Source code for fluids.fittings

# -*- 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
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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_foot_valve_Crane(fd, style=0): r'''Returns the loss coefficient for a foot valve of either poppet disc (`style` = 0) or hinged-disk (`style` = 1) as shown in [1]_. Both valves are specified include the loss of the attached strainer. .. math:: K = K_1 = K_2 = N\cdot f_d For style 0 (poppet disk), N = 420; for style 1 (hinged disk), N = 75. 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 (poppet disk foot valve) or 1 (hinged disk foot 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_foot_valve_Crane(0.015, style=0) 6.3 References ---------- .. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009. ''' coeffs = {0: 420, 1: 75} try: K = coeffs[style]*fd except KeyError: raise KeyError('Accepted valve styles are 0 (poppet disk) or 1 (hinged disk) 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