How Do Nuclear Fuel Pellets and the Alaska Pipeline Manage Heat Transfer?

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In summary: The energy lost to the ice water during the half hour is approximately 750,000 Joules. This is equivalent to about 180 kilocalories, which means you could eat about 0.15 Big Mac meals without gaining weight. In summary, the conversation discusses the steady state heat conduction equations for nuclear fuel pellets and the Alaska pipeline. It also proposes a plan, known as the Squires DietTM, where one can lose weight by sitting in ice water for half an hour. The calculations for the temperature distribution, energy loss, and resistance to heat transfer are also provided.
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shawn26
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Homework Statement


Any hints would help a lot for any question

3. Nuclear fuel pellets consist of a spherical core of radius R1 of fissionable material of thermal conductivity kn that generates an energy per volume S. To more easily handle this radioactive material, each pellet is encased in aluminum of thermal conductivity ka and outer radius R2 . No heat is generated within the aluminum. The entire fuel pellets are then immersed in a fluid of temperature Tf and heat transfer coefficient h.
a. Using shell balances for energy, write the steady state heat conduction equations for the nuclear core and aluminum shell.
b. Write the boundary conditions for the inner sphere (R1 ) and the outer sphere (R2 ). Write a short explanation of each.
c. Solve for the temperature distribution for the core and the shell. What is the maximum temperature of the core and the outer shell?

4. The Alaska pipeline runs 800 miles (1280 km), from the arctic ocean to Valdez, Alaska. The pipe itself is 48 inches (122 cm) in diameter, with 1/2 inch (1.27 cm) thick steel. It is insulated with a 4 inch (10 cm) layer of fiberglass insulation (k = .04W/mK), and the oil is generally pumped at 145◦ F , in air that can be −40◦ F . Assume the pipe to have a heat transfer coefficient in air h = 24W/m2 K (assuming a 1 mm boundary layer).
a. Compute the energy loss per foot per time due to heat transfer. Compute the total energy loss for the entire pipeline. Compute the general formulas first, then plug in numbers.
b. Where does most of the temperature drop occur? Compare the insulating properties of the steel pipe, the fiberglass insulation, and the convection through the air. Where is the resistance to heat transfer the highest?
c. Does the system exceed the requirement set by the critical insulation thickness, in order that the insulation actually help?

5. The Squires DietTM , which is sure to make me rich if I can just figure out a way to charge for it. Your body is constantly maintained at 98.6 ◦ F, or 37C. You burn food to maintain this temperature. Rather than scrimp on food, or eat healthy, or – gasp – exercise, why not just make your thermostat work a little harder? Here’s the plan: fill your bathtub with ice water (0 C), then lay down in it for one half hour per day, while watching Judge Judy on TV. Assume Newton’s law of cooling (heat transfer coefficient h ≈ 500W/m2 C, which I got assuming a 1 mm boundary layer), and that your body has about 2 m2 of surface area. How much Joules of energy do you lose to the ice water during this time? Convert this energy to kilocalories (which is a food ‘calorie’). Since a Big Mac, super-size fries, and diet coke run about 1200 ‘calories’ (kcal), how many such value meals can you now eat without gaining weight?


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The Attempt at a Solution

 
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3. a. Core: K_n*(dT/dr) = -S Shell: K_a*(dT/dr) = 0 b. Inner Sphere: T(R1) = Tf + (Tc-Tf)*(R1/R2) Outer Sphere: h*(Tc-Tf) = K_a*(Tc-Tf)/R2 The inner sphere boundary condition states that the temperature at the inner surface of the shell is equal to the temperature of the fluid times the fraction of the radius of the inner surface to the outer surface plus the temperature of the core times the fraction of the radius of the inner surface to the outer surface. The outer sphere boundary condition states that the heat transfer coefficient multiplied by the temperature difference between the core and the fluid is equal to the thermal conductivity of the aluminum multiplied by the temperature difference between the core and the fluid divided by the radius of the outer surface. c. Maximum Temperature of Core: Tc = Tf + (h*R1*R2/K_a) Maximum Temperature of Outer Shell: Ta = Tf + (Tc-Tf)*(R2/R1) 4. a. Energy Loss Per Foot Per Time = h*A*(Tc-Tf)/L Total Energy Loss for Entire Pipeline = h*A*(Tc-Tf)*L b. Most of the temperature drop occurs in the air convection since the convection heat transfer coefficient is much higher than the other two. The resistance to heat transfer is highest in the steel pipe since it has the lowest thermal conductivity. c. Yes, the system exceeds the requirement set by the critical insulation thickness.
 

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