NIGERIA LNG LTD

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NLNG SIX PROJECT

Appendix - 1

Valve Tag - 520-FRC-800

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

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NLNG SIX PROJECT

Appendix - 1

Valve Tag - 520-FRC-800

Input Data:

P1 22 bara
P2 20.65 bara
xT 0.694 (@ 100% valve travel for a 4" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 4
d2 16
D 6
D2 36
Cv 251 (@ 100% valve travel for a 4" valve size)
Cv2 63001
k 1.38
g 22.12 Kg/m3
W 4.31 Kg/s
15516 Kg/hr

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(16/36))^2 0.5
K1 0.1543 1
2
From eqn. 5, K2 = 1.0*(1-(16/36))^2 4
K2 0.3086

From eqn.6, KB1 or KB2 = 1-(4/6)^4

KB1 or KB2 = 0.8025

From eqn.3, K = 0.1543+0.3086+KB1-KB2, (KB1 = KB2)

K = 0.4630

From eqn.2, Fp = (1+(0.4630/890)*(251/16)^2)^-0.5

Fp = 0.9415

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.0614

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NLNG SIX PROJECT

Appendix - 1

Valve Tag - 520-FRC-800

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.9568

A 0.00066401

B = 15.6875
B =
2
246.10

C = 1.1634

E= 0.7828

XTP = 0.6729

Calculating Fk in eqn. 10,

Fk = 0.9857

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9692

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 24.912

J = 5.465

Cv = W/(H*J) 113.978

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NLNG SIX PROJECT

Appendix - 2

Valve Tag - 520-FRC-801

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 22.18 bara
P2 21.04 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 6
d2 36
D 6
D2 36
Cv 357 (@ 100% valve travel for a 6" valve size
Cv2 127449
g 600.9 Kg/m3
W 23.76 Kg/s
85536 Kg/hr

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NLNG SIX PROJECT

Appendix - 2

Valve Tag - 520-FRC-801

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(36/36))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(36/36))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(6/6)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 98.3402778

Fp = 1

Solving for eqn. 10,

L = 26.173

Substituting the value of L into eqn. 11 results in:

Cv = 119.71

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NLNG SIX PROJECT

Appendix - 3

Valve Tag - 520-PRCA-803A

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(Pc*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Pc = P1*XT*Fk [14]

Let xTKi/N5 = A [15]

Let Cv/d2 = B [16]

Let (1+((A)*(B)2)) = C [17]

Let (xT/FP2) = E [18]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [19]

Let N6*Fp*Y = H [20]

Let (Pc*g)^0.5 = J [21]

Therefore, Cv = W/(H*J) [22]

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NLNG SIX PROJECT

Appendix - 3

Valve Tag - 520-PRCA-803A

Input Data:

P1 19.5 bara
P2 4.04 bara (Based on back pressure of existing)
xT 0.694 (@ 100% valve travel for a 10" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 10
d2 100
D 10
D2 100
Cv 240 (@ 100% valve travel for a 10" valve size)
Cv2 57600
k 1.25
g 28.38 Kg/m3
W 16.25 Kg/s
58500 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(100/100))^2 4
12
144
K1 0.0

From eqn. 5, K2 = 1.0*(1-(100/144))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(10/12)^4

KB1 or KB2 = 0.518

From eqn.3, K = 0.0+0.0934+KB1-KB2, (KB1 = KB2)

K = -0.5177

From eqn.2, Fp = (1+(-0.424/890)*(570/100)^2)^-0.5

Fp = 1.0017

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.7928

From eqn. 14

Pc = 12.0830

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NLNG SIX PROJECT

Appendix - 3

Valve Tag - 520-PRCA-803A

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.5177

A 0.00035932

B = 2.4
B2 = 5.76

C = 1.0021

E= 0.6917

XTP = 0.6902

Calculating Fk in eqn. 10,

Fk = 0.8929

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.5712

Cv = (W/((N6*Fp*Y)*(Pc*g)^0.5))

H = 15.620

J = 18.518

Cv = W/(H*J) 202.251

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NLNG SIX PROJECT

Appendix - 4

Valve Tag - 520-PRCA-803B

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^--1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

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NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 4

Valve Tag - 520-PRCA-803B

Input Data:

P1 19.5 bara
P2 18.8 bara
xT 0.807 (@ 100% valve travel for a 8" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 8
d2 64
D 10
D2 100
Cv 818 (@ 100% valve travel for a 8" valve size)
Cv2 669124
k 1.25
g 28.55 Kg/m3
W 11.2 Kg/s
40320 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(64/100))^2 4

K1 0.0648

From eqn. 5, K2 = 1.0*(1-(64/100))^2

K2 0.1296

From eqn.6, KB1 or KB2 = 1-(8/10)^4

KB1 or KB2 = 0.5904

From eqn.3, K = 0.0648+0.1296+KB1-KB2, (KB1 = KB2)

K = 0.1944

From eqn.2, Fp = (1+(0.1944/890)*(818/64)^2)^-0.5

Fp = 0.9826

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.0359

Solving from equations 13 to 17 and substituting into 18 results in:

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NLNG SIX PROJECT

Appendix - 4

Valve Tag - 520-PRCA-803B

Ki = 0.6552

A 0.00052875

B = 12.78125
B2 = 163.36

C = 1.0864

E= 0.8358

XTP = 0.7693

Calculating Fk in eqn. 10,

Fk = 0.8929

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9826

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 26.358

J = 4.470

Cv = W/(H*J) 342.177

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NLNG SIX PROJECT

Appendix - 5

Valve Tag - 520-FRC-804

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 19.63 bara
P2 8.63 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 2
d2 4
D 3
D2 9
Cv 59.7 (@ 100% valve travel for a 2" valve size)
Cv2 3564.09
g 650 Kg/m3
W 12.96 Kg/s
46656 Kg/hr

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NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 5

Valve Tag - 520-FRC-804

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(4/9))^2 0.5
K1 0.1543 1
2
From eqn. 5, K2 = 1.0*(1-(4/9))^2 4
K2 0.3086

From eqn.6, KB1 or KB2 = 1-(2/3)^4

KB1 or KB2 = 0.8025

From eqn.3, K = 0.1543+0.3086+KB1-KB2, (KB1 = KB2)

K = 0.4630

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0.00052

R = 222.755625

Fp = 0.9467

Solving for eqn. 10,

L = 84.558

Substituting the value of L into eqn. 11 results in:

Cv = 21.35

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NLNG SIX PROJECT

Appendix - 6

Valve Tag - 520-FICA-805

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 22.9 bara
P2 22.2 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 3
d2 9
D 3
D2 9
Cv 121 (@ 100% valve travel for a 3" valve size)
Cv2 14641
g 505.2 Kg/m3
W 8.54 Kg/s
30744 Kg/hr

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NLNG SIX PROJECT

Appendix - 6

Valve Tag - 520-FICA-805

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(9/9))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(9/9))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(3/3)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 180.753086

Fp = 1

Solving for eqn. 10,

L = 19.072

Substituting the value of L into eqn. 11 results in:

Cv = 59.05

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NLNG SIX PROJECT

Appendix - 7

Valve Tag - 520-FRC-807

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 9.15 bara (HOLD 1)

P2 8.62 bara (HOLD 1)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 10
d2 100
D 12
D2 144
Cv 946 (@ 100% valve travel for a 10" valve size)
Cv2 894916
g 724.7 Kg/m3
W 122.9 Kg/s
442440 Kg/hr

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NLNG SIX PROJECT

Appendix - 7

Valve Tag - 520-FRC-807

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(10/144))^2 0.5
K1 0.0467 1
2
From eqn. 5, K2 = 1.0*(1-(100/144))^2 4
K2 0.0934

From eqn.6, KB1 or KB2 = 1-(10/12)^4

KB1 or KB2 = 0.5177

From eqn.3, K = 0.0467+0.0934+KB1-KB2, (KB1 = KB2)

K = 0.1400

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0.00015736

R = 89.492

Fp = 0.993

Solving for eqn. 10,

L = 19.598

Substituting the value of L into eqn. 11 results in:

Cv = 832.743

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NLNG SIX PROJECT

Appendix - 8

Valve Tag - 520-FICA-808

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 23.38 bara
P2 20.29 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 1.5
d2 2.25
D 2
D2 4
Cv 21 (@ 100% valve travel for a 1.5" valve size)
Cv2 441
g 508 Kg/m3
W 3.34 Kg/s
12024 Kg/hr

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NLNG SIX PROJECT

Appendix - 8

Valve Tag - 520-FICA-808

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(2.25/4))^2 0.5
K1 0.0957 1
2
From eqn. 5, K2 = 1.0*(1-(2.25/4))^2 4
K2 0.1914

From eqn.6, KB1 or KB2 = 1-(1.5/2)^4

KB1 or KB2 = 0.6836

From eqn.3, K = 0.0957+0.1914+KB1-KB2, (KB1 = KB2)

K = 0.2871

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0.00032259

R = 87.111

Fp = 0.986

Solving for eqn. 10,

L = 39.620

Substituting the value of L into eqn. 11 results in:

Cv = 11.272

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NLNG SIX PROJECT

Appendix - 9

Valve Tag - 520-LICA-809

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 10.93 bara
P2 1.38 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 2
d2 4
D 3
D2 9
Cv 13.9 (@ 100% valve travel for a 2" valve size)
Cv2 193.21
g 505.2 Kg/m3
W 1.528 Kg/s
5500.8 Kg/hr

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 9

Valve Tag - 520-LICA-809

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(4/9))^2 0.5
K1 0.1543 1
2
From eqn. 5, K2 = 1.0*(1-(4/9))^2 4
K2 0.3086

From eqn.6, KB1 or KB2 = 1-(2/3)^4

KB1 or KB2 = 0.8025

From eqn.3, K = 0.1543+0.3086+KB1-KB2, (KB1 = KB2)

K = 0.4630

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0.00052018

R = 12.075625

Fp = 0.997

Solving for eqn. 10,

L = 69.460

Substituting the value of L into eqn. 11 results in:

Cv = 2.910

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 10

Valve Tag - 520-FRC-809

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 10

Valve Tag - 520-FRC-809

Input Data:

P1 24.9 bara
P2 19.4 bara
xT 0.694 (@ 100% valve travel for a 8" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 8
d2 64
D 8
D2 64
Cv 570 (@ 100% valve travel for a 8" valve size)
Cv2 324900
k 1.24
g 17.14 Kg/m3
W 8.73 Kg/s
31428 Kg/hr

Calculations:
Control Point
Solving for equations 4, 5 & 6 0.5
1
From eqn. 4, K1 = 0.5*(1-(64/64))^2 2
K1 0.0000 4

From eqn. 5, K2 = 1.0*(1-(64/64))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(8/8)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0/890)*(570/64)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.2209

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 10

Valve Tag - 520-FRC-809

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.0000

A 0

B = 8.90625
B2 = 79.32

C = 1.0000

E= 0.6940

XTP = 0.6940

Calculating Fk in eqn. 10,

Fk = 0.8857

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.8802

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 24.030

J = 9.709

Cv = W/(H*J) 134.703

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NLNG SIX PROJECT

Appendix - 11

Valve Tag - 520-FIC-812

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 22.32 bara
P2 21.47 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 3
d2 9
D 3
D2 9
Cv 136 (@ 100% valve travel for a 3" valve size)
Cv2 18496
g 645 Kg/m3
W 10 Kg/s
36000 Kg/hr

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 11

Valve Tag - 520-FIC-812

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(9/9))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(9/9))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(3/3)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 228.345679

Fp = 1

Solving for eqn. 10,

L = 23.415

Substituting the value of L into eqn. 11 results in:

Cv = 56.318

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 12

Valve Tag - 520-FRC-815

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTP = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTP = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 12

Valve Tag - 520-FRC-815

Input Data:

P1 24.9 bara
P2 21.65 bara
xT 0.724 (@ 100% valve travel for a 8" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 8
d2 64
D 8
D2 64
Cv 567 (@ 100% valve travel for a 8" valve size)
Cv2 321489
k 1.24
g 17.14 Kg/m3
W 8.73 Kg/s
31428 Kg/hr

Calculations:
Control Point
Solving for equations 4, 5 & 6 0.5
1
From eqn. 4, K1 = 0.5*(1-(64/64))^2 2
K1 0.0000 4

From eqn. 5, K2 = 1.0*(1-(64/64))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(8/8)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0/890)*(570/64)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.1305

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NLNG SIX PROJECT

Appendix - 12

Valve Tag - 520-FRC-815

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.0000

A 0

B = 8.859375
B2 = 78.49

C = 1.0000

E= 0.7240

XTP = 0.7240

Calculating Fk in eqn. 10,

Fk = 0.8857

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9322

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 25.448

J = 7.464

Cv = W/(H*J) 165.470

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 13

Valve Tag - 441-FIC-782

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 9.8 bara (Ref. 18)

P2 6.95 bara (Ref. 18)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 6
d2 36
D 6
D2 36
Cv 322 (@ 100% valve travel for a 6" valve size)
Cv2 103684
g 724.7 Kg/m3
W 44.29 Kg/s (Ref. 18, Pg.6)
159444 Kg/hr

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 13

Valve Tag - 441-FIC-782

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(36/36))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(36/36))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(6/6)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 80.003

Fp = 1.000

Solving for eqn. 10,

L = 45.447

Substituting the value of L into eqn. 11 results in:

Cv = 128.512

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 14

Valve Tag - 541-FIC-782

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 9.8 bara (Ref. 18)

P2 6.95 bara (Ref. 18)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 6
d2 36
D 6
D2 36
Cv 322 (@ 100% valve travel for a 6" valve size)
Cv2 103684
g 724.7 Kg/m3
W 44.29 Kg/s (Ref. 18, Pg. 6)
159444 Kg/hr

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 14

Valve Tag - 541-FIC-782

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(36/36))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(36/36))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(6/6)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 80.003

Fp = 1.000

Solving for eqn. 10,

L = 45.447

Substituting the value of L into eqn. 11 results in:

Cv = 128.51

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 15

Valve Tag - 441-HIC-782

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 10.2 bara
P2 7.97 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 6
d2 36
D 6
D2 36
Cv 322 (@ 100% valve travel for a 6" valve size)
Cv2 103684
g 724.7 Kg/m3
W 44.29 Kg/s (Ref. 18, Pg.6)
159444 Kg/hr

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 15

Valve Tag - 441-HIC-782

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(36/36))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(36/36))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(6/6)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 80.003

Fp = 1.000

Solving for eqn. 10,

L = 40.201

Substituting the value of L into eqn. 11 results in:

Cv = 145.3

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 16

Valve Tag - 541-HC-782

Sizing Equations:

Cv = W/(N6*Fp*((P1-P2)g)^0.5) [1]

Fp = (1+K/N2(Cv/d2)^2)^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Let K/N2 = Q [7]

Let (Cv/d2)^2 = R [8]

Substituting equations 7 & 8 into 2 results in:

Fp = (1+(Q*R))^-0.5 [9]

Let ((P1-P2)g)^0.5 = L [10]

Substituting equation 10 into 1 results in:

Cv = W/N6*FP*L [11]

Input Data:

P1 10.2 bara
P2 7.97 bara
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 6
d2 36
D 6
D2 36
Cv 322 (@ 100% valve travel for a 6" valve size)
Cv2 103684
g 724.7 Kg/m3
W 44.29 Kg/s (Ref. 18, Pg.6)
159444 Kg/hr

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
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NLNG SIX PROJECT

Appendix - 16

Valve Tag - 541-HC-782

Calculations:

Solving for equations 4, 5 & 6

Control Point
From eqn. 4, K1 = 0.5*(1-(36/36))^2 0.5
K1 0.0000 1
2
From eqn. 5, K2 = 1.0*(1-(36/36))^2 4
K2 0.0000

From eqn.6, KB1 or KB2 = 1-(6/6)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

Solving for eqn. 7 & 8 and substituting into eqn. 9 results in:

Q = 0

R = 80.003

Fp = 1.000

Solving for eqn. 10,

L = 40.201

Substituting the value of L into eqn. 11 results in:

Cv = 145.3

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 17

Valve Tag - 520-HIC-811

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^--1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 17

Valve Tag - 520-HIC-811

Input Data:

P1 19.5 bara
P2 19.06 bara
xT 0.807 (@ 100% valve travel for a 10" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 10
d2 100
D 10
D2 100
Cv 810 (@ 100% valve travel for a 10" valve size)
Cv2 656100
k 1.13
g 22.13 Kg/m3
W 17 Kg/s
61200 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(64/100))^2 4

K1 0.0000

From eqn. 5, K2 = 1.0*(1-(64/100))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(8/10)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0.0648+0.1296+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0.1944/890)*(818/64)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.0226

Solving from equations 13 to 17 and substituting into 18 results in:

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
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NLNG SIX PROJECT

Appendix - 17

Valve Tag - 520-HIC-811

Ki = 0.0000

A 0

B = 8.1
B =
2
65.61

C = 1.0000

E= 0.8070

XTP = 0.8070

Calculating Fk in eqn. 10,

Fk = 0.8071

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9885

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 26.985

J = 3.120

Cv = W/(H*J) 726.801

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 18

Valve Tag - 520-HIC-812

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^--1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 18

Valve Tag - 520-HIC-812

Input Data:

P1 19.5 bara
P2 16.21 bara
xT 0.694 (@ 100% valve travel for a 2" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 2
d2 4
D 2
D2 4
Cv 33.3 (@ 100% valve travel for a 2" valve size)
Cv2 1108.89
k 1.25
g 20.5
W 1.2 Kg/s
4320 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(4/4))^2 4

K1 0.0000

From eqn. 5, K2 = 1.0*(1-(4/4))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(2/2)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0.1944/890)*(818/64)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.1687

Solving from equations 13 to 17 and substituting into 18 results in:

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 18

Valve Tag - 520-HIC-812

Ki = 0.0000

A 0

B = 8.325
B =
2
69.31

C = 1.0000

E= 0.6940

XTP = 0.6940

Calculating Fk in eqn. 10,

Fk = 0.8929

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9092

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 24.822

J = 8.212

Cv = W/(H*J) 21.192

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
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NLNG SIX PROJECT

Appendix - 19

Valve Tag - 520-HZ-806

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^--1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 19

Valve Tag - 520-HZ-806

Input Data:

P1 24.5 bara
P2 3.8 bara
xT 0.839 (@ 100% valve travel for a 8" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 8
d2 64
D 10
D2 100
Cv 688 (@ 100% valve travel for a 8" valve size)
Cv2 473344
k 1.063
g 50.92 Kg/m3
W 28.9 Kg/s
104040 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(64/100))^2 4

K1 0.0648

From eqn. 5, K2 = 1.0*(1-(64/100))^2

K2 0.1296

From eqn.6, KB1 or KB2 = 1-(8/10)^4

KB1 or KB2 = 0.5904

From eqn.3, K = 0.0648+0.1296+KB1-KB2, (KB1 = KB2)

K = 0.1944

From eqn.2, Fp = (1+(0.1944/890)*(818/64)^2)^-0.5

Fp = 0.9876

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.8449

Solving from equations 13 to 17 and substituting into 18 results in:

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
Page 46 of 55
NLNG SIX PROJECT

Appendix - 19

Valve Tag - 520-HZ-806

Ki = 0.6552

A 0.00054971

B = 10.75
B =
2
115.56

C = 1.0635

E= 0.8602

XTP = 0.8088

Calculating Fk in eqn. 10,

Fk = 0.7593

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.5414

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 14.597

J = 32.466

Cv = W/(H*J) 219.536

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 20

Valve Tag - 520-HZ-814

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^--1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
Page 48 of 55
NLNG SIX PROJECT

Appendix - 20

Valve Tag - 520-HZ-814

Input Data:

P1 27.61 bara
P2 2.1 bara
xT 0.813 (@ 100% valve travel for a 4" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 4
d2 16
D 6
D2 36
Cv 112 (@ 100% valve travel for a 4" valve size)
Cv2 12544
k 1.13
g 40.79 Kg/m3
W 3.99 Kg/s
14364 Kg/hr

Calculations: Control Point

0.5
Solving for equations 4, 5 & 6 1
2
From eqn. 4, K1 = 0.5*(1-(64/100))^2 4

K1 0.1543

From eqn. 5, K2 = 1.0*(1-(64/100))^2

K2 0.3086

From eqn.6, KB1 or KB2 = 1-(8/10)^4

KB1 or KB2 = 0.8025

From eqn.3, K = 0.0648+0.1296+KB1-KB2, (KB1 = KB2)

K = 0.4630

From eqn.2, Fp = (1+(0.1944/890)*(818/64)^2)^-0.5

Fp = 0.9875

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.9239

Solving from equations 13 to 17 and substituting into 18 results in:

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 20

Valve Tag - 520-HZ-814

Ki = 0.9568

A 0.00077787

B = 7
B =
2
49.00

C = 1.0381

E= 0.8337

XTP = 0.8031

Calculating Fk in eqn. 10,

Fk = 0.8071

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.5249

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 14.150

J = 32.258

Cv = W/(H*J) 31.47

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 21

Valve Tag - 47-PIC-077

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 21

Valve Tag - 47-PIC-077

Input Data:

P1 8 bara
P2 7.5 bara
xT 0.667 (@ 100% valve travel for a 1" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 2
d2 4
D 2
D2 4
Cv 17.2 (@ 100% valve travel for a 1" valve size)
Cv2 295.84
k 1.4
g 8.66 Kg/m3
W 0.051 Kg/s
183.6 Kg/hr

Calculations:
Control Point
Solving for equations 4, 5 & 6 0.5
1
From eqn. 4, K1 = 0.5*(1-(1/1))^2 2
K1 0.0000 4

From eqn. 5, K2 = 1.0*(1-(1/1))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(1/1)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0/890)*(17.2/1)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.0625

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.0000

Document: NE-CSU-PRO-F-6053
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NLNG SIX PROJECT

Appendix - 21

Valve Tag - 47-PIC-077

A 0

B = 4.3
B2 = 18.49

C = 1.0000

E= 0.6670

XTP = 0.6670

Calculating Fk in eqn. 10,

Fk = 1.0000

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9688

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 26.447

J = 2.081

Cv = W/(H*J) 3.336

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
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NLNG SIX PROJECT

Appendix - 22

Valve Tag - 47-PIC-078

Sizing Equations:

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5)) [1]

Fp = [1+K/N2(Cv/d2)^2]^-0.5 [2]

K = K1+K2+KB1-KB2 [3]

K1 = 0.5(1-d2/D2)^2 [4]

K2 = 1.0(1-d2/D2)^2 [5]

KB1 or KB2 = 1-(d/D)^4 [6]

Y = 1-(x/(3Fk*xTP)) [7]

x = dP/P1 [8]

dP = P1-P2 [9]

Therefore, x = (P1-P2)/P1 [10]

Fk = k/1.40 [11]

XTp = (xT/FP2)*(1+(xTKi/N5)*(Cv/d2)2)^-1 [12]

Ki = K1 + KB1 [13]

Let xTKi/N5 = A [14]

Let Cv/d2 = B [15]

Let (1+((A)*(B)2)) = C [16]

Let (xT/FP2) = E [17]

Substituting equations 16 & 17 into 12 results into:

XTp = E*(C)^-1 [18]

Let N6*Fp*Y = H [19]

Let (x*P1*g)^0.5 = J [20]

Therefore, Cv = W/(H*J) [21]

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06
NIGERIA LNG LTD
Page 54 of 55
NLNG SIX PROJECT

Appendix - 22

Valve Tag - 47-PIC-078

Input Data:

P1 8 bara
P2 7.5 bara
xT 0.667 (@ 100% valve travel for a 1" valve size)
N2 890 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N5 1000 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
N6 27.3 (Ref. 1 Catalog 12 Section 2 tab pg. 2-3, Equation constants)
d 2
d2 4
D 2
D2 4
Cv 17.2 (@ 100% valve travel for a 1" valve size)
Cv2 295.84
k 1.4
g 8.66 Kg/m3
W 0.051 Kg/s
183.6 Kg/hr

Calculations:
Control Point
Solving for equations 4, 5 & 6 0.5
1
From eqn. 4, K1 = 0.5*(1-(1/1))^2 2
K1 0.0000 4

From eqn. 5, K2 = 1.0*(1-(1/1))^2

K2 0.0000

From eqn.6, KB1 or KB2 = 1-(1/1)^4

KB1 or KB2 = 0.0000

From eqn.3, K = 0+0+KB1-KB2, (KB1 = KB2)

K = 0.0000

From eqn.2, Fp = (1+(0/890)*(17.2/1)^2)^-0.5

Fp = 1.0000

From eqn.8, x = dP/P1

Therefore x = (P1-P2)/P1
x= 0.0625

Solving from equations 13 to 17 and substituting into 18 results in:

Ki = 0.0000

Document: NE-CSU-PRO-F-6053
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Date :23/11/06
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NLNG SIX PROJECT

Appendix - 22

Valve Tag - 47-PIC-078

A 0

B = 4.3
B2 = 18.49

C = 1.0000

E= 0.6670

XTP = 0.6670

Calculating Fk in eqn. 10,

Fk = 1.0000

Substituting the value of Fk, xTP, x into eqn. 7 results in:

Y = 0.9688

Cv = (W/((N6*Fp*Y)*(x*P1*g)^0.5))

H = 26.447

J = 2.081

Cv = W/(H*J) 3.336

Document: NE-CSU-PRO-F-6053
NLNG SIX PROJECT file:///conversion/tmp/scratch/359804461.xls Revision : C
Date :23/11/06