Heat Transfer Boundary Conditions

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SUMMARY

The discussion focuses on deriving the temperature profile for a high-temperature gas-cooled nuclear reactor with a composite cylindrical wall consisting of thorium fuel (thermal conductivity Ka=57 W/m*K) and graphite (Kb=3 W/m*K). The helium coolant at 600 K and a convection coefficient of h=2000 W/m^2*K are critical parameters. The governing equation for heat transfer is established as 1/r d/dr (r dT/dr) + q/k = 0, where q=10^8 W/m^3 represents the uniform thermal energy generation in the fuel element. The boundary conditions applied include dT/dr=0 at the center (r=0) and T(r0)=Ts at the outer surface of the fuel element.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction and convection.
  • Familiarity with cylindrical coordinates in thermal analysis.
  • Knowledge of thermal conductivity values for materials (e.g., thorium and graphite).
  • Ability to apply boundary conditions in differential equations.
NEXT STEPS
  • Study the derivation of temperature profiles in cylindrical coordinates using the heat conduction equation.
  • Learn about thermal boundary layers and their impact on heat transfer in gas-cooled reactors.
  • Explore numerical methods for solving heat transfer problems in composite materials.
  • Investigate the effects of varying convection coefficients on thermal performance in nuclear reactors.
USEFUL FOR

Engineers and physicists involved in nuclear reactor design, thermal management specialists, and students studying heat transfer in engineering applications.

luca131
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Homework Statement


A high temperature, gas cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element (Ka=57 W/m*K) is encased in graphite (Kb= 3 W/m*K) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is T=600k and the convection coefficient at the outer surface of the graphite is h= 2000 W/m^2*K. If thermal energy is uniformly generated in the fuel elemnt at a rate of q=10^8 W/m^3 what are the temperaturesT1 and T2 at the inner and outer surfaces, respectively, of the fuel element?

Homework Equations


1/r d/dr (r dT/dr)+q/k=0

The Attempt at a Solution


Not really sure how to derive this formula. The boundary conditions I have chosen was
at r=0 dT/dr=0 because the max temperature should be in the center. My second boundary condition is the T(r0)= Ts. If someone could help derive this formula that would be great been trying to derive the temperature profile for sometime now.
If someone could so me a crude derivation of the temperature profile that would be great.
 
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If you do a heat balance on the section of one of the layers between r and r + delta r, you obtain: $$\left[-2\pi r k\frac{d T}{dr}\right]_{r+dr}-\left[-2\pi r k\frac{dT}{d r}\right]_{r}=(2\pi r dr)q$$or$$\frac{1}{r}\frac{d}{dr}\left(rk\frac{dT}{dr}\right)=-q$$Heat is generated only in the thorium fuel element, and the temperature of the surrounding helium is 600 K, although this is not the outer temperature of the graphite (because of the thermal boundary layer between the outer surface and the bulk helium).
 

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