Integrated methodology for heavy-duty gas turbine modeling, performance simulation and verification

(1)

• • • •

(2)

• • • • • •

(3)

𝜂𝑖𝑠 𝑐 = (ℎ_{𝑐 𝑜𝑢𝑡 𝑖𝑠}− ℎ_{𝑐 𝑖𝑛}) (ℎ𝑐 𝑜𝑢𝑡− ℎ𝑐 𝑖𝑛) ℎ_{𝑐 𝑜𝑢𝑡 𝑖𝑠} ℎ𝑐 𝑜𝑢𝑡 𝑖𝑠= ℎ(𝑠𝑐 𝑖𝑛, 𝑝𝑐 𝑜𝑢𝑡) ℎ_{𝑐 𝑜𝑢𝑡} = ℎ(𝑇_{𝑐 𝑜𝑢𝑡}, 𝑝_{𝑐 𝑜𝑢𝑡}) 𝑚̇𝑐 𝑖𝑛 =𝑃𝐺𝑇 𝑒𝑙+ 𝑃𝑙𝑜𝑠𝑠+ 𝑚̇𝑓∙ (ℎ𝑡 𝑜𝑢𝑡− (𝐿𝐻𝑉 + ℎ𝑓) ∙ 𝜂𝐶𝐶) − ∑ 𝑚̇𝑙𝑒𝑎𝑘∙ (ℎ𝑙𝑒𝑎𝑘− ℎ𝑡 𝑜𝑢𝑡) ℎ𝑐 𝑖𝑛− ℎ𝑡 𝑜𝑢𝑡 o 𝑚̇_𝑓 o 𝑃𝐺𝑇 𝑒𝑙 o 𝑃_{𝑙𝑜𝑠𝑠} o ℎ𝑐 𝑖𝑛 ℎ𝑡 𝑜𝑢𝑡 o 𝑚̇_{𝑙𝑒𝑎𝑘} o ℎ_{𝑙𝑒𝑎𝑘}

(4)

ℎ(𝑇𝐼𝑆𝑂) =𝑚̇𝑐 𝑖𝑛∙ ℎ𝑐 𝑖𝑛+ 𝑚̇𝑓∙ (𝐿𝐻𝑉 + ℎ𝑓) ∙ 𝜂𝐶𝐶 + 𝑃𝑐 𝑚̇_{𝑐 𝑖𝑛}+ 𝑚̇_𝑓 o ℎ(𝑇𝐼𝑆𝑂) o 𝑚̇_{𝑐 𝑖𝑛} o ℎ𝑐 𝑜𝑢𝑡 o 𝑚̇_𝑓 o 𝑃𝑐 𝑚̇_𝑒𝑞 𝑃𝑐 = (𝑚̇𝑐 𝑖𝑛− 𝑚̇𝑒𝑞) ∙ (ℎ𝑐 𝑜𝑢𝑡− ℎ𝑐 𝑖𝑛) 𝜂_𝑡 = ℎ(𝑇𝐼𝑇) − ℎ(𝑇𝐸𝑇) ℎ(𝑇𝐼𝑇) − ℎ(𝑇𝐸𝑇, 𝑖𝑠)

(5)

• • • • • • • • • • • • • •

(6)

• 𝑂2 = 2𝑂 𝑁₂+ 𝑂 = 𝑁𝑂 + 𝑁 𝑁 + 𝑂2 = 𝑁𝑂 + 𝑂 𝑁 + 𝑂𝐻 = 𝑁𝑂 + 𝐻 •

(7)

• • •

(8)

(9)

• • • • • • • • •

(10)

(11)

𝑚̇_𝐵𝑂 =𝑚̇𝑏𝑙 𝑜𝑢𝑡∙ (ℎ𝑓𝑙 𝑜𝑢𝑡− ℎ𝑏𝑙 𝑜𝑢𝑡) ℎ_{𝑏𝑙 𝑜𝑢𝑡}− ℎ_{𝑓𝑙 𝑜𝑢𝑡}

• •

(12)

• • •

(13)

• •

(14)

(15)

𝜂_{𝑡 𝑓𝑙 𝑚𝑖𝑥} = ℎ(𝑇𝐼𝑇) − ℎ(𝑇𝑏𝑙 𝑜𝑢𝑡, 𝑝𝑓𝑙 𝑜𝑢𝑡) ℎ(𝑇𝐼𝑇) − ℎ(𝑇_{𝑏𝑙 𝑜𝑢𝑡}, 𝑝_{𝑓𝑙 𝑜𝑢𝑡}, 𝑖𝑠)

(16)

• • •

(17)

• • • •

(18)

(19)

• • • • • • • • • • •

(20)

(21)

(22)

(23)

• • •

(24)

(25)

(26)

• • • •

(27)

•

(28)

(29)

(30)

•

(31)

(32)

(33)

(34)

• ⁓

(35)

(36)

(37)

(38)

(39)

(40)

(41)

- - 𝑝𝑣 - - 𝑝1 𝑇1 - 𝑝₄ - 𝐷 - 𝐷 / 𝐷0 - 𝐿₁ 𝐿₂ 𝑦 𝐶𝑉∗_{= 𝑓(𝑦)}

(42)

𝑋𝑡 = 𝑓(𝑦) 𝐶𝐺∗= 40 ∙ 𝐶𝑉∗∙ √𝑋𝑡 𝐶𝑉∗ _𝐶𝐺∗ 𝐶𝐺 = 4.02195 ∙ 10−7∙ 𝐶𝐺∗ 𝐶𝑉 = 2.3837 ∙ 10−7_{∙ 𝐶𝑉}∗ 𝐾𝑣 = 𝜋2 _{∙ 𝐷}4 8 ∙ 𝐶𝑉2 𝜈1= 𝑅𝑔𝑎𝑠∙ 𝑇1/𝑝1 𝐴 =𝜋 ∙ 𝐷 2 4 𝐾𝑡1= 4 ∙ 𝑓 ∙ 𝐿1 𝐷 + 𝐾𝑐𝑜𝑛+ 𝐾𝑏𝑓1 𝐾𝑡2 = 4 ∙ 𝑓 ∙ 𝐿2 𝐷 + 𝐾𝑏𝑓2

(43)

𝐾𝑐𝑜𝑛 = −2.604 ∙ ( 𝐷 𝐷0 )5 + 7.552 ∙ (𝐷 𝐷0 )4− 6.875 ∙ (𝐷 𝐷0 )3 + 1.823 ∙ (𝐷 𝐷0 )2 − 0.3958 ∙ (𝐷 𝐷0 ) + 0.5 𝐾_𝑏𝑓1 𝐾_𝑏𝑓2 𝐾_{𝑐𝑜𝑛𝑐} 𝐾_{𝑐𝑜𝑛𝑐} 𝐶𝐿1 = 𝐴 ∙ √ 2 𝐾𝑡1

(44)

ln(𝐾𝑡) 𝑝3𝑐 𝑝1 = 𝜑0+ 𝜑1∙ ln(𝐾𝑡) + 𝜑2∙ [ln (𝐾𝑡)]2+ 𝜑3∙ [ln (𝐾𝑡)]3+ 𝜑4∙ [ln (𝐾𝑡)]4 𝑏₀ 𝑚_𝑏 𝑏0(𝐾𝑡, 0.95) = 𝜎0+ 𝜎1∙ ln(𝐾𝑡) + 𝜎2∙ [ln (𝐾𝑡)]2+ 𝜎3∙ [ln (𝐾𝑡)]3+ 𝜎4 ∙ [ln (𝐾𝑡)]4 𝑚𝑏(𝐾𝑡) = 𝜓0+ 𝜓1∙ ln(𝐾𝑡) + 𝜓2∙ [ln(𝐾𝑡)]2+ 𝜓3∙ [ln(𝐾𝑡)]3+ 𝜓4∙ [ln(𝐾𝑡)]4 + 𝜓5∙ [ln (𝐾𝑡)]5 𝜑_𝑖 𝛽_𝑖 𝜓_𝑖 𝛾 = 𝐾𝑡 ≤ 1 1 < 𝐾𝑡≤ 2048 𝐾_𝑡 𝛾 = 1.4 𝑝_𝑣 𝑏_0,1 𝑏0,1(𝐾𝑡1, 0.95) = 𝜎0+ 𝜎1∙ ln(𝐾𝑡1) + 𝜎2∙ [ln (𝐾𝑡1)]2+ 𝜎3∙ [ln (𝐾𝑡1)]3+ 𝜎4 ∙ [ln (𝐾𝑡1)]4 𝑚𝑏,1(𝐾𝑡1) = 𝜓0+ 𝜓1∙ ln(𝐾𝑡1) + 𝜓2∙ [ln(𝐾𝑡1)]2+ 𝜓3∙ [ln(𝐾𝑡1)]3+ 𝜓4 ∙ [ln(𝐾𝑡1)]4+ 𝜓5∙ [ln (𝐾𝑡1)]5 𝜎_𝑖 𝜓_𝑖 𝐾𝑡 𝐾𝑡1 𝐾 ≤ 1 1 < 𝐾 ≤ 2048

(45)

𝑣_𝑎𝑣𝑔= 𝑣₁∙𝛾 + 1 𝛾 ∙ (1 − 𝑝3 𝑝1 ) / [1 − (𝑝3 𝑝1 ) 𝛾+1 𝛾 ] 𝑣𝑎𝑣𝑔 𝑏0 = 𝑏0(𝐾𝑡, 0.95) + 𝑚𝑏(𝐾𝑡) ∙ ( 𝑝₃ 𝑝1 − 0.95) 1 √𝜆𝐷 + 2𝑙𝑜𝑔10( 1 3.7 𝜀 𝐷+ 1.256 𝑅𝑒√𝜆𝐷 ) = 0 𝜆𝐷= 4 ∙ 𝑓



𝑓 = 0.0045 𝑣̂ = 𝑣_{𝑎𝑣𝑔,1} _𝑎𝑣𝑔 𝑏̂ = 𝑏_0,1 ₀ (𝑝𝑣 𝑝1 ) ̂ = 1 −𝑣̂𝑎𝑣𝑔,1 𝑣𝑎𝑣𝑔 ∙ (𝑏0 𝑏̂_0,1) 2 ∙ (𝐶𝑡 𝐶𝐿1 ) ∙ (1 −𝑝3 𝑝1 )

(46)

𝑣̂ = 𝑣_{𝑎𝑣𝑔,1} ₁∙𝛾 + 1 𝛾 ∙ [1 − ( 𝑝𝑣 𝑝₁) ̂ ] / [1 − (𝑝𝑣 𝑝₁) ̂𝛾+1𝛾 ] 𝑏̂ = 𝑏0,1 0(𝐾𝑡1, 0.95) + 𝑚𝑏(𝐾𝑡1) ∙ [( 𝑝_𝑣 𝑝1 ) ̂ − 0.95] 𝑝_𝑣⁄𝑝₁ 𝜈𝑣 = 𝜈1∙ (1/ 𝑝_𝑣 𝑝1 ) 1/𝛾 𝑚̇_{𝑙𝑜𝑛𝑔 𝑝𝑖𝑝𝑒} = 𝑏₀∙ 𝐶_𝑡∙ √𝑝1∙ √ 1 − 𝑝3/𝑝1 𝜈_𝑎𝑣𝑔 𝑚̇_{𝑎𝑐𝑡𝑢𝑎𝑙} = 𝑚𝑖𝑛(𝑚̇_{𝑠𝑜𝑛𝑖𝑐} , 𝑚̇_{𝑙𝑜𝑛𝑔 𝑝𝑖𝑝𝑒}) {_{𝑖𝑓 𝑚̇}𝑖𝑓 𝑝3𝑐 > 𝑝4 → 𝑐ℎ𝑜𝑘𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑠𝑜𝑛𝑖𝑐 > 𝑚̇𝑙𝑜𝑛𝑔 𝑝𝑖𝑝𝑒 → 𝑐ℎ𝑜𝑘𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑣𝑒} 𝑅𝑒 =𝑚̇𝑎𝑐𝑡𝑢𝑎𝑙∙ 𝐷 𝜇 ∙ 𝐴

(47)

𝜇 =(1.458 ∙ 10 −6_{) ∙ 𝑇} 13/2 𝑇₁+ 110.4 [ 𝑘𝑔 𝑚𝑠] 𝑇1



(48)

(49)

(50)

(51)

(52)

(53)

(54)

(55)