# -*- coding: utf-8 -*-
'''Chemical Engineering Design Library (ChEDL). Utilities for process modeling.
Copyright (C) 2016, 2017 Caleb Bell <Caleb.Andrew.Bell@gmail.com>
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
SOFTWARE.'''
from __future__ import division
from math import cos, sin, tan, atan, pi, radians
import numpy as np
from scipy.constants import inch
from fluids.friction import friction_factor
__all__ = ['contraction_sharp', 'contraction_round',
'contraction_conical', 'contraction_beveled', 'diffuser_sharp',
'diffuser_conical', 'diffuser_conical_staged', 'diffuser_curved',
'diffuser_pipe_reducer',
'entrance_sharp', 'entrance_distance', 'entrance_angled',
'entrance_rounded', 'entrance_beveled', 'entrance_beveled_orifice',
'exit_normal', 'bend_rounded',
'bend_miter', 'helix', 'spiral','Darby3K', 'Hooper2K', 'Kv_to_Cv', 'Cv_to_Kv',
'Kv_to_K', 'K_to_Kv', 'Cv_to_K', 'K_to_Cv', 'change_K_basis', 'Darby',
'Hooper', 'K_gate_valve_Crane', 'K_angle_valve_Crane', 'K_globe_valve_Crane',
'K_swing_check_valve_Crane', 'K_lift_check_valve_Crane',
'K_tilting_disk_check_valve_Crane', 'K_globe_stop_check_valve_Crane',
'K_angle_stop_check_valve_Crane', 'K_ball_valve_Crane',
'K_diaphragm_valve_Crane', 'K_foot_valve_Crane', 'K_butterfly_valve_Crane',
'K_plug_valve_Crane', 'K_branch_converging_Crane', 'K_run_converging_Crane',
'K_branch_diverging_Crane', 'K_run_diverging_Crane', 'v_lift_valve_Crane']
[docs]def change_K_basis(K1, D1, D2):
r'''Converts a loss coefficient `K1` from the basis of one diameter `D1`
to another diameter, `D2`. This is necessary when dealing with pipelines
of changing diameter.
.. math::
K_2 = K_1\frac{D_2^4}{D_1^4} = K_1 \frac{A_2^2}{A_1^2}
Parameters
----------
K1 : float
Loss coefficient with respect to diameter `D`, [-]
D1 : float
Diameter of pipe for which `K1` has been calculated, [m]
D2 : float
Diameter of pipe for which `K2` will be calculated, [m]
Returns
-------
K2 : float
Loss coefficient with respect to the second diameter, [-]
Notes
-----
This expression is shown in [1]_ and can easily be derived:
.. math::
\frac{\rho V_{1}^{2}}{2} \cdot K_{1} = \frac{\rho V_{2}^{2} }{2}
\cdot K_{2}
Substitute velocities for flow rate divided by area:
.. math::
\frac{8 K_{1} Q^{2} \rho}{\pi^{2} D_{1}^{4}} = \frac{8 K_{2} Q^{2}
\rho}{\pi^{2} D_{2}^{4}}
From here, simplification and rearrangement is all that is required.
Examples
--------
>>> change_K_basis(K1=32.68875692997804, D1=.01, D2=.02)
523.0201108796487
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
return K1*(D2/D1)**4
### Entrances
[docs]def entrance_sharp():
r'''Returns loss coefficient for a sharp entrance to a pipe
as shown in [1]_.
.. math::
K = 0.57
.. figure:: fittings/flush_mounted_sharp_edged_entrance.png
:scale: 30 %
:alt: flush mounted sharp edged entrance; after [1]_
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Other values used have been 0.5.
Examples
--------
>>> entrance_sharp()
0.57
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
return 0.57
[docs]def entrance_distance(Di, t):
r'''Returns loss coefficient for a sharp entrance to a pipe at a distance
from the wall of a reservoir, as shown in [1]_.
.. math::
K = 1.12 - 22\frac{t}{d} + 216\left(\frac{t}{d}\right)^2 +
80\left(\frac{t}{d}\right)^3
.. figure:: fittings/sharp_edged_entrace_extended_mount.png
:scale: 30 %
:alt: sharp edged entrace, extended mount; after [1]_
Parameters
----------
Di : float
Inside diameter of pipe, [m]
t : float
Thickness of pipe wall, [m]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Recommended for cases where the length of the inlet pipe extending into a
tank divided by the inner diameter of the pipe is larger than 0.5.
If the pipe is 10 cm in diameter, the pipe should extend into the tank
at least 5 cm. This type of inlet is also known as a Borda's mouthpiece.
It is not of practical interest according to [1]_.
If the pipe wall thickness to diameter ratio `t`/`Di` is larger than 0.05,
it is rounded to 0.05; the effect levels off at that ratio and K=0.57.
Examples
--------
>>> entrance_distance(Di=0.1, t=0.0005)
1.0154100000000001
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
ratio = t/Di
if ratio > 0.05:
ratio = 0.05
return 1.12 - 22.*ratio + 216.*ratio**2 + 80*ratio**3
[docs]def entrance_angled(angle):
r'''Returns loss coefficient for a sharp, angled entrance to a pipe
flush with the wall of a reservoir, as shown in [1]_.
.. math::
K = 0.57 + 0.30\cos(\theta) + 0.20\cos(\theta)^2
.. figure:: fittings/entrance_mounted_at_an_angle.png
:scale: 30 %
:alt: entrace mounted at an angle; after [1]_
Parameters
----------
angle : float
Angle of inclination (90=straight, 0=parallel to pipe wall) [degrees]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Not reliable for angles under 20 degrees.
Loss coefficient is the same for an upward or downward angled inlet.
Examples
--------
>>> entrance_angled(30)
0.9798076211353316
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
angle = angle/(180/pi)
return 0.57 + 0.30*cos(angle) + 0.20*cos(angle)**2
[docs]def entrance_rounded(Di, rc):
r'''Returns loss coefficient for a rounded entrance to a pipe
flush with the wall of a reservoir, as shown in [1]_.
.. math::
K = 0.0696\left(1 - 0.569\frac{r}{d}\right)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622\left(1 - 0.30\sqrt{\frac{r}{d}}
- 0.70\frac{r}{d}\right)^4
.. figure:: fittings/flush_mounted_rounded_entrance.png
:scale: 30 %
:alt: rounded entrace mounted straight and flush; after [1]_
Parameters
----------
Di : float
Inside diameter of pipe, [m]
rc : float
Radius of curvature of the entrance, [m]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
For generously rounded entrance (rc/Di >= 1), the loss coefficient converges
to 0.03.
Examples
--------
>>> entrance_rounded(Di=0.1, rc=0.0235)
0.09839534618360923
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
if rc/Di > 1:
return 0.03
lbd = 1. + 0.622*(1. - 0.30*(rc/Di)**0.5 - 0.70*(rc/Di))**4
return 0.0696*(1. - 0.569*rc/Di)*lbd**2 + (lbd - 1.)**2
[docs]def entrance_beveled(Di, l, angle):
r'''Returns loss coefficient for a beveled or chamfered entrance to a pipe
flush with the wall of a reservoir, as shown in [1]_.
.. math::
K = 0.0696\left(1 - C_b\frac{l}{d}\right)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622\left[1-1.5C_b\left(\frac{l}{d}
\right)^{\frac{1-(l/d)^{1/4}}{2}}\right]
C_b = \left(1 - \frac{\theta}{90}\right)\left(\frac{\theta}{90}
\right)^{\frac{1}{1+l/d}}
.. figure:: fittings/flush_mounted_beveled_entrance.png
:scale: 30 %
:alt: Beveled entrace mounted straight; after [1]_
Parameters
----------
Di : float
Inside diameter of pipe, [m]
l : float
Length of bevel measured parallel to the pipe length, [m]
angle : float
Angle of bevel with respect to the pipe length, [degrees]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
A cheap way of getting a lower pressure drop.
Little credible data is available.
Examples
--------
>>> entrance_beveled(Di=0.1, l=0.003, angle=45)
0.45086864221916984
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
Cb = (1-angle/90.)*(angle/90.)**(1./(1 + l/Di ))
lbd = 1 + 0.622*(1 - 1.5*Cb*(l/Di)**((1 - (l/Di)**0.25)/2.))
return 0.0696*(1 - Cb*l/Di)*lbd**2 + (lbd - 1.)**2
[docs]def entrance_beveled_orifice(Di, do, l, angle):
r'''Returns loss coefficient for a beveled or chamfered orifice entrance to
a pipe flush with the wall of a reservoir, as shown in [1]_.
.. math::
K = 0.0696\left(1 - C_b\frac{l}{d_o}\right)\lambda^2 + \left(\lambda
-\left(\frac{d_o}{D_i}\right)^2\right)^2
\lambda = 1 + 0.622\left[1-C_b\left(\frac{l}{d_o}\right)^{\frac{1-
(l/d_o)^{0.25}}{2}}\right]
C_b = \left(1 - \frac{\Psi}{90}\right)\left(\frac{\Psi}{90}
\right)^{\frac{1}{1+l/d_o}}
.. figure:: fittings/flush_mounted_beveled_orifice_entrance.png
:scale: 30 %
:alt: Beveled orifice entrace mounted straight; after [1]_
Parameters
----------
Di : float
Inside diameter of pipe, [m]
do : float
Inside diameter of orifice, [m]
l : float
Length of bevel measured parallel to the pipe length, [m]
angle : float
Angle of bevel with respect to the pipe length, [degrees]
Returns
-------
K : float
Loss coefficient [-]
Examples
--------
>>> entrance_beveled_orifice(Di=0.1, do=.07, l=0.003, angle=45)
1.2987552913818574
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
Cb = (1-angle/90.)*(angle/90.)**(1./(1 + l/do ))
lbd = 1 + 0.622*(1 - Cb*(l/do)**((1 - (l/do)**0.25)/2.))
return 0.0696*(1 - Cb*l/do)*lbd**2 + (lbd - (do/Di)**2)**2
### Exits
[docs]def exit_normal():
r'''Returns loss coefficient for any exit to a pipe
as shown in [1]_ and in other sources.
.. math::
K = 1
.. figure:: fittings/flush_mounted_exit.png
:scale: 28 %
:alt: Exit from a flush mounted wall; after [1]_
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
It has been found on occasion that K = 2.0 for laminar flow, and ranges
from about 1.04 to 1.10 for turbulent flow.
Examples
--------
>>> exit_normal()
1.0
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
return 1.0
### Bends
[docs]def bend_rounded(Di, angle, fd, rc=None, bend_diameters=5):
r'''Returns loss coefficient for any rounded bend in a pipe
as shown in [1]_.
.. math::
K = f\alpha\frac{r}{d} + (0.10 + 2.4f)\sin(\alpha/2)
+ \frac{6.6f(\sqrt{\sin(\alpha/2)}+\sin(\alpha/2))}
{(r/d)^{\frac{4\alpha}{\pi}}}
.. figure:: fittings/bend_rounded.png
:scale: 30 %
:alt: rounded bend; after [1]_
Parameters
----------
Di : float
Inside diameter of pipe, [m]
angle : float
Angle of bend, [degrees]
fd : float
Darcy friction factor [-]
rc : float, optional
Radius of curvature of the entrance, optional [m]
bend_diameters : float, optional (used if rc not provided)
Number of diameters of pipe making up the bend radius [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
When inputting bend diameters, note that manufacturers often specify
this as a multiplier of nominal diameter, which is different than actual
diameter. Those require that rc be specified.
First term represents surface friction loss; the second, secondary flows;
and the third, flow separation.
Encompasses the entire range of elbow and pipe bend configurations.
This was developed for bend angles between 0 and 180 degrees; and r/D
ratios above 0.5.
Note the loss coefficient includes the surface friction of the pipe as if
it was straight.
Examples
--------
>>> bend_rounded(Di=4.020, rc=4.0*5, angle=30, fd=0.0163)
0.10680196344492195
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
angle = angle/(180/pi)
if not rc:
rc = Di*bend_diameters
return (fd*angle*rc/Di + (0.10 + 2.4*fd)*sin(angle/2.)
+ 6.6*fd*(sin(angle/2.)**0.5 + sin(angle/2.))/(rc/Di)**(4.*angle/pi))
[docs]def bend_miter(angle):
r'''Returns loss coefficient for any single-joint miter bend in a pipe
as shown in [1]_.
.. math::
K = 0.42\sin(\alpha/2) + 2.56\sin^3(\alpha/2)
.. figure:: fittings/bend_mitre.png
:scale: 25 %
:alt: Miter bend, one joint only; after [1]_
Parameters
----------
angle : float
Angle of bend, [degrees]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Applies for bends from 0 to 150 degrees. One joint only.
Examples
--------
>>> bend_miter(150)
2.7128147734758103
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
angle = angle/(180/pi)
return 0.42*sin(angle*0.5) + 2.56*sin(angle*0.5)**3
[docs]def helix(Di, rs, pitch, N, fd):
r'''Returns loss coefficient for any size constant-pitch helix
as shown in [1]_. Has applications in immersed coils in tanks.
.. math::
K = N \left[f\frac{\sqrt{(2\pi r)^2 + p^2}}{d} + 0.20 + 4.8 f\right]
Parameters
----------
Di : float
Inside diameter of pipe, [m]
rs : float
Radius of spiral, [m]
pitch : float
Distance between two subsequent coil centers, [m]
N : float
Number of coils in the helix [-]
fd : float
Darcy friction factor [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Formulation based on peak secondary flow as in two 180 degree bends per
coil. Flow separation ignored. No f, Re, geometry limitations.
Source not compared against others.
Examples
--------
>>> helix(Di=0.01, rs=0.1, pitch=.03, N=10, fd=.0185)
14.525134924495514
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
return N*(fd*((2*pi*rs)**2 + pitch**2)**0.5/Di + 0.20 + 4.8*fd)
[docs]def spiral(Di, rmax, rmin, pitch, fd):
r'''Returns loss coefficient for any size constant-pitch spiral
as shown in [1]_. Has applications in immersed coils in tanks.
.. math::
K = \frac{r_{max} - r_{min}}{p} \left[ f\pi\left(\frac{r_{max}
+r_{min}}{d}\right) + 0.20 + 4.8f\right]
+ \frac{13.2f}{(r_{min}/d)^2}
Parameters
----------
Di : float
Inside diameter of pipe, [m]
rmax : float
Radius of spiral at extremity, [m]
rmin : float
Radius of spiral at end near center, [m]
pitch : float
Distance between two subsequent coil centers, [m]
fd : float
Darcy friction factor [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Source not compared against others.
Examples
--------
>>> spiral(Di=0.01, rmax=.1, rmin=.02, pitch=.01, fd=0.0185)
7.950918552775473
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
return (rmax-rmin)/pitch*(fd*pi*(rmax+rmin)/Di + 0.20 + 4.8*fd) + 13.2*fd/(rmin/Di)**2
### Contractions
[docs]def contraction_sharp(Di1, Di2):
r'''Returns loss coefficient for any sharp edged pipe contraction
as shown in [1]_.
.. math::
K = 0.0696(1-\beta^5)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622(1-0.215\beta^2 - 0.785\beta^5)
\beta = d_2/d_1
.. figure:: fittings/contraction_sharp.png
:scale: 40 %
:alt: Sharp contraction; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe, [m]
Di2 : float
Inside diameter of following pipe, [m]
Returns
-------
K : float
Loss coefficient in terms of the following pipe [-]
Notes
-----
A value of 0.506 or simply 0.5 is often used.
Examples
--------
>>> contraction_sharp(Di1=1, Di2=0.4)
0.5301269161591805
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di2/Di1
lbd = 1 + 0.622*(1-0.215*beta**2 - 0.785*beta**5)
return 0.0696*(1-beta**5)*lbd**2 + (lbd-1)**2
[docs]def contraction_round(Di1, Di2, rc):
r'''Returns loss coefficient for any round edged pipe contraction
as shown in [1]_.
.. math::
K = 0.0696\left(1 - 0.569\frac{r}{d_2}\right)\left(1-\sqrt{\frac{r}
{d_2}}\beta\right)(1-\beta^5)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622\left(1 - 0.30\sqrt{\frac{r}{d_2}}
- 0.70\frac{r}{d_2}\right)^4 (1-0.215\beta^2-0.785\beta^5)
\beta = d_2/d_1
.. figure:: fittings/contraction_round.png
:scale: 30 %
:alt: Cirucular round contraction; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe, [m]
Di2 : float
Inside diameter of following pipe, [m]
rc : float
Radius of curvature of the contraction, [m]
Returns
-------
K : float
Loss coefficient in terms of the following pipe [-]
Notes
-----
Rounding radius larger than 0.14Di2 prevents flow separation from the wall.
Further increase in rounding radius continues to reduce loss coefficient.
Examples
--------
>>> contraction_round(Di1=1, Di2=0.4, rc=0.04)
0.1783332490866574
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di2/Di1
lbd = 1 + 0.622*(1 - 0.30*(rc/Di2)**0.5 - 0.70*rc/Di2)**4*(1-0.215*beta**2 - 0.785*beta**5)
return 0.0696*(1-0.569*rc/Di2)*(1-(rc/Di2)**0.5*beta)*(1-beta**5)*lbd**2 + (lbd-1)**2
[docs]def contraction_conical(Di1, Di2, fd, l=None, angle=None):
r'''Returns loss coefficient for any conical pipe contraction
as shown in [1]_.
.. math::
K = 0.0696[1+C_B(\sin(\alpha/2)-1)](1-\beta^5)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622(\alpha/180)^{0.8}(1-0.215\beta^2-0.785\beta^5)
\beta = d_2/d_1
.. figure:: fittings/contraction_conical.png
:scale: 30 %
:alt: contraction conical; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe, [m]
Di2 : float
Inside diameter of following pipe, [m]
fd : float
Darcy friction factor [-]
l : float
Length of the contraction, optional [m]
angle : float
Angle of contraction, optional [degrees]
Returns
-------
K : float
Loss coefficient in terms of the following pipe [-]
Notes
-----
Cheap and has substantial impact on pressure drop.
Examples
--------
>>> contraction_conical(Di1=0.1, Di2=0.04, l=0.04, fd=0.0185)
0.15779041548350314
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di2/Di1
if angle:
angle = angle/(180/pi)
l = (Di1 - Di2)/(2*tan(angle/2))
elif l:
angle = 2*atan((Di1-Di2)/2/l)
else:
raise Exception('Either l or angle is required')
lbd = 1 + 0.622*(angle/pi)**0.8*(1-0.215*beta**2 - 0.785*beta**5)
return fd*(1-beta**4)/(8*sin(angle/2)) + 0.0696*sin(angle/2)*(1-beta**5)*lbd**2 + (lbd-1)**2
[docs]def contraction_beveled(Di1, Di2, l=None, angle=None):
r'''Returns loss coefficient for any sharp beveled pipe contraction
as shown in [1]_.
.. math::
K = 0.0696[1+C_B(\sin(\alpha/2)-1)](1-\beta^5)\lambda^2 + (\lambda-1)^2
\lambda = 1 + 0.622\left[1+C_B\left(\left(\frac{\alpha}{180}
\right)^{0.8}-1\right)\right](1-0.215\beta^2-0.785\beta^5)
C_B = \frac{l}{d_2}\frac{2\beta\tan(\alpha/2)}{1-\beta}
\beta = d_2/d_1
.. figure:: fittings/contraction_beveled.png
:scale: 30 %
:alt: contraction beveled; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe, [m]
Di2 : float
Inside diameter of following pipe, [m]
l : float
Length of the bevel along the pipe axis ,[m]
angle : float
Angle of bevel, [degrees]
Returns
-------
K : float
Loss coefficient in terms of the following pipe [-]
Notes
-----
Examples
--------
>>> contraction_beveled(Di1=0.5, Di2=0.1, l=.7*.1, angle=120)
0.40946469413070485
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
angle = angle/(180/pi)
beta = Di2/Di1
CB = l/Di2*2*beta*tan(angle/2)/(1-beta)
lbd = 1 + 0.622*(1 + CB*((angle/pi)**0.8-1))*(1-0.215*beta**2-0.785*beta**5)
return 0.0696*(1 + CB*(sin(angle/2)-1))*(1-beta**5)*lbd**2 + (lbd-1)**2
### Expansions (diffusers)
[docs]def diffuser_sharp(Di1, Di2):
r'''Returns loss coefficient for any sudden pipe diameter expansion
as shown in [1]_ and in other sources.
.. math::
K_1 = (1-\beta^2)^2
Parameters
----------
Di1 : float
Inside diameter of original pipe (smaller), [m]
Di2 : float
Inside diameter of following pipe (larger), [m]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Highly accurate.
Examples
--------
>>> diffuser_sharp(Di1=.5, Di2=1)
0.5625
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di1/Di2
return (1. - beta*beta)**2
[docs]def diffuser_conical(Di1, Di2, l=None, angle=None, fd=None):
r'''Returns loss coefficient for any conical pipe expansion
as shown in [1]_. Five different formulas are used, depending on
the angle and the ratio of diameters.
For 0 to 20 degrees, all aspect ratios:
.. math::
K_1 = 8.30[\tan(\alpha/2)]^{1.75}(1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}
For 20 to 60 degrees, beta < 0.5:
.. math::
K_1 = \left\{1.366\sin\left[\frac{2\pi(\alpha-15^\circ)}{180}\right]^{0.5}
- 0.170 - 3.28(0.0625-\beta^4)\sqrt{\frac{\alpha-20^\circ}{40^\circ}}\right\}
(1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}
For 20 to 60 degrees, beta >= 0.5:
.. math::
K_1 = \left\{1.366\sin\left[\frac{2\pi(\alpha-15^\circ)}{180}\right]^{0.5}
- 0.170 \right\}(1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}
For 60 to 180 degrees, beta < 0.5:
.. math::
K_1 = \left[1.205 - 3.28(0.0625-\beta^4)-12.8\beta^6\sqrt{\frac
{\alpha-60^\circ}{120^\circ}}\right](1-\beta^2)^2
For 60 to 180 degrees, beta >= 0.5:
.. math::
K_1 = \left[1.205 - 0.20\sqrt{\frac{\alpha-60^\circ}{120^\circ}}
\right](1-\beta^2)^2
.. figure:: fittings/diffuser_conical.png
:scale: 60 %
:alt: diffuser conical; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe (smaller), [m]
Di2 : float
Inside diameter of following pipe (larger), [m]
l : float
Length of the contraction along the pipe axis, optional[m]
angle : float
Angle of contraction, [degrees]
fd : float
Darcy friction factor [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
For angles above 60 degrees, friction factor is not used.
Examples
--------
>>> diffuser_conical(Di1=1/3., Di2=1, angle=50, fd=0.03)
0.8081340270019336
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di1/Di2
if angle is not None:
angle_rad = angle/(180/pi)
l = (Di2 - Di1)/(2*tan(angle_rad/2))
elif l is not None:
angle_rad = 2*atan((Di2-Di1)/2/l)
angle = angle_rad*(180/pi)
else:
raise Exception('Either `l` or `angle` must be specified')
if 0 < angle <= 20:
K = 8.30*tan(angle_rad/2)**1.75*(1-beta**2)**2 + fd*(1-beta**4)/8./sin(angle_rad/2)
elif 20 < angle <= 60 and 0 <= beta < 0.5:
K = (1.366*sin(2*pi*(angle-15)/180.)**0.5-0.170
- 3.28*(0.0625-beta**4)*((angle-20)/40.)**0.5)*(1-beta**2)**2 + fd*(1-beta**4)/8./sin(angle_rad/2)
elif 20 < angle <= 60 and beta >= 0.5:
K = (1.366*sin(2*pi*(angle-15)/180.)**0.5-0.170)*(1-beta**2)**2 + fd*(1-beta**4)/8./sin(angle_rad/2)
elif 60 < angle <= 180 and 0 <= beta < 0.5:
K = (1.205 - 3.28*(0.0625-beta**4) - 12.8*beta**6*((angle-60)/120.)**0.5)*(1-beta**2)**2
elif 60 < angle <= 180 and beta >= 0.5:
K = (1.205 - 0.20*((angle-60)/120.)**0.5)*(1-beta**2)**2
else:
raise Exception('Conical diffuser inputs incorrect')
return K
[docs]def diffuser_conical_staged(Di1, Di2, DEs, ls, fd=None):
r'''Returns loss coefficient for any series of staged conical pipe expansions
as shown in [1]_. Five different formulas are used, depending on
the angle and the ratio of diameters. This function calls diffuser_conical.
Parameters
----------
Di1 : float
Inside diameter of original pipe (smaller), [m]
Di2 : float
Inside diameter of following pipe (larger), [m]
DEs : array
Diameters of intermediate sections, [m]
ls : array
Lengths of the various sections, [m]
fd : float
Darcy friction factor [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Only lengths of sections currently allowed. This could be changed
to understand angles also.
Formula doesn't make much sense, as observed by the example comparing
a series of conical sections. Use only for small numbers of segments of
highly differing angles.
Examples
--------
>>> diffuser_conical(Di1=1., Di2=10.,l=9, fd=0.01)
0.973137914861591
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
K = 0
DEs.insert(0, Di1)
DEs.append(Di2)
for i in range(len(ls)):
K += diffuser_conical(Di1=float(DEs[i]), Di2=float(DEs[i+1]), l=float(ls[i]), fd=fd)
return K
[docs]def diffuser_curved(Di1, Di2, l):
r'''Returns loss coefficient for any curved wall pipe expansion
as shown in [1]_.
.. math::
K_1 = \phi(1.43-1.3\beta^2)(1-\beta^2)^2
\phi = 1.01 - 0.624\frac{l}{d_1} + 0.30\left(\frac{l}{d_1}\right)^2
- 0.074\left(\frac{l}{d_1}\right)^3 + 0.0070\left(\frac{l}{d_1}\right)^4
.. figure:: fittings/curved_wall_diffuser.png
:scale: 25 %
:alt: diffuser curved; after [1]_
Parameters
----------
Di1 : float
Inside diameter of original pipe (smaller), [m]
Di2 : float
Inside diameter of following pipe (larger), [m]
l : float
Length of the curve along the pipe axis, [m]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Beta^2 should be between 0.1 and 0.9.
A small mismatch between tabulated values of this function in table 11.3
is observed with the equation presented.
Examples
--------
>>> diffuser_curved(Di1=.25**0.5, Di2=1., l=2.)
0.2299781250000002
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
beta = Di1/Di2
phi = 1.01 - 0.624*l/Di1 + 0.30*(l/Di1)**2 - 0.074*(l/Di1)**3 + 0.0070*(l/Di1)**4
return phi*(1.43 - 1.3*beta**2)*(1 - beta**2)**2
[docs]def diffuser_pipe_reducer(Di1, Di2, l, fd1, fd2=None):
r'''Returns loss coefficient for any pipe reducer pipe expansion
as shown in [1]. This is an approximate formula.
.. math::
K_f = f_1\frac{0.20l}{d_1} + \frac{f_1(1-\beta)}{8\sin(\alpha/2)}
+ f_2\frac{0.20l}{d_2}\beta^4
\alpha = 2\tan^{-1}\left(\frac{d_1-d_2}{1.20l}\right)
Parameters
----------
Di1 : float
Inside diameter of original pipe (smaller), [m]
Di2 : float
Inside diameter of following pipe (larger), [m]
l : float
Length of the pipe reducer along the pipe axis, [m]
fd1 : float
Darcy friction factor at inlet diameter [-]
fd2 : float
Darcy friction factor at outlet diameter, optional [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Industry lack of standardization prevents better formulas from being
developed. Add 15% if the reducer is eccentric.
Friction factor at outlet will be assumed the same as at inlet if not specified.
Doubt about the validity of this equation is raised.
Examples
--------
>>> diffuser_pipe_reducer(Di1=.5, Di2=.75, l=1.5, fd1=0.07)
0.06873244301714816
References
----------
.. [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
if fd2 is None:
fd2 = fd1
beta = Di1/Di2
angle = -2*atan((Di1-Di2)/1.20/l)
K = fd1*0.20*l/Di1 + fd1*(1-beta)/8./sin(angle/2) + fd2*0.20*l/Di2*beta**4
return K
### TODO: Tees
### 3 Darby 3K Method (with valves)
Darby = {}
Darby['Elbow, 90°, threaded, standard, (r/D = 1)'] = {'K1': 800, 'Ki': 0.14, 'Kd': 4}
Darby['Elbow, 90°, threaded, long radius, (r/D = 1.5)'] = {'K1': 800, 'Ki': 0.071, 'Kd': 4.2}
Darby['Elbow, 90°, flanged, welded, bends, (r/D = 1)'] = {'K1': 800, 'Ki': 0.091, 'Kd': 4}
Darby['Elbow, 90°, (r/D = 2)'] = {'K1': 800, 'Ki': 0.056, 'Kd': 3.9}
Darby['Elbow, 90°, (r/D = 4)'] = {'K1': 800, 'Ki': 0.066, 'Kd': 3.9}
Darby['Elbow, 90°, (r/D = 6)'] = {'K1': 800, 'Ki': 0.075, 'Kd': 4.2}
Darby['Elbow, 90°, mitered, 1 weld, (90°)'] = {'K1': 1000, 'Ki': 0.27, 'Kd': 4}
Darby['Elbow, 90°, 2 welds, (45°)'] = {'K1': 800, 'Ki': 0.068, 'Kd': 4.1}
Darby['Elbow, 90°, 3 welds, (30°)'] = {'K1': 800, 'Ki': 0.035, 'Kd': 4.2}
Darby['Elbow, 45°, threaded standard, (r/D = 1)'] = {'K1': 500, 'Ki': 0.071, 'Kd': 4.2}
Darby['Elbow, 45°, long radius, (r/D = 1.5)'] = {'K1': 500, 'Ki': 0.052, 'Kd': 4}
Darby['Elbow, 45°, mitered, 1 weld, (45°)'] = {'K1': 500, 'Ki': 0.086, 'Kd': 4}
Darby['Elbow, 45°, mitered, 2 welds, (22.5°)'] = {'K1': 500, 'Ki': 0.052, 'Kd': 4}
Darby['Elbow, 180°, threaded, close-return bend, (r/D = 1)'] = {'K1': 1000, 'Ki': 0.23, 'Kd': 4}
Darby['Elbow, 180°, flanged, (r/D = 1)'] = {'K1': 1000, 'Ki': 0.12, 'Kd': 4}
Darby['Elbow, 180°, all, (r/D = 1.5)'] = {'K1': 1000, 'Ki': 0.1, 'Kd': 4}
Darby['Tee, Through-branch, (as elbow), threaded, (r/D = 1)'] = {'K1': 500, 'Ki': 0.274, 'Kd': 4}
Darby['Tee, Through-branch,(as elbow), (r/D = 1.5)'] = {'K1': 800, 'Ki': 0.14, 'Kd': 4}
Darby['Tee, Through-branch, (as elbow), flanged, (r/D = 1)'] = {'K1': 800, 'Ki': 0.28, 'Kd': 4}
Darby['Tee, Through-branch, (as elbow), stub-in branch'] = {'K1': 1000, 'Ki': 0.34, 'Kd': 4}
Darby['Tee, Run-through, threaded, (r/D = 1)'] = {'K1': 200, 'Ki': 0.091, 'Kd': 4}
Darby['Tee, Run-through, flanged, (r/D = 1)'] = {'K1': 150, 'Ki': 0.05, 'Kd': 4}
Darby['Tee, Run-through, stub-in branch'] = {'K1': 100, 'Ki': 0, 'Kd': 0}
Darby['Valve, Angle valve, 45°, full line size, β = 1'] = {'K1': 950, 'Ki': 0.25, 'Kd': 4}
Darby['Valve, Angle valve, 90°, full line size, β = 1'] = {'K1': 1000, 'Ki': 0.69, 'Kd': 4}
Darby['Valve, Globe valve, standard, β = 1'] = {'K1': 1500, 'Ki': 1.7, 'Kd': 3.6}
Darby['Valve, Plug valve, branch flow'] = {'K1': 500, 'Ki': 0.41, 'Kd': 4}
Darby['Valve, Plug valve, straight through'] = {'K1': 300, 'Ki': 0.084, 'Kd': 3.9}
Darby['Valve, Plug valve, three-way (flow through)'] = {'K1': 300, 'Ki': 0.14, 'Kd': 4}
Darby['Valve, Gate valve, standard, β = 1'] = {'K1': 300, 'Ki': 0.037, 'Kd': 3.9}
Darby['Valve, Ball valve, standard, β = 1'] = {'K1': 300, 'Ki': 0.017, 'Kd': 3.5}
Darby['Valve, Diaphragm, dam type'] = {'K1': 1000, 'Ki': 0.69, 'Kd': 4.9}
Darby['Valve, Swing check'] = {'K1': 1500, 'Ki': 0.46, 'Kd': 4}
Darby['Valve, Lift check'] = {'K1': 2000, 'Ki': 2.85, 'Kd': 3.8}
[docs]def Darby3K(NPS=None, Re=None, name=None, K1=None, Ki=None, Kd=None):
r'''Returns loss coefficient for any various fittings, depending
on the name input. Alternatively, the Darby constants K1, Ki and Kd
may be provided and used instead. Source of data is [1]_.
Reviews of this model are favorable.
.. math::
K_f = \frac{K_1}{Re} + K_i\left(1 + \frac{K_d}{D_{\text{NPS}}^{0.3}}
\right)
Note this model uses nominal pipe diameter in inches.
Parameters
----------
NPS : float
Nominal diameter of the pipe, [in]
Re : float
Reynolds number, [-]
name : str
String from Darby dict representing a fitting
K1 : float
K1 parameter of Darby model, optional [-]
Ki : float
Ki parameter of Darby model, optional [-]
Kd : float
Kd parameter of Darby model, optional [in]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Also described in Albright's Handbook and Ludwig's Applied Process Design.
Relatively uncommon to see it used.
The possibility of combining these methods with those above are attractive.
Examples
--------
>>> Darby3K(NPS=2., Re=10000., name='Valve, Angle valve, 45°, full line size, β = 1')
1.1572523963562353
>>> Darby3K(NPS=12., Re=10000., K1=950, Ki=0.25, Kd=4)
0.819510280626355
References
----------
.. [1] Silverberg, Peter, and Ron Darby. "Correlate Pressure Drops through
Fittings: Three Constants Accurately Calculate Flow through Elbows,
Valves and Tees." Chemical Engineering 106, no. 7 (July 1999): 101.
.. [2] Silverberg, Peter. "Correlate Pressure Drops Through Fittings."
Chemical Engineering 108, no. 4 (April 2001): 127,129-130.
'''
if name:
if name in Darby:
d = Darby[name]
K1, Ki, Kd = d['K1'], d['Ki'], d['Kd']
else:
raise Exception('Name of fitting not in list')
elif K1 and Ki and Kd:
pass
else:
raise Exception('Name of fitting or constants are required')
return K1/Re + Ki*(1. + Kd/NPS**0.3)
### 2K Hooper Method
Hooper = {}
Hooper['Elbow, 90°, Standard (R/D = 1), Screwed'] = {'K1': 800, 'Kinfty': 0.4}
Hooper['Elbow, 90°, Standard (R/D = 1), Flanged/welded'] = {'K1': 800, 'Kinfty': 0.25}
Hooper['Elbow, 90°, Long-radius (R/D = 1.5), All types'] = {'K1': 800, 'Kinfty': 0.2}
Hooper['Elbow, 90°, Mitered (R/D = 1.5), 1 weld (90° angle)'] = {'K1': 1000, 'Kinfty': 1.15}
Hooper['Elbow, 90°, Mitered (R/D = 1.5), 2 weld (45° angle)'] = {'K1': 800, 'Kinfty': 0.35}
Hooper['Elbow, 90°, Mitered (R/D = 1.5), 3 weld (30° angle)'] = {'K1': 800, 'Kinfty': 0.3}
Hooper['Elbow, 90°, Mitered (R/D = 1.5), 4 weld (22.5° angle)'] = {'K1': 800, 'Kinfty': 0.27}
Hooper['Elbow, 90°, Mitered (R/D = 1.5), 5 weld (18° angle)'] = {'K1': 800, 'Kinfty': 0.25}
Hooper['Elbow, 45°, Standard (R/D = 1), All types'] = {'K1': 500, 'Kinfty': 0.2}
Hooper['Elbow, 45°, Long-radius (R/D 1.5), All types'] = {'K1': 500, 'Kinfty': 0.15}
Hooper['Elbow, 45°, Mitered (R/D=1.5), 1 weld (45° angle)'] = {'K1': 500, 'Kinfty': 0.25}
Hooper['Elbow, 45°, Mitered (R/D=1.5), 2 weld (22.5° angle)'] = {'K1': 500, 'Kinfty': 0.15}
Hooper['Elbow, 45°, Standard (R/D = 1), Screwed'] = {'K1': 1000, 'Kinfty': 0.7}
Hooper['Elbow, 180°, Standard (R/D = 1), Flanged/welded'] = {'K1': 1000, 'Kinfty': 0.35}
Hooper['Elbow, 180°, Long-radius (R/D = 1.5), All types'] = {'K1': 1000, 'Kinfty': 0.3}
Hooper['Elbow, Used as, Standard, Screwed'] = {'K1': 500, 'Kinfty': 0.7}
Hooper['Elbow, Elbow, Long-radius, Screwed'] = {'K1': 800, 'Kinfty': 0.4}
Hooper['Elbow, Elbow, Standard, Flanged/welded'] = {'K1': 800, 'Kinfty': 0.8}
Hooper['Elbow, Elbow, Stub-in type branch'] = {'K1': 1000, 'Kinfty': 1}
Hooper['Tee, Run, Screwed'] = {'K1': 200, 'Kinfty': 0.1}
Hooper['Tee, Through, Flanged or welded'] = {'K1': 150, 'Kinfty': 0.05}
Hooper['Tee, Tee, Stub-in type branch'] = {'K1': 100, 'Kinfty': 0}
Hooper['Valve, Gate, Full line size, Beta = 1'] = {'K1': 300, 'Kinfty': 0.1}
Hooper['Valve, Ball, Reduced trim, Beta = 0.9'] = {'K1': 500, 'Kinfty': 0.15}
Hooper['Valve, Plug, Reduced trim, Beta = 0.8'] = {'K1': 1000, 'Kinfty': 0.25}
Hooper['Valve, Globe, Standard'] = {'K1': 1500, 'Kinfty': 4}
Hooper['Valve, Globe, Angle or Y-type'] = {'K1': 1000, 'Kinfty': 2}
Hooper['Valve, Diaphragm, Dam type'] = {'K1': 1000, 'Kinfty': 2}
Hooper['Valve, Butterfly,'] = {'K1': 800, 'Kinfty': 0.25}
Hooper['Valve, Check, Lift'] = {'K1': 2000, 'Kinfty': 10}
Hooper['Valve, Check, Swing'] = {'K1': 1500, 'Kinfty': 1.5}
Hooper['Valve, Check, Tilting-disc'] = {'K1': 1000, 'Kinfty': 0.5}
[docs]def Hooper2K(Di, Re, name=None, K1=None, Kinfty=None):
r'''Returns loss coefficient for any various fittings, depending
on the name input. Alternatively, the Hooper constants K1, Kinfty
may be provided and used instead. Source of data is [1]_.
Reviews of this model are favorable less favorable than the Darby method
but superior to the constant-K method.
.. math::
K = \frac{K_1}{Re} + K_\infty\left(1 + \frac{1\text{ inch}}{D_{in}}\right)
Note this model uses actual inside pipe diameter in inches.
Parameters
----------
Di : float
Actual inside diameter of the pipe, [in]
Re : float
Reynolds number, [-]
name : str, optional
String from Hooper dict representing a fitting
K1 : float, optional
K1 parameter of Hooper model, optional [-]
Kinfty : float, optional
Kinfty parameter of Hooper model, optional [-]
Returns
-------
K : float
Loss coefficient [-]
Notes
-----
Also described in Ludwig's Applied Process Design.
Relatively uncommon to see it used.
No actual example found.
Examples
--------
>>> Hooper2K(Di=2., Re=10000., name='Valve, Globe, Standard')
6.15
>>> Hooper2K(Di=2., Re=10000., K1=900, Kinfty=4)
6.09
References
----------
.. [1] Hooper, W. B., "The 2-K Method Predicts Head Losses in Pipe
Fittings," Chem. Eng., p. 97, Aug. 24 (1981).
.. [2] Hooper, William B. "Calculate Head Loss Caused by Change in Pipe
Size." Chemical Engineering 95, no. 16 (November 7, 1988): 89.
.. [3] Kayode Coker. Ludwig's Applied Process Design for Chemical and
Petrochemical Plants. 4E. Amsterdam ; Boston: Gulf Professional
Publishing, 2007.
'''
if name:
if name in Hooper:
d = Hooper[name]
K1, Kinfty = d['K1'], d['Kinfty']
else:
raise Exception('Name of fitting not in list')
elif K1 and Kinfty:
pass
else:
raise Exception('Name of fitting or constants are required')
return K1/Re + Kinfty*(1. + 1./Di)
### Valves
[docs]def Kv_to_Cv(Kv):
r'''Convert valve flow coefficient from imperial to common metric units.
.. math::
C_v = 1.156 K_v
Parameters
----------
Kv : float
Metric Kv valve flow coefficient (flow rate of water at a pressure drop
of 1 bar) [m^3/hr]
Returns
-------
Cv : float
Imperial Cv valve flow coefficient (flow rate of water at a pressure
drop of 1 psi) [gallons/minute]
Notes
-----
Kv = 0.865 Cv is in the IEC standard 60534-2-1.
It has also been said that Cv = 1.17Kv; this is wrong by current standards.
The conversion factor does not depend on the density of the fluid or the
diameter of the valve. It is calculated with the definition of a US gallon
as 231 cubic inches, and a psi as a pound-force per square inch.
The exact conversion coefficient between Kv to Cv is 1.1560992283536566;
it is rounded in the formula above.
Examples
--------
>>> Kv_to_Cv(2)
2.3121984567073133
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return 1.1560992283536566*Kv
[docs]def Cv_to_Kv(Cv):
r'''Convert valve flow coefficient from imperial to common metric units.
.. math::
K_v = C_v/1.156
Parameters
----------
Cv : float
Imperial Cv valve flow coefficient (flow rate of water at a pressure
drop of 1 psi) [gallons/minute]
Returns
-------
Kv : float
Metric Kv valve flow coefficient (flow rate of water at a pressure drop
of 1 bar) [m^3/hr]
Notes
-----
Kv = 0.865 Cv is in the IEC standard 60534-2-1.
It has also been said that Cv = 1.17Kv; this is wrong by current standards.
The conversion factor does not depend on the density of the fluid or the
diameter of the valve. It is calculated with the definition of a US gallon
as 231 cubic inches, and a psi as a pound-force per square inch.
The exact conversion coefficient between Kv to Cv is 1.1560992283536566;
it is rounded in the formula above.
Examples
--------
>>> Cv_to_Kv(2.312)
1.9998283393826013
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return Cv/1.1560992283536566
[docs]def Kv_to_K(Kv, D):
r'''Convert valve flow coefficient from common metric units to regular
loss coefficients.
.. math::
K = 1.6\times 10^9 \frac{D^4}{K_v^2}
Parameters
----------
Kv : float
Metric Kv valve flow coefficient (flow rate of water at a pressure drop
of 1 bar) [m^3/hr]
D : float
Inside diameter of the valve [m]
Returns
-------
K : float
Loss coefficient, [-]
Notes
-----
Crane TP 410 M (2009) gives the coefficient of 0.04 (with diameter in mm).
It also suggests the density of water should be found between 5-40°C.
Older versions specify the density should be found at 60 °F, which is
used here, and the pessure for the appropriate density is back calculated.
.. math::
\Delta P = 1 \text{ bar} = \frac{1}{2}\rho V^2\cdot K
V = \frac{\frac{K_v\cdot \text{ hour}}{3600 \text{ second}}}{\frac{\pi}{4}D^2}
\rho = 999.29744568 \;\; kg/m^3 \text{ at } T=60° F, P = 703572 Pa
The value of density is calculated with IAPWS-95; it is chosen as it makes
the coefficient a very convenient round number. Others constants that have
been used are 1.604E9, and 1.60045E9.
Examples
--------
>>> Kv_to_K(2.312, .015)
15.153374600399898
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return 1.6E9*D**4*Kv**-2
[docs]def K_to_Kv(K, D):
r'''Convert regular loss coefficient to valve flow coefficient.
.. math::
K_v = 4\times 10^4 \sqrt{ \frac{D^4}{K}}
Parameters
----------
K : float
Loss coefficient, [-]
D : float
Inside diameter of the valve [m]
Returns
-------
Kv : float
Metric Kv valve flow coefficient (flow rate of water at a pressure drop
of 1 bar) [m^3/hr]
Notes
-----
Crane TP 410 M (2009) gives the coefficient of 0.04 (with diameter in mm).
It also suggests the density of water should be found between 5-40°C.
Older versions specify the density should be found at 60 °F, which is
used here, and the pessure for the appropriate density is back calculated.
.. math::
\Delta P = 1 \text{ bar} = \frac{1}{2}\rho V^2\cdot K
V = \frac{\frac{K_v\cdot \text{ hour}}{3600 \text{ second}}}{\frac{\pi}{4}D^2}
\rho = 999.29744568 \;\; kg/m^3 \text{ at } T=60° F, P = 703572 Pa
The value of density is calculated with IAPWS-95; it is chosen as it makes
the coefficient a very convenient round number. Others constants that have
been used are 1.604E9, and 1.60045E9.
Examples
--------
>>> K_to_Kv(15.15337460039990, .015)
2.312
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return D*D*(1.6E9/K)**0.5
[docs]def K_to_Cv(K, D):
r'''Convert regular loss coefficient to imperial valve flow coefficient.
.. math::
K_v = 1.156 \cdot 4\times 10^4 \sqrt{ \frac{D^4}{K}}
Parameters
----------
K : float
Loss coefficient, [-]
D : float
Inside diameter of the valve [m]
Returns
-------
Cv : float
Imperial Cv valve flow coefficient (flow rate of water at a pressure
drop of 1 psi) [gallons/minute]
Notes
-----
The conversion factor does not depend on the density of the fluid or the
diameter of the valve. It is calculated with the definition of a US gallon
as 231 cubic inches, and a psi as a pound-force per square inch.
The exact conversion coefficient between Kv to Cv is 1.1560992283536566;
it is rounded in the formula above.
Examples
--------
>>> K_to_Cv(16, .015)
2.601223263795727
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return 1.1560992283536566*D*D*(1.6E9/K)**0.5
[docs]def Cv_to_K(Cv, D):
r'''Convert imperial valve flow coefficient from imperial units to regular
loss coefficients.
.. math::
K = 1.6\times 10^9 \frac{D^4}{\left(\frac{C_v}{1.56}\right)^2}
Parameters
----------
Cv : float
Imperial Cv valve flow coefficient (flow rate of water at a pressure
drop of 1 psi) [gallons/minute]
D : float
Inside diameter of the valve [m]
Returns
-------
K : float
Loss coefficient, [-]
Notes
-----
The exact conversion coefficient between Kv to Cv is 1.1560992283536566;
it is rounded in the formula above.
Examples
--------
>>> Cv_to_K(2.712, .015)
14.719595348352552
References
----------
.. [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft
'''
return 1.6E9*D**4*(Cv/1.1560992283536566)**-2
[docs]def K_gate_valve_Crane(D1, D2, angle, fd):
r'''Returns loss coefficient for a gate valve of types wedge disc, double
disc, or plug type, as shown in [1]_.
If β = 1 and θ = 0:
.. math::
K = K_1 = K_2 = 8f_d
If β < 1 and θ <= 45°:
.. math::
K_2 = \frac{K + \sin \frac{\theta}{2} \left[0.8(1-\beta^2)
+ 2.6(1-\beta^2)^2\right]}{\beta^4}
If β < 1 and θ > 45°:
.. math::
K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2)
+ (1-\beta^2)^2}{\beta^4}
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
angle : float
Angle formed by the reducer in the valve, [degrees]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions [2]_.
Examples
--------
Example 7-4 in [1]_; a 150 by 100 mm glass 600 steel gate valve, conically
tapered ports, length 550 mm, back of sear ring ~150 mm. The valve is
connected to 146 mm schedule 80 pipe. The angle can be calculated to be
13 degrees. The valve is specified to be operating in turbulent conditions.
>>> K_gate_valve_Crane(D1=.1, D2=.146, angle=13.115, fd=0.015)
1.145830368873396
The calculated result is lower than their value of 1.22; the difference is
due to intermediate rounding.
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
.. [2] Harvey Wilson. "Pressure Drop in Pipe Fittings and Valves |
Equivalent Length and Resistance Coefficient." Katmar Software. Accessed
July 28, 2017. http://www.katmarsoftware.com/articles/pipe-fitting-pressure-drop.htm.
'''
angle = radians(angle)
beta = D1/D2
K1 = 8*fd # This does not refer to upstream loss per se
if beta == 1 or angle == 0:
return K1 # upstream and down
else:
if angle <= pi/4:
K = (K1 + sin(angle/2)*(0.8*(1-beta**2) + 2.6*(1-beta**2)**2))/beta**4
else:
K = (K1 + 0.5*(sin(angle/2))**0.5 * (1 - beta**2) + (1-beta**2)**2)/beta**4
return K
[docs]def K_globe_valve_Crane(D1, D2, fd):
r'''Returns the loss coefficient for all types of globe valve, (reduced
seat or throttled) as shown in [1]_.
If β = 1:
.. math::
K = K_1 = K_2 = 340 f_d
Otherwise:
.. math::
K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_globe_valve_Crane(.01, .02, fd=.015)
87.1
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = D1/D2
K1 = 340*fd
if beta == 1:
return K1 # upstream and down
else:
return (K1 + beta*(0.5*(1-beta)**2 + (1-beta**2)**2))/beta**4
[docs]def K_angle_valve_Crane(D1, D2, fd, style=0):
r'''Returns the loss coefficient for all types of angle valve, (reduced
seat or throttled) as shown in [1]_.
If β = 1:
.. math::
K = K_1 = K_2 = N\cdot f_d
Otherwise:
.. math::
K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}
For style 0 and 2, N = 55; for style 1, N=150.
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
One of 0, 1, or 2; refers to three different types of angle valves
as shown in [1]_ [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_angle_valve_Crane(.01, .02, fd=.016)
19.58
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = D1/D2
if style not in [0, 1, 2]:
raise Exception('Valve style should be 0, 1, or 2')
if style == 0 or style == 2:
K1 = 55*fd
else:
K1 = 150*fd
if beta == 1:
return K1 # upstream and down
else:
return (K1 + beta*(0.5*(1-beta)**2 + (1-beta**2)**2))/beta**4
[docs]def K_swing_check_valve_Crane(fd, angled=True):
r'''Returns the loss coefficient for a swing check valve as shown in [1]_.
.. math::
K_2 = N\cdot f_d
For angled swing check valves N = 100; for straight valves, N = 50.
Parameters
----------
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
angled : bool, optional
If True, returns a value 2x the unangled value; the style of the valve
[-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_swing_check_valve_Crane(fd=.016)
1.6
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
if angled:
return 100.*fd
return 50.*fd
[docs]def K_lift_check_valve_Crane(D1, D2, fd, angled=True):
r'''Returns the loss coefficient for a lift check valve as shown in [1]_.
If β = 1:
.. math::
K = K_1 = K_2 = N\cdot f_d
Otherwise:
.. math::
K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}
For angled lift check valves N = 55; for straight valves, N = 600.
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
angled : bool, optional
If True, returns a value 2x the unangled value; the style of the valve
[-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_lift_check_valve_Crane(.01, .02, fd=.016)
21.58
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = D1/D2
if angled:
K1 = 55*fd
if beta == 1:
return K1
else:
return (K1 + beta*(0.5*(1 - beta**2) + (1 - beta**2)**2))/beta**4
else:
K1 = 600.*fd
if beta == 1:
return K1
else:
return (K1 + beta*(0.5*(1 - beta**2) + (1 - beta**2)**2))/beta**4
[docs]def K_tilting_disk_check_valve_Crane(D, angle, fd):
r'''Returns the loss coefficient for a tilting disk check valve as shown in
[1]_. Results are specified in [1]_ to be for the disk's resting position
to be at 5 or 25 degrees to the flow direction. The model is implemented
here so as to switch to the higher loss 15 degree coefficients at 10
degrees, and use the lesser coefficients for any angle under 10 degrees.
.. math::
K = N\cdot f_d
N is obtained from the following table:
+--------+-------------+-------------+
| | angle = 5 ° | angle = 15° |
+========+=============+=============+
| 2-8" | 40 | 120 |
+--------+-------------+-------------+
| 10-14" | 30 | 90 |
+--------+-------------+-------------+
| 16-48" | 20 | 60 |
+--------+-------------+-------------+
The actual change of coefficients happen at <= 9" and <= 15".
Parameters
----------
D : float
Diameter of the pipe section the valve in mounted in; the
same as the line size [m]
angle : float
Angle of the tilting disk to the flow direction; nominally 5 or 15
degrees [degrees]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_tilting_disk_check_valve_Crane(.01, 5, fd=.016)
0.64
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
if angle < 10:
# 5 degree case
if D <= 0.2286:
# 2-8 inches, split at 9 inch
return 40*fd
elif D <= 0.381:
# 10-14 inches, split at 15 inch
return 30*fd
else:
# 16-18 inches
return 20*fd
else:
# 15 degree case
if D < 0.2286:
# 2-8 inches
return 120*fd
elif D < 0.381:
# 10-14 inches
return 90*fd
else:
# 16-18 inches
return 60*fd
[docs]def K_globe_stop_check_valve_Crane(D1, D2, fd, style=0):
r'''Returns the loss coefficient for a globe stop check valve as shown in
[1]_.
If β = 1:
.. math::
K = K_1 = K_2 = N\cdot f_d
Otherwise:
.. math::
K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}
Style 0 is the standard form; style 1 is angled, with a restrition to force
the flow up through the valve; style 2 is also angled but with a smaller
restriction forcing the flow up. N is 400, 300, and 55 for those cases
respectively.
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
One of 0, 1, or 2; refers to three different types of angle valves
as shown in [1]_ [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_globe_stop_check_valve_Crane(.1, .02, .0165, style=1)
4.51992
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
coeffs = {0: 400, 1: 300, 2: 55}
try:
K = coeffs[style]*fd
except KeyError:
raise KeyError('Accepted valve styles are 0, 1, and 2 only')
beta = D1/D2
if beta == 1:
return K
else:
return (K + beta*(0.5*(1 - beta**2) + (1 - beta**2)**2))/beta**4
[docs]def K_angle_stop_check_valve_Crane(D1, D2, fd, style=0):
r'''Returns the loss coefficient for a angle stop check valve as shown in
[1]_.
If β = 1:
.. math::
K = K_1 = K_2 = N\cdot f_d
Otherwise:
.. math::
K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}
Style 0 is the standard form; style 1 has a restrition to force
the flow up through the valve; style 2 is has the clearest flow area with
no guides for the angle valve. N is 200, 350, and 55 for those cases
respectively.
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be smaller or equal to `D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
One of 0, 1, or 2; refers to three different types of angle valves
as shown in [1]_ [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_angle_stop_check_valve_Crane(.1, .02, .0165, style=1)
4.52124
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
coeffs = {0: 200, 1: 350, 2: 55}
try:
K = coeffs[style]*fd
except KeyError:
raise KeyError('Accepted valve styles are 0, 1, and 2 only')
beta = D1/D2
if beta == 1:
return K
else:
return (K + beta*(0.5*(1 - beta**2) + (1 - beta**2)**2))/beta**4
[docs]def K_ball_valve_Crane(D1, D2, angle, fd):
r'''Returns the loss coefficient for a ball valve as shown in [1]_.
If β = 1:
.. math::
K = K_1 = K_2 = 3f_d
If β < 1 and θ <= 45°:
.. math::
K_2 = \frac{K + \sin \frac{\theta}{2} \left[0.8(1-\beta^2)
+ 2.6(1-\beta^2)^2\right]} {\beta^4}
If β < 1 and θ > 45°:
.. math::
K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2)
+ (1-\beta^2)^2}{\beta^4}
Parameters
----------
D1 : float
Diameter of the valve seat bore (must be equal to or smaller than
`D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
angle : float
Angle formed by the reducer in the valve, [degrees]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_ball_valve_Crane(.01, .02, 50, .025)
14.100545785228675
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = D1/D2
K1 = 3*fd
angle = radians(angle)
if beta == 1:
return K1
else:
if angle <= pi/4:
return (K1 + sin(angle/2)*(0.8*(1-beta**2) + 2.6*(1-beta**2)**2))/beta**4
else:
return (K1 + 0.5*(sin(angle/2))**0.5 * (1 - beta**2) + (1-beta**2)**2)/beta**4
[docs]def K_diaphragm_valve_Crane(fd, style=0):
r'''Returns the loss coefficient for a diaphragm valve of either weir
(`style` = 0) or straight-through (`style` = 1) as shown in [1]_.
.. math::
K = K_1 = K_2 = N\cdot f_d
For style 0 (weir), N = 149; for style 1 (straight through), N = 39.
Parameters
----------
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
Either 0 (weir type valve) or 1 (straight through weir valve) [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_diaphragm_valve_Crane(0.015, style=0)
2.235
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
coeffs = {0: 149, 1: 39}
try:
K = coeffs[style]*fd
except KeyError:
raise KeyError('Accepted valve styles are 0 (weir) or 1 (straight through) only')
return K
[docs]def K_butterfly_valve_Crane(D, fd, style=0):
r'''Returns the loss coefficient for a butterfly valve as shown in
[1]_. Three different types are supported; Centric (`style` = 0),
double offset (`style` = 1), and triple offset (`style` = 2).
.. math::
K = N\cdot f_d
N is obtained from the following table:
+------------+---------+---------------+---------------+
| Size range | Centric | Double offset | Triple offset |
+============+=========+===============+===============+
| 2" - 8" | 45 | 74 | 218 |
+------------+---------+---------------+---------------+
| 10" - 14" | 35 | 52 | 96 |
+------------+---------+---------------+---------------+
| 16" - 24" | 25 | 43 | 55 |
+------------+---------+---------------+---------------+
The actual change of coefficients happen at <= 9" and <= 15".
Parameters
----------
D : float
Diameter of the pipe section the valve in mounted in; the
same as the line size [m]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
Either 0 (centric), 1 (double offset), or 2 (triple offset) [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_butterfly_valve_Crane(.01, .016, style=2)
3.488
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
coeffs = {0: (45, 35, 25), 1: (74, 52, 43), 2: (218, 96, 55)}
try:
c1, c2, c3 = coeffs[style]
except KeyError:
raise KeyError('Accepted valve styles are 0 (centric), 1 (double offset), or 2 (triple offset) only.')
if D <= 0.2286:
# 2-8 inches, split at 9 inch
return c1*fd
elif D <= 0.381:
# 10-14 inches, split at 15 inch
return c2*fd
else:
# 16-18 inches
return c3*fd
[docs]def K_plug_valve_Crane(D1, D2, angle, fd, style=0):
r'''Returns the loss coefficient for a plug valve or cock valve as shown in
[1]_.
If β = 1:
.. math::
K = K_1 = K_2 = Nf_d
Otherwise:
.. math::
K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2)
+ (1-\beta^2)^2}{\beta^4}
Three types of plug valves are supported. For straight-through plug valves
(`style` = 0), N = 18. For 3-way, flow straight through (`style` = 1)
plug valves, N = 30. For 3-way, flow 90° valves (`style` = 2) N = 90.
Parameters
----------
D1 : float
Diameter of the valve plug bore (must be equal to or smaller than
`D2`), [m]
D2 : float
Diameter of the pipe attached to the valve, [m]
angle : float
Angle formed by the reducer in the valve, [degrees]
fd : float
Darcy friction factor calculated for the actual pipe flow in clean
steel (roughness = 0.0018 inch) in the fully developed turbulent
region [-]
style : int, optional
Either 0 (straight-through), 1 (3-way, flow straight-through), or 2
(3-way, flow 90°) [-]
Returns
-------
K : float
Loss coefficient with respect to the pipe inside diameter [-]
Notes
-----
This method is not valid in the laminar regime and the pressure drop will
be underestimated in those conditions.
Examples
--------
>>> K_plug_valve_Crane(.01, .02, 50, .025)
20.100545785228675
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
coeffs = {0: 18, 1: 30, 2: 90}
beta = D1/D2
try:
K = coeffs[style]*fd
except KeyError:
raise KeyError('Accepted valve styles are 0 (straight-through), 1 (3-way, flow straight-through), or 2 (3-way, flow 90°)')
angle = radians(angle)
if beta == 1:
return K
else:
return (K + 0.5*(sin(angle/2))**0.5 * (1 - beta**2) + (1-beta**2)**2)/beta**4
branch_converging_Crane_Fs = np.array([1.74, 1.41, 1, 0])
branch_converging_Crane_angles = np.array([30, 45, 60, 90])
[docs]def K_branch_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
r'''Returns the loss coefficient for the branch of a converging tee or wye
according to the Crane method [1]_.
.. math::
K_{branch} = C\left[1 + D\left(\frac{Q_{branch}}{Q_{comb}\cdot
\beta_{branch}^2}\right)^2 - E\left(1 - \frac{Q_{branch}}{Q_{comb}}
\right)^2 - \frac{F}{\beta_{branch}^2} \left(\frac{Q_{branch}}
{Q_{comb}}\right)^2\right]
\beta_{branch} = \frac{D_{branch}}{D_{comb}}
In the above equation, D = 1, E = 2. See the notes for definitions of F and
C.
Parameters
----------
D_run : float
Diameter of the straight-through inlet portion of the tee or wye [m]
D_branch : float
Diameter of the pipe attached at an angle to the straight-through, [m]
Q_run : float
Volumetric flow rate in the straight-through inlet of the tee or wye,
[m^3/s]
Q_branch : float
Volumetric flow rate in the pipe attached at an angle to the straight-
through, [m^3/s]
angle : float, optional
Angle the branch makes with the straight-through (tee=90, wye<90)
[degrees]
Returns
-------
K : float
Loss coefficient of branch with respect to the velocity and inside
diameter of the combined flow outlet [-]
Notes
-----
F is linearly interpolated from the table of angles below. There is no
cutoff to prevent angles from being larger or smaller than 30 or 90
degrees.
+-----------+------+
| Angle [°] | |
+===========+======+
| 30 | 1.74 |
+-----------+------+
| 45 | 1.41 |
+-----------+------+
| 60 | 1 |
+-----------+------+
| 90 | 0 |
+-----------+------+
If :math:`\beta_{branch}^2 \le 0.35`, C = 1
If :math:`\beta_{branch}^2 > 0.35` and :math:`Q_{branch}/Q_{comb} > 0.4`,
C = 0.55.
If neither of the above conditions are met:
.. math::
C = 0.9\left(1 - \frac{Q_{branch}}{Q_{comb}}\right)
Note that there is an error in the text of [1]_; the errata can be obtained
here: http://www.flowoffluids.com/publications/tp-410-errata.aspx
Examples
--------
Example 7-35 of [1]_. A DN100 schedule 40 tee has 1135 liters/minute of
water passing through the straight leg, and 380 liters/minute of water
converging with it through a 90° branch. Calculate the loss coefficient in
the branch. The calculated value there is -0.04026.
>>> K_branch_converging_Crane(0.1023, 0.1023, 0.018917, 0.00633)
-0.04044108513625682
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = (D_branch/D_run)
beta2 = beta*beta
Q_comb = Q_run + Q_branch
Q_ratio = Q_branch/Q_comb
if beta2 <= 0.35:
C = 1.
elif Q_ratio <= 0.4:
C = 0.9*(1 - Q_ratio)
else:
C = 0.55
D, E = 1., 2.
F = np.interp(angle, branch_converging_Crane_angles, branch_converging_Crane_Fs)
K = C*(1. + D*(Q_ratio/beta2)**2 - E*(1. - Q_ratio)**2 - F/beta2*Q_ratio**2)
return K
run_converging_Crane_Fs = np.array([1.74, 1.41, 1])
run_converging_Crane_angles = np.array([30, 45, 60])
[docs]def K_run_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
r'''Returns the loss coefficient for the run of a converging tee or wye
according to the Crane method [1]_.
.. math::
K_{branch} = C\left[1 + D\left(\frac{Q_{branch}}{Q_{comb}\cdot
\beta_{branch}^2}\right)^2 - E\left(1 - \frac{Q_{branch}}{Q_{comb}}
\right)^2 - \frac{F}{\beta_{branch}^2} \left(\frac{Q_{branch}}
{Q_{comb}}\right)^2\right]
\beta_{branch} = \frac{D_{branch}}{D_{comb}}
In the above equation, C=1, D=0, E=1. See the notes for definitions of F
and also the special case of 90°.
Parameters
----------
D_run : float
Diameter of the straight-through inlet portion of the tee or wye
[m]
D_branch : float
Diameter of the pipe attached at an angle to the straight-through, [m]
Q_run : float
Volumetric flow rate in the straight-through inlet of the tee or wye,
[m^3/s]
Q_branch : float
Volumetric flow rate in the pipe attached at an angle to the straight-
through, [m^3/s]
angle : float, optional
Angle the branch makes with the straight-through (tee=90, wye<90)
[degrees]
Returns
-------
K : float
Loss coefficient of run with respect to the velocity and inside
diameter of the combined flow outlet [-]
Notes
-----
F is linearly interpolated from the table of angles below. There is no
cutoff to prevent angles from being larger or smaller than 30 or 60
degrees. The switch to the special 90° happens at 75°.
+-----------+------+
| Angle [°] | |
+===========+======+
| 30 | 1.74 |
+-----------+------+
| 45 | 1.41 |
+-----------+------+
| 60 | 1 |
+-----------+------+
For the special case of 90°, the formula used is as follows.
.. math::
K_{run} = 1.55\left(\frac{Q_{branch}}{Q_{comb}} \right)
- \left(\frac{Q_{branch}}{Q_{comb}}\right)^2
Examples
--------
Example 7-35 of [1]_. A DN100 schedule 40 tee has 1135 liters/minute of
water passing through the straight leg, and 380 liters/minute of water
converging with it through a 90° branch. Calculate the loss coefficient in
the run. The calculated value there is 0.03258.
>>> K_run_converging_Crane(0.1023, 0.1023, 0.018917, 0.00633)
0.32575847854551254
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = (D_branch/D_run)
beta2 = beta*beta
Q_comb = Q_run + Q_branch
Q_ratio = Q_branch/Q_comb
if angle < 75.:
C = 1
else:
return 1.55*(Q_ratio) - Q_ratio*Q_ratio
D, E = 0, 1
F = np.interp(angle, run_converging_Crane_angles, run_converging_Crane_Fs)
K = C*(1. + D*(Q_ratio/beta2)**2 - E*(1. - Q_ratio)**2 - F/beta2*Q_ratio**2)
return K
[docs]def K_branch_diverging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
r'''Returns the loss coefficient for the branch of a diverging tee or wye
according to the Crane method [1]_.
.. math::
K_{branch} = G\left[1 + H\left(\frac{Q_{branch}}{Q_{comb}
\beta_{branch}^2}\right)^2 - J\left(\frac{Q_{branch}}{Q_{comb}
\beta_{branch}^2}\right)\cos\theta\right]
\beta_{branch} = \frac{D_{branch}}{D_{comb}}
See the notes for definitions of H, J, and G.
Parameters
----------
D_run : float
Diameter of the straight-through inlet portion of the tee or wye [m]
D_branch : float
Diameter of the pipe attached at an angle to the straight-through, [m]
Q_run : float
Volumetric flow rate in the straight-through outlet of the tee or wye,
[m^3/s]
Q_branch : float
Volumetric flow rate in the pipe attached at an angle to the straight-
through, [m^3/s]
angle : float, optional
Angle the branch makes with the straight-through (tee=90, wye<90)
[degrees]
Returns
-------
K : float
Loss coefficient of branch with respect to the velocity and inside
diameter of the combined flow inlet [-]
Notes
-----
If :math:`\beta_{branch} = 1, \theta = 90^\circ`, H = 0.3 and J = 0.
Otherwise H = 1 and J = 2.
G is determined according to the following pseudocode:
.. code-block:: python
if angle < 75:
if beta2 <= 0.35:
if Q_ratio <= 0.4:
G = 1.1 - 0.7*Q_ratio
else:
G = 0.85
else:
if Q_ratio <= 0.6:
G = 1.0 - 0.6*Q_ratio
else:
G = 0.6
else:
if beta2 <= 2/3.:
G = 1
else:
G = 1 + 0.3*Q_ratio*Q_ratio
Note that there are several errors in the text of [1]_; the errata can be
obtained here: http://www.flowoffluids.com/publications/tp-410-errata.aspx
Examples
--------
Example 7-36 of [1]_. A DN150 schedule 80 wye has 1515 liters/minute of
water exiting the straight leg, and 950 liters/minute of water
exiting it through a 45° branch. Calculate the loss coefficient in
the branch. The calculated value there is 0.4640.
>>> K_branch_diverging_Crane(0.146, 0.146, 0.02525, 0.01583, angle=45)
0.4639895627496694
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = (D_branch/D_run)
beta2 = beta*beta
Q_comb = Q_run + Q_branch
Q_ratio = Q_branch/Q_comb
if angle < 60 or beta <= 2/3.:
H, J = 1., 2.
else:
H, J = 0.3, 0
if angle < 75:
if beta2 <= 0.35:
if Q_ratio <= 0.4:
G = 1.1 - 0.7*Q_ratio
else:
G = 0.85
else:
if Q_ratio <= 0.6:
G = 1.0 - 0.6*Q_ratio
else:
G = 0.6
else:
if beta2 <= 2/3.:
G = 1
else:
G = 1 + 0.3*Q_ratio*Q_ratio
angle_rad = radians(angle)
K_branch = G*(1 + H*(Q_ratio/beta2)**2 - J*(Q_ratio/beta2)*cos(angle_rad))
return K_branch
[docs]def K_run_diverging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
r'''Returns the loss coefficient for the run of a converging tee or wye
according to the Crane method [1]_.
.. math::
K_{run} = M \left(\frac{Q_{branch}}{Q_{comb}}\right)^2
\beta_{branch} = \frac{D_{branch}}{D_{comb}}
See the notes for the definition of M.
Parameters
----------
D_run : float
Diameter of the straight-through inlet portion of the tee or wye [m]
D_branch : float
Diameter of the pipe attached at an angle to the straight-through, [m]
Q_run : float
Volumetric flow rate in the straight-through outlet of the tee or wye,
[m^3/s]
Q_branch : float
Volumetric flow rate in the pipe attached at an angle to the straight-
through, [m^3/s]
angle : float, optional
Angle the branch makes with the straight-through (tee=90, wye<90)
[degrees]
Returns
-------
K : float
Loss coefficient of run with respect to the velocity and inside
diameter of the combined flow inlet [-]
Notes
-----
M is calculated according to the following pseudocode:
.. code-block:: python
if beta*beta <= 0.4:
M = 0.4
elif Q_branch/Q_comb <= 0.5:
M = 2*(2*Q_branch/Q_comb - 1)
else:
M = 0.3*(2*Q_branch/Q_comb - 1)
Examples
--------
Example 7-36 of [1]_. A DN150 schedule 80 wye has 1515 liters/minute of
water exiting the straight leg, and 950 liters/minute of water
exiting it through a 45° branch. Calculate the loss coefficient in
the branch. The calculated value there is -0.06809.
>>> K_run_diverging_Crane(0.146, 0.146, 0.02525, 0.01583, angle=45)
-0.06810067607153049
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
beta = (D_branch/D_run)
beta2 = beta*beta
Q_comb = Q_run + Q_branch
Q_ratio = Q_branch/Q_comb
if beta2 <= 0.4:
M = 0.4
elif Q_ratio <= 0.5:
M = 2.*(2.*Q_ratio - 1.)
else:
M = 0.3*(2.*Q_ratio - 1.)
return M*Q_ratio*Q_ratio
[docs]def v_lift_valve_Crane(rho, D1=None, D2=None, style='swing check angled'):
r'''Calculates the approximate minimum velocity required to lift the disk
or other controlling element of a check valve to a fully open, stable,
position according to the Crane method [1]_.
.. math::
v_{min} = N\cdot \text{m/s} \cdot \sqrt{\frac{\text{kg/m}^3}{\rho}}
v_{min} = N\beta^2 \cdot \text{m/s} \cdot \sqrt{\frac{\text{kg/m}^3}{\rho}}
See the notes for the definition of values of N and which check valves use
which formulas.
Parameters
----------
rho : float
Density of the fluid [kg/m^3]
D1 : float, optional
Diameter of the valve bore (must be equal to or smaller than
`D2`), [m]
D2 : float, optional
Diameter of the pipe attached to the valve, [m]
style : str
The type of valve; one of ['swing check angled', 'swing check straight',
'swing check UL', 'lift check straight', 'lift check angled',
'tilting check 5°', 'tilting check 15°', 'stop check globe 1',
'stop check angle 1', 'stop check globe 2', 'stop check angle 2',
'stop check globe 3', 'stop check angle 3', 'foot valve poppet disc',
'foot valve hinged disc'], [-]
Returns
-------
v_min : float
Approximate minimum velocity required to keep the disc fully lifted,
preventing chattering and wear [m/s]
Notes
-----
This equation is not dimensionless.
+--------------------------+-----+------+
| Name/string | N | Full |
+==========================+=====+======+
| 'swing check angled' | 45 | No |
+--------------------------+-----+------+
| 'swing check straight' | 75 | No |
+--------------------------+-----+------+
| 'swing check UL' | 120 | No |
+--------------------------+-----+------+
| 'lift check straight' | 50 | Yes |
+--------------------------+-----+------+
| 'lift check angled' | 170 | Yes |
+--------------------------+-----+------+
| 'tilting check 5°' | 100 | No |
+--------------------------+-----+------+
| 'tilting check 15°' | 40 | No |
+--------------------------+-----+------+
| 'stop check globe 1' | 70 | Yes |
+--------------------------+-----+------+
| 'stop check angle 1' | 95 | Yes |
+--------------------------+-----+------+
| 'stop check globe 2' | 75 | Yes |
+--------------------------+-----+------+
| 'stop check angle 2' | 75 | Yes |
+--------------------------+-----+------+
| 'stop check globe 3' | 170 | Yes |
+--------------------------+-----+------+
| 'stop check angle 3' | 170 | Yes |
+--------------------------+-----+------+
| 'foot valve poppet disc' | 20 | No |
+--------------------------+-----+------+
| 'foot valve hinged disc' | 45 | No |
+--------------------------+-----+------+
Examples
--------
>>> v_lift_valve_Crane(rho=998.2, D1=0.0627, D2=0.0779, style='lift check straight')
1.0252301935349286
References
----------
.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
2009.
'''
specific_volume = 1./rho
if D1 is not None and D2 is not None:
beta = D1/D2
beta2 = beta*beta
if style == 'swing check angled':
return 45*specific_volume**0.5
elif style == 'swing check straight':
return 75*specific_volume**0.5
elif style == 'swing check UL':
return 120*specific_volume**0.5
elif style == 'lift check straight':
return 50.*beta2*specific_volume**0.5
elif style == 'lift check angled':
return 170.*beta2*specific_volume**0.5
elif style == 'tilting check 5°':
return 100*specific_volume**0.5
elif style == 'tilting check 15°':
return 40*specific_volume**0.5
elif style == 'stop check globe 1':
return 70*beta2*specific_volume**0.5
elif style == 'stop check angle 1':
return 95*beta2*specific_volume**0.5
elif style in ['stop check globe 2', 'stop check angle 2']:
return 75*beta2*specific_volume**0.5
elif style in ['stop check globe 3', 'stop check angle 3']:
return 170*beta2*specific_volume**0.5
elif style == 'foot valve poppet disc':
return 20*specific_volume**0.5
elif style == 'foot valve hinged disc':
return 45*specific_volume**0.5