Introduction to Nuclear System - Chapter 2-2

Introduction to Nuclear System and Operation - Chapter 2-2

• This chapter focuses on heat removal from nuclear reactors using conduction.
• Conduction is heat transfer caused by temperature differences, without macroscopic matter movement.
• Convection is heat transfer due to temperature differences involving moving fluids (liquid or gas).
• Radiation is heat transfer via thermal radiation.
• Fourier's law describes heat conduction: q" = -k∇T, where q" is heat flux, k is thermal conductivity, and ∇T is the temperature gradient.
• This law is similar to Fick's law, comparing the mechanisms of heat and mass transfer.

Energy Conservation

• In a unit volume, the net rate of heat flow out equals the heat production rate (steady state).
• The equation derived is ∇ ⋅ q" + q" = 0.
• This is the steady state heat conduction equation or Poisson's equation.
• If there is no internal heat generation, it reduces to Laplace's equation (∇²T = 0).

Plate-type Fuel Element Solutions

• Fuel elements are plates with a central fueled strip.
• Cladding surrounds the fuel strip.
• Heat generation density is uniform (q").
• Heat flow is only in the x-direction.
• Poisson's equation is used to model heat transfer: ∇²T + q" / k = 0.
• The general solution for temperature is T = (q"/2k)x² + C₁x + C₂.
• Boundary conditions determine constants: T(x=0) = Tm, dT/dx|x=0 = 0.

Temperature Distribution and Heat Flow

• The temperature distribution is described by T = (q"/2k)x² + Tm .
• Heat flow out of the fuel surface is calculated as q = q"A.
• The temperature at the surface of fuel can be expressed as Ts = (q"/2k)a² + Tm.
• Thermal resistance can be used to simplify calculations.

Temperature Distribution in Cladding

• Cladding temperature distribution is determined without internal heat generation (∇²T = 0).
• Cladding general solution is T = C₁ x + C₂.
• Boundary conditions are used to determine constants: T(x=a) = Ts, T(x=a+b) = Tc.
• Heat balance ensures the heat flow into the cladding equals the heat generated in the fuel.

Temperature Difference (Fuel and Cladding)

• The heat flow in fuel meat (Tm - Ts) is calculated by q=a/2kA.
• Heat flow in cladding (Ts - Tc) is given by q = k A (Ts - Tc)/ b.
• The total temperature difference (Tm-Tc) can be simplified by adding the thermal resistance of the fuel and cladding.

Cylindrical Fuel Rod Solutions

• A cylindrical fuel rod has a central fueled rod of radius 'a' surrounded by cladding of radius 'b'.
• Uniform heat generation density (q").
• Heat flow is steady in the radial direction only.
• Poisson's equation is used to model the heat flow: ∇²T + q" / k = 0.
• The general solution for temperature is T = (q"/4k)r² + C₁lnr + C₂.
• Boundary conditions determine constants: T(r=a) = Ts, T(r=b) = Tc.

Temperature Distribution (Cylindrical)

• The general solution for a cylindrical fuel rod is given, and the expressions for temperature distribution and heat flow are determined, incorporating natural logs and thermal resistances.

Space Dependent Heat Source

• The discussion involves systems with varying heat generation across the component, emphasizing radial non-uniformity within a given fuel rod.
• The formulation focuses on a specific z-coordinate within a finite core and calculates the heat flow in a more generalized form.

Exponential Heat Source

• The heat source attenuates exponentially, like in a shield or structural material.
• The heat flow out of the surface of the fuel is calculated for exponential heat sources in slab structures using energy balance and boundary conditions.
• The temperature distribution can exceed the surface temperatures.

Homework Assignment

• The assignment is due one week before the next class.