Skalak, R. (1956), Ref. 4 Wave speeds ( ) Nomenclature 11 CASA_06-27.mcd Appendix 1 γs 76972.857 kg m2s2 = pcf 157.087464 kg m2s2 ρs γs g:= ρs 7849.047053 kg m3 A := 8 A 8= …

Mathcad 11 CASA_06-27.mcd Appendix 1

γs 76972.857kg

m2

s2

= pcf 157.087464kg

m2

s2

=

ρs

γs

g:= ρs 7849.047053

kg

m3

=

A 8:= A 8=

m1

ρ0 a⋅

A:= m1 38.09371

kg

m2

=

ha

62.752:= h 0.004857 m=

m2 ρs h⋅:= m2 38.124515kg

m2

=m2 m1−

m1

0.000809=

m3

ρs π a h+( )2

⋅ π a2

⋅− ⋅

2 π⋅ a1

2h⋅+

⋅

:= m3 38.124515kg

m2

=m3 m1−

m1

0.000809=

R1 12.478:=E

K

ρ0

ρs 1 ν2

−( )⋅

⋅ 12.468436=

Skalak, R. (1956), Ref. 4"An extension of the theory of water hammer."Transactions of the ASME 78, 105-116.(PhD Thesis, Columbia University, New York, USA, 1954, Ref. 2)

Table 1

g 32.174049ft

s2

= g 9.80665m

s2

=

a 1 ft⋅:= a 0.3048m=

c 5000ft

s⋅:= c 1524

m

s=

ρ0 1.94slug

ft3

⋅:= ρ0 999.834908kg

m3

=

K ρ0 c2

⋅:= K 2.322193 109

× Pa=

E 30 106

⋅ psi⋅:= E 2.068427 1011

× Pa=

ν 0.3:= ν 0.3= pcflbf

ft3

:=

γs 490 pcf⋅:=


There is something wrong with Skalak's equations [74] and [76]: the -c02 term in

the last factor must be deleted in [74] and [76] misses a square root.

c2

c0

0.981061=c2

c3.464192=

c1

ce

0.999004=c1

c0.643647=

ce

c0.644288=

c0

c3.531067=

Check with Table 1

Eq. (76)c2 5279.429013m

s=

c2 c2 A⋅ R⋅ R+ R

21 ν

2−( )⋅+ 2 A⋅ R⋅ R+ R

21 ν

2−( )⋅+

2

4 R2

⋅ 1 ν2

−( )⋅ 2 A⋅ R+( )⋅−+

2 2 A⋅ R+( )⋅

0.5

⋅:=

Eq. (76)c1 980.917501m

s=

c1 c2 A⋅ R⋅ R+ R

21 ν

2−( )⋅+ 2 A⋅ R⋅ R+ R

21 ν

2−( )⋅+

2

4 R2

⋅ 1 ν2

−( )⋅ 2 A⋅ R+( )⋅−−

2 2 A⋅ R+( )⋅

0.5

⋅:=

Eq. (1)ce 981.895414m

s=ce

c

12 K⋅ a⋅

E h⋅+

:=

Korteweg - Joukowsky

A 7.993536=Aρ0 a⋅

m2

:=

R R2:=R2 12.468436=R2

c02

c2

:=

Nomenclaturec0 5381.346495m

s=c0

E h⋅

m2 1 ν2

−( )⋅

:=

Wave speeds


ct 5133.477378m

s=

ct

c0

0.953939=

With FSI:

γ2 c2F c2t+ ν2

ρf

ρt

⋅D

ee⋅ c2F⋅+:= γ2 2.883457 10

7×

m2

s2

=

λ211

2γ2 γ2

24 c2F⋅ c2t⋅−−

⋅:= λ21 9.622 10

5×

m2

s2

=

λ231

2γ2 γ2

24 c2F⋅ c2t⋅−+

⋅:= λ23 2.787 10

7×

m2

s2

=

λ1 λ21:= λ1 980.917501m

s= cF ct⋅ 5.17868431352064 10

6×

m2

s2

=

λ3 λ23:= λ3 5279.429013m

s= λ1 λ3⋅ 5.17868431352065 10

6×

m2

s2

=

λ3

ct

1.028431=Check with Table 1

λ1

c0.643647=

λ1

ce

0.999004=λ3

c3.464192=

λ3

c0

0.981061=

ρf ρ0:= ρf 999.834908kg

m3

= Tijsseling

D 2 a⋅:= D 0.6096 m=

ee h:= ee 0.004857m=

ρt ρs:= ρt 7849.047053kg

m3

=

Longitudinal wave speeds

Classical:

c2FK

ρf

1K D⋅

E ee⋅1 ν

2−( )⋅+

1−

⋅:= cF c2F:= cF 1008.806299m

s=

cF

ce

1.027407=

c2fK

ρf

1K D⋅

E ee⋅+

1−

⋅:= cf c2f:= cf 981.895414m

s=

cf

ce

1=

c2tE

ρt

:= ct c2t:=


d2 11.477473m

3

s=d1 2.925721

m3

s=

d2 c a2

⋅

A 4+( )c2

c

5c2

c

3

1 R+( )⋅−c2

c

R⋅+

16−c2

c

2

⋅ 2 A⋅ R+( )⋅ 8 R⋅ 2 A⋅ 1+( )⋅+ 8 R2

⋅ 1 ν2

−( )+

⋅:=Eq. (78)

d1 c a2

⋅

A 4+( )c1

c

5c1

c

3

1 R+( )⋅−c1

c

R⋅+

16−c1

c

2

⋅ 2 A⋅ R+( )⋅ 8 R⋅ 2 A⋅ 1+( )⋅+ 8 R2

⋅ 1 ν2

−( )+

⋅:=Eq. (78)

Nominal length of the wave front according to Tijsseling

d2old 11.477844m

3

s=d1old 2.925652

m3

s=

d2old c a2

⋅

A 4+( )c2

c

5c2

c

3

1 R+( )⋅−c2

c

R⋅+

c2

c

h2

12 a2

⋅

⋅ 1 2 R2

⋅ ν⋅+( )⋅−

16−c2

c

2

⋅ 2 A⋅ R+( )⋅ 8 R⋅ 2 A⋅ 1+( )⋅+ 8 R2

⋅ 1 ν2

−( )+

⋅:=Eq. (78)

d1old c a2

⋅

A 4+( )c1

c

5c1

c

3

1 R+( )⋅−c1

c

R⋅+

c1

c

h2

12 a2

⋅

⋅ 1 2 R2

⋅ ν⋅+( )⋅−

16−c1

c

2

⋅ 2 A⋅ R+( )⋅ 8 R⋅ 2 A⋅ 1+( )⋅+ 8 R2

⋅ 1 ν2

−( )+

⋅:=Eq. (78)

Nominal length of the wave front according to Skalak

λ3

c2

1=λ1

c1

1=

Check Skalak - Tijsseling


L2 5 s⋅( ) 15.669309 m=L2 5 s⋅( ) 51.408494 ft=

L2 1 s⋅( ) 9.163467 m=L2 1 s⋅( ) 30.06387 ft=

L1 5 s⋅( ) 9.935248 m=L1 5 s⋅( ) 32.595957 ft=

L1 1 s⋅( ) 5.810168 m=L1 1 s⋅( ) 19.062231 ft=

Check with Table 1

Eq. (64)L2 t( )3 π⋅

3d2 t⋅⋅

Γ1

3

sinπ

3

⋅

:=L1 t( )3 π⋅

3d1 t⋅⋅

Γ1

3

sinπ

3

⋅

:=

3 π⋅

Γ1

3

sinπ

3

⋅

4.062354=Γ1

3

2

3π

3

1

2

Γ2

3

⋅⋅→sinπ

3

1

23

1

2⋅→

c2

c

h2

12 a2

⋅

⋅ 1 2 R2

⋅ ν⋅+( )⋅

A 4+( )c2

c

5c2

c

3

1 R+( )⋅−c2

c

R⋅+

0.000032−=

c1

c

h2

12 a2

⋅

⋅ 1 2 R2

⋅ ν⋅+( )⋅

A 4+( )c1

c

5c1

c

3

1 R+( )⋅−c1

c

R⋅+

0.000024=

Error because of "wrong" (h/a)2 term in Eq. (78)

h

a

2

0.000254=R 12.468436=A 7.993536=


0 1 2 3 4 5 6 7 8 9 100

5

10

15

20

L1 t( )

L2 t( )

t

TOL 1012−

:=even function to be integrated

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 51

0

1

2

sin η η3

+( )η

η


0

9.3

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.840919869378856=β 10:=

0

20

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.11015525927976=β 1:=

0

43

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.42315940650369=β1

10:=

0

93

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.56376138708481=β1

100:=

0

201

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57083144591109=β1

1000:=

0

430

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57077652953821=β1

10000:=

0

907

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57076569930658=β1

100000:=

0

1843

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57082809089641=β1

1000000:=

π

21.570796=

Eq. (49)

0

6600

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57093027532399=β 0:=

Determination of the maximum possible upper boundaries of integrals:


0

20

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.432399540407618=β 1−:=

0

43

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.73656497765972=β1−

10:=

0

94

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.71751324086010=β1−

100:=

0

202

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.18218831662671=β1−

1000:=

0

441

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.56865597358518=β1−

10000:=

0

962

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.70193923364200=β1−

100000:=

0

2115

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.62570169385300=β1−

1000000:=

π

21.570796=

0

6600

ηsin η β η

3⋅+( )

η

⌠⌡

d 1.57093027532399=β 0:=

0

2.0

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.599018563818040=β 1000:=

0

4.3

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.681123567409668=β 100:=


β 10−:=

0

9.3

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.123095673983503−=

β 100−:=

0

4.3

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.347949976012880−=

β 1000−:=

0

2.0

ηsin η β η

3⋅+( )

η

⌠⌡

d 0.444274408294637−=

Maximum possible upper boundaries b of integrals as function of β:

b β( ) 1843 β 0.000001≤if

907 0.000001 β< 0.00001≤if

430 0.00001 β< 0.0001≤if

201 0.0001 β< 0.001≤if

93 0.001 β< 0.01≤if

43 0.01 β< 0.1≤if

20 0.1 β< 1≤if

9.3 1 β< 10≤if

4.3 10 β< 100≤if

2.0 100 β< 1000≤if

0 otherwise

:=


Define range of n values for β:

n 1000:=

i 0 n..:= j i( )10

ni⋅ 7−:= β i( ) 10

j i( ):= β i( )

110 -7

1.02310 -7

1.04710 -7

1.07210 -7

1.09610 -7

1.12210 -7

1.14810 -7

1.17510 -7

1.20210 -7

1.2310 -7

1.25910 -7

1.28810 -7

1.31810 -7

1.34910 -7

1.3810 -7

1.41310 -7

= b β i( )( )1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

1843

=

i

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

= j i( )

-7

-6.99

-6.98

-6.97

-6.96

-6.95

-6.94

-6.93

-6.92

-6.91

-6.9

-6.89

-6.88

-6.87

-6.86

-6.85

=

Maximum possible upper boundaries b of integrals as function of β:

1 .107

1 .106

1 .105

1 .104

1 .103

0.01 0.1 1 10 100 1 .103

0

500

1000

1500

2000

b β i( )( )

β i( )


Eq. (49) for negative (f=wn) and positive (g=wp) values of β:

f(x) is the integral from 0 to minus b(x) OR f(x) is minus the integral from 0 to b(x)

f x( )

0

b x( )

ηsin η x η

3⋅−( )

η

⌠⌡

d−:= g x( )

0

b x( )

ηsin η x η

3⋅+( )

η

⌠⌡

d:=

wni f β i( )( ):= wpi g β i( )( ):=

wp

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

1.571

=wn

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

-1.550

-1.591

-1.616

-1.570

-1.544

-1.525

-1.537

-1.547

-1.556

-1.564

-1.565

-1.559

-1.547

-1.534

-1.526

-1.536

=

ββi β i( ):= zni1−

3β i( )

:= zpi1

3β i( )

:=

ββ

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

110 -7

1.02310 -7

1.04710 -7

1.07210 -7

1.09610 -7

1.12210 -7

1.14810 -7

1.17510 -7

1.20210 -7

1.2310 -7

1.25910 -7

1.28810 -7

1.31810 -7

1.34910 -7

1.3810 -7

1.41310 -7

= zn

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

-215.443

-213.796

-212.162

-210.539

-208.93

-207.332

-205.747

-204.174

-202.613

-201.064

-199.526

-198.001

-196.487

-194.984

-193.494

-192.014

= zp

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

215.443

213.796

212.162

210.539

208.93

207.332

205.747

204.174

202.613

201.064

199.526

198.001

196.487

194.984

193.494

192.014

=


Skalak, R. (1956), Ref. 4"An extension of the theory of water hammer."Transactions of the ASME 78, 105-116.

Figure 4, not scaled

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 62

1

0

1

2

3

wni−

wpi−

π

2

π

2−

π

6−

zni zpi, zni, zpi, zpi,

Figure 4, scaled and vertically shifted

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 60.4

0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1

π− wni⋅

1

2+

1

π− wpi⋅

1

2+

1

0

1

3



1000 800 600 400 200 0 200 400 600 800 10003

2

1

0

1

2

wni

wpi

π

2−

π

2

π

6

π

6

ββ i− ββ i, 0, 0, ββ i, ββ i−,

1 .107

1 .106

1 .105

1 .104

1 .103

0.01 0.1 1 10 100 1 .103

3

2

1

0

1

2

wni

wpi

π

2−

π

2

π

6

ββ i


Eq. (1-8), thesisw0 9.247031 106−

× m=w0

p0 a2

⋅

E h⋅:=

Eq. (3-24), thesiswI0 1.268253− 106−

× m=wI0

p0

6 m2⋅c1

c1−

⋅ D1⋅

:=

Eq. (3-22), thesiswIpinf 3.804758− 106−

× m=wIpinf

p0

2 m2⋅c1

c1−

⋅ D1⋅

:=

Eq. (3-22), thesiswIninf 3.804758 106−

× m=wIninf

p0−

2 m2⋅c1

c1−

⋅ D1⋅

:=

displacements

Eq. (56)CP 1.077752 1010

×Pa

m=CP

2 ρ0⋅ c12

⋅

ac1

2

c2

1−

⋅

−:=

Eqs. (47), (49)C 2.422184− 106−

× m=Cp0

π m2⋅c1

c1−

⋅ D1⋅

:=

p0 100000 Pa⋅:=h

2

12 a2

⋅

0.000021=D1 9.672924 108

×1

s2

=

Eq. (45),Eq. (3-14), thesis

D1

4 ρ0⋅ c14

⋅

m2 c2

⋅ a⋅c1

2

c2

1−

2

⋅

2 c02

⋅

a2

ν2

1c0

2

c12

−

21+ ν

2−

h2

12 a2

⋅

+

⋅+:=

Constants for "positive" waterhammer wave


Scaling

cPn CP C⋅ wn⋅1

2p0⋅+:= cPp CP C⋅ wp⋅

1

2p0⋅+:= Eqs. (47), (49), (56)

PIninf CP wIninf⋅1

2p0⋅+:= PIninf 91005.873668Pa= NOT equal to p0

because .....

PIpinf CP wIpinf⋅1

2p0⋅+:= PIpinf 8994.126332Pa= NOT equal to 0

because .....

PI0 CP wI0⋅1

2p0⋅+:= PI0 36331.375444Pa=

Dimensional pressure for p0 100000 Pa=

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 60

1 .104

2 .104

3 .104

4 .104

5 .104

6 .104

7 .104

8 .104

9 .104

1 .105

1.1 .105

1.2 .105

cPni

cPpi

PIninf

PIpinf

PI0



Dimensionless pressure for a

h62.752=

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

cPni

p0

cPpi

p0

PIninf

p0

PIpinf

p0

PI0

p0


PIninf

p0

0.910059=PIpinf

p0

0.089941=

0.089941 0.0046− 0.085341= off set add precursor, see Figure 5.

PIninf

p0

0.085341− 0.824718=PIpinf

p0

0.085341− 0.0046=

contributions of constant and other three solutions

PngIpinf Png2Ipinf+ Pg2Ininf+

p0

0+ 0.085332−= off set

osw


p0

0+:= osw 0.085332−=


t 1 s⋅:=

zzni

c1 t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zzpi

c1 t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)


h62.752= as function of z \ (c

1t)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

cPni

p0

cPpi

p0

PIninf

p0

PIpinf

p0

PI0

p0

zzni zzpi, zzni, zzpi, zzpi,

L1 t( ) 5.810168m= 0.005 c1⋅ t⋅ 4.904588 m=


t 5 s⋅:=

zzni

c1 t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zzpi

c1 t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)



1t)

0.97 0.975 0.98 0.985 0.99 0.995 1 1.005 1.01 1.015 1.020

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

cPni

p0

cPpi

p0

PIninf

p0

PIpinf

p0

PI0

p0


L1 t( ) 9.935248m= 0.002 c1⋅ t⋅ 9.809175 m=


t 10 s⋅:=

zzni

c1 t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zzpi

c1 t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)



1t)

0.99 0.992 0.994 0.996 0.998 1 1.002 1.004 1.006 1.008 1.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

cPni

p0

cPpi

p0

PIninf

p0

PIpinf

p0

PI0

p0


L1 t( ) 12.517628m= 0.001 c1⋅ t⋅ 9.809175 m=


Eqs. (47), (49), (56)cPp2 CP2 C2⋅ wp⋅:=cPn2 CP2 C2⋅ wn⋅:=

Scaling

Eq. (3-24), thesis,Eq. (56)

P2I0 76.712762− Pa=P2I0

p0

6 m2⋅c2

c1−

⋅ D2⋅

CP2⋅:=

NOT equal to 0

because .....

Eq. (3-22), thesis,Eq. (56)

P2Ipinf 230.138286− Pa=P2Ipinf

p0

2 m2⋅c2

c1−

⋅ D2⋅

CP2⋅:=

Eq. (3-22), thesis,Eq. (56)

P2Ininf 230.138286Pa=P2Ininf

p0−

2 m2⋅c2

c1−

⋅ D2⋅

CP2⋅:=

Eq. (56)CP2 1.662263− 1010

×Pa

m=CP2

2 ρ0⋅ c22

⋅

ac2

2

c2

1−

⋅

−:=

Eqs. (47), (49)C2 8.813923 109−

× m=C2

p0

π m2⋅c2

c1−

⋅ D2⋅

:=

p0 100000 Pa=h

2

12 a2

⋅

0.000021=D2 3.844165 1010

×1

s2

=

Eq. (45),Eq. (3-14), thesis

D2

4 ρ0⋅ c24

⋅

m2 c2

⋅ a⋅c2

2

c2

1−

2

⋅

2 c02

⋅

a2

ν2

1c0

2

c22

−

21+ ν

2−

h2

12 a2

⋅

+

⋅+:=

Constants for "positive" precursor wave



30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 6300

200

100

0

100

200

300

400

cPn2i

cPp2i

P2Ininf

P2Ipinf

P2I0




h62.752=

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 2 4 60.003

0.002

0.001

0

0.001

0.002

0.003

0.004

cPn2i

p0

cPp2i

p0

P2Ininf

p0

P2Ipinf

p0

P2I0

p0


P2Ininf

p0

0.002301=P2Ipinf

p0

0.002301−= off set

P2Ininf

p0

0.002301+ 0.004602=P2Ipinf

p0

0.002301+ 3.828584− 107−

×=


PgIpinf PngIpinf+ Png2Ipinf+

p0

1

2+ 0.002308= off set

osp


p0

1

2+:= osp 0.002308=


t 1 s⋅:=

zzn2i

c2 t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zzp2i

c2 t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)



2t)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.003

0.002

0.001

0

0.001

0.002

0.003

0.004

cPn2i

p0

cPp2i

p0

P2Ininf

p0

P2Ipinf

p0

P2I0

p0

zzn2i zzp2i, zzn2i, zzp2i, zzp2i,

L2 t( ) 9.163467m= 0.002 c2⋅ t⋅ 10.558858 m=


t 5 s⋅:=

zzn2i

c2 t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zzp2i

c2 t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)

Dimensionless pressure displacement for a


2t)

0.995 0.996 0.997 0.998 0.999 1 1.001 1.002 1.003 1.004 1.0050.003

0.002

0.001

0

0.001

0.002

0.003

0.004

cPn2i

p0

cPp2i

p0

P2Ininf

p0

P2Ipinf

p0

P2I0

p0


L2 t( ) 15.669309m= 0.0005 c2⋅ t⋅ 13.198573 m=


t 10 s⋅:=

zzn2i

c2 t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zzp2i

c2 t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)



2t)

0.997 0.9975 0.998 0.9985 0.999 0.9995 1 1.0005 1.001 1.0015 1.0020.003

0.002

0.001

0

0.001

0.002

0.003

0.004

cPn2i

p0

cPp2i

p0

P2Ininf

p0

P2Ipinf

p0

P2I0

p0


L2 t( ) 19.742092m= 0.0004 c2⋅ t⋅ 21.117716 m=


Eq. (1-8), thesisw0 9.247031 106−

× m=w0

p0 a2

⋅

E h⋅:=

Eq. (3-24), thesiswnI0 2.749655− 107−

× m=wnI0

p0−

6 m2⋅c1

c1+

⋅ D1⋅

:=

Eq. (3-22), thesiswnIpinf 8.248965 107−

× m=wnIpinf

p0

2 m2⋅c1

c1+

⋅ D1⋅

:=

Eq. (3-22), thesiswnIninf 8.248965− 107−

× m=wnIninf

p0−

2 m2⋅c1

c1+

⋅ D1⋅

:=

displacements

Eqs. (47), (49)Cn 5.251454 107−

× m=Cnp0

π m2⋅c1

c1+

⋅ D1⋅

:=

p0 100000 Pa=h

2

12 a2

⋅

0.000021=D1 9.672924 108

×1

s2

=

Eq. (45),Eq. (3-14), thesis

D1

4 ρ0⋅ c14

⋅

m2 c2

⋅ a⋅c1

2

c2

1−

2

⋅

2 c02

⋅

a2

ν2

1c0

2

c12

−

21+ ν

2−

h2

12 a2

⋅

+

⋅+:=

Constants for "negative" waterhammer wave


6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 303.5 .10

4

4 .104

4.5 .104

5 .104

5.5 .104

6 .104

cPnni

cPnpi

PnIninf

PnIpinf

PnI0

zni zpi, zni, zpi, zni,


Eq. (56)PnI0 52963.447212Pa=PnI0 CP wnI0−⋅1

2p0⋅+:=

Eq. (56)PnIpinf 41109.658365Pa=PnIpinf CP wnIpinf−⋅1

2p0⋅+:=

NOT equal to 0

because .....PnIninf 58890.341635Pa=PnIninf CP wnIninf−⋅

1

2p0⋅+:=

Eqs. (47), (49), (56)cPnp CP Cn−⋅ wnp⋅1

2p0⋅+:=cPnn CP Cn−⋅ wnn⋅

1

2p0⋅+:=

wnp wn−:=wnn wp−:=

Scaling



h62.752=

6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 300.35

0.4

0.45

0.5

0.55

0.6

cPnni

p0

cPnpi

p0

PnIninf

p0

PnIpinf

p0

PnI0

p0


PnIninf

p0

0.588903=PnIpinf

p0

0.411097=

0.588903− 1.0025+ 0.413597= off set add precursor, see Figure 5.

PnIninf

p0

0.413597+ 1.0025=PnIpinf

p0

0.413597+ 0.824694=


PgIninf Pg2Ininf+ Png2Ipinf+

p0

0+ 0.41363= off set

oswn


p0

0+:= oswn 0.41363=


t 1 s⋅:=

zznni

c1− t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zznpi

c1− t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)



1t)

1.1 1.08 1.06 1.04 1.02 1 0.98 0.96 0.94 0.92 0.90.35

0.4

0.45

0.5

0.55

0.6

cPnni

p0

cPnpi

p0

PnIninf

p0

PnIpinf

p0

PnI0

p0

zznni zznpi, zznni, zznpi, zznni,

L1 t( ) 5.810168m= 0.005 c1⋅ t⋅ 4.904588 m=


t 5 s⋅:=

zznni

c1− t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zznpi

c1− t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)



1t)

1.02 1.015 1.01 1.005 1 0.995 0.99 0.985 0.98 0.975 0.970.35

0.4

0.45

0.5

0.55

0.6

cPnni

p0

cPnpi

p0

PnIninf

p0

PnIpinf

p0

PnI0

p0


L1 t( ) 9.935248m= 0.002 c1⋅ t⋅ 9.809175 m=


t 10 s⋅:=

zznni

c1− t⋅ zni3

d1 t⋅⋅+

c1 t⋅:= zznpi

c1− t⋅ zpi3

d1 t⋅⋅+

c1 t⋅:= Eqs. (40), (48), (51)



1t)

1.01 1.008 1.006 1.004 1.002 1 0.998 0.996 0.994 0.992 0.990.35

0.4

0.45

0.5

0.55

0.6

cPnni

p0

cPnpi

p0

PnIninf

p0

PnIpinf

p0

PnI0

p0


L1 t( ) 12.517628m= 0.001 c1⋅ t⋅ 9.809175 m=


Pn2I0 42.344725Pa= Eq. (3-24), thesis

Scaling

wnn wp−:= wnp wn−:=

cPnn2 CP2 Cn2−⋅ wnn⋅:= cPnp2 CP2 Cn2−⋅ wnp⋅:= Eqs. (47), (49), (56)

Pn2Ininf Pn2Ininf−:= Pn2Ininf 127.034176− Pa= NOT equal to 0

because .....

Pn2Ipinf Pn2Ipinf−:= Pn2Ipinf 127.034176Pa= Eq. (56)

Pn2I0 Pn2I0−:= Pn2I0 42.344725− Pa= Eq. (56)

Constants for "negative" precursor wave

D2

4 ρ0⋅ c24

⋅

m2 c2

⋅ a⋅c2

2

c2

1−

2

⋅

2 c02

⋅

a2

ν2

1c0

2

c22

−

21+ ν

2−

h2

12 a2

⋅

+

⋅+:= Eq. (45),Eq. (3-14), thesis

D2 3.844165 1010

×1

s2

=h

2

12 a2

⋅

0.000021= p0 100000 Pa=

Cn2

p0

π m2⋅c2

c1+

⋅ D2⋅

:= Cn2 4.865203 109−

× m= Eqs. (47), (49)

Pn2Ininf

p0−

2 m2⋅c2

c1+

⋅ D2⋅

CP2⋅:= Pn2Ininf 127.034176Pa= Eq. (3-22), thesis

Pn2Ipinf

p0

2 m2⋅c2

c1+

⋅ D2⋅

CP2⋅:= Pn2Ipinf 127.034176− Pa= Eq. (3-22), thesis

NOT equal to 0

because .....

Pn2I0

p0−

6 m2⋅c2

c1+

⋅ D2⋅

CP2⋅:=



6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30150

100

50

0

50

100

150

200

250

cPnn2i

cPnp2i

Pn2Ininf

Pn2Ipinf

Pn2I0




h62.752=

6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 300.0015

0.001

5 .104

0

5 .104

0.001

0.0015

0.002

0.0025

cPnn2i

p0

cPnp2i

p0

Pn2Ininf

p0

Pn2Ipinf

p0

Pn2I0

p0


Pn2Ininf

p0

0.00127034−=Pn2Ipinf

p0

0.00127034=

0.00127034 1+ 1.00127= off set see Figure 5

Pn2Ininf

p0

1.00127+ 1=Pn2Ipinf

p0

1.00127+ 1.00254=


PgIninf PngIninf+ Pg2Ininf+

p0

1

2+ 1.001264= off set

ospn


p0

1

2+:= ospn 1.001264=


t 1 s⋅:=

zznn2i

c2− t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zznp2i

c2− t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)



2t)

1.02 1.015 1.01 1.005 1 0.995 0.99 0.985 0.98 0.975 0.970.0015

0.001

5 .104

0

5 .104

0.001

0.0015

0.002

0.0025

cPnn2i

p0

cPnp2i

p0

Pn2Ininf

p0

Pn2Ipinf

p0

Pn2I0

p0

zznn2i zznp2i, zznn2i, zznp2i, zznn2i,

L2 t( ) 9.163467m= 0.002 c2⋅ t⋅ 10.558858 m=


t 5 s⋅:=

zznn2i

c2− t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zznp2i

c2− t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)



2t)

1.005 1.004 1.003 1.002 1.001 1 0.999 0.998 0.997 0.996 0.9950.0015

0.001

5 .104

0

5 .104

0.001

0.0015

0.002

0.0025

cPnn2i

p0

cPnp2i

p0

Pn2Ininf

p0

Pn2Ipinf

p0

Pn2I0

p0


L2 t( ) 15.669309m= 0.0005 c2⋅ t⋅ 13.198573 m=


t 10 s⋅:=

zznn2i

c2− t⋅ zni3

d2 t⋅⋅+

c2 t⋅:= zznp2i

c2− t⋅ zpi3

d2 t⋅⋅+

c2 t⋅:= Eqs. (40), (48), (51)



2t)

1.002 1.0015 1.001 1.0005 1 0.9995 0.999 0.9985 0.998 0.9975 0.9970.0015

0.001

5 .104

0

5 .104

0.001

0.0015

0.002

0.0025

cPnn2i

p0

cPnp2i

p0

Pn2Ininf

p0

Pn2Ipinf

p0

Pn2I0

p0


L2 t( ) 19.742092m= 0.0004 c2⋅ t⋅ 21.117716 m=


H2(∞)Pn2Ipinf 127.034176Pa=

Png2Ipinf 127.034176Pa≡Png2Ininf 127.034176− Pa≡

H1(∞)PnIpinf1

2p0⋅− 8890.341635− Pa=

PngIpinf 8890.341635− Pa≡PngIninf 8890.341635Pa≡

I2(∞)P2Ipinf 230.138286− Pa=

Pg2Ipinf 230.138286− Pa≡Pg2Ininf 230.138286Pa≡

I1(∞)PIpinf1

2p0⋅− 41005.873668− Pa=

PgIpinf 41005.873668− Pa≡PgIninf 41005.873668Pa≡

Global definition of values at infinity

p0 CP⋅

2 m2⋅c1

c1−

⋅ D1⋅

p0 CP⋅

2 m2⋅c1

c1+

⋅ D1⋅

−p0 CP2⋅

2 m2⋅c2

c1−

⋅ D2⋅

+p0 CP2⋅

2 m2⋅c2

c1+

⋅ D2⋅

−

p0

0.499993−=

PIninf PnIninf+ P2Ininf+ Pn2Ininf+

p0

1.499993=Solution for z = negative infinity,see Eq. (56)

PIpinf PnIpinf+ P2Ipinf+ Pn2Ipinf+

p0

0.500007=Solution for z = positive infinity,see Eq. (56)



p0

0+ 0.085332−= p. 16


p0

1

2+ 0.002308= p. 22


p0

0+ 0.41363= p. 28


p0

1

2+ 1.001264= p. 34

Solution for z = positive infinity,see Eq. (56)

PgIpinf PngIpinf+ Pg2Ipinf+ Png2Ipinf+

p0

0.499993−=

Solution for z = negative infinity,see Eq. (56)

PgIninf PngIninf+ Pg2Ininf+ Png2Ininf+

p0

0.499993=



zzJ0 PIninf:= zzJ1 PIpinf:= Joukowsky

30 26 22 18 14 10 6 2 2 60

2 .104

4 .104

6 .104

8 .104

1 .105

1.2 .105

scaled axial distance (-)

pre

ssu

re (

Pa)

Factors for conversion from radial wall displacement to pressure

E h⋅

a2

1.081428 1010

×Pa

m=

E h⋅

a2

w0

zzJ0

⋅ 1.09883=

CP 1.077752 1010

×Pa

m= CP

w0

zzJ0

⋅ 1.095095=


zzznp2i c2− t⋅ zpi3

d2 t⋅⋅+:= Eqs. (40), (48), (51)

off sets

osw osw p0⋅:= osp osp p0⋅:=

PIpinf osw+ 460.957159 Pa= P2Ininf osp+ 460.957159 Pa=

oswn oswn p0⋅:= ospn ospn p0⋅:=

PnIninf oswn+ 100253.387765 Pa= Pn2Ipinf ospn+ 100253.387765 Pa=

axial coordinates of end points of connecting horizontal lines

z0w0

700

m⋅:= zwp1200

3000

m⋅:= zpw3000

4800

m⋅:= zinfp5800

6000

m⋅:=

Positive and negative waterhammer and precursor wavesin one figure

t 1 s⋅:= c1 t⋅ 980.917501 m= zn1 213.796209−= zp1 213.796209=

3d1 t⋅ 1.430247 m= znn 0.1−= zpn 0.1=

zzzni c1 t⋅ zni3

d1 t⋅⋅+:= zzzpi c1 t⋅ zpi3

d1 t⋅⋅+:= Eqs. (40), (48), (51)

zzzn2i c2 t⋅ zni3

d2 t⋅⋅+:= zzzp2i c2 t⋅ zpi3

d2 t⋅⋅+:= Eqs. (40), (48), (51)

zzznni c1− t⋅ zni3

d1 t⋅⋅+:= zzznpi c1− t⋅ zpi3

d1 t⋅⋅+:= Eqs. (40), (48), (51)

zzznn2i c2− t⋅ zni3

d2 t⋅⋅+:=


Dimensional pressure for a

h62.752= and p0 100000Pa=

as function of z, WATERHAMMER + precursor, upstream

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 60002 .10

4

0

2 .104

4 .104

6 .104

8 .104

1 .105

1.2 .105

cPni osw+( )

cPpi osw+( )

PIninf osw+( )

PIpinf osw+( )

cPn2i osp+( )

cPp2i osp+( )

P2Ininf osp+( )

P2Ipinf osp+( )

zzzni zzzpi, z0wi, zwpi, zzzn2i, zzzp2i, zpwi, zinfpi,

L1 t( ) 5.810168m= 0.005 c1⋅ t⋅ 4.904588 m=



h62.752= and p0 100000Pa=

as function of z, PRECURSOR, upstream

3000 3500 4000 4500 5000 5500 6000100

0

100

200

300

400

500

600

cPn2i osp+( )

cPp2i osp+( )

P2Ininf osp+( )

P2Ipinf osp+( )

zzzn2i zzzp2i, zpwi, zinfpi,

L2 t( ) 9.163467m= 0.002 c2⋅ t⋅ 10.558858 m=



h62.752= and p0 100000Pa=

as function of z, WATERHAMMER + precursor, downstream

6000 5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 02 .10

4

0

2 .104

4 .104

6 .104

8 .104

1 .105

1.2 .105

cPnni oswn+( )

cPnpi oswn+( )

PnIpinf oswn+( )

PnIninf oswn+( )

cPnn2i ospn+( )

cPnp2i ospn+( )

Pn2Ipinf ospn+( )

Pn2Ininf ospn+( )

zzznni zzznpi, z0w−( )i

, zwp−( )i

, zzznn2i, zzznp2i, zpw−( )i

, zinfp−( )i

,



h62.752= and p0 100000Pa=

as function of z, precursor, downstream

6000 5750 5500 5250 5000 4750 4500 4250 4000 3750 3500 3250 30009.99 .10

4

1 .105

1.001 .105

1.002 .105

1.003 .105

1.004 .105

cPnni oswn+( )

cPnpi oswn+( )

PnIpinf oswn+( )

PnIninf oswn+( )

cPnn2i ospn+( )

cPnp2i ospn+( )

Pn2Ipinf ospn+( )

Pn2Ininf ospn+( )

zzznni zzznpi, z0w−( )i

, zwp−( )i

, zzznn2i, zzznp2i, zpw−( )i

, zinfp−( )i

,


fstatic 5215.453765Hz=fstatic

c1

λstatic

:=

λstatic 188.079033 mm=λstatic 2 π⋅

4

a2

h2

⋅

3 1 ν2

−( )⋅

⋅:=

h 1.5 mm⋅:=

a 40 mm⋅:=

Asselman (1969), eq (2.2.6), Fig. 2.10, static solution,MSc Thesis, Dept. of Applied Physics, TU Eindhoven

1f3ring

f0ring

− 0.422494=

f3ring 1548.007982Hz=f3ring1

2 π⋅

4 E⋅

D2

ρs⋅ 1D

8 ee⋅

ρf

ρs

⋅+

⋅

⋅:=

Barez et al (1979), part II, eq (2), Ref. 16

1f2ring

f0ring

− 0.477614=


2 π⋅

4 E⋅

D2

ρs⋅ 1D

6 ee⋅

ρf

ρs

⋅+

⋅

⋅:=

Kellner and Schoenfelder (1982), Ref. 17

1f1ring

f0ring

− 0.552642=


2 π⋅

4 E⋅

D2

ρs⋅ 1D

4 ee⋅

ρf

ρs

⋅+

⋅

⋅:=

Walker and Phillips (1977), Ref. 18


π D⋅

E

ρs

⋅:=

Ring frequencies


flobar 6( ) 279.339293Hz=flobar 3( ) 50.508256Hz=

flobar 2( ) 16.070848Hz=flobar 5( ) 181.179919Hz=

flobar 1( ) 0Hz=flobar 4( ) 105.158248Hz=

Lobar modes

c0

ee1.108 10

6× Hz=flobar n( ) f0ring Ω2 n( )⋅:=

(2-32)Ω2 n( ) β2n

2n

21−( )2

⋅

1 n2

+ 2 n⋅ µ⋅+

⋅:=

c0

D8827.668135Hz=

f00ring

f0ring

1.048285=f00ring

c0

πD:=

Lobar frequencies

aee

2+

R0.024641m=β2 2.082909 10

5−×=β2

ee2

12 aee

2+

2

⋅

:=

1

4

ρf

ρt

⋅D

ee⋅ 3.996768=µ 3.965174=µ

ρf Af⋅

ρt At⋅:=

Constants

It 44253.676749 cm4

=At 93.762442 cm2

=Af 0.29186351 m2

=

Itπ

4a ee+( )

4a

4− ⋅:=At π a ee+( )

2a2

− ⋅:=Af π a2

⋅:=

Cross-sectional areas

Input data

Lobar frequenciesDe Jong (1994) Section 2.4PhD Thesis, Dept. of Mechanical Engineering, TU Eindhoven


Ovalizing frequency

flobar 2( ) 16.070847864527Hz=

foval2 ee⋅

Dρf

ρt

D

ee⋅ 5+⋅

f0ring⋅:= PS9, Time Scales, Eq. (14), wrong

foval 9.324237 Hz=

fcoval2 3⋅ ee⋅

Dρf

ρt

D

ee⋅ 5+⋅

f0ring⋅:= PS9, Time Scales, Eq. (14), corrected

fcoval 16.150053 Hz=

fcoval2 3⋅ ee⋅

D 4 µ⋅ 5+⋅f0ring⋅:= PS9, Time Scales, Eq. (14), corrected, more accurate

fcoval 16.1988983478942 Hz= fcovala

aee

2+

⋅ 16.070847864527Hz=



h62.752= and p0 100000Pa=

as function of z, WATERHAMMER + precursor, upstream

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 60002 .10

4

0

2 .104

4 .104

6 .104

8 .104

1 .105

1.2 .105

connects to figure on (Ap)p. 44

Skalak, R. (1956), Ref. 4 Wave speeds ( ) Nomenclature 11 CASA_06-27.mcd Appendix 1 γs 76972.857 kg m2s2 = pcf 157.087464 kg m2s2 ρs γs g:= ρs 7849.047053 kg m3 A := 8 A 8= …

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