This notebook contains course material from FAT0413156 by Gilberto Xavier (gmxavier at fat.uerj.br); the content is available on Github. The text is released under the CC-BY-NC-SA-4.0 license, and code is released under the MIT license.
Control and safety valves¶
Control valves¶
from IPython.display import YouTubeVideo
YouTubeVideo('DVd0TW7HffU') # globe
Sliding-stem valves (Kuphaldt, 2019, p. 2084)¶
YouTubeVideo('8hfvHXhLvGY') # globe
YouTubeVideo('C5ZMLWujKGs') # gate
YouTubeVideo('dxelSWY6bhg') # diaphragm
Rotary stem valves (Kuphaldt, 2019, p. 2096)¶
YouTubeVideo('qOvbgVLjoK8') # ball
YouTubeVideo('wPr3BCE9BgM') # butterfly
YouTubeVideo('xVrNx-nrZA4') # plug
YouTubeVideo('ubTcwCh9I4w') # disk
Control valve actuators (Kuphaldt, 2019, p. 2112)¶
YouTubeVideo('BdmNFyN8tB8') # spring-diaphragm
YouTubeVideo('xo8DPGACE-o') # piston
YouTubeVideo('WAI5Lcu_QXE') # rotary actuators
Actuator bench-set (Kuphaldt, 2019, p. 2137)¶
YouTubeVideo('xGGk88FMyyk') # bench-set
YouTubeVideo('BPJUGkdKCJ8') # bench-set
Valve positioners (Kuphaldt, 2019, p. 2146)¶
YouTubeVideo('dNq4H9WfrfE')
Control valve sizing (Kuphaldt, 2019, p. 2179)¶
YouTubeVideo('CNTs3xH_WHw')
Incompressible fluids by hand (ISA, 2007, p. 14)¶
YouTubeVideo('ldLkbV3W7Pk') # incompressible
Incompressible fluids using Python¶
The function size_control_valve_l from the great Caleb Bell's fluids package calculates flow coefficient of a control valve passing a liquid according to ISA-75.01.01-2007.
Let's try to solve ISA-75.01.01-2007 (IEC 60534-2-1 Mod) Example 1.
In this case, for a globe, parabolic plug, flow-to-open valve passing water, we have:
try:
import fluids
except:
!pip install fluids
from fluids.control_valve import size_control_valve_l, Kv_to_Cv
Kv = size_control_valve_l(rho=965.4, Psat=70.1E3, Pc=22120E3, mu=3.1472E-4, P1=680E3, P2=220E3, Q=0.1,
D1=0.15, D2=0.15, d=0.15, FL=0.9, Fd=0.46)
print('Kv = %.0f mˆ3/h' % Kv) # hand calculation gives Kv = 165 mˆ3/h
Kv = 165 mˆ3/h
Now, with ISA-75.01.01-2007 (IEC 60534-2-1 Mod) Example 2 for a ball, segmented ball, flow-to-open valve passing water, we have:
Kv = size_control_valve_l(rho=965.4, Psat=70.1E3, Pc=22120E3, mu=3.1472E-4, P1=680E3, P2=220E3, Q=0.1,
D1=0.1, D2=0.1, d=0.1, FL=0.6, Fd=0.98)
print('Kv = %.0f mˆ3/h' % Kv) # hand calculation gives Kv = 238 mˆ3/h
Kv = 238 mˆ3/h
To illustrate a more complex case, we borrow an example from Emerson's Control Valve Handbook. It involves a flow of 800 gpm of liquid propane. The inlet and outlet pipe size is 8 inches, but the size of the valve itself is unknown. The desired pressure drop is 25 psi.
Using the scipy package to converting this problem to SI units and the other great Caleb Bell's thermo package to obtain the necessary properties of the fluid, we calculate the necessary Kv of the valve based on an assumed valve size of 3 inches:
try:
import thermo
except:
!pip install thermo
from scipy.constants import *
from thermo.chemical import Chemical
P1 = 300*psi + 14.7*psi # to Pa
P2 = 275*psi + 14.7*psi # to Pa
T = 273.15 + 21 # to K
propane = Chemical('propane', P=(P1+P2)/2, T=T)
rho = propane.rho
Psat = propane.Psat
Pc = propane.Pc
mu = propane.mu
Q = 800*gallon/minute # to m^3/s
D1 = D2 = 8*inch # to m
d = 3*inch # to m
The standard specifies two more parameters specific to a valve:
- FL, Liquid pressure recovery factor of a control valve without attached fittings
- Fd, Valve style modifier
Both of these are factors between 0 and 1. In the borrowed example, they are not considered in the sizing procedure and set to 1. These factors are also a function of the diameter of the valve and are normally tabulated next to the values of Cv or Kv for a valve. Again, using size_control_valve_l to solve for the flow coefficient:
Kv = size_control_valve_l(rho, Psat, Pc, mu, P1, P2, Q, D1, D2, d, FL=1, Fd=1)
print('Cv = %.1f' % Kv_to_Cv(Kv=Kv)) # hand calculation gives Cv = 125.7
Cv = 126.4
As this value exceeds the capacity of the assumed valve, which has a Cv of 121, we will need to try a larger valve size.
d = 4*inch # to m
Kv = size_control_valve_l(rho, Psat, Pc, mu, P1, P2, Q, D1, D2, d, FL=1, Fd=1)
print('Cv = %.1f' % Kv_to_Cv(Kv=Kv)) # hand calculation gives Cv = 125.7
Cv = 116.1
As the new calculated Cv is well under the larger valve’s maximum rated Cv, which has a value of 203, we can select this larger valve size.
This model requires a vapor pressure and a critical pressure of the fluid as inputs. There is no clarification in the standard about how to handle mixtures, which do not have these values. It is reasonable to calculate vapor pressure as the bubble pressure, and the mixture’s critical pressure through a mole-weighted average.
For actual values of Cv, Fl, Fd, and available diameters, an excellent resource is the Fisher Catalog #12.
Compressible fluids by hand (ISA, 2007, p. 16)¶
YouTubeVideo('yQypZR-ocoQ') # compressible
Compressible fluids using Python¶
The function size_control_valve_g from the great Caleb Bell's fluids package calculates flow coefficient of a control valve passing a liquid according to ISA-75.01.01-2007.
From ISA-75.01.01-2007 (IEC 60534-2-1 Mod) Example 3 for non-choked gas flow with attached fittings and a rotary, eccentric plug, flow-to-open control valve passing carbon dioxide, we have:
from fluids.control_valve import size_control_valve_g
Kv = size_control_valve_g(T=433., MW=44.01, mu=1.4665E-4, gamma=1.30, Z=0.988, P1=680E3, P2=310E3, Q=38/36.,
D1=0.08, D2=0.1, d=0.05,FL=0.85, Fd=0.42, xT=0.60)
print('Kv = %.1f mˆ3/h' % Kv) # hand calculation gives Kv = 72.1 m3/h
Kv = 72.6 mˆ3/h
From ISA-75.01.01-2007 (IEC 60534-2-1 Mod) Example 4 for a small flow trim sized tapered needle plug valve passing water, we have:
Kv = size_control_valve_g(T=320., MW=39.95, mu=5.625E-5, gamma=1.67, Z=1.0, P1=2.8E5, P2=1.3E5, Q=0.46/3600.,
D1=0.015, D2=0.015, d=0.015, FL=0.98, Fd=0.07, xT=0.8)
print('Kv = %.4f mˆ3/h' % Kv) # hand calculation gives Cv = 0.0184
Kv = 0.0165 mˆ3/h
To illustrate a more complex case, we borrow an example from Emerson's Control Valve Handbook. It involves a flow of 6 million scfh of natural gas. The inlet and outlet pipe size is 8 inches, but the size of the valve itself is unknown. The desired pressure drop is 150 psi.
Using the scipy package to converting this problem to SI units and the other great Caleb Bell's thermo package to obtain the necessary properties of the fluid, we calculate the necessary Kv of the valve based on an assumed valve size of 8 inches:
P1 = 200*psi + 14.7*psi # to Pa
P2 = 50*psi + 14.7*psi # to Pa
T = 273.15 + 16 # to K
natural_gas = Chemical('natural gas')
natural_gas.calculate(T=T, P=(P1+P2)/2)
Z = natural_gas.Z
MW = natural_gas.MW
mu = natural_gas.mul
gamma = natural_gas.isentropic_exponent
Q = 6E6*foot**3/hour # to m^3/s
D1 = D2 = d = 8*inch # 8-inch Fisher Design V250
The standard specifies three more parameters specific to a valve:
- FL, Liquid pressure recovery factor of a control valve without attached fittings
- Fd, Valve style modifier
- xT, Pressure difference ratio factor of a valve without fittings at choked flow
All three of these are factors between 0 and 1. In the borrowed example, FL and Fd are not considered in the sizing procedure and set to 1. xT is specified as 0.137 at full opening. These factors are also a function of the diameter of the valve and are normally tabulated next to the values of Cv or Kv for a valve. Again using size_control_valve_g to solve for the flow coefficient:
Kv = size_control_valve_g(T, MW, mu, gamma, Z, P1, P2, Q, D1, D2, d, FL=1, Fd=1, xT=.137)
print('Cv = %.0f' % Kv_to_Cv(Kv=Kv))
Cv = 1522
As the calculated Cv is well under the 8-inch valve maximum rated Cv, which has a value of 2190, the valve is adequate to provide the desired flow.
The calculated Cv value in the borrowed example is 1504, differing slightly due to the natural gas properties used (look at the molar mass difference!).
print('The hand calculation assumes a molar mass value of 17.38 kg/kmol and here we used a value of %.2f kg/kmol'% natural_gas.MW)
The hand calculation assumes a molar mass value of 17.38 kg/kmol and here we used a value of 16.04 kg/kmol
The example next goes on to determine the actual opening position the valve should be set at to provide the required flow. Their conclusion is approximately 83% open; we can do better using a numerical solver. The values of opening at different positions are obtained in this example from the Fisher Catalog #12.
Loading the data and creating interpolation functions so FL, Fd, and xT are all smooth functions:
# loading some libraries
from scipy.interpolate import interp1d
from scipy.optimize import newton
# taking values from the catalog, page 325
openings = [20, 30, 40, 50, 60, 70, 80, 90] # in degrees
Fds = [0.59, 0.75, 0.85, 0.92, 0.96, 0.98, 0.99, 0.99]
Fls = [0.9, 0.9, 0.9, 0.85, 0.78, 0.68, 0.57, 0.45]
xTs = [0.92, 0.81, 0.85, 0.63, 0.58, 0.48, 0.29, 0.14]
Kvs = [24.1, 79.4, 153, 266, 413, 623, 1060, 1890]
# creating the smooth interpolation functions to each coefficient
Fd_interp = interp1d(openings, Fds, kind='cubic')
Fl_interp = interp1d(openings, Fls, kind='cubic')
xT_interp = interp1d(openings, xTs, kind='cubic')
Kv_interp = interp1d(openings, Kvs, kind='cubic')
Creating and solving the objective function:
# creating the objective function
def to_solve(opening):
Fd = float(Fd_interp(opening))
Fl = float(Fl_interp(opening))
xT = float(xT_interp(opening))
Kv_lookup = float(Kv_interp(opening))
Kv_calc = size_control_valve_g(T, MW, mu, gamma, Z, P1, P2, Q, D1, D2, d, FL=Fl, Fd=Fd, xT=xT)
return Kv_calc - Kv_lookup
# solving with Newton's method
Xv = newton(to_solve, 60) # initial guess of 60 degrees
print('Cv = %.0f' % Kv_to_Cv(Kv=float(Kv_interp(Xv)))) # hand calculation gives Cv = 923
print('xT = %.3f' % float(xT_interp(Xv))) # hand calculation gives xT = 0.372
print('Xv = %.0f degrees' % Xv) # hand calculation gives Xv = 74 degrees
Cv = 883 xT = 0.407 Xv = 74 degrees
The hand calculated Xv value in the borrowed example is 74 degrees, which agrees exactly with our result.
Again, the Cv values differing due to the difference between the natural gas composition as shown earlier.
Control valve problems (Kuphaldt, 2019, p. 2207)¶
YouTubeVideo('gRsvO4Gpnf0') # flashing/cavitation
YouTubeVideo('6gZqvykrxJ0') # noise