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In a nuclear reactor, heat is generated uniformly in the 5 -cm-diameter cylindrical uranium rods at a rate of \(2 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\). If the length of the rods is \(1 \mathrm{~m}\), determine the rate of heat generation in each rod. Answer: \(393 \mathrm{~kW}\)

Short Answer

Expert verified
Answer: The rate of heat generation in each uranium rod is 393 kW.

Step by step solution

01

Calculate the volume of each rod

To find the rate of heat generation in each rod, we first need to calculate the volume of each rod. The rods have a cylindrical shape and we are given the diameter and the length. The formula for the volume of a cylinder is: \[V = \pi r^2 h\] where 'V' is the volume, 'r' is the radius, and 'h' is the height (length) of the cylinder. Since we have the diameter (5 cm) and length (1 m), we can find the radius by dividing the diameter by 2: \[r = \frac{d}{2} = \frac{5\mathrm{~cm}}{2} = 2.5\mathrm{~cm}\] Now, convert the radius and length to meters: \[r = 2.5\mathrm{~cm} \times \frac{1\mathrm{~m}}{100\mathrm{~cm}} = 0.025\mathrm{~m}\] \[h = 1\mathrm{~m}\] Now plug in the values for radius and height into the volume formula: \[V = \pi (0.025\mathrm{~m})^2 (1\mathrm{~m})\]
02

Calculate the volume

Now we have to calculate the actual volume of each cylindrical rod. Using the formula and the values we just found, we can find the volume: \[V = \pi (0.025\mathrm{~m})^2 (1\mathrm{~m}) = 0.001963\mathrm{~m}^3\]
03

Calculate the rate of heat generation

We are given the heat generation rate per cubic meter, which is \(2 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\). To find the rate of heat generation in each rod, we just need to multiply the rate in \(\mathrm{W} / \mathrm{m}^{3}\) by the volume of the rod: \[\textit{Rate of heat generation} = 2 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3} \times 0.001963\mathrm{~m}^3 = 393000\mathrm{~W}\]
04

Express the answer in kilowatts

Lastly, we need to express our answer in kilowatts. To do this, divide the watts by 1000: \[\textit{Rate of heat generation} = \frac{393000\mathrm{~W}}{1000} = 393\mathrm{~kW}\] Hence, the rate of heat generation in each uranium rod is 393 kW.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Volume Calculation
Understanding how to calculate the volume of a cylinder is fundamental, especially in fields like thermal engineering and nuclear reactor design. The volume of a cylinder is determined using the formula:\[ V = \pi r^2 h \]where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height (or length) of the cylinder. In the context of the given exercise, these cylindrical uranium rods in a nuclear reactor contribute to heat generation based on their volume.To perform this calculation:
  • First, determine the radius from the diameter. In this case, the diameter of 5 cm is converted to 2.5 cm (or 0.025 m) when halved.
  • Use the height of 1 m, as given.
  • Substitute these values into the volume formula to get \( V = \pi (0.025)^2 (1) \) or approximately 0.001963 m3.
This calculation is pivotal, as it will directly affect the thermal calculations involved in subsequent steps.
Thermal Engineering
In thermal engineering, understanding heat generation and dissipation is crucial for designing effective energy systems. Heat generation inside a nuclear reactor is a complex process that requires a precise calculation to ensure safe and efficient operation. Here, we are dealing with a uniform heat generation rate inside cylindrical uranium rods.Key considerations:
  • The uniform heat generation rate is specified as \( 2 \times 10^{8} \, \mathrm{W/m^3} \). This indicates the amount of heat produced per unit volume of the rods.
  • Using the volume calculated previously, we multiply this rate by the volume of each rod, yielding the absolute heat generation rate.
  • The heat generated is crucial for understanding the reactor's energy output and for maintaining the necessary temperature conditions inside the reactor.
This calculation ensures engineers can predict the thermal behavior of reactor materials and avoid overheating by designing appropriate cooling systems.
Nuclear Reactor Design
Nuclear reactor design involves careful planning to balance heat production with safety and efficiency. The uranium rods are central to this, as they generate the necessary heat through nuclear reactions. The calculation of heat generation rate is vital within this design process. Core principles include:
  • The need to precisely calculate the heat output per rod to design adequate cooling systems, which prevent overheating.
  • The physical dimensions of the rods influence the reactor design as their volume affects the rate of heat generation.
  • Converting heat generation data into a usable form (e.g., from watts to kilowatts) helps engineers and scientists assess and optimize reactor performance.
This exemplary case of cylindrical uranium rods provides crucial insights into the safe and efficient design of nuclear reactors, emphasizing precision in calculations to ensure both energy maximization and safety.

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Most popular questions from this chapter

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

The temperature of a plane wall during steady onedimensional heat conduction varies linearly when the thermal conductivity is constant. Is this still the case when the thermal conductivity varies linearly with temperature?

The heat conduction equation in a medium is given in its simplest form as $$ \frac{1}{r} \frac{d}{d r}\left(r k \frac{d T}{d r}\right)+\dot{e}_{\text {gen }}=0 $$ Select the wrong statement below. (a) The medium is of cylindrical shape. (b) The thermal conductivity of the medium is constant. (c) Heat transfer through the medium is steady. (d) There is heat generation within the medium. (e) Heat conduction through the medium is one-dimensional.

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius \(r_{1}=15 \mathrm{~cm}\), outer radius \(r_{2}=20 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

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