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compressible.py
1895 lines (1610 loc) · 73.4 KB
/
compressible.py
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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, isinf
from fluids.constants import R
from fluids.numerics import secant, newton, ridder, 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']
__numba_additional_funcs__ = ['isothermal_gas_err_P1', 'isothermal_gas_err_P2',
'isothermal_gas_err_P2_basis', 'isothermal_gas_err_D',
'_to_solve_Spitzglass_high', '_to_solve_Spitzglass_low',
'_to_solve_Oliphant']
def isothermal_work_compression(P1, P2, T, Z=1.0):
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
.. math::
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.
This is the best possible case for compression; all actual compresssors
require more work to do the compression.
By making the compression take a large number of stages and cooling the gas
between stages, this can be approached reasonable closely. Integrally
geared compressors are often used for this purpose.
Examples
--------
>>> isothermal_work_compression(1E5, 1E6, 300)
5743.427304244769
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=1.0, 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 efficiency 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}}
.. math::
W = \int_{P_1}^{P_2} V dP = \int_{P_1}^{P_2}\frac{P_1^{1/k}V_1}
{P^{1/k}}dP
.. math::
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)
10416.876986384483
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 eta is not None and P1 is not None and P2 is not None:
return k/(k - 1.0)*Z*R*T1*((P2/P1)**((k-1.)/k) - 1.0)/eta
elif P1 is None and eta is not None and W is not None and P2 is not None:
return P2*(1.0 + W*eta/(R*T1*Z) - W*eta/(R*T1*Z*k))**(-k/(k - 1.0))
elif P2 is None and eta is not None and W is not None and P1 is not None:
return P1*(1.0 + W*eta/(R*T1*Z) - W*eta/(R*T1*Z*k))**(k/(k - 1.0))
elif eta is None and P1 is not None and P2 is not None and W is not None:
return R*T1*Z*k*((P2/P1)**((k - 1.0)/k) - 1.0)/(W*(k - 1.0))
else:
raise ValueError('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 - 1.0)/k) - 1.0)/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}
.. math::
\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) [-]
eta_s : float, optional
Isentropic (adiabatic) 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 is not None:
return ((P2/P1)**((k-1.0)/k)-1.0)/((P2/P1)**((k-1.0)/(k*eta_p))-1.0)
elif eta_p is None and eta_s is not None:
return (k - 1.0)*log(P2/P1)/(k*log(
(eta_s + (P2/P1)**((k - 1.0)/k) - 1.0)/eta_s))
else:
raise ValueError('Either eta_s or eta_p is required')
def polytropic_exponent(k, n=None, eta_p=None):
r'''Calculates one of:
* Polytropic exponent from polytropic efficiency
* Polytropic efficiency from the polytropic exponent
.. math::
n = \frac{k\eta_p}{1 - k(1-\eta_p)}
.. math::
\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) [-]
n : float, optional
Polytropic exponent of the process [-]
eta_p : float, optional
Polytropic efficiency of the process, [-]
Returns
-------
n or eta_p : float
Polytropic 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 is not None:
return k*eta_p/(1.0 - k*(1.0 - eta_p))
elif eta_p is None and n is not None:
return n*(k - 1.0)/(k*(n - 1.0))
else:
raise ValueError('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*2.0/(k + 1.0)
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*(2.0/(k + 1.))**(k/(k - 1.0))
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 [-]
D : float
Diameter of pipe, [m]
L : float
Length 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 + 1.0) + L*fd)/(2.0*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 - 1.0))
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 - 1.0)/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 temperature `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_err_P1(P1, fd, rho, P2, L, D, m):
return m - isothermal_gas(rho, fd, P1=P1, P2=P2, L=L, D=D)
def isothermal_gas_err_P2(P2, rho, fd, P1, L, D, m):
return m - isothermal_gas(rho, fd, P1=P1, P2=P2, L=L, D=D)
def isothermal_gas_err_P2_basis(P1, P2, rho, fd, m, L, D):
return abs(P2 - isothermal_gas(rho, fd, m=m, P1=P1, P2=None, L=L, D=D))
def isothermal_gas_err_D(D, m, rho, fd, P1, P2, L):
return m - isothermal_gas(rho, fd, P1=P1, P2=P2, L=L, D=D)
def isothermal_gas(rho, fd, 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]
fd : 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}}
.. math::
B = \frac{\pi^2 D^4}{4^2} \rho_{avg}
.. math::
C = f_d \frac{L}{D}
A wide range of conditions are impossible due to choked 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.
A number of limitations exist for this model:
* Density dependence is that of an ideal gas.
* If calculating the pressure drop, the average gas density cannot
be known immediately; iteration must be used to correct this.
* The friction factor depends on both the gas density and velocity,
so it should be solved for iteratively as well. It changes throughout
the pipe as the gas expands and velocity increases.
* The model is not easily adapted to include elevation effects due to
the acceleration term included in it.
* As the gas expands, it will change temperature slightly, further
altering the density and friction factor.
There are many commercial packages which perform the actual direct
integration of the flow, such as OLGA Dynamic Multiphase Flow Simulator,
or ASPEN Hydraulics.
This expression has also been presented with the ideal gas assumption
directly incorporated into it [4]_ (note R is the specific gas constant, in
units of J/kg/K):
.. math::
\dot m^2 = \frac{\left(\frac{\pi D^2}{4}\right)^2
\left(P_1^2-P_2^2\right)}{RT\left(f_d\frac{L}{D} + 2\ln\frac{P_1}{P_2}
\right)}
Examples
--------
>>> isothermal_gas(rho=11.3, fd=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.
.. [4] Rennels, Donald C., and Hobart M. Hudson. Pipe Flow: A Practical
and Comprehensive Guide. 1st edition. Hoboken, N.J: Wiley, 2012.
'''
if m is None and P1 is not None and P2 is not None and L is not None and D is not None:
Pcf = P_isothermal_critical_flow(P=P1, fd=fd, D=D, L=L)
if P2 < Pcf:
raise ValueError('Given outlet pressure is not physically possible ' # numba: delete
'due to the formation of choked flow at P2=%f, specified outlet pressure was %f' % (Pcf, P2)) # numba: delete
# raise ValueError("Not possible") # numba: uncomment
if P2 > P1:
raise ValueError('Specified outlet pressure is larger than the '
'inlet pressure; fluid will flow backwards.')
return (0.0625*pi*pi*D**4*rho/(P1*(fd*L/D + 2.0*log(P1/P2)))*(P1*P1 - P2*P2))**0.5
elif L is None and P1 is not None and P2 is not None and D is not None and m is not None:
return D*(pi*pi*D**4*rho*(P1*P1 - P2*P2) - 32.0*P1*m*m*log(P1/P2))/(16.0*P1*fd*m*m)
elif P1 is None and L is not None and P2 is not None and D is not None and m is not None:
Pcf = P_upstream_isothermal_critical_flow(P=P2, fd=fd, D=D, L=L)
try:
# Use the explicit solution for P2 with different P1 guesses;
# newton doesn't like solving for m.
P1 = secant(isothermal_gas_err_P2_basis, (P2+Pcf)/2., args=(P2, rho, fd, m, L, D))
if not (P2 <= P1):
raise ValueError("Failed")
return P1
except:
try:
return ridder(isothermal_gas_err_P1, a=P2, b=Pcf, args=(fd, rho, P2, L, D, m))
except:
m_max = isothermal_gas(rho, fd, P1=Pcf, P2=P2, L=L, D=D)
raise ValueError('The desired mass flow rate of %f kg/s cannot ' # numba: delete
'be achieved with the specified downstream pressure; the maximum flowrate is ' # numba: delete
'%f kg/s at an upstream pressure of %f Pa' %(m, m_max, Pcf)) # numba: delete
# raise ValueError("Failed") # numba: uncomment
elif P2 is None and L is not None and P1 is not None and D is not None and m is not None:
try:
Pcf = P_isothermal_critical_flow(P=P1, fd=fd, D=D, L=L)
m_max = isothermal_gas(rho, fd, P1=P1, P2=Pcf, L=L, D=D)
if not (m <= m_max):
raise ValueError("Failed")
C = fd*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.
if isinf(lambert_ans):
raise ValueError("Should not be infinity")
P2 = P1/exp((-C*m**2+lambert_ans*m**2+B*P1)/m**2/2.)
if not (P2 < P1):
raise ValueError("Should not be the case")
return P2
except:
Pcf = P_isothermal_critical_flow(P=P1, fd=fd, D=D, L=L)
try:
return ridder(isothermal_gas_err_P2, a=Pcf, b=P1, args=(rho, fd, P1, L, D, m))
except:
m_max = isothermal_gas(rho, fd, P1=P1, P2=Pcf, L=L, D=D)
raise ValueError('The desired mass flow rate cannot be achieved ' # numba: delete
'with the specified upstream pressure of %f Pa; the maximum flowrate is %f ' # numba: delete
'kg/s at a downstream pressure of %f' %(P1, m_max, Pcf)) # numba: delete
# raise ValueError("Failed") # numba: uncomment
# 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 P2 is not None and P1 is not None and L is not None and m is not None:
return secant(isothermal_gas_err_D, 0.1, args=(m, rho, fd, P1, P2, L))
else:
raise ValueError('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=1.0, 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 at `Ts` and `Ps`, [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,