Unformatted text preview: Chapter 9 Chapter 9 Dynamic Measurement (vs. steadystate)
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 )
(913)
dt
where U is the heat transfer coefficient and A is the heat transfer
area. Rearranging gives
⎛ mC ⎞ dTm
+ Tm = T
⎜
⎟
⎝ UA ⎠ dt (914) 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. (915) (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. AirtoOpen vs. AirtoClose Control Valves Figure 9.7 A pneumatic control valve (airtoopen). • Normally, the choice of AO (airtoopen) or AC (airtoclose) 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 (92) 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 socalled 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 (93) 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. 92, 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 ⎞
(95)
=⎜
⎟
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 (96) 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 (97) 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 (92) (98) We will “pick” Cv = 125. For a linear characteristic valve, use
the relation between f and q from Eq. 92 with f = :
q
=
where ΔPv depends on q (99)
Cv ΔPv / g s
Using Eq. 99 and values of ΔPv ( q ) from Eq. 97 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 “ys = q values” and
calculate “xs = 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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 Winter '08
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 pressure drop, Valve, ΔPv

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