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Derivation Of Nusselt Number For Flow Of Liquid Metal Over An Isothermal Flat Plate
- 1. Derivation Of Nusselt Number For Flow Of Liquid Metal Over An Isothermal Flat Plate Consider the 2D laminar boundary-layer, stedy and incompressible flow Let the free-stream temperature T∞ be constant. fluid has constant thermophysical properties. velocity distribution becomes independent of the temperature distribution when the properties are constant. thermal energy generation due to viscous dissipation is negligible. As in the use of the momentum integral equation and the energy integral equation for an approximation is third order polynomial for velocity and temperature profile is becomes temperature profile is 𝜃 𝜃∞ = 𝑇−𝑇 𝑤 𝑇∞−𝑇 𝑤 = 3𝑦 2𝛿 − 1 2 𝑦 𝛿 3 ------(1) velocity profile is, 𝑢 𝑈∞ = 3𝑦 2𝛿 − 1 2 𝑦 𝛿 3 -- -------(2)
- 2. Hydrodynamic and thermal boundary layers for flow of liquid metal over on a flat plate Flat plate kept at a constant temperature Tw over its entire length. The thermal boundary- layer thickness δT is thicker than the velocity boundary-layer thickness δ.
- 3. The energy integral Equation is 𝑑 𝑑𝑥 0 𝛿 𝑇 𝑇∞ − 𝑇 𝑢𝑑𝑦 = 𝛼( 𝜕𝑇 𝜕𝑦 ) 𝑦=0 When the thermal boundary layer is thicker than the velocity boundary layer, the integral in Equation above must be split up into two parts: one from y = 0 to y = δ and the other from y = δ to y = δT, that is, 𝑑 𝑑𝑥 [ 0 𝛿. 𝑇∞ − 𝑇 𝑢𝑑𝑦 + 𝑈∞ 𝛿 𝛿 𝑇 𝑇∞ − 𝑇 𝑑𝑦] = 𝛼( 𝜕𝑇 𝜕𝑦 ) 𝑦=0 Put value of u and T from equation number (1) &(2) Then we get 𝑈∞ 𝑇∞ − 𝑇 𝑤 𝑑 𝑑𝑥 0 𝛿 1 − 3 2 𝑦 𝛿 𝑇 + 1 2 𝑦 𝛿 𝑇 3 3𝑦 2𝛿 − 1 2 𝑦 𝛿 3 𝑑𝑦 + 𝛿 𝛿 𝑇 1 − 3 2 𝑦 𝛿 𝑇 + 1 2 𝑦 𝛿 𝑇 3 𝑑𝑦 = 𝛼( 𝜕𝑇 𝜕𝑦 ) 𝑦=0
- 4. 𝑈∞ 𝑇∞ − 𝑇 𝑤 𝑑 𝑑𝑥 3 2𝛿 𝑦2 2 − 1 2𝛿3 𝑦4 4 − 9 4𝛿𝛿 𝑇 𝑦3 3 + 3 4𝛿3 𝛿 𝑇 𝑦5 5 + 3 4𝛿𝛿 𝑇 3 𝑦5 5 −
- 5. For liquid metal have 𝑃𝑟 ≪ 1 and also TBL ≫ VBL 𝛿 𝛿 𝑇 ≪ 1 since we can neglect terms which have 𝛿 𝛿 𝑇 Then we get 𝑈∞ 𝑇∞ − 𝑇 𝑤 𝑑 𝑑𝑥 5 8 𝛿 + 3 8 𝛿 𝑇 − 𝛿 =𝛼( 𝜕𝑇 𝜕𝑦 ). 𝑦=0 𝑈∞ 𝑇∞ − 𝑇 𝑤 3 8 𝑑 𝑑𝑥 𝛿 𝑇 − 𝛿 =𝛼(𝑇 𝑤-𝑇∞) 3 2𝛿 𝑇 𝑑 𝑑𝑥 𝛿 𝑇(1 − 𝛿 𝛿 𝑇 ) = 𝛼4 𝑈∞ 𝛿 𝑇 put 𝛿 𝑇=𝑍 ∗ 𝛿 Z 𝛿 𝑑 𝑑𝑥 (Z 𝛿) = 4𝛼 𝑈∞
- 6. . Z 𝛿 𝑑 𝑑𝑥 (Z 𝛿) = 4𝛼 𝑈∞ By product rule 𝑍2 𝛿 𝑑 𝑑𝑥 (𝛿) +Z𝛿2 𝑑𝑍 𝑑𝑥 = 4𝛼 𝑈∞ put 𝛿 𝑑 𝑑𝑥 (𝛿) = 140𝜗 13𝑈∞ and 𝛿2 = 280𝜗𝑥 13𝑈∞ 𝑍2 140 13 𝜗 𝑈∞ +Z 280 13 𝜗𝑥 𝑈∞ 𝑑𝑍 𝑑𝑥 = 4𝛼 𝑈∞ 140 13 (𝑍2+2Zx 𝑑𝑍 𝑑𝑥 ) = 4𝛼 𝜗 𝑍2 +x 𝑑𝑍2 𝑑𝑥 = 13∗4𝛼 140𝑥𝜗 𝑑𝑍2 𝑑𝑥 + 𝑍2 𝑥 = 13𝛼 35𝑥𝜗 Here 𝑃𝑟= 𝜗 𝛼
- 7. It is in the form of 𝑑𝑦 𝑑𝑥 + Py =Q IF=𝑒 𝑝𝑑𝑥 = 𝑒 1 𝑥 𝑑𝑥 = x Therefore solution is in the form of y *IF = 𝑄 ∗ 𝐼𝐹 𝑍2 ∗ x = 13 35 1 𝑃𝑟 𝑥 ∗ 𝑥
- 8. After integration we get 𝑍2 = 0.3714 𝑃𝑟 That is Z= 𝛿 𝑇 𝑥 𝛿 𝑥 = 0.609449 𝑃𝑟 .5 We know the local heat flux per unit area is From newton’s law of cooling From above two equation we get,
- 9. Nusselt number is, Nu= ℎ 𝑥 𝑘 = 1.5 𝛿 𝑇 𝑥 Nu= 1.5 𝛿 𝑥 0.609449 𝑃 𝑟 .5 Here we know 𝛿 𝑥 = 4.64 𝑅 𝑒 .5 Nu=0.5314 𝑃𝑟 .5 𝑅 𝑒 .5