# Fittings pressure drop (fluids.fittings)¶

fluids.fittings.contraction_sharp(Di1, Di2)[source]

Returns loss coefficient for any sharp edged pipe contraction as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
Parameters: Di1 : float Inside diameter of original pipe, [m] Di2 : float Inside diameter of following pipe, [m] K : float Loss coefficient in terms of the following pipe [-]

Notes

A value of 0.506 or simply 0.5 is often used.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> contraction_sharp(Di1=1, Di2=0.4)
0.5301269161591805
fluids.fittings.contraction_round(Di1, Di2, rc)[source]

Returns loss coefficient for any round edged pipe contraction as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
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] 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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> contraction_round(Di1=1, Di2=0.4, rc=0.04)
0.1783332490866574
fluids.fittings.contraction_conical(Di1, Di2, fd, l=None, angle=None)[source]

Returns loss coefficient for any conical pipe contraction as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
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] K : float Loss coefficient in terms of the following pipe [-]

Notes

Cheap and has substantial impact on pressure drop.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> contraction_conical(Di1=0.1, Di2=0.04, l=0.04, fd=0.0185)
0.15779041548350314
fluids.fittings.contraction_beveled(Di1, Di2, l=None, angle=None)[source]

Returns loss coefficient for any sharp beveled pipe contraction as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
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] K : float Loss coefficient in terms of the following pipe [-]

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> contraction_beveled(Di1=0.5, Di2=0.1, l=.7*.1, angle=120)
0.40946469413070485
fluids.fittings.diffuser_sharp(Di1, Di2)[source]

Returns loss coefficient for any sudden pipe diameter expansion as shown in [1] and in other sources.

$K_1 = (1-\beta^2)^2$
Parameters: Di1 : float Inside diameter of original pipe (smaller), [m] Di2 : float Inside diameter of following pipe (larger), [m] K : float Loss coefficient [-]

Notes

Highly accurate.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> diffuser_sharp(Di1=.5, Di2=1)
0.5625
fluids.fittings.diffuser_conical(Di1, Di2, l=None, angle=None, fd=None)[source]

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:

$K_1 = 8.30[\tan(\alpha/2)]^{1.75}(1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}$

For 20 to 60 degrees, beta < 0.5:

$K_1 = \left\{1.366\sin\left[\frac{2\pi(\alpha-15^\circ)}{180}\right]^{0.5} - 0.170 - 3.28(0.0625-\beta^4)\sqrt{\frac{\alpha-20^\circ}{40^\circ}}\right\} (1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}$

For 20 to 60 degrees, beta >= 0.5:

$K_1 = \left\{1.366\sin\left[\frac{2\pi(\alpha-15^\circ)}{180}\right]^{0.5} - 0.170 \right\}(1-\beta^2)^2 + \frac{f(1-\beta^4)}{8\sin(\alpha/2)}$

For 60 to 180 degrees, beta < 0.5:

$K_1 = \left[1.205 - 3.28(0.0625-\beta^4)-12.8\beta^6\sqrt{\frac {\alpha-60^\circ}{120^\circ}}\right](1-\beta^2)^2$

For 60 to 180 degrees, beta >= 0.5:

$K_1 = \left[1.205 - 0.20\sqrt{\frac{\alpha-60^\circ}{120^\circ}} \right](1-\beta^2)^2$
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 [-] K : float Loss coefficient [-]

Notes

For angles above 60 degrees, friction factor is not used.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> diffuser_conical(Di1=1/3., Di2=1, angle=50, fd=0.03)
0.8081340270019336
fluids.fittings.diffuser_conical_staged(Di1, Di2, DEs, ls, fd=None)[source]

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 [-] 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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> diffuser_conical(Di1=1., Di2=10.,l=9, fd=0.01)
0.973137914861591
fluids.fittings.diffuser_curved(Di1, Di2, l)[source]

Returns loss coefficient for any curved wall pipe expansion as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
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] 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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> diffuser_curved(Di1=.25**0.5, Di2=1., l=2.)
0.2299781250000002
fluids.fittings.diffuser_pipe_reducer(Di1, Di2, l, fd1, fd2=None)[source]

Returns loss coefficient for any pipe reducer pipe expansion as shown in [1]. This is an approximate formula.

\begin{align}\begin{aligned}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)\end{aligned}\end{align}
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 [-] 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.

References

 [1] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> diffuser_pipe_reducer(Di1=.5, Di2=.75, l=1.5, fd1=0.07)
0.06873244301714816
fluids.fittings.entrance_sharp()[source]

Returns loss coefficient for a sharp entrance to a pipe as shown in [1].

$K = 0.57$
Returns: K : float Loss coefficient [-]

Notes

Other values used have been 0.5.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_sharp()
0.57
fluids.fittings.entrance_distance(Di, t)[source]

Returns loss coefficient for a sharp entrance to a pipe at a distance from the wall of a reservoir, as shown in [1].

$K = 1.12 - 22\frac{t}{d} + 216\left(\frac{t}{d}\right)^2 + 80\left(\frac{t}{d}\right)^3$
Parameters: Di : float Inside diameter of pipe, [m] t : float Thickness of pipe wall, [m] 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.

References

 [1] (1, 2, 3) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_distance(Di=0.1, t=0.0005)
1.0154100000000001
fluids.fittings.entrance_angled(angle)[source]

Returns loss coefficient for a sharp, angled entrance to a pipe flush with the wall of a reservoir, as shown in [1].

$K = 0.57 + 0.30\cos(\theta) + 0.20\cos(\theta)^2$
Parameters: angle : float Angle of inclination (90=straight, 0=parallel to pipe wall) [degrees] K : float Loss coefficient [-]

Notes

Not reliable for angles under 20 degrees. Loss coefficient is the same for an upward or downward angled inlet.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_angled(30)
0.9798076211353316
fluids.fittings.entrance_rounded(Di, rc)[source]

Returns loss coefficient for a rounded entrance to a pipe flush with the wall of a reservoir, as shown in [1].

\begin{align}\begin{aligned}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\end{aligned}\end{align}
Parameters: Di : float Inside diameter of pipe, [m] rc : float Radius of curvature of the entrance, [m] K : float Loss coefficient [-]

Notes

For generously rounded entrance (rc/Di >= 1), the loss coefficient converges to 0.03.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_rounded(Di=0.1, rc=0.0235)
0.09839534618360923
fluids.fittings.entrance_beveled(Di, l, angle)[source]

Returns loss coefficient for a beveled or chamfered entrance to a pipe flush with the wall of a reservoir, as shown in [1].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}
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] K : float Loss coefficient [-]

Notes

A cheap way of getting a lower pressure drop. Little credible data is available.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_beveled(Di=0.1, l=0.003, angle=45)
0.45086864221916984
fluids.fittings.entrance_beveled_orifice(Di, do, l, angle)[source]

Returns loss coefficient for a beveled or chamfered orifice entrance to a pipe flush with the wall of a reservoir, as shown in [1].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}
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] K : float Loss coefficient [-]

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> entrance_beveled_orifice(Di=0.1, do=.07, l=0.003, angle=45)
1.2987552913818574
fluids.fittings.exit_normal()[source]

Returns loss coefficient for any exit to a pipe as shown in [1] and in other sources.

$K = 1$
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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> exit_normal()
1.0
fluids.fittings.bend_rounded(Di, angle, fd, rc=None, bend_diameters=5)[source]

Returns loss coefficient for any rounded bend in a pipe as shown in [1].

$K = f\alpha\frac{r}{d} + (0.10 + 2.4f)\sin(\alpha/2) + \frac{6.6f(\sqrt{\sin(\alpha/2)}+\sin(\alpha/2))} {(r/d)^{\frac{4\alpha}{\pi}}}$
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 [-] 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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> bend_rounded(Di=4.020, rc=4.0*5, angle=30, fd=0.0163)
0.10680196344492195
fluids.fittings.bend_miter(angle)[source]

Returns loss coefficient for any single-joint miter bend in a pipe as shown in [1].

$K = 0.42\sin(\alpha/2) + 2.56\sin^3(\alpha/2)$
Parameters: angle : float Angle of bend, [degrees] K : float Loss coefficient [-]

Notes

Applies for bends from 0 to 150 degrees. One joint only.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> bend_miter(150)
2.7128147734758103
fluids.fittings.helix(Di, rs, pitch, N, fd)[source]

Returns loss coefficient for any size constant-pitch helix as shown in [1]. Has applications in immersed coils in tanks.

$K = N \left[f\frac{\sqrt{(2\pi r)^2 + p^2}}{d} + 0.20 + 4.8 f\right]$
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 [-] 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.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> helix(Di=0.01, rs=0.1, pitch=.03, N=10, fd=.0185)
14.525134924495514
fluids.fittings.spiral(Di, rmax, rmin, pitch, fd)[source]

Returns loss coefficient for any size constant-pitch spiral as shown in [1]. Has applications in immersed coils in tanks.

$K = \frac{r_{max} - r_{min}}{p} \left[ f\pi\left(\frac{r_{max} +r_{min}}{d}\right) + 0.20 + 4.8f\right] + \frac{13.2f}{(r_{min}/d)^2}$
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 [-] K : float Loss coefficient [-]

Notes

Source not compared against others.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> spiral(Di=0.01, rmax=.1, rmin=.02, pitch=.01, fd=0.0185)
7.950918552775473
fluids.fittings.Darby3K(NPS=None, Re=None, name=None, K1=None, Ki=None, Kd=None)[source]

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.

$K_f = \frac{K_1}{Re} + K_i\left(1 + \frac{K_d}{D_{\text{NPS}}^{0.3}} \right)$

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] 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.

References

 [1] (1, 2) 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.

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
fluids.fittings.Hooper2K(Di, Re, name=None, K1=None, Kinfty=None)[source]

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.

$K = \frac{K_1}{Re} + K_\infty\left(1 + \frac{1\text{ inch}}{D_{in}}\right)$

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 [-] K : float Loss coefficient [-]

Notes

Also described in Ludwig’s Applied Process Design. Relatively uncommon to see it used. No actual example found.

References

 [1] (1, 2) 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.

Examples

>>> Hooper2K(Di=2., Re=10000., name='Valve, Globe, Standard')
6.15
>>> Hooper2K(Di=2., Re=10000., K1=900, Kinfty=4)
6.09
fluids.fittings.Kv_to_Cv(Kv)[source]

Convert valve flow coefficient from imperial to common metric units.

$C_v = 1.156 K_v$
Parameters: Kv : float Metric Kv valve flow coefficient (flow rate of water at a pressure drop of 1 bar) [m^3/hr] 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.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> Kv_to_Cv(2)
2.3121984567073133
fluids.fittings.Cv_to_Kv(Cv)[source]

Convert valve flow coefficient from imperial to common metric units.

$K_v = C_v/1.156$
Parameters: Cv : float Imperial Cv valve flow coefficient (flow rate of water at a pressure drop of 1 psi) [gallons/minute] 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.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> Cv_to_Kv(2.312)
1.9998283393826013
fluids.fittings.Kv_to_K(Kv, D)[source]

Convert valve flow coefficient from common metric units to regular loss coefficients.

$K = 1.6\times 10^9 \frac{D^4}{K_v^2}$
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] 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.

\begin{align}\begin{aligned}\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\end{aligned}\end{align}

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.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> Kv_to_K(2.312, .015)
15.153374600399898
fluids.fittings.K_to_Kv(K, D)[source]

Convert regular loss coefficient to valve flow coefficient.

$K_v = 4\times 10^4 \sqrt{ \frac{D^4}{K}}$
Parameters: K : float Loss coefficient, [-] D : float Inside diameter of the valve [m] 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.

\begin{align}\begin{aligned}\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\end{aligned}\end{align}

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.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> K_to_Kv(15.15337460039990, .015)
2.312
fluids.fittings.Cv_to_K(Cv, D)[source]

Convert imperial valve flow coefficient from imperial units to regular loss coefficients.

$K = 1.6\times 10^9 \frac{D^4}{\left(\frac{C_v}{1.56}\right)^2}$
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] K : float Loss coefficient, [-]

Notes

The exact conversion coefficient between Kv to Cv is 1.1560992283536566; it is rounded in the formula above.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> Cv_to_K(2.712, .015)
14.719595348352552
fluids.fittings.K_to_Cv(K, D)[source]

Convert regular loss coefficient to imperial valve flow coefficient.

$K_v = 1.156 \cdot 4\times 10^4 \sqrt{ \frac{D^4}{K}}$
Parameters: K : float Loss coefficient, [-] D : float Inside diameter of the valve [m] 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.

References

 [1] ISA-75.01.01-2007 (60534-2-1 Mod) Draft

Examples

>>> K_to_Cv(16, .015)
2.601223263795727
fluids.fittings.change_K_basis(K1, D1, D2)[source]

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.

$K_2 = K_1\frac{D_2^4}{D_1^4} = K_1 \frac{A_2^2}{A_1^2}$
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] K2 : float Loss coefficient with respect to the second diameter, [-]

Notes

This expression is shown in [1] and can easily be derived:

$\frac{\rho V_{1}^{2}}{2} \cdot K_{1} = \frac{\rho V_{2}^{2} }{2} \cdot K_{2}$

Substitute velocities for flow rate divided by area:

$\frac{8 K_{1} Q^{2} \rho}{\pi^{2} D_{1}^{4}} = \frac{8 K_{2} Q^{2} \rho}{\pi^{2} D_{2}^{4}}$

From here, simplification and rearrangement is all that is required.

References

 [1] (1, 2) Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.

Examples

>>> change_K_basis(K1=32.68875692997804, D1=.01, D2=.02)
523.0201108796487
fluids.fittings.K_gate_valve_Crane(D1, D2, angle, fd)[source]

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:

$K = K_1 = K_2 = 8f_d$

If β < 1 and θ <= 45°:

$K_2 = \frac{K + \sin \frac{\theta}{2} \left[0.8(1-\beta^2) + 2.6(1-\beta^2)^2\right]}{\beta^4}$

If β < 1 and θ > 45°:

$K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2) + (1-\beta^2)^2}{\beta^4}$
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 [-] 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].

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.
 [2] (1, 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.

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.

fluids.fittings.K_angle_valve_Crane(D1, D2, fd, style=0)[source]

Returns the loss coefficient for all types of angle valve, (reduced seat or throttled) as shown in [1].

If β = 1:

$K = K_1 = K_2 = N\cdot f_d$

Otherwise:

$K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}$

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] [-] 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.

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_angle_valve_Crane(.01, .02, fd=.016)
19.58
fluids.fittings.K_globe_valve_Crane(D1, D2, fd)[source]

Returns the loss coefficient for all types of globe valve, (reduced seat or throttled) as shown in [1].

If β = 1:

$K = K_1 = K_2 = 340 f_d$

Otherwise:

$K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}$
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 [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_globe_valve_Crane(.01, .02, fd=.015)
87.1
fluids.fittings.K_swing_check_valve_Crane(fd, angled=True)[source]

Returns the loss coefficient for a swing check valve as shown in [1].

$K_2 = N\cdot f_d$

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 [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_swing_check_valve_Crane(fd=.016)
1.6
fluids.fittings.K_lift_check_valve_Crane(D1, D2, fd, angled=True)[source]

Returns the loss coefficient for a lift check valve as shown in [1].

If β = 1:

$K = K_1 = K_2 = N\cdot f_d$

Otherwise:

$K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}$

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 [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_lift_check_valve_Crane(.01, .02, fd=.016)
21.58
fluids.fittings.K_tilting_disk_check_valve_Crane(D, angle, fd)[source]

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.

$K = N\cdot f_d$

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 [-] 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.

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_tilting_disk_check_valve_Crane(.01, 5, fd=.016)
0.64
fluids.fittings.K_globe_stop_check_valve_Crane(D1, D2, fd, style=0)[source]

Returns the loss coefficient for a globe stop check valve as shown in [1].

If β = 1:

$K = K_1 = K_2 = N\cdot f_d$

Otherwise:

$K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}$

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] [-] 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.

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_globe_stop_check_valve_Crane(.1, .02, .0165, style=1)
4.51992
fluids.fittings.K_angle_stop_check_valve_Crane(D1, D2, fd, style=0)[source]

Returns the loss coefficient for a angle stop check valve as shown in [1].

If β = 1:

$K = K_1 = K_2 = N\cdot f_d$

Otherwise:

$K_2 = \frac{K + \left[0.5(1-\beta^2) + (1-\beta^2)^2\right]}{\beta^4}$

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] [-] 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.

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_angle_stop_check_valve_Crane(.1, .02, .0165, style=1)
4.52124
fluids.fittings.K_ball_valve_Crane(D1, D2, angle, fd)[source]

Returns the loss coefficient for a ball valve as shown in [1].

If β = 1:

$K = K_1 = K_2 = 3f_d$

If β < 1 and θ <= 45°:

$K_2 = \frac{K + \sin \frac{\theta}{2} \left[0.8(1-\beta^2) + 2.6(1-\beta^2)^2\right]} {\beta^4}$

If β < 1 and θ > 45°:

$K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2) + (1-\beta^2)^2}{\beta^4}$
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 [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_ball_valve_Crane(.01, .02, 50, .025)
14.100545785228675
fluids.fittings.K_diaphragm_valve_Crane(fd, style=0)[source]

Returns the loss coefficient for a diaphragm valve of either weir (style = 0) or straight-through (style = 1) as shown in [1].

$K = K_1 = K_2 = N\cdot f_d$

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) [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_diaphragm_valve_Crane(0.015, style=0)
2.235
fluids.fittings.K_foot_valve_Crane(fd, style=0)[source]

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.

$K = K_1 = K_2 = N\cdot f_d$

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) [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_foot_valve_Crane(0.015, style=0)
6.3
fluids.fittings.K_butterfly_valve_Crane(D, fd, style=0)[source]

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).

$K = N\cdot f_d$

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) [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_butterfly_valve_Crane(.01, .016, style=2)
3.488
fluids.fittings.K_plug_valve_Crane(D1, D2, angle, fd, style=0)[source]

Returns the loss coefficient for a plug valve or cock valve as shown in [1].

If β = 1:

$K = K_1 = K_2 = Nf_d$

Otherwise:

$K_2 = \frac{K + 0.5\sqrt{\sin\frac{\theta}{2}}(1-\beta^2) + (1-\beta^2)^2}{\beta^4}$

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°) [-] 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.

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> K_plug_valve_Crane(.01, .02, 50, .025)
20.100545785228675
fluids.fittings.K_branch_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90)[source]

Returns the loss coefficient for the branch of a converging tee or wye according to the Crane method [1].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}

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] 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 $$\beta_{branch}^2 \le 0.35$$, C = 1

If $$\beta_{branch}^2 > 0.35$$ and $$Q_{branch}/Q_{comb} > 0.4$$, C = 0.55.

If neither of the above conditions are met:

$C = 0.9\left(1 - \frac{Q_{branch}}{Q_{comb}}\right)$

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

References

 [1] (1, 2, 3, 4) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

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
fluids.fittings.K_run_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90)[source]

Returns the loss coefficient for the run of a converging tee or wye according to the Crane method [1].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}

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] 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.

$K_{run} = 1.55\left(\frac{Q_{branch}}{Q_{comb}} \right) - \left(\frac{Q_{branch}}{Q_{comb}}\right)^2$

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

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
fluids.fittings.K_branch_diverging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90)[source]

Returns the loss coefficient for the branch of a diverging tee or wye according to the Crane method [1].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}

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] K : float Loss coefficient of branch with respect to the velocity and inside diameter of the combined flow inlet [-]

Notes

If $$\beta_{branch} = 1, \theta = 90^\circ$$, H = 0.3 and J = 0. Otherwise H = 1 and J = 2.

G is determined according to the following pseudocode:

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

References

 [1] (1, 2, 3, 4) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

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
fluids.fittings.K_run_diverging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90)[source]

Returns the loss coefficient for the run of a converging tee or wye according to the Crane method [1].

\begin{align}\begin{aligned}K_{run} = M \left(\frac{Q_{branch}}{Q_{comb}}\right)^2\\\beta_{branch} = \frac{D_{branch}}{D_{comb}}\end{aligned}\end{align}

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] 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:

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)

References

 [1] (1, 2, 3) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

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
fluids.fittings.v_lift_valve_Crane(rho, D1=None, D2=None, style='swing check angled')[source]

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].

\begin{align}\begin{aligned}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}}\end{aligned}\end{align}

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’], [-] 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

References

 [1] (1, 2) Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 2009.

Examples

>>> v_lift_valve_Crane(rho=998.2, D1=0.0627, D2=0.0779, style='lift check straight')
1.0252301935349286