are normally not operated to cut off flow completely! To “Design” / Select an Equal Percentage Valve: a) Plot the pump characteristic curve and ΔPs , the system pressure drop curve without pressure drop curve without the valve, as shown in Fig. 9.10. valve, as shown in Fig. 9.10. The difference between these two curves is ΔPv . The pump should be sized to obtain the desired ratio of ΔPv / ΔPs , for example, a ratio of 0.25 – 0.33 = the pressure drop across the valve is 25 to 33% of the drop across the equipment, at the design flow rate qd. Chapter 9 Chapter 9 Rangeability Figure 9.10 Calculation of the valve pressure drop ( ΔPv ) from the pump characteristic curve and the system pressure drop without the valve ( ΔPs ) .As flow increases, pressure required to force fluid through equipment increases, but pressure available from pump often decreases ! 14 15 How would you calculate the numbers for these curves ? For case (a) linear valve, q is NOT linear with lift ! c) Compute q as a function of using Eq. 9-2, the rated Cv, and ΔPv from (a). A plot of the valve characteristic (q vs. ) sho should be “reasonably” linear in the operating region of be “reasonabl linear in the operating region interest (at least around the design flow rate). If it is not suitably linear, adjust the rated Cv and repeat. Example 9.2 – pg. 219 SEM2 A pump furnishes a constant head of 40 psi over the entire flow rate range of interest The heat exchanger pressure drop 30 psig rate range of interest. The heat exchanger pressure drop is 30 psig at 200 gal/min (qd) and can be assumed to be proportional to q 2. Select the rated Cv of the valve and plot the installed characteristic for the following case: a) A linear valve that is half open (l = 0.5) at the design flow rate. 16 Chapter 9 Chapter 9 b) Calculate the valve’s rated Cv, the value that yields at least 100% of qd with the available pressure drop at that higher flow rate. Figure 9.12 Installed valve characteristics for Example 9.2. 17 Solution This should be Chapter 9 Chapter 9 First we write an expression for the pressure drop across the heat exchanger (assumes pressure drop is proportional to square of flow rate) 2 ΔPhe ⎛ q ⎞ (9-5) =⎜ ⎟ 30 ⎝ 200 ⎠ ΔPv Figure 9.9 A control valve placed in series with a pump and a heat exchanger. Pump discharge pressure is constant. di ⎛q⎞ ΔPs = ΔPhe = 30 ⎜ ⎟ ⎝ 200 ⎠ 2 (9-6) Because the pump head is constant at 40 psi, the pressure drop available for the valve is ⎛q⎞ ΔPv = 40 − ΔPhe = 40 − 30 ⎜ ⎟ ⎝ 200 ⎠ 2 (9-7) Figure 9.11 illustrates these relations. Note that in all four design cases ΔPv / ΔPs = 10 / 30 = 33% at qd. 18 19 a) First calculate the rated Cv from q = Cv f for f = 0.5, gs = 1.0. Chapter 9 Chapter 9 Cv = Figure 9.11 Pump characteristic and system pressure drop for Example 9.2. As valve is opened to allow more flow, the pressure drop across the valve drops and you get less flow than if ΔPv had stayed constant. () 200 gpm = 126.5 0.5 (10 psig/1.0) ΔPv gs (9-2) (9-8) We will “pick” Cv = 125. For a linear characteristic valve, use the relation between f and q from Eq. 9-2 with f = : q = where ΔPv depends on q (9-9) Cv ΔPv / g s Using Eq. 9-9 and values of ΔPv ( q ) from Eq. 9-7 for a set of values of flow q, the installed valve characteristic curve (q vs. ) can be plotted from values of for a set of q values. 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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