October, 21st, 2014 Des. Ind. Plant, 2

Compressible Flow along pipelines

a.

Consider a pipe (diameter D, lenght L) used to transport gas (molar mass M , ratio of specific heat evaluated atconstant pressure and at constant volume given by γ = cp/cv) from point 1 to point 2.

1. Derive the relation between specific mass flow rate G and pressure at the upstream/downstream section of thepipeline. Assume isothermal transformation for the gas (p/ρ = C) along the pipeline.Since the pipeline works in steady state conditions, m = cost and if the pipe diameter is constant along thelength of pipe, also G = const. The flow along the line is described by Benrnoulli equation:

dv2

2+

dp

ρ+ gdh = dws −

dlvρ

(1)

where gdh ≃ 0, dws = 0 (no compressor along the line) and dlv/ρ = 2fdxv2/D as for incompressible flow.Bernoulli should be integrated between point 1 and 2, but it can not be integrated in the form of Equation 1because we do not know the law of variation of velocity along the pipe, v(x), which is required to quantify theviscous losses. To integrate that term, we need to rewrite Bernoulli as:

ρ2(

dv2

2+

dp

ρ

)

= −2f(v2ρ2)dx

D(2)

which, considering G = ρv, becomes

ρ2G2

2d1

ρ2+ ρdp = −2fG2dx

D(3)

Integration of the first term gives

∫

2

1

ρ2G2

2d1

ρ2=

∫

2

1

ρ2G2

2(−2)

1

ρ3dρ = −G2

dρ

ρ= −G2 ln

ρ2ρ1

= −G2 lnp2p1

= G2 lnp1p2

(4)

where we considered the relationship between p and ρ in isothermal flow. Integration of the second term gives

∫

2

1

ρdp =

∫

2

1

pM

RTdp =

M

2RT(p2

2− p2

1) (5)

Integration of the right hand side gives

∫

2

1

2fG2dx

D= 2f

L

DG2 (6)

The integral version of Bernoulli equation for isothermal flow along the pipeline is:

G2 lnp1p2

+M

2RT(p2

2− p2

1) + 2f

L

DG2 = 0 = F (G, p1, p2) (7)

from which we calculate

G =

√

√

√

√

M2RT

(p21− p2

2)

ln p1

p2

+ 2f LD

= f(p1, p2) (8)

the relationship we were looking for. If we assume that the upstream pressure p1 is fixed, the specific flow rateis a function of the downstream pressure p2 only. p2 can varies in the range [0 : p1]. G is positive definite, andis null for both p2 = p1 and p2 = 0. For the Weierstass theorem it has maximum in the range [0 : p1]. Tofind the maximu we should calculate dG(p2)/dp2 = 0. Since the function G(p2) is rather complex, we can useEquation 7 to derive the derivative of G in a simpler way. Considering the differential of F (G, p1, p2) we get

1 ©··⌣ 2014 marina.campolo@uniud.it

October, 21st, 2014 Des. Ind. Plant, 2

dF (G, p2) =∂F

∂GdG+

∂F

∂p2dp2 = 0 → dG

dp2= −∂F/∂p2

∂F/∂G(9)

from which dGdp2

= 0 only if ∂F/∂p2 = 0. Therefore:

∂F

∂p2= −G2

p2+

M

RTp2 = 0 → G =

√

M

RTp22=

√ρ2p2 (10)

and the velocity of gas at the outlet section is v2 = G/ρ2 =√

p2/ρ2 = vsound. When G is maximum, the velocityof gas at the outlet section is the sound speed (evaluated in isothermal conditions). Substituting the value ofGmax in Equation 7 we obtain

M

RTp22ln

p1p2

+Mp2

2

2RT

[

1−(

p1p2

)2]

+ 2fL

D

M

RTp22= 0 (11)

which can be simplified as

lnp1p2

+1

2

[

1−(

p1p2

)2]

+ 2fL

D= 0 (12)

This is a funtion of (p1/p2) only and gives the value of the critical pressure ratio corresponding to G = Gmax.As already discussed for the efflux from a tank, when the gas velocity at the outlet section equals the speed ofsound, if the outer pressure is lowered there is no way for this information to propagate upstream along the pipetoward the inlet section. Therefore, once the critical flow is established, no further variation of G is expectedeven if p2 is reduced and G = Gmax.

2. Derive the formula for the specific mass flow rate at critical conditions when the gas undergoes adiabatic (irre-versible, i.e. non isoentropic) transformation moving along the pipe.

In this case, the integral form of Bernoulli will be given by:

G2 lnp1p2

+

∫

2

1

ρdp+ 2fL

DG2 = 0 (13)

where we need a relationship between ρ and p along the pipeline which is valid for irreversible adiabatic tran-sformation. We cannot use p/ργ = cost which is only valid for revesible adiabatic. Nevertheless, we can derivean alternative relationship from the total energy conservation equation:

d(e+p

ρ+

1

2v2 + gh) = dq + dws (14)

e is the internal energy, q is the heat flux added to the control volume of gas and ws is the mechanical workmade on the control volume. If we define the enthalpy as H = e+ p/ρ, the total energy equation becomes

d(H +1

2v2 + gh) = dq + dws (15)

where we can neglect gh, dws = 0 (there is no compressor on the pipeline) and dq = 0 (for adiabatic flow). Weobtain

d(H +1

2v2) = 0 → cpdT +

dv2

2= 0 (16)

where, according to Mayer equation, cp is given by

cp =γ

γ − 1

R

M(17)

2

October, 21st, 2014 Des. Ind. Plant, 2

and according to ideal gas law dT = M/Rd(p/ρ) and

γ

γ − 1d

(

p

ρ

)

+1

2dv2 = 0 (18)

2γ

γ − 1

(

p

ρ

)

+G2

ρ2= C = cost (19)

Equation 19 gives a relationship between p and ρ which is valid for each point along the pipe and can be usedto integrate the term in Equation 13. From 19 we caan write

p =

(

C − G2

ρ2

)

γ − 1

2γρ (20)

and

dp =

[

−(−2)G2

ρ3ρ+

(

C − G2

ρ2

)]

γ − 1

2γdρ =

[

2G2

ρ2+ C − G2

ρ2

]

γ − 1

2γdρ =

[

G2

ρ2+ C

]

γ − 1

2γdρ (21)

The integral in Bernoulli becomes

∫

2

1

ρdp =

∫

2

1

ρ

[

G2

ρ2+ C

]

γ − 1

2γdρ =

γ − 1

2γ

[

G2 lnρ2ρ1

+C

2(ρ2

2− ρ2

1)

]

(22)

Further simplifications led to

∫

2

1

ρdp =γ − 1

2γ

[

G2 lnρ2ρ1

+

(

γ

γ − 1p1ρ1 +

G2

2

)

(

(

ρ2ρ1

)2

− 1

)]

(23)

and the final form for integral Bernoulli equation for adiabatic irreversible flow:

G2 lnρ1ρ2

γ + 1

γ+

(

p1ρ1 +γ − 1

γG2

)

[

(

ρ2ρ1

)2

− 1

]

+ 4fL

DG2 = 0 (24)

which is again an implicit function F (G, ρ1, ρ2) = 0. If we consider the upstream condition fixed, F (G, ρ2) = 0describes the variation of G as the downstream condition change. Following the same procedure as before, wecan find the condition for which G = Gmax from

dG

dρ2= 0 → dG

dρ2= −∂F/dρ2

∂F/∂G→ ∂F

dρ2= 0 (25)

which gives

∂F

dρ2= −G2

ρ2

γ + 1

γ+

(

ρ1p1 +γ − 1

2γG2

)

2

ρ21

ρ2 = 0 (26)

and using Equation 19 to rewrite the term in parentesis, we get

G2

ρ2

γ + 1

γ=

γ − 1

2γ

[

G2

ρ22

+2γ

γ − 1

p2ρ2

]

2ρ2 (27)

and

G2

ρ2

[

γ + 1

γ− γ − 1

γ

]

= 2p2 → 2G2

γ= 2ρ2p2 → Gmax =

√γp2ρ2 (28)

3

October, 21st, 2014 Des. Ind. Plant, 2

The last equation indicates that when the specific flow rate is maximum, G is a function of pressure and densityat the outlet section and the formula is the same as that calculated for the flow exiting from a tank underadiabatic conditions. The velocity of gas at the outlet section turns out to be v2 = G/ρ2 =

√

γp2/ρ2 which isthe speed of sound. Substituting the value of Gmax into Equation 24 we obtain the equation which gives thelink between densities at the upstream and downstream section of the pipe at critical flow conditions:

lnρ1ρ2

+1

2

[

1−(

ρ1ρ2

)2]

+ 4fL

D

γ

γ + 1= 0 (29)

which together with

p1p2

=ρ1ρ2

γ + 1

2− γ − 1

2

ρ2ρ1

(30)

allows to link upstream and downstream conditions of the pipe. Equation 29 is the analogous of the criticalpressure ratio calculated for isothermal transport. Equation 30 is derived from Equation 19 using the result ofEquation 29.

4