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Modeling and simulation of gas networks coupled to power grids

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In this paper, a mathematical framework for the coupling of gas networks to electric grids is presented to describe in particular the transition from gas to power. The dynamics of the gas flow are given by the isentropic Euler equations, while the power flow equations are used to model the power grid. We derive pressure laws for the gas flow that allow for the well-posedness of the coupling and a rigorous treatment of solutions. For simulation purposes, we apply appropriate numerical methods and show in a experimental study how gas-to-power might influence the dynamics of the gas and power network, respectively.
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Modeling and simulation of gas networks coupled to power
E. Fokken
, S. Göttlich
, O. Kolb
January 1, 2019
In this paper, a mathematical framework for the coupling of gas networks to electric grids
is presented to describe in particular the transition from gas to power. The dynamics of the
gas flow are given by the isentropic Euler equations, while the power flow equations are used
to model the power grid. We derive pressure laws for the gas flow that allow for the well-
posedness of the coupling and a rigorous treatment of solutions. For simulation purposes, we
apply appropriate numerical methods and show in a experimental study how gas-to-power
might influence the dynamics of the gas and power network, respectively.
AMS Classification: 35L65, 65M08
Keywords: Gas networks, pressure laws, power flow equations, simulation
1 Introduction
Power generation is changing drastically with more and more available energy coming from renew-
able sources. Most of these, like wind and solar power, are much more volatile than traditional
sources. The problems of incorporating them into the power system have sparked much research.
Due to their flexibility in comparison to other power plants, gas turbines and gas engine power
plants are often proposed as a means to balance power demands that cannot be met with re-
newable power sources at a given time. The repercussions on the gas pipeline networks by this
balancing have been studied [6] and also joint optimal control of gas and power networks are
investigated [24,27]. But so far only steady-state flow of the gas dynamics is examined. Although
this is often sufficient, the comparably low acoustic speed allows for short term effects that cannot
be seen with steady state methods.
We propose a fully time-dependent gas model on a pipeline network coupled to a power network.
Gas networks have been studied during the last two decades and questions of node conditions have
been analyzed from a theoretical point of view, see for example [1,2,4,5,7]. Therein, a system
of conservation or balance laws based on the Euler equations is used and the concept of Lax
curves [22] allows for a discussion on well-posedness of solutions. Finite volume or finite difference
methods are typically applied to solve the network problem numerically [15,19,21]. Power grids
are mainly described by the power flow equations consisting of a nonlinear system of equations to
determine power and voltage, see for example [3,17].
Although the power network could also be modeled with full time-dependency [9], the high
signal speed of electricity renders this unnecessary. Coupling of the networks happens at nodes
common to both, where a gas-fired generator is placed such that it can convert gas to electricity,
see [16,27]. The coupling is modeled by an equality constraint relating generated power to
consumed gas flow. In section 2and 3, we derive conditions on the gas pressure and power
demand, respectively, that guarantee that the coupling conditions lead to well-posedness of the
gas-power coupling. For the numerical study in section 4, we first give some validation results for
University of Mannheim, Department of Mathematics, 68131 Mannheim, Germany
arXiv:1812.11438v1 [math.NA] 29 Dec 2018
the proposed discretizations, then we present the influence of different pressure laws and lastly we
showcase the solution of a coupled gas-power system.
2 Gas networks
Gas networks have been investigated very intensively during the last ten years, see for example
[2,5,21]. Coupling conditions at nodes have been established to ensure well-posedness of the
network solution [7] and a rigorous numerical treatment [8,15,21,25]. As underlying structure
for gas networks we consider a directed graph with nodes VGas and arcs EGas . Nodes as well as
arcs may have state variables, which are in general time-dependent. We briefly recall the common
In each pipe eEGas, the gas flow is described by the isentropic Euler equations in one space
+ q
p(ρ) + q2
S(ρ, q),(1)
where tIR+is the time, x[0, le]is the position along the pipeline of length le,ρis the density
of the gas, qis the momentum of the gas, pis the pressure and Sincludes source terms that
influence the momentum of the gas. The pressure law must be a function of the form
pC1(IR+,IR+), p0(ρ)>0for all ρIR+,
to ensure strict hyperbolicity. Usual examples are the isothermal pressure law
p(ρ) = c2ρ, (2)
where cis the acoustic speed of the gas, or, more general, the γ-law
p(ρ) = κργ,(3)
for suitable constants κand γ, which we examine in Proposition 2. In fact, we show that the
class of possible pressure laws can be enlarged, leading to non-standard pressure functions. From
literature we know that the eigenvalues λand the eigenvectors rfor the system (1) are given by
λ1(ρ, q) = q
ρpp0(ρ), λ2(ρ, q) = q
r1(ρ, q) = 1
λ1(ρ, q), r2(ρ, q) = 1
λ2(ρ, q).
At the (gas) network inflow boundaries, typically the pressure pin is prescribed and at the outflow
boundaries the flow qout is given. Further, the following coupling conditions, an introduction to
which can be found in [2], are used at all inner nodes of VGas :
equality of pressure: the pressure values at the ends of all arcs connected to the same node
must be equal, that is, there is a coupling pressure pcoupling such that
pe=pcoupling (4)
at the end of all arcs econnected to the junction. This condition is equivalent to a condition
ρe=ρcoupling ,
because pas a strictly increasing function is one-to-one.
conservation of mass: the sum of all incoming fluxes must equal the sum of all outgoing
fluxes (including eventual source/sink terms), i.e.,
incoming pipes
qpipe =X
outgoing pipes
qpipe (5)
2.1 Well-posedness of the Riemann problem for the isentropic Euler
In this section, we investigate conditions for the pressure function (3) and the flows that guarantee
well-posedness of the Riemann problem for the isentropic Euler equations without a source term,
see [4] for an introduction. By the front-tracking technique this is sufficient for the well-posedness
of the isentropic Euler equations, see [12]. Assuming strict hyperbolicity, we start with the as-
sumption p0>0. We also assume that the initial data Ul= (ρl, ql)and Ur= (ρr, qr)of the
Riemann problem is sub-sonic, i.e.,
< c(ρ) = pp0(ρ).(6)
A solution to the Riemann problem with left state Uland right state Uris given by an intersection
point of the Lax curves through these two points, again see [4]. The Lax curve of all states
U= (ρ, q)reachable via 1-rarefaction or 1-shock (that is, waves corresponding to λ1above) from
a left state Ul= (ρl, ql)is given by (taken from [7])
Ll(ρ;ρl, ql) =
sdsfor ρρl(rarefaction)
ρlpf(ρ, ρl)for ρlρ(shock),
where fis defined by
f(ρ, ρl) = a(ρ, ρl)∆p(ρ, ρl)with a(ρ, ρl) = ρ
(ρρl),p(ρ, ρl) = p(ρ)p(ρl)(8)
such that U= (ρ, Ll(ρ)) holds.
Analogously, the Lax curve of all states U= (ρ, q)reachable via 2-rarefaction or 2-shock from
a right state Ur= (ρr, qr)is given by
Lr(ρ;ρr, qr) =
sdsfor ρρr(rarefaction)
ρr+pf(ρ, ρr)for ρrρ(shock).
Whenever we write Ll(ρ)and Lr(ρ), it will mean Ll(ρ, ρl, ql)and Lr(ρ, ρr, qr), respectively. Beware
that later in section 2.2 different second and third arguments will appear and we will write them
out again.
A solution is then at a point ρ, where
An example of this is shown in figure 1.
Note that
Lr(ρ;ρr, qr) = Ll(ρ;ρr,qr).(11)
So Lland Lrshare most properties. Therefore equation (10) is best understood not as a
difference of two functions but as a sum of two very similar functions. Similar to [16], we show
certain properties of the Lax curves that are used to define the new pressure laws. There will
appear three propositions, which are numbered by A, B and C. These propositions present lists of
conditions that are closely related. To make things tractable, each list bears the same numbering
but has the letter of the corresponding proposition in front. For example, the conditions A1, B1
and C1 all govern concavity of the Lax curves.
Proposition A (Proposition 1).Let LC2(R+,R+)and let it fulfill conditions A1 through A3
(where only one of A3(a) and A3(b) must hold).
Figure 1: Typical Lax curves with pressure law p(ρ) = c2ρ. The zero of the purple curve is the
desired solution.
(A1) L00 0,
(A2) ρ > 0:L(ρ)<0,
(A3) (a) 0<limρ0L(ρ)<or
(b) limρ0L(ρ)=0and limρ0L0(ρ)>0.
Then there is a unique ρ > 0with L(ρ)=0such that for all ˆρ>ρholds L(ˆρ)<0and L(ρ)→ −∞
for ρ→ ∞.
Proof. Condition A3 ensures positive values near 0, condition A2 ensures negative values for some
ρ > 0, together yielding a zero in between and condition A1 makes Lconcave, guaranteeing
uniqueness of the zero and implying then that L(ρ)→ −∞ for ρ→ ∞.
Note if two functions L1,L2satisfy the prerequisites of Proposition A, so does L1+L2. There-
fore we search for conditions on pressure functions that make Lland Lrsatisfy the conditions
of Proposition A, as then their sum in equation (10) will, too, and so it will have a unique zero.
Our main result is the following. It generalizes the usual notion of γ-laws to allow for γ < 1.
Further, as we will show below (Proposition C), positive linear combinations of such valid pressure
laws again lead to valid pressure laws.
Proposition 2 (generalized γ-laws).Let the derivative of the pressure be given by
p0(ρ) = αρδ.
This translates to pressure functions given by
p(ρ) = α
for γ6= 0 and
p(ρ) = αlog(ρ) + const .
For these pressure functions there holds
The Lax curves through any sub-sonic initial states Ul,Urhave a unique intersection point
at some ρ > 0if and only if |δ| ≤ 2and α > 0.
For every δwith |δ|>2there are sub-sonic states Ul,Ursuch that the Lax curves have no
intersection at all.
Expressed in the usual form of γ-laws, this result means p(ρ) = κργis a valid pressure function
if and only if 0< γ < 3and κ > 0or 1< γ < 0and κ < 0. The proof of Proposition 2will be
given at the end of this section. To prove Proposition 2we will reformulate conditions A1 through
A3 in Proposition Ain terms of the pressure function.
Because of equation (11), both Lax curves behave essentially identically and therefore we prove
our findings for Llonly as the proofs for Lrare the same. Along the way the derivatives of the
Lax curves will be important and so we provide them here,
l(ρ) =
sdsc(ρ)for ρρl
2ffor ρlρ
l(ρ) =
ρ+c0(ρ)) for ρρl
4ffor ρlρ ,
r(ρ) =
sds+c(ρ)for ρρr
2ffor ρrρ
r(ρ) =
ρ+c0(ρ)for ρρr
4ffor ρrρ .
Next, we intend to give conditions under which Proposition Ais applicable to Ll. Before we do
so, we provide the following lemma:
Lemma 3. Let gC1(R+,R+)be a non-negative function, g0, such that also (ρ2g(ρ))00
for all ρ > 0and let Gbe given by G(ρ) = Rρl
ρg(s) ds. Then, there holds ρ2g(ρ)ρ0
0if and only
if ρG(ρ)ρ0
Proof. The proof can be found in Appendix A.
The role of gin Lemma 3will be played by p0(ρ)and c(ρ)
ρin the following. Let us now put the
focus on the Lax curve Llagain. The following Proposition Bwill turn conditions A1 through A3
in Proposition Ainto conditions on the pressure function.
Proposition B (Proposition 4).Let pC2(R+)with p0>0. Then conditions A1,A2,A3 hold
for Llfor all ρl,qlwith
ρl<pp0(ρl)if and only if conditions B1 through B3 hold (again with
only one of B3(a) and B3(b) fulfilled).
(B1) These inequalities hold:
2p0(ρ) + ρp00(ρ)0ρ > 0(14)
p2+a22∆pp00 (p0)2+1
2(a2)0(∆p2)00ρl>0, ρ > ρl,(15)
where the arguments ρand ρlhave been omitted for readability, see equation (8).
(B2) Let p= limρ→∞ p(ρ)R∪ {∞}. For all ρ > 0there holds
(B3) (a) There is p0>0such that p(ρ) = p0
(b) p(ρ)o
ρand limρ0Rρl
sdsc(ρ)c(ρl)0for all ρl>0.
B1 A1: The equivalence B1 A1 is immediate from the definitions, inequality (14) is for the rar-
efaction part, inequality (15) for the shock part. Note that
2ff 00 (f0)2= ∆p2+a22∆pp00 (p0)2+1
A1,A3(a)B1,B3(a): Assume conditions A1 and A3(a). Let F(ρ) = Rρl
0< l = lim
ρ0Ll(ρ) = lim
sds= lim
ρ0ρF (ρ).(16)
Therefore F(ρ) = l
ρ. Note now that
2pp0(ρ)(2p0(ρ) + ρp00(ρ)) 0
due to condition A1. Lemma 3therefore shows that
ρ=F0(ρ) = l
and hence
p0(ρ) = c(ρ)2=l2
which again with lemma 3yields
p(ρ) = l2
A1,A3(a)B1,B3(a): Assume now conditions B1 and B3(a). We now note that ρ2p0(ρ)0=ρ2p0(ρ) + ρp00(ρ)
0, because of B1 and use the lemmas to arrive at A3(a).
A1,A3(b)B1,B3(b): We assume condition B3(b). The last proof (here we need again conditions A1 and B1
respectively) also shows that
For the derivative we find
l(ρ) = lim
due to the sub-sonic condition. This is non-negative due to condition B3(b).
A1,A3(b)B1,B3(b): Assume that B3(b) does not hold, that is
sdsc(ρ)c(ρl)≤ −δ < 0.
Choosing ρl, qlsuch that ql
ρl=c(ρl) + 1
l(ρ)≤ −1
2δ < 0
A2B2: First of all we note that the limit exists, as p0>0. For ρ>ρlwe find
Ll(ρ) = ρ
ρlsf(ρ, ρl)
This is less than zero if and only if the bracket is less then zero. If now p=, this is
fulfilled for some ρ>ρl. If p<, we have the limit
ρ→∞ s1
Let now δand ρbe defined by
Then δ > 0as the state is sub-sonic and ρ0. Therefore let ρbe so great that δρ>0.
Then, we find
which implies A2.
A2B2: In case p=condition B2 is fulfilled. Let now p<and let there be for all sub-sonic
(ρl, ql)aρ>0such that
that is, let A2 be true. Note that the right-hand-side is increasing in ρbecause its derivative
is positive. Therefore taking suprema of (18) yields
With Proposition Bwe have a list of conditions B1,B2,B3 which is equivalent to conditions
A1,A2,A3 from Proposition Abut now only involves the pressure function. This is fortunate as
we want to prove Proposition 2, which only contains pressure functions. In principle, we could just
plug the pressure functions into our conditions Band check, what generalized γ-laws are allowed.
But there is more insight to be gained by simplifying the conditions further. We will do so in
Propositions 5,6and 7. Yet this comes at a price, the next list of conditions in Proposition C
below will only be sufficient conditions for Propositions Aand Bto hold.
Proposition 5. Let pC3(R+). Condition B1 holds if
2p0(ρ) + ρp00(ρ)0
6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)0
holds for all ρ > 0.
Proof. We note that 2ff00 (f0)2(ρl, ρl)=0and that
2ff 00 (f0)20= 2f f 000 .
As f0, we see that 2ff00 (f0)20for all ρρlif f000(ρ, ρl)0for all ρρl. This was also
proved in [13]. This is the case if and only if ρlf000(ρ, ρl)0. For this we find
ρlf000(ρ, ρl) = 6p0(ρ) + 3(2ρρl)p00(ρ) + ρ(ρρl)p000(ρ)
=h6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)iρl3p00(ρ) + ρp000(ρ),
which is an affine function in ρl. Hence for given ρ > 0this function takes its minimum in one of
the edges of the simplex {ρl|0ρlρ}. The values on these are given by
6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)and
32p0(ρ) + ρp00(ρ).
The next condition we will replace is condition B3(b).
Proposition 6. Condition B3(b) is fulfilled if either of these conditions is satisfied:
C3(b) (i) limρ0c(ρ)=0and 2p0(ρ)ρp00(ρ)0for all ρ > 0.
(ii) 0<limρ0c(ρ)<.
(iii) There is a η(0,1) such that limρ0ρηc(ρ)exists and 0<limρ0ρηc(ρ)<.
(i) Here we have
sdsc(ρ)c(ρl) = lim
= lim
= lim
2spp0(s)(2p0(s)sp00(s)) ds
(ii) In this case the integral is unbounded near zero, but c(ρ) + c(ρl)is finite.
(iii) In this case there is a > 0such that ρηc(ρ) = a1 + r(ρ)with r(ρ)0. Then, for every
mNthere is ρm>0such that for all ρ < ρmthere holds c(ρ)a11
mρη. Choose
now mso great that
and then ρm,0so small that for all ρ<ρm,0there holds
ηr(ρ) = 1 + θ
with θ > 0. Define also
Then we find for ρ < ρm,0
sdsc(ρ)c(ρl) = Zρm
=a 11
> aθρηa11
Proposition 7. All conditions only depend on differences in pressures, that is, if cRis a
constant then if a pressure function psatisfies our conditions, so does p+c. This means that if
p<in condition B2, we can instead choose p= 0, yielding the clearer condition
p0(ρ)≤ −p(ρ)
Having all we need, we now sum up our findings in the following Proposition C:
Proposition C (Proposition 8).The Riemann problem for arbitrary sub-sonic left- and right-hand
states is well-posed if the pressure function psatisfies the following conditions:
(C1) both inequalities of Proposition 5:
2p0(ρ) + ρp00(ρ)0,(19)
6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)0,(20)
(C2) one of the conditions according to Proposition 7:
(a) p→ ∞ for ρ→ ∞ or
(b) p0(ρ)≤ −p(ρ)
ρfor all ρ > 0,
(C3) according to Proposition 6:
(a) There is p0>0such that p(ρ) = p0
ρfor ρ0or
(b) one of the following conditions is fulfilled:
(i) limρ0c(ρ)=0and 2p0(ρ)ρp00(ρ)0for all ρ > 0,
(ii) 0<limρ0c(ρ)<,
(iii) there is a η(0,1) such that limρ0ρηc(ρ)exists and 0<limρ0ρηc(ρ)<.
These conditions are positively linear in pand hence all pressure functions satisfying them define
a convex cone. In addition, because the integral is monotone and linear, we can also integrate over
pressure functions to find new pressure functions.
With Proposition Cwe have a useful list of conditions which can be checked easily for any
given candidate for a pressure function. The proposition now makes it easy to prove our central
result Proposition 2and we will do so now.
Proof of Proposition 2.
For the first part of the claim we note that for |δ| ≤ 2and α > 0the generalized γ-law
fulfills the conditions in C. Note that at δ= 0 the conditions in C3 switch and at δ=1
the conditions in C2 switch.
For the second part of the claim we observe: For δ > 2, choose c(ρl)<ql
which is obviously sub-sonic. Then Llis strictly negative for ρ > 0, as is easily computed.
A similar range for qr
ρrshows the same for Lr. For δ < 2, choose q1
δ+1 c(ρl)<ql
ρl< c(ρl)
and Llis strictly positive for ρ > 0and similarly for Lr.
Example 9. As just proved, the functions
p(ρ) = ρ3
p(ρ) = ρ
p(ρ) = ρ1
p(ρ) = ρ1
p(ρ) = ln(ρ)
are all valid pressure functions. Even a and more exotic
p(ρ) = ρ3ρ
might be possible.
2.2 Extension to junctions
Our findings guarantee well-posedness of Riemann problems with sub-sonic initial conditions under
certain conditions on the pressure function. We now examine the coupling conditions (4) and (5)
and do so by considering the generalized Riemann problem at the junction in accordance with
[1,2,7]. Consider a junction with incoming pipes indexed by eIand outgoing pipes indexed by
fO. At the junction-facing end of each pipe there are initial states Ui= (ρi, qi)for each iI
and iO, respectively. To make things tractable, we restrict the solution of the junction Riemann
problem to be of this form: In each pipe iIO, there appears exactly one new state Vinext to
the junction such that the Visatisfy the junction conditions and are connected to their respective
Uiby a shock or a rarefaction wave. On ingoing pipes these must be 1-waves, on outgoing pipes
these must be 2-waves. A sketch of this is shown in figure 2.
(a) Junction at t= 0.
(b) Junction after some time.
Figure 2: A junction with initial states. i= 1,2,3are incoming pipes, i= 4,5are outgoing.
This is not a great restriction as this is the only solution structure found in pipelines with low
Mach number, yet for different solutions, see [13]. For these new states to appear at all, the wave
speed between Viand Uimust be negative on ingoing pipes and positive on outgoing ones. To
keep things simple, we examine a single ingoing pipe iwith initial condition Uiand new state
Vi= (ρ, Ll(ρ;Ui)).
Proposition 10. Let Proposition Band hence Abe fulfilled for the pressure. Let ρi,min be such
that L0
l(ρi,min;Ui) = 0, if it exists, otherwise let ρi,min = 0. The wave speed of the wave between
Uiand Viis negative if and only if ρ>ρi,min.
Proof. One easily sees that λ1(ρ, Ll(ρ;Ui)) = L0
l(ρ;Ui). Also using the definitions (7) and (12)
and the concavity of Ll, one can see that ρi,min is always unique. Because Llis concave, L0
decreasing, so λ1(ρ, Ll(ρ;Ui)) <0if and only if ρ>ρi,min .
: Let ρ>ρi,min. Then λ1(Vi)<0. If Viis in the rarefaction part of Ll(ρ;Ui)the wave speed
is negative. If Viis in the shock part of Ll(ρ;Ui), we compute:
s0(ρ) = Ll(ρ;Ui)L(ρi;Ui)
since Llis concave. So
s(ρ)s(ρi) = λ1(ρi)<0
because Uiis sub-sonic.
: Let ρρi,min . Because Uiis sub-sonic, we have ρρi,min < ρiand are dealing with a
rarefaction wave. But then the wave speed is just given by λ1(ρ, Ll(ρ;Ui)which is non-
For outgoing pipes one defines ρi,min 0such that L0
r(ρi,min;Ui)=0, if possible, or ρi,min = 0
otherwise and obtains in the same way
Proposition 11. Let Proposition Bbe fulfilled for the pressure. On an outgoing pipe the wave
speed between Uiand Vi= (ρ, Lr(ρ;Ui)) is positive if and only if ρ>ρi,min.
Coming back to junctions we define the junction minimal density as
ρmin = max
iIOρi,min .(21)
A solution (Vi)iIOto the junction Riemann problem is admissible if and only if the density ρat
the junction fulfills ρ>ρmin . If it is not admissible, then in at least one pipe there is a super-sonic
gas flow.
Note that a usual Riemann problem with sub-sonic initial conditions can be treated as a
junction with one ingoing and one outgoing pipe. In this case only one new state Vis created and
the admissibility criterion guarantees that λ1(V)<0< λ2(V). For the sake of completeness, we
classify the solutions to the usual Riemann problem by wave types in Table 1.
Table 1: (r)arefaction waves and (s)hocks for different values of the density ρ.
(a) wave types for ρlρr.
value of ρleft wave right wave
ρρmin invalid
ρmin < ρ ρlr r
ρlρρrs r
ρrρs s
(b) wave types for ρrρl.
value of ρleft wave right wave
ρρmin invalid
ρmin < ρ ρrr r
ρrρρlr s
ρlρs s
2.3 Additional constraints for consistency
We have found a complete list of conditions for a usual Riemann problem and a junction Riemann
problem to be well-posed. For the usual Riemann problem this is enough, as just noted. Yet our
conditions do not guarantee that the solution states at the junction are sub-sonic, as only one of
each eigenvalue is restricted. Since sub-sonic states are necessary at the junction, we introduce
an additional condition by hand. We again focus on one incoming pipe iwith initial condition Ui
and new state Vi= (ρ, Ll(ρ;Ui)). We define
ρi,max = min(ρ > 0|λ2(ρ, Ll(ρ;Ui)) <0∪ {∞})(22)
and further
ρmax = min
Although there is no knowledge about λ2(Vi)as it is not conveniently given by the derivative of a
Lax curve, we still have λ2(Vi) = λ1(Vi) + 2c(ρ)> λ1(Vi), so there holds at least ρi,max > ρi,min
in every pipe. But there is no guarantee that ρmax > ρmin.
3 Coupling to power grids
In this section, we focus on the coupling of the gas network to power grids. Following the ideas
in [16,27], the coupling is done extending the already existing coupling conservation of fluxes
(5) by a gas power plant. To model the power grid, we use the well-known powerflow equations
[11], which describe real and reactive power nodewise. We have to investigate again under which
assumptions this kind of coupling is well-posed.
3.1 Power flow model
Power grids are usually modeled as a graph, whose nodes (buses) are indexed by kand whose
edges (transmission lines) are indexed by the indices of the nodes connected by the edge. For each
node kVPG of the power grid (PG) there are four (time-dependent) state variables, namely real
power Pk(t), reactive power Qk(t), voltage magnitudeVk(t)and voltage angle φk(t). We consider
three different types of nodes: load/PQ buses (Pund Qare given), generater/PV buses (Pand
|V|are given) and a single slack bus (|V|and φare given). Thus we always have two unknowns
per node kVPG . Accordingly, we apply the powerflow equations for real and reactive power for
each node:
jVPG |Vk||Vj|Gkj cos(φkφj) + Bkj sin(φkφj),(24a)
jVPG |Vk||Vj|Gkj sin(φkφj)Bkj cos(φkφj),(24b)
where Gkj is the real part of the entry ykj in the bus admittance matrix and Bkj is the imaginary
part. For k6=jwe consider Gkj and Bkj as properties of the arc/transmission line connecting
nodes kand j.Gkk and Bkk are considered as node properties.
Figure 3: Sketch of a gas network (purple) coupled to a power grid (black).
3.2 Gas-to-power coupling
We focus on a junction with one incoming pipeline, one outgoing pipeline and an outlet that draws
a set amount of flow ε, which is converted to electric power. The coupling condition is taken from
[16] and reads
pin =pout (25a)
qin =qout +ε , (25b)
where the outflow εis non-negative. The fuel is converted to gas via the following heat rate
formula, taken from [27],
ε(P) = a0+a1P+a2P2,(26)
where Pis the real power and a0,a1,a2are constants.
To find what εare allowed, we examine another generalized Riemann problem at the junction.
We start with two constant states on the ingoing and outgoing pipes and solve the Riemann
problem where the flow qjumps by an amount εat the junction. We again demand the waves to
leave the junction, giving rise to a picure like that in Figure 4, where next to the junction in red
21 0 1 2
Figure 4: Solution after a short time.
two new states Vl,Vrappear that fulfill the coupling conditions (25). The analysis of this setting
is similar to a three-way junction but with the flow on one outgoing pipe fixed to ε. We must now
solve the equation
Ll(ρ)Lr(ρ) = ε , (27)
which is similar to equation (10) of a usual (two-way) Riemann problem but with a non-zero
right-hand side. We even can forego the additional consistency constraints of section 2.3: One half
of the eigenvalues is on the right side of zero due to the ρmin-criterion, which we need here again
to make the waves have the right direction.
λ1(Vl)0and λ2(Vr)0,
Instead of invoking ρmax, we note that ε0and compute for the remaining two eigenvalues
λ2(Vl) = Ll(ρ)
ρ+c(ρ) = Lr(ρ) + ε
ρ+c(ρ) = λ2(Vr) + ε
λ1(Vr) = Lr(ρ)
ρc(ρ) = Ll(ρ)ε
Remark 12. In our setting this is sufficient. If we also considered power-to-gas plants, we would
need to also treat negative e. In this case, we would have to adhere to the ρmax-criterion (23)
Solution structure for different outflows ε
For the usual Riemann problem we had a unique non-zero solution due to our findings in Section
2.1. For ε > 0we now have two solutions to Ll(ρ)Lr(ρ) = ε, one of which is not admissible as
it lies to the left of the maximum and hence has non-negative derivative (which means it would
be super-sonic). As LlLris decreasing in the admissible regime, greater εresult in smaller ρ.
One of the two solution structures is given in table 2.
Table 2: Solution structure for ρlρr.
value of εleft wave right wave
(LlLr)(ρmin)> ε (LlLr)(ρl)r r
(LlLr)(ρl)ε(LlLr)(ρr)s r
(LlLr)(ρr)εs s
4 Numerical results
Within the following numerical examples, we consider two different discretization schemes. The
first is a third-order CWENO scheme (CWENO3) with suitable boundary treatment [20,23],
which relies on a local Lax-Friedrichs flux function for the inner discretization points of each pipe
and handles coupling points by explicitly solving equation (27).
The second scheme is an implicit box scheme (IBOX) [21], suitable for sub-sonic flows. For a
general system of balance laws
the considered scheme reads
j1)+ ∆tG(Un+1
j) + G(Un+1
Here, tand xare the temporal and spatial mesh size, respectively, and the numerical approx-
imation is thought in the following sense:
jU(x, t)for x(j1
2)∆x, (j+1
2)∆x, t nt, (n+ 1)∆t.(30)
The implicit box scheme has to obey an inverse CFL condition [21], which is beneficial for
problems with large characteristic speeds whereas the solution is merely quasi-stationary. This is
usually the case for daily operation tasks in gas networks and therefore motivates the choice of
this scheme for the real-world scenario below.
The first test example in section 4.1 is supposed to demonstrate the different cases revealed in
the analysis above (section 3.2). Further, since the applied implicit box scheme does not explicitly
make use of any Riemann solver, this scenario is also considered as a numerical validation of its
applicability, where the CWENO3 scheme with Riemann solver at the junction serves as reference.
Within the second example (section 4.2) we briefly demonstrate the differences resulting from
various pressure functions, which are all covered by our theoretical results above.
The third example (section 4.3) considers a more complex scenario: An increasing power
demand within the power grid leads to an increasing fuel demand of a gas-to-power generator and
further to a significant pressure drop in the gas network.
4.1 Validation
We consider the isentropic Euler equations with pressure law p(ρ) = κργand parameters κ= 1.0,
γ= 1.4, and a Riemann problem with left state Ul=4.0
1.0and right state Ur=3.0
Further we assume a gas demand εat the coupling point of the two states (here x= 0). Then,
from table 2, we get the following solution structure:
s-s solution for ε0.57877,
r-s solution for 0.57877 ε3.0594,
r-r solution for 3.0594 ε.
Further, one can easily compute ρ1,min 1.8819,ρ2,min 1.5041, and therewith ρmin 1.8819
and maximum gas demand ε4.3892. We will consider the numerical simulation of the described
setting until time t= 0.1for ε∈ {0.25,1.75,3.25}and the following discretization parameters:
CWENO3: t= 5 ·105,x= 5 ·104,
IBOX: t= 5 ·104,x= 5 ·105.
The different choices result from the (usual) CFL condition the explicit CWENO3 scheme has
to obey, in contrast to the inverse CFL condition of the IBOX scheme. Figures 5to 7show
the computed densities at the final time. Both schemes show the correct solution structure
(shock/rarefaction waves), where CWENO3 expectably achieves the sharper resolution. 0 0.1 0.2 0.3 0.4 0.5
Figure 5: Density profile at t= 0.1for ε= 0.25. (s-s solution) 0 0.1 0.2 0.3 0.4 0.5
Figure 6: Density profile at t= 0.1for ε= 1.75. (r-s solution)
4.2 Different pressure laws
In our second test case, we apply various pressure laws, which are all covered by our theoretical
study, and are interested in the different dynamics one may observe even on a single pipeline.
Therefore, we consider a single pipe with length l= 0.1and the following pressure laws:
p(ρ) = 1
γργwith γ= 1.4(“gamma law”),
p(ρ) = 1
ρ(“inverse”, corresponding to γ=1),
p(ρ) = ln(ρ)(“logarithmic”),
15 0 0.1 0.2 0.3 0.4 0.5
Figure 7: Density profile at t= 0.1for ε= 3.25. (r-r solution)
p(ρ) = 1
1+i/5(“sum of gamma laws”).
Note that all considered pressure functions are scaled in such a way that p0(ρ=1)=1. Initially,
we have ρ= 1 and q= 0 in the whole pipe. Further, we fix ρ= 1 on the left-hand boundary,
whereas qat the right-hand boundary linearly increases from 0to 0.2until time t= 0.1and stays
constant afterwards until the final time t= 0.5. We approximate the solution to this problem
by CWENO3 with discretization parameters t= 5 ·104and x= 103. The variety of the
resulting dynamics is demonstrated in figures 8and 9, which show the density in the pipeline at
times t= 0.25 and t= 0.5, respectively.
0 0.025 0.05 0.075 0.1
gamma law
sum of gamma laws
Figure 8: Simulation result at time t= 0.25 for different pressure laws.
0 0.025 0.05 0.075 0.1
gamma law
sum of gamma laws
Figure 9: Simulation result at time t= 0.5for different pressure laws.
4.3 Coupled gas and power grid
We consider the network(s) depicted in figure 10, containing a power grid from the example
“case9” of the MATPOWER Matlab programming suite [26] and a small part of the GasLib-40
network [18], extended by a compressor station in front of node S17 and a gas-to-power generator
between S4 and N1, providing the necessary power at the latter node. The parameters of the gas
network are given in table 3, the parameters of the power grid in table 4.
N4 N5 N6 N7 N8
S5 S17
S20 S25
Figure 10: Gas network connected to a power grid. Red nodes are PQ/demand nodes, green
nodes are generators (PV nodes) and the blue node is the slack bus (also a generator, with gas
consumption of the form ε(P) = a0+a1P+a2P2).
Table 3: Parameters of the gas network.
Pipe From To Length [km] Diameter [mm] Roughness [mm]
P10 S4 S20 20.322 600 0.05
P20 S5 S17 20.635 600 0.05
P21 S17 S4 10.586 600 0.05
P22 S17 S8 10.452 600 0.05
P24 S8 S20 19.303 600 0.05
P25 S20 S25 66.037 600 0.05
P99 S4 S8 5.000 600 0.05
Within each pipe eEGas of the gas network (with length le, diameter de, cross section Ae,
roughness ke) we consider the isothermal Euler equations with acoustic speed c= 340m
s. The
source term in the momentum equation is given by
S(ρ, q) = λ(q)
with friction factor λ(q), which is determined by the Prandtl-Colebrook formula:
λ=2 log10 2.51
with Reynolds number
Re(q) = de
and dynamic viscosity η= 105kg
ms .
Initially, the gas network is in a stationary state: The pressure at S5 is fixed at 60bar, the
ouflow at S25 is q= 100m3
Aewith ρ0= 0.785 kg
m3, and there is an additional gas consumption
at S4 resulting from the gas to power transformation (a0= 2,a1= 5,a2= 10) due to the power
demand at the slack bus N1. The initial (stationary) state of the power grid is determined by
boundary conditions given in table 5.
Table 4: Parameters of the power grid.
(a) Busses
Node G B
N1 0.0000 -17.3611
N2 0.0000 -16.0000
N3 0.0000 -17.0648
N4 3.3074 -39.3089
N5 3.2242 -15.8409
N6 2.4371 -32.1539
N7 2.7722 -23.3032
N8 2.8047 -35.4456
N9 2.5528 -17.3382
(b) Transmission lines
Edge From To G B
TL14 N1 N4 0.0000 17.3611
TL45 N4 N5 -1.9422 10.5107
TL56 N5 N6 -1.2820 5.5882
TL36 N3 N6 0.0000 17.0648
TL67 N6 N7 -1.1551 9.7843
TL78 N7 N8 -1.6171 13.6980
TL82 N8 N2 0.0000 16.0000
TL89 N8 N9 -1.1876 5.9751
TL94 N9 N4 -1.3652 11.6041
Table 5: Initial boundary conditions of the power grid.
Node P Q |V|φ
N1 - - 1 0
N2 163 - 1 -
N3 85 - 1 -
N4 0 0 - -
N5 -90 -30 - -
N6 0 0 - -
N7 -100 -35 - -
N8 0 0 - -
N9 -125 -50 - -
In the course of the simulation, the power (and reactive power) demand at N5 is(/are) linearly
increased between t= 1 hour and t= 1.5hours from 0.9p.u. to 1.8p.u. (reactive power from 0.3
p.u. to 0.6p.u.) - see also figure 11 (left). Accordingly, the power demand at the slack bus N1
increases (see figure 11 (right)) and therewith the gas consumption at S4, which also results in an
increase of the inflow at S5 (see figure 12). Due to the increased flow values, the pressure in the
gas network decreases (see figure 13 for the pressure at the nodes S20 and S25).
0 2 4 6 8 10 12
time [hours]
power [p.u.]
Pdemanded at N5
Qdemanded at N5
0 2 4 6 8 10 12
time [hours]
power [p.u.]
Pat slack
Qat slack
Figure 11: Power and reactive power at demand node N5 (left) and the slack bus (right).
0 2 4 6 8 10 12
time [hours]
flow [m3/s]
inflow at node S5
Figure 12: Inflow at node S5.
0 2 4 6 8 10 12
time [hours]
pressure [bar]
pressure at node S20
pressure at node S25
Figure 13: Pressure at nodes S20 and S25.
5 Conclusion and future work
We have presented a coupled model for gas and power allowing for a mathematically well-defined
transition from gas to power. The framework involves also the consideration of non-standard
pressure functions. Various simulation results show the properties of the presented approach.
Future work includes the investigation of optimal control problems for the coupled model as for
example the control of compressor stations [14] or the inclusion of uncertain customer demands [10].
This will immediately lead to the class of nonlinear (stochastic) optimization problems and tailored
solution techniques.
The authors gratefully thank the BMBF project ENets (05M18VMA) for the financial support.
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A Proof of Lemma 3
The following two technical results are the key ingredients to prove Lemma 3.
Lemma 13. Let gC1(R+,R+)be a non-negative function, g0, and let Gbe given by G(ρ) =
ρg(s) ds. Then There holds
1. If ρ2g(ρ)ρ0
0, then ρG(ρ)ρ0
2. If ρG(ρ)ρ0
0, then lim infρ0ρ2g(ρ) = 0.
1. By assumption ρ2g(ρ)ρ0
0. For mNchoose ρm>0such that ρ2g(ρ)1
2for ρ<ρm.
Now choose ρm,0< ρmso small that
ρm,0 Zρl
g(s) ds1
Then, for ρ<ρm,0, there holds
g(s) dsρZρl
g(s) ds+1
s2ds=ρ Zρl
g(s) ds1
for small enough ρ. Therefore limρ0ρRρl
ρg(s) ds2
mfor all mN. As g0, we also have
ρg(s) ds. Summarizing, we get limρ0ρG(ρ) = limρ0ρRρl
ρg(s) ds= 0.
2. We prove by contradiction: Assume there are ρ0>0and a > 0such that ρ2g(ρ)afor
ρ<ρ0. Then g(ρ)a
ρ2for such ρ. Therefore
g(s) dsρZρl
g(s) ds+ρa
s2ds0 + a
which contradicts ρG(ρ)ρ0
Lemma 14. Let gC1(R+,R+),g0and lim infρ0ρ2g(ρ)=0. Let also ρ2g(ρ)00. Then,
lim supρ0ρ2g(ρ) = lim infρ0ρ2g(ρ)=0.
Proof. We prove by contradiction. Let lim supρ0ρ2g(ρ)>lim inf ρ2g(ρ). Then it is easily seen
that lim infρ0ρ2g(ρ)0=−∞ <0resulting in a contradiction.
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In this paper, we introduce a stationary model for gas flow based on simplified isothermal Euler equations in a non-cycled pipeline network. Especially the problem of the feasibility of a random load vector is analyzed. Feasibility in this context means the existence of a flow vector meeting these loads, which satisfies the physical conservation laws with box constraints for the pressure. An important aspect of the model is the support of compressor stations, which counteract the pressure loss caused by friction in the pipes. The network is assumed to have only one influx node, all other nodes are efflux nodes. With these assumptions the set of feasible loads can be characterized analytically. In addition we show the existence of optimal solutions for some optimization problems with probabilistic constraints. A numerical example based on real data completes this paper.
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