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Modeling and simulation of gas networks coupled to power

grids

E. Fokken∗

, S. Göttlich∗

, O. Kolb∗

January 1, 2019

Abstract

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

({fokken,goettlich,kolb}@uni-mannheim.de).

1

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

setting.

In each pipe e∈EGas, the gas ﬂow is described by the isentropic Euler equations in one space

dimension:

ρ

qt

+ q

p(ρ) + q2

ρ!x

=0

S(ρ, q),(1)

where t∈IR+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

p∈C1(IR+,IR+), p0(ρ)>0for all ρ∈IR+,

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

ρ−pp0(ρ), λ2(ρ, q) = q

ρ+pp0(ρ),

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.,

X

incoming pipes

qpipe =X

outgoing pipes

qpipe (5)

2

2.1 Well-posedness of the Riemann problem for the isentropic Euler

equation

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.,

q

ρ

< 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) =

ρql

ρl+Rρl

ρ

c(s)

sdsfor ρ≤ρl(rarefaction)

ρql

ρl−pf(ρ, ρl)for ρl≤ρ(shock),

(7)

where fis deﬁned by

f(ρ, ρl) = a(ρ, ρl)∆p(ρ, ρl)with a(ρ, ρl) = ρ

ρl

(ρ−ρl),∆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) =

ρqr

ρr−Rρr

ρ

c(s)

sdsfor ρ≤ρr(rarefaction)

ρqr

ρr+pf(ρ, ρr)for ρr≤ρ(shock).

(9)

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

Ll(ρ)−Lr(ρ)=0.(10)

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 L∈C2(R+,R+)and let it fulﬁll conditions A1 through A3

(where only one of A3(a) and A3(b) must hold).

3

ρ

q

Ll

Lr

Ll−Lr

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<limρ→0L(ρ)<∞or

(b) limρ→0L(ρ)=0and limρ→0L0(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

γργ+const

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.

4

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,

L0

l(ρ) =

ql

ρl+Rρl

ρ

c(s)

sds−c(ρ)for ρ≤ρl

ql

ρl−f0

2√ffor ρl≤ρ

L00

l(ρ) =

−(c(ρ)

ρ+c0(ρ)) for ρ≤ρl

−2f00f−(f0)2

4√ffor ρl≤ρ ,

(12)

and

L0

r(ρ) =

qr

ρr−Rρr

ρ

c(s)

sds+c(ρ)for ρ≤ρr

qr

ρr+f0

2√ffor ρr≤ρ

L00

r(ρ) =

c(ρ)

ρ+c0(ρ)for ρ≤ρr

2f00f−(f0)2

4√ffor ρr≤ρ .

(13)

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 g∈C1(R+,R+)be a non-negative function, g≥0, such that also (ρ2g(ρ))0≥0

for all ρ > 0and let Gbe given by G(ρ) = Rρl

ρg(s) ds. Then, there holds ρ2g(ρ)ρ→0

−→ 0if and only

if ρG(ρ)ρ→0

−→ 0.

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 p∈C2(R+)with p0>0. Then conditions A1,A2,A3 hold

for Llfor all ρl,qlwith

ql

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)≥0∀ρ > 0(14)

∆p2+a22∆pp00 −(p0)2+1

2(a2)0(∆p2)0≥0∀ρl>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

p∞−ρp0(ρ)−p(ρ)≥0.

(B3) (a) There is p0>0such that p(ρ) = −p0

ρ+o

ρ→01

ρor

5

(b) p(ρ)∈o

ρ→01

ρand limρ→0Rρl

ρ

c(s)

sds−c(ρ)−c(ρl)≥0for all ρl>0.

Proof.

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

2(a2)0(∆p2)0.

A1,A3(a)⇒B1,B3(a): Assume conditions A1 and A3(a). Let F(ρ) = Rρl

ρ

c(s)

sds.

0< l = lim

ρ→0Ll(ρ) = lim

ρ→0ρZρl

ρ

c(s)

sds= lim

ρ→0ρF (ρ).(16)

Therefore F(ρ) = l

ρ+o

ρ→01

ρ. Note now that

ρ2c(ρ)

ρ0=1

2pp0(ρ)(2p0(ρ) + ρp00(ρ)) ≥0

due to condition A1. Lemma 3therefore shows that

c(ρ)

ρ=−F0(ρ) = l

ρ2+o

ρ→01

ρ2(17)

and hence

p0(ρ) = c(ρ)2=l2

ρ2+o

ρ→01

ρ2,

which again with lemma 3yields

p(ρ) = −l2

ρ+o

ρ→01

ρ.

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

lim

ρ→0Ll(ρ)=0⇔p∈o

ρ→01

ρ.

For the derivative we ﬁnd

lim

ρ→0L0

l(ρ) = lim

ρ→0

ql

ρl

+Zρl

ρ

c(s)

sds−c(ρ)>lim

ρ→0Zρl

ρ

c(s)

sds−c(ρ)−c(ρl)

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

lim

ρ→0Zρl

ρ

c(s)

sds−c(ρ)−c(ρl)≤ −δ < 0.

Choosing ρl, qlsuch that ql

ρl=−c(ρl) + 1

2δyields

lim

ρ→0L0

l(ρ)≤ −1

2δ < 0

6

A2⇐B2: First of all we note that the limit exists, as p0>0. For ρ>ρlwe ﬁnd

Ll(ρ) = ρ

ql

ρl−sf(ρ, ρl)

ρ2

=ρ

ql

ρl−s1

ρl1−ρl

ρp(ρ)−p(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

lim

ρ→∞ s1

ρl1−ρl

ρp(ρ)−p(ρl)=sp∞−p(ρl)

ρl

.

Let now δand ρbe deﬁned by

ql

ρl

+δ=c(ρl)

s1

ρl1−ρl

ρp(ρ)−p(ρl)=sp∞−p(ρl)

ρl−ρ.

Then δ > 0as the state is sub-sonic and ρ→0. Therefore let ρbe so great that δ−ρ>0.

Then, we ﬁnd

ql

ρl

<ql

ρl

+δ−ρ=c(ρl)−ρ≤sp∞−p(ρl)

ρl−ρ

=s1

ρl1−ρl

ρp(ρ)−p(ρl)

which implies A2.

A2⇒B2: In case p∞=∞condition B2 is fulﬁlled. Let now p∞<∞and let there be for all sub-sonic

(ρl, ql)aρ−>0such that

ql

ρl

<s1

ρl1−ρl

ρ−p(ρ−)−p(ρl),(18)

that is, let A2 be true. Note that the right-hand-side is increasing in ρ−because its derivative

is positive. Therefore taking suprema of (18) yields

c(ρl)≤sp∞−p(ρl)

ρl⇒p0(ρl)≤p∞−p(ρl)

ρl

.

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 p∈C3(R+). Condition B1 holds if

2p0(ρ) + ρp00(ρ)≥0

6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)≥0

holds for all ρ > 0.

7

Proof. We note that 2ff00 −(f0)2(ρl, ρl)=0and that

2ff 00 −(f0)20= 2f f 000 .

As f≥0, we see that 2ff00 −(f0)2≥0for all ρ≥ρlif f000(ρ, ρl)≥0for all ρ≥ρl. 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(2ρ−ρl)p00(ρ) + ρ(ρ−ρl)p000(ρ)

=h6p0(ρ)+6ρp00(ρ) + ρ2p000(ρ)i−ρl3p00(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|0≤ρl≤ρ}. 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) limρ→0c(ρ)=0and 2p0(ρ)−ρp00(ρ)≥0for all ρ > 0.

(ii) 0<limρ→0c(ρ)<∞.

(iii) There is a η∈(0,1) such that limρ→0ρηc(ρ)exists and 0<limρ→0ρηc(ρ)<∞.

Proof.

(i) Here we have

lim

ρ→0Zρl

ρ

c(s)

sds−c(ρ)−c(ρl) = lim

ρ→0Zρl

ρc(s)

s−c0(s)ds−2c(ρ)

= lim

ρ→0Zρl

ρc(s)

s−c0(s)ds

= lim

ρ→0Zρl

ρ

1

2spp0(s)(2p0(s)−sp00(s)) ds

≥0.

(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

m∈Nthere is ρm>0such that for all ρ < ρmthere holds c(ρ)≥a1−1

mρ−η. Choose

now mso great that

1−1

m1

η>1

and then ρm,0so small that for all ρ<ρm,0there holds

1−1

m1

η−r(ρ) = 1 + θ

with θ > 0. Deﬁne also

Cm=Zρl

ρm

c(s)

sds.

8

Then we ﬁnd for ρ < ρm,0

Zρl

ρ

c(s)

sds−c(ρ)−c(ρl) = Zρm

ρ

c(s)

sds+Zρl

ρm

c(s)

sds−c(ρ)−c(ρl)

≥a1−1

mZρm

ρ

s−η−1ds+Cm−c(ρ)−c(ρl)

=a 1−1

m1

η−1−r(ρ)!ρ−η−a1−1

m1

ηρ−η

m+Cm−c(ρl)

> aθρ−η−a1−1

m1

ηρ−η

m+Cm−c(ρl)

ρ→0

→+∞>0.

Proposition 7. All conditions only depend on diﬀerences in pressures, that is, if c∈Ris 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)

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

ρ+o1

ρfor ρ→0or

(b) one of the following conditions is fulﬁlled:

(i) limρ→0c(ρ)=0and 2p0(ρ)−ρp00(ρ)≥0for all ρ > 0,

(ii) 0<limρ→0c(ρ)<∞,

(iii) there is a η∈(0,1) such that limρ→0ρηc(ρ)exists and 0<limρ→0ρηc(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.

9

•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

ρl<−2

δc(ρl),

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 q−1

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) = −ρ−1

p(ρ) = −ρ−1

2

p(ρ) = ln(ρ)

are all valid pressure functions. Even a and more exotic

p(ρ) = ρ3−ρ

ln(ρ)=Z3

1

ργdγ

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 e∈Iand outgoing pipes indexed by

f∈O. At the junction-facing end of each pipe there are initial states Ui= (ρi, qi)for each i∈I

and i∈O, respectively. To make things tractable, we restrict the solution of the junction Riemann

problem to be of this form: In each pipe i∈I∪O, 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.

U1

U2

U3

U4

U5

(a) Junction at t= 0.

U1

V1

U2V2

U3

V3

U4

V4

U5

V5

(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)).

10

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

lis

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)

ρ−ρi0=L0

l(ρ;Ui)(ρ−ρi)−(Ll(ρ;Ui)−Ll(ρi;Ui))

(ρ−ρi)2≤0,

since Llis concave. So

s(ρ)≤s(ρi) = λ1(ρi)<0

because Uiis sub-sonic.

⇒: Let ρ≤ρi,min . Because Uiis sub-sonic, we have ρ≤ρi,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-

negative.

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

i∈I∪Oρi,min .(21)

A solution (Vi)i∈I∪Oto 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 ρl≤ρr.

value of ρleft wave right wave

ρ≤ρmin invalid

ρmin < ρ ≤ρlr r

ρl≤ρ≤ρrs r

ρr≤ρs s

(b) wave types for ρr≤ρl.

value of ρleft wave right wave

ρ≤ρmin invalid

ρmin < ρ ≤ρrr r

ρr≤ρ≤ρlr s

ρl≤ρs 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

11

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

i∈I∪Oρi,max.(23)

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 k∈VPG 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 k∈VPG . Accordingly, we apply the powerﬂow equations for real and reactive power for

each node:

Pk=X

j∈VPG |Vk||Vj|Gkj cos(φk−φj) + Bkj sin(φk−φj),(24a)

Qk=X

j∈VPG |Vk||Vj|Gkj sin(φk−φj)−Bkj cos(φk−φj),(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).

12

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

−2−1 0 1 2

0

0.5

1

1.5

2

Ul

Vl

Vr

Ur

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 ε≥0and compute for the remaining two eigenvalues

λ2(Vl) = Ll(ρ)

ρ+c(ρ) = Lr(ρ) + ε

ρ+c(ρ) = λ2(Vr) + ε

ρ≥λ2(Vr)≥0,

and

λ1(Vr) = Lr(ρ)

ρ−c(ρ) = Ll(ρ)−ε

ρ−c(ρ)λ1(Vl)−ε

ρ≤λ1(Vl)≤0.

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)

again.

13

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 Ll−Lris 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 ρl≤ρr.

value of εleft wave right wave

ε≥(Ll−Lr)(ρmin)invalid

(Ll−Lr)(ρmin)> ε ≥(Ll−Lr)(ρl)r r

(Ll−Lr)(ρl)≥ε≥(Ll−Lr)(ρr)s r

(Ll−Lr)(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

Ut+F(U)x=G(U),(28)

the considered scheme reads

Un+1

j−1+Un+1

j

2=Un

j−1+Un

j

2−∆t

∆xF(Un+1

j)−F(Un+1

j−1)+ ∆tG(Un+1

j) + G(Un+1

j−1)

2.(29)

Here, ∆tand ∆xare the temporal and spatial mesh size, respectively, and the numerical approx-

imation is thought in the following sense:

Un

j≈U(x, t)for x∈(j−1

2)∆x, (j+1

2)∆x, t ∈n∆t, (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

−1.0.

14

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 ε≤0.57877,

•r-s solution for 0.57877 ≤ε≤3.0594,

•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 ε≈4.3892. 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 ·10−5,∆x= 5 ·10−4,

•IBOX: ∆t= 5 ·10−4,∆x= 5 ·10−5.

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.

−0.5−0.4−0.3−0.2−0.1 0 0.1 0.2 0.3 0.4 0.5

3

3.5

4

x

density

CWENO3

IBOX

Figure 5: Density proﬁle at t= 0.1for ε= 0.25. (s-s solution)

−0.5−0.4−0.3−0.2−0.1 0 0.1 0.2 0.3 0.4 0.5

3

3.5

4

x

density

CWENO3

IBOX

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

−0.5−0.4−0.3−0.2−0.1 0 0.1 0.2 0.3 0.4 0.5

3

3.5

4

x

density

CWENO3

IBOX

Figure 7: Density proﬁle at t= 0.1for ε= 3.25. (r-r solution)

•p(ρ) = 1

10

10

P

i=1

ρ1+i/5

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 ·10−4and ∆x= 10−3. 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

0.9

1

1.1

1.2

x

density

gamma law

sum of gamma laws

logarithmic

inverse

Figure 8: Simulation result at time t= 0.25 for diﬀerent pressure laws.

0 0.025 0.05 0.075 0.1

0.6

0.8

1

1.2

x

density

gamma law

sum of gamma laws

logarithmic

inverse

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.

16

N1

N2N3

N4 N5 N6 N7 N8

N9

S5 S17

S4

S8

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 e∈EGas 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)

2de

q|q|

ρ

with friction factor λ(q), which is determined by the Prandtl-Colebrook formula:

1

√λ=−2 log10 2.51

Re(q)√λ+ke

3.71de!

with Reynolds number

Re(q) = de

ηq

and dynamic viscosity η= 10−5kg

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

s·ρ0

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.

17

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

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

time [hours]

power [p.u.]

Pdemanded at N5

Qdemanded at N5

0 2 4 6 8 10 12

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

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).

18

0 2 4 6 8 10 12

110

120

130

140

time [hours]

ﬂow [m3/s]

inflow at node S5

Figure 12: Inﬂow at node S5.

0 2 4 6 8 10 12

30

40

50

60

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.

Acknowledgments

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 g∈C1(R+,R+)be a non-negative function, g≥0, and let Gbe given by G(ρ) =

Rρl

ρg(s) ds. Then There holds

1. If ρ2g(ρ)ρ→0

→0, then ρG(ρ)ρ→0

→0.

2. If ρG(ρ)ρ→0

→0, then lim infρ→0ρ2g(ρ) = 0.

Proof.

1. By assumption ρ2g(ρ)ρ→0

→0. For m∈Nchoose ρm>0such that ρ2g(ρ)≤1

mρ2for ρ<ρm.

Now choose ρm,0< ρmso small that

ρm,0 Zρl

ρm

g(s) ds−1

mρm!≤1

m

Then, for ρ<ρm,0, there holds

ρZρl

ρ

g(s) ds≤ρZρl

ρm

g(s) ds+1

mρZρm

ρ

1

s2ds=ρ Zρl

ρm

g(s) ds−1

mρm!+1

m≤2

m

for small enough ρ. Therefore limρ→0ρRρl

ρg(s) ds≤2

mfor all m∈N. As g≥0, we also have

0≤limρ→0ρRρl

ρg(s) ds. Summarizing, we get limρ→0ρG(ρ) = limρ→0ρRρl

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

2

1

ρ2for such ρ. Therefore

ρZρl

ρ

g(s) ds≥ρZρl

ρ0

g(s) ds+ρa

2Zρ0

ρ

1

s2ds→0 + a

2ρ−1

sρ0

ρ→a

2,

which contradicts ρG(ρ)ρ→0

−→ 0.

Lemma 14. Let g∈C1(R+,R+),g≥0and lim infρ→0ρ2g(ρ)=0. Let also ρ2g(ρ)0≥0. Then,

lim supρ→0ρ2g(ρ) = lim infρ→0ρ2g(ρ)=0.

Proof. We prove by contradiction. Let lim supρ→0ρ2g(ρ)>lim inf ρ2g(ρ). Then it is easily seen

that lim infρ→0ρ2g(ρ)0=−∞ <0resulting in a contradiction.

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