Example – Convection – Problem with Solution

Example – Convection – Cladding Surface Temperature

Convection - Convective Heat Transfer

Example – Convection – Problem with Solution 

Cladding is the outer layer of the fuel rods, standing between the reactor coolant and the nuclear fuel (i.e. fuel pellets). It is made of a corrosion-resistant material with low absorption cross section for thermal neutrons, usually zirconium alloy. Cladding prevents radioactive fission products from escaping the fuel matrix into the reactor coolant and contaminating it. Cladding constitute one of barriers in ‘defence-in-depth‘ approach, therefore its coolability is one of key safety aspects.

Consider the fuel cladding of inner radius rZr,2 = 0.408 cm and outer radius rZr,1 = 0.465 cm. In comparison to fuel pellet, there is almost no heat generation in the fuel cladding (cladding is slightly heated by radiation). All heat generated in the fuel must be transferred via conduction through the cladding and therefore the inner surface is hotter than the outer surface.

Assume that:

  • the outer diameter of the cladding is: d = 2 x rZr,1 = 9,3 mm
  • the pitch of fuel pins is: p = 13 mm
  • the thermal conductivity of saturated water at 300°C is: kH2O = 0.545 W/m.K
  • the dynamic viscosity of saturated water at 300°C is: μ = 0.0000859 N.s/m2
  • the fluid density is: ρ = 714 kg/m3
  • the specific heat is: cp = 5.65 kJ/kg.K
  • the core flow velocity is constant and equal to Vcore = 5 m/s
  • the temperature of reactor coolant at this axial coordinate is: Tbulk = 296°C
  • the linear heat rate of the fuel is qL = 300 W/cm (FQ ≈ 2.0) and thus the volumetric heat rate is qV = 597 x 106 W/m3

Hydraulic Diameter - Fuel ChannelCalculate the Prandtl, Reynolds and Nusselt number for this flow regime (internal forced turbulent flow) inside the rectangular fuel lattice (fuel channel), then calculate the heat transfer coefficient and finally the cladding surface temperature, TZr,1.

To calculate the cladding surface temperature, we have to calculate the Prandtl, Reynolds and Nusselt number, because the heat transfer for this flow regime can be described by the Dittus-Boelter equation, which is:

Dittus-Boelter Equation - Formula

Calculation of the Prandtl number

To calculate the Prandtl number, we have to know:

  • the thermal conductivity of saturated water at 300°C is: kH2O = 0.545 W/m.K
  • the dynamic viscosity of saturated water at 300°C is: μ = 0.0000859 N.s/m2
  • the specific heat is: cp = 5.65 kJ/kg.K

Note that, all these parameters significantly differs for water at 300°C from those at 20°C.  Prandtl number for water at 20°C is around 6.91. Prandtl number for reactor coolant at 300°C is then:

prandtl number - example

Calculation of the Reynolds number

To calculate the Reynolds number, we have to know:

  • the outer diameter of the cladding is: d = 2 x rZr,1 = 9,3 mm (to calculate the hydraulic diameter)
  • the pitch of fuel pins is: p = 13 mm  (to calculate the hydraulic diameter)
  • the dynamic viscosity of saturated water at 300°C is: μ = 0.0000859 N.s/m2
  • the fluid density is: ρ = 714 kg/m3

The hydraulic diameter, Dh, is a commonly used term when handling flow in non-circular tubes and channels. The hydraulic diameter of the fuel channel, Dh, is equal to 13,85 mm.

See also: Hydraulic Diameter

The Reynolds number inside the fuel channel is then equal to:

reynolds number - example

This fully satisfies the turbulent conditions.

Calculation of the Nusselt number using Dittus-Boelter equation

For fully developed (hydrodynamically and thermally) turbulent flow in a smooth circular tube, the local Nusselt number may be obtained from the well-known Dittus–Boelter equation.

To calculate the Nusselt number, we have to know:

The Nusselt number for the forced convection inside the fuel channel is then equal to:

nusselt number - example

Calculation of the heat transfer coefficient and the cladding surface temperature, TZr,1

Detailed knowledge of geometry, fluid parameters, outer radius of cladding, linear heat rate, convective heat transfer coefficient allows us to calculate the temperature difference ∆T between the coolant (Tbulk) and the cladding surface (TZr,1).

To calculate the the cladding surface temperature, we have to know:

  • the outer diameter of the cladding is: d = 2 x rZr,1 = 9,3 mm
  • the Nusselt number, which is NuDh = 890
  • the hydraulic diameter of the fuel channel is: Dh = 13,85 mm
  • the thermal conductivity of reactor coolant (300°C) is: kH2O = 0.545 W/m.K
  • the bulk temperature of reactor coolant at this axial coordinate is: Tbulk = 296°C
  • the linear heat rate of the fuel is: qL = 300 W/cm (FQ ≈ 2.0)

The convective heat transfer coefficient, h, is given directly by the definition of Nusselt number:

convective heat transfer coefficient - example

Finally, we can calculate the cladding surface temperature (TZr,1) simply using the Newton’s Law of Cooling:

Newton law of cooling - example

For PWRs at normal operation, there is a compressed liquid water inside the reactor core, loops and steam generators.  The pressure is maintained at approximately 16MPa. At this pressure water boils at approximately 350°C(662°F). As can be seen, the surface temperature TZr,1 = 325°C ensures, that even subcooled boiling does not occur. Note that, subcooled boiling requires TZr,1 = Tsat. Since the inlet temperatures of the water are usually about 290°C (554°F), it is obvious this example corresponds to the lower part of the core. At higher elevations of the core the bulk temperature may reach up to 330°C. The temperature difference of 29°C causes the subcooled boiling may occur (330°C + 29°C > 350°C). On the other hand, nucleate boiling at the surface effectively disrupts the stagnant layer and therefore nucleate boiling significantly increases the ability of a surface to transfer thermal energy to bulk fluid. As a result, the convective heat transfer coefficient significantly increases and therefore at higher elevations, the temperature difference (TZr,1 – Tbulk) significantly decreases.

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. Fundamentals of Heat and Mass Transfer. C. P. Kothandaraman. New Age International, 2006, ISBN: 9788122417722.
  4. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2. 
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See above:

Convection