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# -*- coding: utf-8 -*- 

'''Chemical Engineering Design Library (ChEDL). Utilities for process modeling. 

Copyright (C) 2016, Caleb Bell <Caleb.Andrew.Bell@gmail.com> 

 

Permission is hereby granted, free of charge, to any person obtaining a copy 

of this software and associated documentation files (the "Software"), to deal 

in the Software without restriction, including without limitation the rights 

to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 

copies of the Software, and to permit persons to whom the Software is 

furnished to do so, subject to the following conditions: 

 

The above copyright notice and this permission notice shall be included in all 

copies or substantial portions of the Software. 

 

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 

IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 

FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 

AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 

LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 

OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 

SOFTWARE.''' 

 

from __future__ import division 

from math import log, pi, exp 

import numpy as np 

from scipy.optimize import newton, ridder 

from scipy.constants import R 

from scipy.special import lambertw 

 

__all__ = ['Panhandle_A', 'Panhandle_B', 'Weymouth', 'Spitzglass_high', 

'Spitzglass_low', 'Oliphant', 'Fritzsche', 'Muller', 'IGT', 'isothermal_gas', 

'isothermal_work_compression', 'polytropic_exponent', 

'isentropic_work_compression', 'isentropic_efficiency', 

'isentropic_T_rise_compression', 'T_critical_flow', 

'P_critical_flow', 'P_isothermal_critical_flow', 

'is_critical_flow', 'stagnation_energy', 'P_stagnation', 

'T_stagnation', 'T_stagnation_ideal'] 

 

 

def isothermal_work_compression(P1, P2, T, Z=1): 

r'''Calculates the work of compression or expansion of a gas going through 

an isothermal process. 

 

.. math:: 

W = zRT\ln\left(\frac{P_2}{P_1}\right) 

 

Parameters 

---------- 

P1 : float 

Inlet pressure, [Pa] 

P2 : float 

Outlet pressure, [Pa] 

T : float 

Temperature of the gas going through an isothermal process, [K] 

Z : float 

Constant compressibility factor of the gas, [-] 

 

Returns 

------- 

W : float 

Work performed per mole of gas compressed/expanded [J/mol] 

 

Notes 

----- 

The full derivation with all forms is as follows: 

 

.. math:: 

W = \int_{P_1}^{P_2} V dP = zRT\int_{P_1}^{P_2} \frac{1}{P} dP 

 

W = zRT\ln\left(\frac{P_2}{P_1}\right) = P_1 V_1 \ln\left(\frac{P_2} 

{P_1}\right) = P_2 V_2 \ln\left(\frac{P_2}{P_1}\right) 

 

The substitutions are according to the ideal gas law with compressibility: 

 

.. math: 

PV = ZRT 

 

The work of compression/expansion is the change in enthalpy of the gas. 

Returns negative values for expansion and positive values for compression. 

 

An average compressibility factor can be used where Z changes. For further 

accuracy, this expression can be used repeatedly with small changes in 

pressure and the work from each step summed. 

 

Examples 

-------- 

>>> isothermal_work_compression(1E5, 1E6, 300) 

5743.425357533477 

 

References 

---------- 

.. [1] Couper, James R., W. Roy Penney, and James R. Fair. Chemical Process 

Equipment: Selection and Design. 2nd ed. Amsterdam ; Boston: Gulf 

Professional Publishing, 2009. 

''' 

return Z*R*T*log(P2/P1) 

 

 

def isentropic_work_compression(T1, k, Z=1, P1=None, P2=None, W=None, eta=None): 

r'''Calculation function for dealing with compressing or expanding a gas 

going through an isentropic, adiabatic process assuming constant Cp and Cv.  

The polytropic model is the same equation; just provide `n` instead of `k`  

and use a polytropic efficienty for `eta` instead of a isentropic  

efficiency. Can calculate any of the following, given all the other inputs: 

 

* W, Work of compression 

* P2, Pressure after compression  

* P1, Pressure before compression 

* eta, isentropic efficiency of compression 

 

.. math:: 

W = \left(\frac{k}{k-1}\right)ZRT_1\left[\left(\frac{P_2}{P_1} 

\right)^{(k-1)/k}-1\right]/\eta_{isentropic} 

 

Parameters 

---------- 

T1 : float 

Initial temperature of the gas, [K] 

k : float 

Isentropic exponent of the gas (Cp/Cv) or polytropic exponent `n` to 

use this as a polytropic model instead [-] 

Z : float, optional 

Constant compressibility factor of the gas, [-] 

P1 : float, optional 

Inlet pressure, [Pa] 

P2 : float, optional 

Outlet pressure, [Pa] 

W : float, optional 

Work performed per mole of gas compressed/expanded [J/mol] 

eta : float, optional 

Isentropic efficiency of the process or polytropic efficiency of the 

process to use this as a polytropic model instead [-] 

 

Returns 

------- 

W, P1, P2, or eta : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

For the same compression ratio, this is always of larger magnitude than the 

isothermal case. 

 

The full derivation is as follows: 

 

For constant-heat capacity "isentropic" fluid, 

 

.. math:: 

V = \frac{P_1^{1/k}V_1}{P^{1/k}} 

 

W = \int_{P_1}^{P_2} V dP = \int_{P_1}^{P_2}\frac{P_1^{1/k}V_1} 

{P^{1/k}}dP 

 

W = \frac{P_1^{1/k} V_1}{1 - \frac{1}{k}}\left[P_2^{1-1/k} - 

P_1^{1-1/k}\right] 

 

After performing the integration and substantial mathematical manipulation 

we can obtain: 

 

.. math:: 

W = \left(\frac{k}{k-1}\right) P_1 V_1 \left[\left(\frac{P_2}{P_1} 

\right)^{(k-1)/k}-1\right] 

 

Using PV = ZRT: 

 

.. math:: 

W = \left(\frac{k}{k-1}\right)ZRT_1\left[\left(\frac{P_2}{P_1} 

\right)^{(k-1)/k}-1\right] 

 

The work of compression/expansion is the change in enthalpy of the gas. 

Returns negative values for expansion and positive values for compression. 

 

An average compressibility factor should be used as Z changes. For further 

accuracy, this expression can be used repeatedly with small changes in 

pressure and new values of isentropic exponent, and the work from each step 

summed. 

 

For the polytropic case this is not necessary, as `eta` corrects for the 

simplification. 

 

Examples 

-------- 

>>> isentropic_work_compression(P1=1E5, P2=1E6, T1=300, k=1.4, eta=0.78) 

10416.873455626454 

 

References 

---------- 

.. [1] Couper, James R., W. Roy Penney, and James R. Fair. Chemical Process 

Equipment: Selection and Design. 2nd ed. Amsterdam ; Boston: Gulf 

Professional Publishing, 2009. 

''' 

if W is None and (None not in [eta, P1, P2]): 

return k/(k-1)*Z*R*T1*((P2/P1)**((k-1.)/k) - 1)/eta 

elif P1 is None and (None not in [eta, W, P2]): 

return P2*(1 + W*eta/(R*T1*Z) - W*eta/(R*T1*Z*k))**(-k/(k - 1)) 

elif P2 is None and (None not in [eta, W, P1]): 

return P1*(1 + W*eta/(R*T1*Z) - W*eta/(R*T1*Z*k))**(k/(k - 1)) 

elif eta is None and (None not in [P1, P2, W]): 

return R*T1*Z*k*((P2/P1)**((k - 1)/k) - 1)/(W*(k - 1)) 

else: 

raise Exception('Three of W, P1, P2, and eta must be specified.') 

 

 

def isentropic_T_rise_compression(T1, P1, P2, k, eta=1): 

r'''Calculates the increase in temperature of a fluid which is compressed 

or expanded under isentropic, adiabatic conditions assuming constant 

Cp and Cv. The polytropic model is the same equation; just provide `n` 

instead of `k` and use a polytropic efficienty for `eta` instead of a 

isentropic efficiency. 

 

.. math:: 

T_2 = T_1 + \frac{\Delta T_s}{\eta_s} = T_1 \left\{1 + \frac{1} 

{\eta_s}\left[\left(\frac{P_2}{P_1}\right)^{(k-1)/k}-1\right]\right\} 

 

Parameters 

---------- 

T1 : float 

Initial temperature of gas [K] 

P1 : float 

Initial pressure of gas [Pa] 

P2 : float 

Final pressure of gas [Pa] 

k : float 

Isentropic exponent of the gas (Cp/Cv) or polytropic exponent `n` to 

use this as a polytropic model instead [-] 

eta : float 

Isentropic efficiency of the process or polytropic efficiency of the 

process to use this as a polytropic model instead [-] 

 

Returns 

------- 

T2 : float 

Final temperature of gas [K] 

 

Notes 

----- 

For the ideal case (`eta`=1), the model simplifies to: 

 

.. math:: 

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k} 

 

Examples 

-------- 

>>> isentropic_T_rise_compression(286.8, 54050, 432400, 1.4) 

519.5230938217768 

 

References 

---------- 

.. [1] Couper, James R., W. Roy Penney, and James R. Fair. Chemical Process 

Equipment: Selection and Design. 2nd ed. Amsterdam ; Boston: Gulf 

Professional Publishing, 2009. 

.. [2] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

''' 

dT = T1*((P2/P1)**((k-1)/k)-1)/eta 

return T1 + dT 

 

 

def isentropic_efficiency(P1, P2, k, eta_s=None, eta_p=None): 

r'''Calculates either isentropic or polytropic efficiency from the other 

type of efficiency. 

 

.. math:: 

\eta_s = \frac{(P_2/P_1)^{(k-1)/k}-1} 

{(P_2/P_1)^{\frac{k-1}{k\eta_p}}-1} 

 

\eta_p = \frac{\left(k - 1\right) \log{\left (\frac{P_{2}}{P_{1}} 

\right )}}{k \log{\left (\frac{1}{\eta_{s}} \left(\eta_{s} 

+ \left(\frac{P_{2}}{P_{1}}\right)^{\frac{1}{k} \left(k - 1\right)} 

- 1\right) \right )}} 

 

Parameters 

---------- 

P1 : float 

Initial pressure of gas [Pa] 

P2 : float 

Final pressure of gas [Pa] 

k : float 

Isentropic exponent of the gas (Cp/Cv) or polytropic exponent `n` to 

use this as a polytropic model instead [-] 

eta_s : float, optional 

Isentropic efficiency of the process, [-] 

eta_p : float, optional 

Polytropic efficiency of the process, [-] 

 

Returns 

------- 

eta_s or eta_p : float 

Isentropic or polytropic efficiency, depending on input, [-] 

 

Notes 

----- 

The form for obtained `eta_p` from `eta_s` was derived with SymPy. 

 

Examples 

-------- 

>>> isentropic_efficiency(1E5, 1E6, 1.4, eta_p=0.78) 

0.7027614191263858 

 

References 

---------- 

.. [1] Couper, James R., W. Roy Penney, and James R. Fair. Chemical Process 

Equipment: Selection and Design. 2nd ed. Amsterdam ; Boston: Gulf 

Professional Publishing, 2009. 

''' 

if eta_s is None and eta_p: 

return ((P2/P1)**((k-1)/k)-1)/((P2/P1)**((k-1)/(k*eta_p))-1) 

elif eta_p is None and eta_s: 

return (k - 1.)*log(P2/P1)/(k*log( 

(eta_s + (P2/P1)**((k - 1.)/k) - 1.)/eta_s)) 

else: 

raise Exception('Either eta_s or eta_p is required') 

 

 

def polytropic_exponent(k, n=None, eta_p=None): 

r'''Calculates either the polytropic exponent from polytropic efficiency 

or polytropic efficiency from the polytropic exponent. 

 

.. math:: 

n = \frac{k\eta_p}{1 - k(1-\eta_p)} 

 

\eta_p = \frac{\left(\frac{n}{n-1}\right)}{\left(\frac{k}{k-1} 

\right)} = \frac{n(k-1)}{k(n-1)} 

 

Parameters 

---------- 

k : float 

Isentropic exponent of the gas (Cp/Cv) or polytropic exponent `n` to 

use this as a polytropic model instead [-] 

eta_p : float, optional 

Polytropic efficiency of the process, [-] 

n : float, optional 

Polytropic exponent of the process [-] 

 

Returns 

------- 

n or eta_p : float 

isentropic exponent or polytropic efficiency, depending on input, [-] 

 

Notes 

----- 

 

Examples 

-------- 

>>> polytropic_exponent(1.4, eta_p=0.78) 

1.5780346820809246 

 

References 

---------- 

.. [1] Couper, James R., W. Roy Penney, and James R. Fair. Chemical Process 

Equipment: Selection and Design. 2nd ed. Amsterdam ; Boston: Gulf 

Professional Publishing, 2009. 

''' 

if n is None and eta_p: 

return k*eta_p/(1 - k*(1-eta_p)) 

elif eta_p is None and n: 

return n*(k-1)/(k*(n-1)) 

else: 

raise Exception('Either n or eta_p is required') 

 

 

def T_critical_flow(T, k): 

r'''Calculates critical flow temperature `Tcf` for a fluid with the 

given isentropic coefficient. `Tcf` is in a flow (with Ma=1) whose 

stagnation conditions are known. Normally used with converging/diverging 

nozzles. 

 

.. math:: 

\frac{T^*}{T_0} = \frac{2}{k+1} 

 

Parameters 

---------- 

T : float 

Stagnation temperature of a fluid with Ma=1 [K] 

k : float 

Isentropic coefficient [] 

 

Returns 

------- 

Tcf : float 

Critical flow temperature at Ma=1 [K] 

 

Notes 

----- 

Assumes isentropic flow. 

 

Examples 

-------- 

Example 12.4 in [1]_: 

 

>>> T_critical_flow(473, 1.289) 

413.2809086937528 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return T*2/(k+1.) 

 

 

def P_critical_flow(P, k): 

r'''Calculates critical flow pressure `Pcf` for a fluid with the 

given isentropic coefficient. `Pcf` is in a flow (with Ma=1) whose 

stagnation conditions are known. Normally used with converging/diverging 

nozzles. 

 

.. math:: 

\frac{P^*}{P_0} = \left(\frac{2}{k+1}\right)^{k/(k-1)} 

 

Parameters 

---------- 

P : float 

Stagnation pressure of a fluid with Ma=1 [Pa] 

k : float 

Isentropic coefficient [] 

 

Returns 

------- 

Pcf : float 

Critical flow pressure at Ma=1 [Pa] 

 

Notes 

----- 

Assumes isentropic flow. 

 

Examples 

-------- 

Example 12.4 in [1]_: 

 

>>> P_critical_flow(1400000, 1.289) 

766812.9022792266 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return P*(2/(k+1.))**(k/(k-1)) 

 

 

def P_isothermal_critical_flow(P, fd, D, L): 

r'''Calculates critical flow pressure `Pcf` for a fluid flowing 

isothermally and suffering pressure drop caused by a pipe's friction factor. 

 

.. math:: 

P_2 = P_{1} e^{\frac{1}{2 D} \left(D \left(\operatorname{LambertW} 

{\left (- e^{\frac{1}{D} \left(- D - L f_d\right)} \right )} + 1\right) 

+ L f_d\right)} 

 

Parameters 

---------- 

P : float 

Inlet pressure [Pa] 

fd : float 

Darcy friction factor for flow in pipe [-] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

 

Returns 

------- 

Pcf : float 

Critical flow pressure of a compressible gas flowing from `P1` to `Pcf` 

in a tube of length L and friction factor `fd` [Pa] 

 

Notes 

----- 

Assumes isothermal flow. Developed based on the `isothermal_gas` model, 

using SymPy. 

 

The isothermal gas model is solved for maximum mass flow rate; any pressure 

drop under it is impossible due to the formation of a shock wave. 

 

Examples 

-------- 

>>> P_isothermal_critical_flow(P=1E6, fd=0.00185, L=1000., D=0.5) 

389699.7317645518 

 

References 

---------- 

.. [1] Wilkes, James O. Fluid Mechanics for Chemical Engineers with 

Microfluidics and CFD. 2 edition. Upper Saddle River, NJ: Prentice Hall, 

2005. 

''' 

# Correct branch of lambertw found by trial and error 

lambert_term = float(lambertw(-exp((-D - L*fd)/D), -1).real) 

return P*exp((D*(lambert_term + 1) + L*fd)/(2.*D)) 

 

 

def P_upstream_isothermal_critical_flow(P, fd, D, L): 

'''Not part of the public API. Reverses `P_isothermal_critical_flow`. 

 

Examples 

-------- 

>>> P_upstream_isothermal_critical_flow(P=389699.7317645518, fd=0.00185, 

... L=1000., D=0.5) 

1000000.0000000001 

''' 

lambertw_term = float(lambertw(-exp(-(fd*L+D)/D), -1).real) 

return exp(-0.5*(D*lambertw_term+fd*L+D)/D)*P 

 

 

def is_critical_flow(P1, P2, k): 

r'''Determines if a flow of a fluid driven by pressure gradient 

P1 - P2 is critical, for a fluid with the given isentropic coefficient. 

This function calculates critical flow pressure, and checks if this is 

larger than P2. If so, the flow is critical and choked. 

 

Parameters 

---------- 

P1 : float 

Higher, source pressure [Pa] 

P2 : float 

Lower, downstream pressure [Pa] 

k : float 

Isentropic coefficient [] 

 

Returns 

------- 

flowtype : bool 

True if the flow is choked; otherwise False 

 

Notes 

----- 

Assumes isentropic flow. Uses P_critical_flow function. 

 

Examples 

-------- 

Examples 1-2 from API 520. 

 

>>> is_critical_flow(670E3, 532E3, 1.11) 

False 

>>> is_critical_flow(670E3, 101E3, 1.11) 

True 

 

References 

---------- 

.. [1] API. 2014. API 520 - Part 1 Sizing, Selection, and Installation of 

Pressure-relieving Devices, Part I - Sizing and Selection, 9E. 

''' 

Pcf = P_critical_flow(P1, k) 

return Pcf > P2 

 

 

def stagnation_energy(V): 

r'''Calculates the increase in enthalpy `dH` which is provided by a fluid's 

velocity `V`. 

 

.. math:: 

\Delta H = \frac{V^2}{2} 

 

Parameters 

---------- 

V : float 

Velocity [m/s] 

 

Returns 

------- 

dH : float 

Incease in enthalpy [J/kg] 

 

Notes 

----- 

The units work out. This term is pretty small, but not trivial. 

 

Examples 

-------- 

>>> stagnation_energy(125) 

7812.5 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return 0.5*V*V 

 

 

def P_stagnation(P, T, Tst, k): 

r'''Calculates stagnation flow pressure `Pst` for a fluid with the 

given isentropic coefficient and specified stagnation temperature and 

normal temperature. Normally used with converging/diverging nozzles. 

 

.. math:: 

\frac{P_0}{P}=\left(\frac{T_0}{T}\right)^{\frac{k}{k-1}} 

 

Parameters 

---------- 

P : float 

Normal pressure of a fluid [Pa] 

T : float 

Normal temperature of a fluid [K] 

Tst : float 

Stagnation temperature of a fluid moving at a certain velocity [K] 

k : float 

Isentropic coefficient [] 

 

Returns 

------- 

Pst : float 

Stagnation pressure of a fluid moving at a certain velocity [Pa] 

 

Notes 

----- 

Assumes isentropic flow. 

 

Examples 

-------- 

Example 12-1 in [1]_. 

 

>>> P_stagnation(54050., 255.7, 286.8, 1.4) 

80772.80495900588 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return P*(Tst/T)**(k/(k-1)) 

 

 

def T_stagnation(T, P, Pst, k): 

r'''Calculates stagnation flow temperature `Tst` for a fluid with the 

given isentropic coefficient and specified stagnation pressure and 

normal pressure. Normally used with converging/diverging nozzles. 

 

.. math:: 

T=T_0\left(\frac{P}{P_0}\right)^{\frac{k-1}{k}} 

 

Parameters 

---------- 

T : float 

Normal temperature of a fluid [K] 

P : float 

Normal pressure of a fluid [Pa] 

Pst : float 

Stagnation pressure of a fluid moving at a certain velocity [Pa] 

k : float 

Isentropic coefficient [] 

 

Returns 

------- 

Tst : float 

Stagnation temperature of a fluid moving at a certain velocity [K] 

 

Notes 

----- 

Assumes isentropic flow. 

 

Examples 

-------- 

Example 12-1 in [1]_. 

 

>>> T_stagnation(286.8, 54050, 54050*8, 1.4) 

519.5230938217768 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return T*(Pst/P)**((k - 1)/k) 

 

 

def T_stagnation_ideal(T, V, Cp): 

r'''Calculates the ideal stagnation temperature `Tst` calculated assuming 

the fluid has a constant heat capacity `Cp` and with a specified 

velocity `V` and tempeature `T`. 

 

.. math:: 

T^* = T + \frac{V^2}{2C_p} 

 

Parameters 

---------- 

T : float 

Tempearture [K] 

V : float 

Velocity [m/s] 

Cp : float 

Ideal heat capacity [J/kg/K] 

 

Returns 

------- 

Tst : float 

Stagnation temperature [J/kg] 

 

Examples 

-------- 

Example 12-1 in [1]_. 

 

>>> T_stagnation_ideal(255.7, 250, 1005.) 

286.79452736318405 

 

References 

---------- 

.. [1] Cengel, Yunus, and John Cimbala. Fluid Mechanics: Fundamentals and 

Applications. Boston: McGraw Hill Higher Education, 2006. 

''' 

return T + 0.5*V*V/Cp 

 

 

def isothermal_gas(rho, f, P1=None, P2=None, L=None, D=None, m=None): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline for the complete isothermal flow equation. Can calculate any of 

the following, given all other inputs: 

 

* Mass flow rate 

* Upstream pressure (numerical) 

* Downstream pressure (analytical or numerical if an overflow occurs) 

* Diameter of pipe (numerical) 

* Length of pipe 

 

A variety of forms of this equation have been presented, differing in their 

use of the ideal gas law and choice of gas constant. The form here uses 

density explicitly, allowing for non-ideal values to be used. 

 

.. math:: 

\dot m^2 = \frac{\left(\frac{\pi D^2}{4}\right)^2 \rho_{avg} 

\left(P_1^2-P_2^2\right)}{P_1\left(f_d\frac{L}{D} + 2\ln\frac{P_1}{P_2} 

\right)} 

 

Parameters 

---------- 

rho : float 

Average density of gas in pipe, [kg/m^3] 

f : float 

Darcy friction factor for flow in pipe [-] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

m : float, optional 

Mass flow rate of gas through pipe, [kg/s] 

 

Returns 

------- 

m, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

The solution for P2 has the following closed form, derived using Maple: 

 

.. math:: 

P_2={P_1 \left( {{ e}^{0.5\cdot{\frac {1}{{m}^{2}} \left( -C{m}^{2} 

+\text{ lambertW} \left(-{\frac {BP_1}{{m}^{2}}{{ e}^{-{\frac {-C{m}^{ 

2}+BP_1}{{m}^{2}}}}}}\right){}{m}^{2}+BP_1 \right) }}} \right) ^{-1}} 

 

B = \frac{\pi^2 D^4}{4^2} \rho_{avg} 

 

C = f_d \frac{L}{D} 

 

A wide range of conditions are impossible due to chocked flow. See 

`P_isothermal_critical_flow` for details. An exception is raised when 

they occur. 

 

The 2 multiplied by the logarithm is often shown as a power of the 

pressure ratio; this is only the case when the pressure ratio is raised to 

the power of 2 before its logarithm is taken. 

 

Examples 

-------- 

>>> isothermal_gas(11.3, 0.00185, P1=1E6, P2=9E5, L=1000, D=0.5) 

145.4847572636031 

 

References 

---------- 

.. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 

2009. 

.. [2] Kim, J. and Singh, N. "A Novel Equation for Isothermal Pipe Flow.". 

Chemical Engineering, June 2012, http://www.chemengonline.com/a-novel-equation-for-isothermal-pipe-flow/?printmode=1 

.. [3] Wilkes, James O. Fluid Mechanics for Chemical Engineers with 

Microfluidics and CFD. 2 edition. Upper Saddle River, NJ: Prentice Hall, 

2005. 

''' 

if m is None and (None not in [P1, P2, L, D]): 

Pcf = P_isothermal_critical_flow(P=P1, fd=f, D=D, L=L) 

if P2 < Pcf: 

raise Exception('Given outlet pressure is not physically possible \ 

due to the formation of choked flow at P2=%f, specified outlet pressure was %f' % (Pcf, P2)) 

if P2 > P1: 

raise Exception('Specified outlet pressure is larger than the \ 

inlet pressure; fluid will flow backwards.') 

return (pi**2/16*D**4*rho/P1/(f*L/D + 2*log(P1/P2))*(P1**2-P2**2))**0.5 

elif L is None and (None not in [P1, P2, D, m]): 

return D*(pi**2*D**4*rho*(P1**2 - P2**2) - 32*P1*m**2*log(P1/P2))/(16*P1*f*m**2) 

elif P1 is None and (None not in [L, P2, D, m]): 

Pcf = P_upstream_isothermal_critical_flow(P=P2, fd=f, D=D, L=L) 

 

def to_solve(P1): 

return m - isothermal_gas(rho, f, P1=P1, P2=P2, L=L, D=D) 

try: 

# Use the explicit solution for P2 with different P1 guesses; 

# newton doesn't like solving for m. 

def to_solve_P2_basis(P1): 

return abs(P2 - isothermal_gas(rho, f, m=m, P1=P1, P2=None, L=L, D=D)) 

P1 = newton(to_solve_P2_basis, (P2+Pcf)/2.) 

assert P2 <= P1 

return P1 

except: 

try: 

return ridder(to_solve, a=P2, b=Pcf) 

except: 

m_max = isothermal_gas(rho, f, P1=Pcf, P2=P2, L=L, D=D) 

raise Exception('The desired mass flow rate cannot be achieved \ 

with the specified downstream pressure; the maximum flowrate is %f at an \ 

upstream pressure of %f' %(m_max, Pcf)) 

elif P2 is None and (None not in [L, P1, D, m]): 

try: 

C = f*L/D 

B = (pi/4*D**2)**2*rho 

arg = -B/m**2*P1*exp(-(-C*m**2+B*P1)/m**2) 

# Consider the two real branches of the lambertw function. 

# The k=-1 branch produces the higher P2 values; the k=0 branch is 

# physically impossible. 

lambert_ans = float(lambertw(arg,k=-1).real) 

# Large overflow problem here; also divide by zero problems! 

# Fail and try a numerical solution if it doesn't work. 

assert np.isfinite(lambert_ans) 

P2 = P1/exp((-C*m**2+lambert_ans*m**2+B*P1)/m**2/2.) 

assert P2 < P1 

return P2 

except: 

Pcf = P_isothermal_critical_flow(P=P1, fd=f, D=D, L=L) 

def to_solve(P2): 

return m - isothermal_gas(rho, f, P1=P1, P2=P2, L=L, D=D) 

# return abs(m - isothermal_gas(rho, f, P1=P1, P2=P2, L=L, D=D)) 

# return fminbound(to_solve, x1=Pcf, x2=P1) 

try: 

return ridder(to_solve, a=Pcf, b=P1) 

except: 

m_max = isothermal_gas(rho, f, P1=P1, P2=Pcf, L=L, D=D) 

raise Exception('The desired mass flow rate cannot be achieved\ 

with the specified upstream pressure; the maximum flowrate is %f at an \ 

downstream pressure of %f' %(m_max, Pcf)) 

# A solver which respects its boundaries is required here. 

# ridder cuts the time down from 2 ms to 200 mircoseconds. 

# Is is believed Pcf and P1 will always bracked the root, however 

# leave the commented code for testing 

elif D is None and (None not in [P2, P1, L, m]): 

def to_solve(D): 

return m - isothermal_gas(rho, f, P1=P1, P2=P2, L=L, D=D) 

return newton(to_solve, 0.1) 

else: 

raise Exception('This function solves for either mass flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Panhandle_A(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=0.92): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Panhandle A formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Panhandle A equation. Here, a new form is developed with all units 

in base SI, based on the work of [1]_. 

 

.. math:: 

Q = 158.02053 E \left(\frac{T_s}{P_s}\right)^{1.0788}\left[\frac{P_1^2 

-P_2^2}{L \cdot {SG}^{0.8539} T_{avg}Z_{avg}}\right]^{0.5394}D^{2.6182} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

[1]_'s original constant was 4.5965E-3, and it has units of km (length), 

kPa, mm (diameter), and flowrate in m^3/day. 

 

The form in [2]_ has the same exponents as used here, units of mm 

(diameter), kPa, km (length), and flow in m^3/hour; its leading constant is 

1.9152E-4. 

 

The GPSA [3]_ has a leading constant of 0.191, a bracketed power of 0.5392, 

a specific gravity power of 0.853, and otherwise the same constants. 

It is in units of mm (diameter) and kPa and m^3/day; length is stated to be 

in km, but according to the errata is in m. 

 

[4]_ has a leading constant of 1.198E7, a specific gravity of power of 0.8541, 

and a power of diameter which is under the root of 4.854 and is otherwise 

the same. It has units of kPa and m^3/day, but is otherwise in base SI 

units. 

 

[5]_ has a leading constant of 99.5211, but its reference correction has no 

exponent; other exponents are the same as here. It is entirely in base SI 

units. 

 

[6]_ has pressures in psi, diameter in inches, length in miles, Q in 

ft^3/day, T in degrees Rankine, and a constant of 435.87. 

Its reference condition power is 1.07881, and it has a specific gravity 

correction outside any other term with a power of 0.4604. 

 

Examples 

-------- 

>>> Panhandle_A(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

42.56082051195928 

 

References 

---------- 

.. [1] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [2] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 

2009. 

.. [3] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

.. [4] Campbell, John M. Gas Conditioning and Processing, Vol. 2: The 

Equipment Modules. 7th edition. Campbell Petroleum Series, 1992. 

.. [5] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

.. [6] Ikoku, Chi U. Natural Gas Production Engineering. Malabar, Fla: 

Krieger Pub Co, 1991. 

''' 

c1 = 1.0788 

c2 = 0.8539 

c3 = 0.5394 

c4 = 2.6182 

c5 = 158.0205328706957220332831680508433862787 # 45965*10**(591/1250)/864 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*E*(Ts/Ps)**c1*((P1**2 - P2**2)/(L*SG**c2*Tavg*Zavg))**c3*D**c4 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Q*(Ts/Ps)**(-c1)*(SG**(-c2)*(P1**2 - P2**2)/(L*Tavg*Zavg))**(-c3)/(E*c5))**(1./c4) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*SG**c2*Tavg*Zavg*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(1./c3) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*SG**c2*Tavg*Zavg*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(1./c3) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return SG**(-c2)*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(-1./c3)*(P1**2 - P2**2)/(Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Panhandle_B(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=0.92): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Panhandle B formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Panhandle B equation. Here, a new form is developed with all units 

in base SI, based on the work of [1]_. 

 

.. math:: 

Q = 152.88116 E \left(\frac{T_s}{P_s}\right)^{1.02}\left[\frac{P_1^2 

-P_2^2}{L \cdot {SG}^{0.961} T_{avg}Z_{avg}}\right]^{0.51}D^{2.53} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

[1]_'s original constant was 1.002E-2, and it has units of km (length), 

kPa, mm (diameter), and flowrate in m^3/day. 

 

The form in [2]_ has the same exponents as used here, units of mm 

(diameter), kPa, km (length), and flow in m^3/hour; its leading constant is 

4.1749E-4. 

 

The GPSA [3]_ has a leading constant of 0.339, and otherwise the same constants. 

It is in units of mm (diameter) and kPa and m^3/day; length is stated to be 

in km, but according to the errata is in m. 

 

[4]_ has a leading constant of 1.264E7, a diameter power of 4.961 which is 

also under the 0.51 power, and is otherwise the same. It has units of kPa 

and m^3/day, but is otherwise in base SI units. 

 

[5]_ has a leading constant of 135.8699, but its reference correction has 

no exponent and its specific gravity has a power of 0.9608; the other 

exponents are the same as here. It is entirely in base SI units. 

 

[6]_ has pressures in psi, diameter in inches, length in miles, Q in 

ft^3/day, T in degrees Rankine, and a constant of 737 with the exponents 

the same as here. 

 

Examples 

-------- 

>>> Panhandle_B(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

42.35366178004172 

 

References 

---------- 

.. [1] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [2] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 

2009. 

.. [3] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

.. [4] Campbell, John M. Gas Conditioning and Processing, Vol. 2: The 

Equipment Modules. 7th edition. Campbell Petroleum Series, 1992. 

.. [5] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

.. [6] Ikoku, Chi U. Natural Gas Production Engineering. Malabar, Fla: 

Krieger Pub Co, 1991. 

''' 

c1 = 1.02 # reference condition power 

c2 = 0.961 # sg power 

c3 = 0.51 # main power 

c4 = 2.53 # diameter power 

c5 = 152.8811634298055458624385985866624419060 # 4175*10**(3/25)/36 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*E*(Ts/Ps)**c1*((P1**2 - P2**2)/(L*SG**c2*Tavg*Zavg))**c3*D**c4 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Q*(Ts/Ps)**(-c1)*(SG**(-c2)*(P1**2 - P2**2)/(L*Tavg*Zavg))**(-c3)/(E*c5))**(1./c4) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*SG**c2*Tavg*Zavg*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(1./c3) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*SG**c2*Tavg*Zavg*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(1./c3) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return SG**(-c2)*(D**(-c4)*Q*(Ts/Ps)**(-c1)/(E*c5))**(-1./c3)*(P1**2 - P2**2)/(Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Weymouth(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=0.92): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Weymouth formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Weymouth equation. Here, a new form is developed with all units 

in base SI, based on the work of [1]_. 

 

.. math:: 

Q = 137.32958 E \frac{T_s}{P_s}\left[\frac{P_1^2 

-P_2^2}{L \cdot {SG} \cdot T_{avg}Z_{avg}}\right]^{0.5}D^{2.667} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

[1]_'s original constant was 3.7435E-3, and it has units of km (length), 

kPa, mm (diameter), and flowrate in m^3/day. 

 

The form in [2]_ has the same exponents as used here, units of mm 

(diameter), kPa, km (length), and flow in m^3/hour; its leading constant is 

1.5598E-4. 

 

The GPSA [3]_ has a leading constant of 0.1182, and otherwise the same constants. 

It is in units of mm (diameter) and kPa and m^3/day; length is stated to be 

in km, but according to the errata is in m. 

 

[4]_ has a leading constant of 1.162E7, a diameter power of 5.333 which is 

also under the 0.50 power, and is otherwise the same. It has units of kPa 

and m^3/day, but is otherwise in base SI units. 

 

[5]_ has a leading constant of 137.2364; the other 

exponents are the same as here. It is entirely in base SI units. 

 

[6]_ has pressures in psi, diameter in inches, length in miles, Q in 

ft^3/hour, T in degrees Rankine, and a constant of 18.062 with the 

exponents the same as here. 

 

Examples 

-------- 

>>> Weymouth(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

32.07729055913029 

 

References 

---------- 

.. [1] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [2] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane, 

2009. 

.. [3] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

.. [4] Campbell, John M. Gas Conditioning and Processing, Vol. 2: The 

Equipment Modules. 7th edition. Campbell Petroleum Series, 1992. 

.. [5] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

.. [6] Ikoku, Chi U. Natural Gas Production Engineering. Malabar, Fla: 

Krieger Pub Co, 1991. 

''' 

c3 = 0.5 # main power 

c4 = 2.667 # diameter power 

c5 = 137.3295809942512546732179684618143090992 # 37435*10**(501/1000)/864 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*E*(Ts/Ps)*((P1**2 - P2**2)/(L*SG*Tavg*Zavg))**c3*D**c4 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Ps*Q*((P1**2 - P2**2)/(L*SG*Tavg*Zavg))**(-c3)/(E*Ts*c5))**(1./c4) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*SG*Tavg*Zavg*(D**(-c4)*Ps*Q/(E*Ts*c5))**(1./c3) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*SG*Tavg*Zavg*(D**(-c4)*Ps*Q/(E*Ts*c5))**(1./c3) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return (D**(-c4)*Ps*Q/(E*Ts*c5))**(-1./c3)*(P1**2 - P2**2)/(SG*Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Spitzglass_high(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=1.): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Spitzglass (high pressure drop) formula. Can calculate 

any of the following, given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe (numerical solution) 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Spitzglass (high pressure drop) formula. Here, the form as in [1]_ 

is used but with a more precise metric conversion from inches to m. 

 

.. math:: 

Q = 125.1060 E \left(\frac{T_s}{P_s}\right)\left[\frac{P_1^2 

-P_2^2}{L \cdot {SG} T_{avg}Z_{avg} (1 + 0.09144/D + \frac{150}{127}D)} 

\right]^{0.5}D^{2.5} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

This equation is often presented without any corection for reference 

conditions for specific gravity. 

 

This model is also presented in [2]_ with a leading constant of 1.0815E-2, 

the same exponents as used here, units of mm (diameter), kPa, km (length), 

and flow in m^3/hour. 

 

Examples 

-------- 

>>> Spitzglass_high(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

29.42670246281681 

 

References 

---------- 

.. [1] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

.. [2] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

''' 

c3 = 1.181102362204724409448818897637795275591 # 0.03/inch or 150/127 

c4 = 0.09144 

c5 = 125.1060 

if Q is None and (None not in [L, D, P1, P2]): 

return (c5*E*Ts/Ps*D**2.5*((P1**2-P2**2) 

/(L*SG*Zavg*Tavg*(1 + c4/D + c3*D)))**0.5) 

elif D is None and (None not in [L, Q, P1, P2]): 

to_solve = lambda D : Q - Spitzglass_high(SG=SG, Tavg=Tavg, L=L, D=D, 

P1=P1, P2=P2, Ts=Ts, Ps=Ps, 

Zavg=Zavg, E=E) 

return newton(to_solve, 0.5) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return ((D**6*E**2*P2**2*Ts**2*c5**2 

+ D**2*L*Ps**2*Q**2*SG*Tavg*Zavg*c3 

+ D*L*Ps**2*Q**2*SG*Tavg*Zavg 

+ L*Ps**2*Q**2*SG*Tavg*Zavg*c4)/(D**6*E**2*Ts**2*c5**2))**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return ((D**6*E**2*P1**2*Ts**2*c5**2 

- D**2*L*Ps**2*Q**2*SG*Tavg*Zavg*c3 

- D*L*Ps**2*Q**2*SG*Tavg*Zavg 

- L*Ps**2*Q**2*SG*Tavg*Zavg*c4)/(D**6*E**2*Ts**2*c5**2))**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return (D**6*E**2*Ts**2*c5**2*(P1**2 - P2**2) 

/(Ps**2*Q**2*SG*Tavg*Zavg*(D**2*c3 + D + c4))) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Spitzglass_low(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=1.): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Spitzglass (low pressure drop) formula. Can calculate 

any of the following, given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe (numerical solution) 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Spitzglass (low pressure drop) formula. Here, the form as in [1]_ 

is used but with a more precise metric conversion from inches to m. 

 

.. math:: 

Q = 125.1060 E \left(\frac{T_s}{P_s}\right)\left[\frac{2(P_1 

-P_2)(P_s+1210)}{L \cdot {SG} \cdot T_{avg}Z_{avg} (1 + 0.09144/D 

+ \frac{150}{127}D)}\right]^{0.5}D^{2.5} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

This equation is often presented without any corection for reference 

conditions for specific gravity. 

 

This model is also presented in [2]_ with a leading constant of 5.69E-2, 

the same exponents as used here, units of mm (diameter), kPa, km (length), 

and flow in m^3/hour. However, it is believed to contain a typo, and gives 

results <1/3 of the correct values. It is also present in [2]_ in imperial 

form; this is believed correct, but makes a slight assumption not done in 

[1]_. 

 

This model is present in [3]_ without reference corrections. The 1210 

constant in [1]_ is an approximation necessary for the reference correction 

to function without a square of the pressure difference. The GPSA version 

is as follows, and matches this formulation very closely: 

 

.. math:: 

Q = 0.821 \left[\frac{(P_1-P_2)D^5}{L \cdot {SG} 

(1 + 91.44/D + 0.0018D)}\right]^{0.5} 

 

The model is also shown in [4]_, with diameter in inches, length in feet, 

flow in MMSCFD, pressure drop in inH2O, and a rounded leading constant of 

0.09; this makes its predictions several percent higher than the model here. 

 

Examples 

-------- 

>>> Spitzglass_low(D=0.154051, P1=6720.3199, P2=0, L=54.864, SG=0.6, Tavg=288.7) 

0.9488775242530617 

 

References 

---------- 

.. [1] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

.. [2] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [3] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

.. [4] PetroWiki. "Pressure Drop Evaluation along Pipelines" Accessed 

September 11, 2016. http://petrowiki.org/Pressure_drop_evaluation_along_pipelines#Spitzglass_equation_2. 

''' 

c3 = 1.181102362204724409448818897637795275591 # 0.03/inch or 150/127 

c4 = 0.09144 

c5 = 125.1060 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*Ts/Ps*D**2.5*E*(((P1-P2)*2*(Ps+1210.))/(L*SG*Tavg*Zavg*(1 + c4/D + c3*D)))**0.5 

elif D is None and (None not in [L, Q, P1, P2]): 

to_solve = lambda D : Q - Spitzglass_low(SG=SG, Tavg=Tavg, L=L, D=D, P1=P1, P2=P2, Ts=Ts, Ps=Ps, Zavg=Zavg, E=E) 

return newton(to_solve, 0.5) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return 0.5*(2.0*D**6*E**2*P2*Ts**2*c5**2*(Ps + 1210.0) + D**2*L*Ps**2*Q**2*SG*Tavg*Zavg*c3 + D*L*Ps**2*Q**2*SG*Tavg*Zavg + L*Ps**2*Q**2*SG*Tavg*Zavg*c4)/(D**6*E**2*Ts**2*c5**2*(Ps + 1210.0)) 

elif P2 is None and (None not in [L, Q, D, P1]): 

return 0.5*(2.0*D**6*E**2*P1*Ts**2*c5**2*(Ps + 1210.0) - D**2*L*Ps**2*Q**2*SG*Tavg*Zavg*c3 - D*L*Ps**2*Q**2*SG*Tavg*Zavg - L*Ps**2*Q**2*SG*Tavg*Zavg*c4)/(D**6*E**2*Ts**2*c5**2*(Ps + 1210.0)) 

elif L is None and (None not in [P2, Q, D, P1]): 

return 2.0*D**6*E**2*Ts**2*c5**2*(P1*Ps + 1210.0*P1 - P2*Ps - 1210.0*P2)/(Ps**2*Q**2*SG*Tavg*Zavg*(D**2*c3 + D + c4)) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Oliphant(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=0.92): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Oliphant formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe (numerical solution) 

* Length of pipe 

 

This model is a more complete conversion to metric of the Imperial version 

presented in [1]_. 

 

.. math:: 

Q = 84.5872\left(D^{2.5} + 0.20915D^3\right)\frac{T_s}{P_s}\left(\frac 

{P_1^2 - P_2^2}{L\cdot {SG} \cdot T_{avg}}\right)^{0.5} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

Recommended in [1]_ for use between vacuum and 100 psi. 

 

The model is simplified by grouping constants here; however, it is presented 

in the imperial unit set inches (diameter), miles (length), psi, Rankine, 

and MMSCFD in [1]_: 

 

.. math:: 

Q = 42(24)\left(D^{2.5} + \frac{D^3}{30}\right)\left(\frac{14.4}{P_s} 

\right)\left(\frac{T_s}{520}\right)\left[\left(\frac{0.6}{SG}\right) 

\left(\frac{520}{T_{avg}}\right)\left(\frac{P_1^2 - P_2^2}{L}\right) 

\right]^{0.5} 

 

Examples 

-------- 

>>> Oliphant(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

28.851535408143057 

 

References 

---------- 

.. [1] GPSA. GPSA Engineering Data Book. 13th edition. Gas Processors 

Suppliers Association, Tulsa, OK, 2012. 

.. [2] F. N. Oliphant, "Production of Natural Gas," Report. USGS, 1902. 

''' 

# c1 = 42*24*Q*foot**3/day*(mile)**0.5*9/5.*(5/9.)**0.5*psi*(1/psi)*14.4/520.*0.6**0.5*520**0.5/inch**2.5 

c1 = 84.587176139918568651410168968141078948974609375000 

c2 = 0.2091519350460528670065940559652517549694 # 1/(30.*0.0254**0.5) 

if Q is None and (None not in [L, D, P1, P2]): 

return c1*(D**2.5 + c2*D**3)*Ts/Ps*((P1**2-P2**2)/(L*SG*Tavg))**0.5 

elif D is None and (None not in [L, Q, P1, P2]): 

to_solve = lambda D : Q - Oliphant(SG=SG, Tavg=Tavg, L=L, D=D, P1=P1, P2=P2, Ts=Ts, Ps=Ps, Zavg=Zavg, E=E) 

return newton(to_solve, 0.5) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*Ps**2*Q**2*SG*Tavg/(Ts**2*c1**2*(D**3*c2 + D**2.5)**2) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*Ps**2*Q**2*SG*Tavg/(Ts**2*c1**2*(D**3*c2 + D**2.5)**2) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return Ts**2*c1**2*(P1**2 - P2**2)*(D**3*c2 + D**2.5)**2/(Ps**2*Q**2*SG*Tavg) 

else: 

raise Exception('This function solves for either flow, upstream \ 

pressure, downstream pressure, diameter, or length; all other inputs \ 

must be provided.') 

 

 

def Fritzsche(SG, Tavg, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, Ps=101325., Zavg=1, E=1): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Fritzsche formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Fritzsche formula. Here, the form as in [1]_ 

is used but with all inputs in base SI units. 

 

.. math:: 

Q = 93.500 \frac{T_s}{P_s}\left(\frac{P_1^2 - P_2^2} 

{L\cdot {SG}^{0.8587} \cdot T_{avg}}\right)^{0.538}D^{2.69} 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

This model is also presented in [1]_ with a leading constant of 2.827, 

the same exponents as used here, units of mm (diameter), kPa, km (length), 

and flow in m^3/hour. 

 

This model is shown in base SI units in [2]_, and with a leading constant 

of 94.2565, a diameter power of 2.6911, main group power of 0.5382 

and a specific gravity power of 0.858. The difference is very small. 

 

Examples 

-------- 

>>> Fritzsche(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, Tavg=277.15) 

39.421535157535565 

 

References 

---------- 

.. [1] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [2] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

''' 

# Rational('2.827E-3')/(3600*24)*(1000)**Rational('2.69')*(1000)**Rational('0.538')*1000/(1000**2)**Rational('0.538') 

c5 = 93.50009798751128188757518688244137811221 # 14135*10**(57/125)/432 

c2 = 0.8587 

c3 = 0.538 

c4 = 2.69 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*E*(Ts/Ps)*((P1**2 - P2**2)/(SG**c2*Tavg*L*Zavg))**c3*D**c4 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Ps*Q*(SG**(-c2)*(P1**2 - P2**2)/(L*Tavg*Zavg))**(-c3)/(E*Ts*c5))**(1./c4) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*SG**c2*Tavg*Zavg*(D**(-c4)*Ps*Q/(E*Ts*c5))**(1./c3) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*SG**c2*Tavg*Zavg*(D**(-c4)*Ps*Q/(E*Ts*c5))**(1./c3) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return SG**(-c2)*(D**(-c4)*Ps*Q/(E*Ts*c5))**(-1./c3)*(P1**2 - P2**2)/(Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream pressure, downstream pressure, diameter, or length; all other inputs must be provided.') 

 

 

def Muller(SG, Tavg, mu, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=1): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the Muller formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the Muller formula. Here, the form as in [1]_ 

is used but with all inputs in base SI units. 

 

.. math:: 

Q = 15.7743\frac{T_s}{P_s}E\left(\frac{P_1^2 - P_2^2}{L \cdot Z_{avg} 

\cdot T_{avg}}\right)^{0.575} \left(\frac{D^{2.725}}{\mu^{0.15} 

SG^{0.425}}\right) 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

mu : float 

Average viscosity of the fluid in the pipeline, [Pa*s] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

This model is presented in [1]_ with a leading constant of 0.4937, the same 

exponents as used here, units of inches (diameter), psi, feet (length), 

Rankine, pound/(foot*second) for viscosity, and 1000 ft^3/hour. 

 

This model is also presented in [2]_ in both SI and imperial form. The 

SI form was incorrectly converted and yields much higher flow rates. The 

imperial version has a leading constant of 85.7368, the same powers as 

used here except with rounded values of powers of viscosity (0.2609) and 

specific gravity (0.7391) rearanged to be inside the bracketed group; 

its units are inches (diameter), psi, miles (length), 

Rankine, pound/(foot*second) for viscosity, and ft^3/day. 

 

This model is shown in base SI units in [3]_, and with a leading constant 

of 15.7650, a diameter power of 2.724, main group power of 0.5747, 

a specific gravity power of 0.74, and a viscosity power of 0.1494. 

 

Examples 

-------- 

>>> Muller(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, mu=1E-5, 

... Tavg=277.15) 

60.45796698148659 

 

References 

---------- 

.. [1] Mohitpour, Mo, Golshan, and Allan Murray. Pipeline Design and 

Construction: A Practical Approach. 3rd edition. New York: Amer Soc 

Mechanical Engineers, 2006. 

.. [2] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [3] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

''' 

# 1000*foot**3/hour*0.4937/inch**2.725*foot**0.575*(5/9.)**0.575*9/5.*(pound/foot)**0.15*psi*(1/psi**2)**0.575 

c5 = 15.77439908642077352939746374951659525108 # 5642991*196133**(17/20)*2**(3/5)*3**(11/40)*5**(7/40)/30645781250 

c2 = 0.575 # main power 

c3 = 2.725 # D power 

c4 = 0.425 # SG power 

c1 = 0.15 # mu power 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*Ts/Ps*E*((P1**2-P2**2)/Tavg/L/Zavg)**c2*D**c3/SG**c4/mu**c1 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Ps*Q*SG**c4*mu**c1*((P1**2 - P2**2)/(L*Tavg*Zavg))**(-c2)/(E*Ts*c5))**(1./c3) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*Tavg*Zavg*(D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(1/c2) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*Tavg*Zavg*(D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(1/c2) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return (D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(-1/c2)*(P1**2 - P2**2)/(Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream pressure, downstream pressure, diameter, or length; all other inputs must be provided.') 

 

 

def IGT(SG, Tavg, mu, L=None, D=None, P1=None, P2=None, Q=None, Ts=288.7, 

Ps=101325., Zavg=1, E=1): 

r'''Calculation function for dealing with flow of a compressible gas in a 

pipeline with the IGT formula. Can calculate any of the following, 

given all other inputs: 

 

* Flow rate 

* Upstream pressure 

* Downstream pressure 

* Diameter of pipe 

* Length of pipe 

 

A variety of different constants and expressions have been presented 

for the IGT formula. Here, the form as in [1]_ 

is used but with all inputs in base SI units. 

 

.. math:: 

Q = 24.6241\frac{T_s}{P_s}E\left(\frac{P_1^2 - P_2^2}{L \cdot Z_{avg} 

\cdot T_{avg}}\right)^{5/9} \left(\frac{D^{8/3}}{\mu^{1/9} 

SG^{4/9}}\right) 

 

Parameters 

---------- 

SG : float 

Specific gravity of fluid with respect to air at the reference 

temperature and pressure `Ts` and `Ps`, [-] 

Tavg : float 

Average temperature of the fluid in the pipeline, [K] 

mu : float 

Average viscosity of the fluid in the pipeline, [Pa*s] 

L : float, optional 

Length of pipe, [m] 

D : float, optional 

Diameter of pipe, [m] 

P1 : float, optional 

Inlet pressure to pipe, [Pa] 

P2 : float, optional 

Outlet pressure from pipe, [Pa] 

Q : float, optional 

Flow rate of gas through pipe, [m^3/s] 

Ts : float, optional 

Reference temperature for the specific gravity of the gas, [K] 

Ps : float, optional 

Reference pressure for the specific gravity of the gas, [Pa] 

Zavg : float, optional 

Average compressibility factor for gas, [-] 

E : float, optional 

Pipeline efficiency, a correction factor between 0 and 1 

 

Returns 

------- 

Q, P1, P2, D, or L : float 

The missing input which was solved for [base SI] 

 

Notes 

----- 

This model is presented in [1]_ with a leading constant of 0.6643, the same 

exponents as used here, units of inches (diameter), psi, feet (length), 

Rankine, pound/(foot*second) for viscosity, and 1000 ft^3/hour. 

 

This model is also presented in [2]_ in both SI and imperial form. Both 

forms are correct. The imperial version has a leading constant of 136.9, 

the same powers as used here except with rounded values of powers of 

viscosity (0.2) and specific gravity (0.8) rearanged to be inside the 

bracketed group; its units are inches (diameter), psi, miles (length), 

Rankine, pound/(foot*second) for viscosity, and ft^3/day. 

 

This model is shown in base SI units in [3]_, and with a leading constant 

of 24.6145, and the same powers as used here. 

 

Examples 

-------- 

>>> IGT(D=0.340, P1=90E5, P2=20E5, L=160E3, SG=0.693, mu=1E-5, Tavg=277.15) 

48.92351786788815 

 

References 

---------- 

.. [1] Mohitpour, Mo, Golshan, and Allan Murray. Pipeline Design and 

Construction: A Practical Approach. 3rd edition. New York: Amer Soc 

Mechanical Engineers, 2006. 

.. [2] Menon, E. Shashi. Gas Pipeline Hydraulics. 1st edition. Boca Raton, 

FL: CRC Press, 2005. 

.. [3] Coelho, Paulo M., and Carlos Pinho. "Considerations about Equations 

for Steady State Flow in Natural Gas Pipelines." Journal of the 

Brazilian Society of Mechanical Sciences and Engineering 29, no. 3 

(September 2007): 262-73. doi:10.1590/S1678-58782007000300005. 

''' 

# 1000*foot**3/hour*0.6643/inch**(8/3.)*foot**(5/9.)*(5/9.)**(5/9.)*9/5.*(pound/foot)**(1/9.)*psi*(1/psi**2)**(5/9.) 

c5 = 24.62412451461407054875301709443930350550 # 1084707*196133**(8/9)*2**(1/9)*6**(1/3)/4377968750 

c2 = 5/9. # main power 

c3 = 8/3. # D power 

c4 = 4/9. # SG power 

c1 = 1/9. # mu power 

if Q is None and (None not in [L, D, P1, P2]): 

return c5*Ts/Ps*E*((P1**2-P2**2)/Tavg/L/Zavg)**c2*D**c3/SG**c4/mu**c1 

elif D is None and (None not in [L, Q, P1, P2]): 

return (Ps*Q*SG**c4*mu**c1*((P1**2 - P2**2)/(L*Tavg*Zavg))**(-c2)/(E*Ts*c5))**(1./c3) 

elif P1 is None and (None not in [L, Q, D, P2]): 

return (L*Tavg*Zavg*(D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(1/c2) + P2**2)**0.5 

elif P2 is None and (None not in [L, Q, D, P1]): 

return (-L*Tavg*Zavg*(D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(1/c2) + P1**2)**0.5 

elif L is None and (None not in [P2, Q, D, P1]): 

return (D**(-c3)*Ps*Q*SG**c4*mu**c1/(E*Ts*c5))**(-1/c2)*(P1**2 - P2**2)/(Tavg*Zavg) 

else: 

raise Exception('This function solves for either flow, upstream pressure, downstream pressure, diameter, or length; all other inputs must be provided.')