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Chapter 2: Problem 17

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 surroundings where the ambient temperature is \(10^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall thickness of the pipe is \(3 \mathrm{~mm}\) and its inner diameter is \(30 \mathrm{~mm}\). The pipe wall has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.23 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.002 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). For safety reasons and to prevent thermal burn to workers, the outer surface temperature of the pipe should be kept below \(50^{\circ} \mathrm{C}\). Determine whether the outer surface temperature of the pipe is at a safe temperature so as to avoid thermal burn.

How is the order of a differential equation determined?

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

The variation of temperature in a plane wall is determined to be \(T(x)=110-60 x\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the thickness of the wall is \(0.75 \mathrm{~m}\), the temperature difference between the inner and outer surfaces of the wall is (a) \(30^{\circ} \mathrm{C}\) (b) \(45^{\circ} \mathrm{C}\) (c) \(60^{\circ} \mathrm{C}\) (d) \(75^{\circ} \mathrm{C}\) (e) \(84^{\circ} \mathrm{C}\)

What is the difference between an ordinary differential equation and a partial differential equation?
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