{"id":20436,"date":"2018-12-02T14:03:21","date_gmt":"2018-12-02T14:03:21","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=20436"},"modified":"2023-02-15T08:12:57","modified_gmt":"2023-02-15T08:12:57","slug":"example-convection-problem-with-solution","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-engineering\/heat-transfer\/convection-convective-heat-transfer\/example-convection-problem-with-solution\/","title":{"rendered":"Example – Convection – Problem with Solution"},"content":{"rendered":"
<\/a><\/strong><\/p>\n Cladding<\/strong> is the outer layer of the fuel rods, standing between the reactor coolant<\/strong> and the nuclear fuel<\/strong> <\/a>(i.e., fuel pellets<\/strong>). It is made of corrosion-resistant material with a low absorption cross-section for thermal neutrons<\/a>, usually zirconium alloy<\/strong>. Cladding<\/strong> prevents radioactive fission products from escaping the fuel matrix into the reactor coolant and contaminating it. Cladding constitutes one of the barriers in the \u2018defense-in-depth <\/b>\u2018 approach; therefore, its coolability<\/strong> is one of the key safety aspects.<\/p>\n Consider the fuel cladding of inner radius r<\/strong>Zr,2<\/sub><\/strong> = 0.408 cm<\/strong> and outer radius r<\/strong>Zr,1<\/sub><\/strong> = 0.465 cm<\/strong>. Compared to fuel pellet, there is almost no heat generation in the fuel cladding (cladding is slightly heated by radiation<\/a>). All heat generated in the fuel must be transferred via conduction<\/strong><\/a> through the cladding, and therefore the inner surface is hotter than the outer surface.<\/p>\n Assume that:<\/p>\n <\/a>Calculate the Prandtl<\/a>, Reynolds<\/a> and Nusselt number for this flow regime (internal forced turbulent flow) inside the rectangular fuel lattice (fuel channel), then calculate the heat transfer coefficient<\/strong> and finally the cladding surface temperature<\/strong>, TZr,1<\/sub><\/strong>.<\/p>\n To calculate the cladding surface temperature<\/strong>, we have to calculate the Prandtl<\/strong>, Reynolds<\/strong> and Nusselt number<\/strong>, because the heat transfer for this flow regime can be described by the Dittus-Boelter equation<\/strong>, which is:<\/p>\n <\/a><\/p><\/div><\/div> To calculate the Prandtl number<\/a>, we have to know:<\/p>\n Note that, all these parameters significantly differs for water at 300\u00b0C from those at 20\u00b0C. \u00a0Prandtl number for water<\/a> at 20\u00b0C is around 6.91. <\/strong>Prandtl number for reactor coolant at 300\u00b0C is then:<\/p>\n <\/a><\/p><\/div><\/div> To calculate the Reynolds number, we have to know:<\/p>\n The hydraulic diameter, D<\/strong>h<\/sub><\/strong>, is a commonly used term when handling flow in non-circular tubes and channels<\/strong>. The hydraulic diameter of the fuel channel<\/strong>, D<\/em>h<\/sub><\/em>, is equal to 13,85 mm.<\/p>\n See also: Hydraulic Diameter<\/a><\/p>\n The Reynolds number <\/strong>inside the fuel channel is then equal to:<\/p>\n <\/a><\/p>\n This fully satisfies the turbulent conditions<\/strong><\/a>.<\/p><\/div><\/div> For fully developed (hydrodynamically and thermally) turbulent flow in a smooth circular tube, the local Nusselt number<\/strong> may be obtained from the well-known Dittus\u0096Boelter equation<\/strong>.<\/p>\n To calculate the Nusselt number<\/strong>, we have to know:<\/p>\n The Nusselt number <\/strong>for the forced convection inside the fuel channel is then equal to:<\/p>\n <\/a><\/p><\/div><\/div> Detailed knowledge of geometry, fluid parameters, the outer radius of cladding, linear heat rate, convective heat transfer coefficient allows us to calculate the temperature difference \u2206T <\/strong>between the coolant (Tbulk<\/sub>) and the cladding surface (TZr,1<\/sub>).<\/p>\n To calculate the cladding surface temperature, we have to know:<\/p>\n The convective heat transfer coefficient, h<\/strong>, is given directly by the definition of Nusselt number:<\/p>\n <\/a><\/p>\n Finally, we can calculate the cladding surface temperature (TZr,1<\/sub>) simply using Newton\u2019s Law of Cooling<\/strong>:<\/p>\n <\/a><\/p>\n For PWRs at normal operation, there is compressed liquid water<\/a> inside the reactor core, loops, and steam generators. The pressure is maintained at approximately 16MPa<\/strong>. At this pressure, water boils at approximately 350\u00b0C<\/strong>(662\u00b0F). As can be seen, the surface temperature TZr,1<\/sub> = 325\u00b0C ensures that even subcooled boiling does not occur. Note that subcooled boiling requires TZr,1<\/sub> = Tsat<\/sub>. Since the inlet temperatures of the water are usually about 290\u00b0C<\/strong> (554\u00b0F), it is obvious this example corresponds to the lower part of the core. At higher core elevations, the bulk temperature may reach up to 330\u00b0C. The temperature difference of 29\u00b0C causes the subcooled boiling may occur (330\u00b0C + 29\u00b0C > 350\u00b0C). On the other hand, nucleate boiling<\/strong> at the surface effectively disrupts the stagnant layer. Therefore, nucleate boiling significantly increases the ability of a surface to transfer thermal energy<\/a> to the bulk fluid. As a result, the convective heat transfer coefficient significantly increases, and therefore at higher elevations, the temperature difference (TZr,1<\/sub> – Tbulk<\/sub>) significantly decreases.<\/p><\/div><\/div>\n
Calculation of the Prandtl number<\/h2>\n
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Calculation of the Reynolds number<\/h2>\n
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Calculation of the Nusselt number using Dittus-Boelter equation<\/h2>\n
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Calculation of the heat transfer coefficient and the cladding surface temperature, TZr,1<\/sub><\/h2>\n
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