/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Uniform internal heat generation... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 28

Uniform internal heat generation at \(\dot{q}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) is occurring in a cylindrical nuclear reactor fuel rod of 50 -mm diameter, and under steady-state conditions the temperature distribution is of the form \(T(r)=a+b r^{2}\), where \(T\) is in degrees Celsius and \(r\) is in meters, while \(a=800^{\circ} \mathrm{C}\) and \(b=-4.167 \times 10^{5}{ }^{\circ} \mathrm{C} / \mathrm{m}^{2}\). The fuel rod properties are \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) What is the rate of heat transfer per unit length of the rod at \(r=0\) (the centerline) and at \(r=25 \mathrm{~mm}\) (the surface)? (b) If the reactor power level is suddenly increased to \(\dot{q}_{2}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what is the initial time rate of temperature change at \(r=0\) and \(r=25 \mathrm{~mm}\) ?

Short Answer

Expert verified
The rate of heat transfer per unit length at \(r=0\) (centerline) is 0 W/m, and at \(r=25 \mathrm{~mm}\) (surface) is \(5.008 \times 10^4 \mathrm{~W/m}\). With the increase in reactor power level, the initial time rate of temperature change at both \(r=0\) and \(r=25 \mathrm{~mm}\) is \(0.114 \mathrm{~K/s}\).

Step by step solution

01

(Step 1: Find the temperature gradient at given positions)

(We are given the temperature distribution T(r) = a + b r^2. Differentiate this expression with respect to r to obtain the temperature gradient, dT/dr.) We have the temperature distribution as \(T(r) = a + b r^{2}\), where a = 800掳C and b = -4.167 x 10鈦� 掳C/m虏. Differentiating with respect to r, \( \frac{dT}{dr} = 2br \)
02

(Step 2: Calculate the temperature gradients at r=0 and r=25 mm for part (a))

(Use the expression obtained in step 1 to calculate the temperature gradients at the given positions.) For r = 0: \( \frac{dT}{dr} = 2b(0) = 0 \) For r = 25 mm (or 0.025 m): \( \frac{dT}{dr} = 2b(0.025) = 2(-4.167*10^5)(0.025)\) Now, calculate the value for the temperature gradient at r = 25 mm.
03

(Step 3: Calculate the heat transfer rates at r=0 and r=25 mm for part (a))

(Using the temperature gradients calculated in step 2 and Fourier's Law of heat conduction, find the heat transfer rates per unit length at the given positions.) Fourier's Law of heat conduction is given by: \( q_r = -k \frac{dT}{dr} \) For r = 0: \( q_r = -k \times 0 = 0 \) For r = 25 mm: \( q_r = -k \times 2b(0.025)\) (Substitute the value of k and the temperature gradient at r=25 mm) Calculate the heat transfer rate at r = 25 mm.
04

(Step 4: Write the energy equation for part (b))

(Write the energy equation for the fuel rod considering internal heat generation, given the properties, 蟻, and cp.) The energy equation is given by: \( \rho c_p \frac{dT}{dt} = k \nabla^2 T + \dot{q} \)
05

(Step 5: Calculate the initial time rate of temperature change at r=0 and r=25 mm)

(Use the energy equation from step 4 and the given values for 蟻, cp, k, and the new heat generation rate to find the initial time rate of temperature changes at the given positions.) Differentiate the temperature distribution with respect to time: \( \frac{dT}{dt} = \frac{k}{\rho c_p}\nabla^2T+\frac{\dot{q}}{\rho c_p} \) For r = 0: \( \frac{dT}{dt}|_{r=0} = 0 + \frac{\dot{q}_{2}}{\rho c_p}\) For r = 25 mm: \( \frac{dT}{dt}|_{r=0.025} = 0 + \frac{\dot{q}_{2}}{\rho c_p}\) Now substitute the values of the new heat generation rate \(\dot{q}_2 = 10^8 W/m鲁\), 蟻, and cp to find the initial time rate of temperature change at r=0 and r=25 mm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Fourier's Law of Heat Conduction
Fourier鈥檚 Law of Heat Conduction is a fundamental principle in thermal physics that describes the transfer of heat within a material. It states that the rate of heat conduction through a substance is proportional to the negative gradient of temperature and the material's ability to conduct heat, known as thermal conductivity. The equation for Fourier鈥檚 Law is given by:

\[ q_r = -k \frac{dT}{dr} \]where:
  • \( q_r \) is the heat transfer rate per unit area,
  • \( k \) is the thermal conductivity of the material,
  • \( \frac{dT}{dr} \) is the temperature gradient.
In the context of the nuclear reactor fuel rod, Fourier's Law helps us understand how heat flows from the interior to the exterior of the rod. By applying it, we calculated the heat transfer at the centerline and the surface of the fuel rod. Without a temperature gradient, as at the center (\( r = 0 \)), no heat flows. However, at the surface (\( r = 0.025 \, \text{m} \)), the gradient exists, resulting in a measurable heat flow. Fourier鈥檚 Law connects temperature distribution to the physical properties of the material, making it a vital concept in heat transfer analysis.
Temperature Distribution
Temperature distribution refers to how temperature varies within an object. For a cylindrical nuclear reactor fuel rod with internal heat generation, understanding temperature distribution is crucial to ensure safe operation. In this scenario, the temperature distribution is given by the equation:

\[ T(r) = a + b r^{2} \] where:
  • \( T(r) \) is the temperature at a distance \( r \) from the center,
  • \( a \) is a constant, representing the maximum temperature,
  • \( b \) determines how temperature changes with \( r^2 \).
Knowing the temperature distribution helps us evaluate where the temperature is highest and how it decreases towards the surface of the rod. At the center (\( r = 0 \)), the temperature is highest due to symmetry and heat generation. The shape of the distribution also tells us the effectiveness of heat dissipation in preventing overheating, vital for maintaining the structural integrity of the reactor's fuel rod under operational conditions.
Nuclear Reactor Fuel Rod
A nuclear reactor fuel rod is a critical component in a nuclear reactor core. It contains the nuclear fuel and is designed to withstand the intense environment of the reactor. The fuel rod not only participates in the fission process but also transfers the generated heat to a coolant, preventing overheating and potential failure.

Key aspects of a fuel rod include:
  • Material Properties: Materials are chosen for their ability to safely contain nuclear fuel and efficiently conduct heat. For our fuel rod, properties include a thermal conductivity of \( k = 30 \, \text{W/m} \cdot \text{K} \).
  • Internal Heat Generation: Heat is generated uniformly within the rod due to nuclear reactions, which must be managed effectively.
  • Temperature Management: The temperature distribution informs us about the thermal conditions throughout the rod, critical for avoiding thermal stresses.
Fuel rods must be meticulously engineered and maintained to ensure the reactor operates safely and efficiently. They play a paramount role in energy production and reactor safety by effectively transferring and managing the heat generated during nuclear reactions.
Internal Heat Generation
Internal heat generation occurs when heat is produced within a substance rather than applied externally. In a nuclear reactor fuel rod, internal heat generation is primarily due to nuclear fission, the process of splitting atomic nuclei to release energy.

For our example, internal heat generation is described by the rate \( \dot{q} = 5 \times 10^{7} \, \text{W/m}^3 \). This figure represents how much heat is produced per unit volume of the fuel rod. The effect of internal heat generation is a central factor in determining the temperature distribution within the rod, influencing the thermal stresses the material undergoes.
  • When this internal heat generation increases, as in the example where reactor power levels suddenly rise, it impacts the time rate of temperature change, potentially hazardously increasing the internal temperature.
  • Engineers must consider these effects to design systems that can handle sudden increases without failure.
Understanding internal heat generation helps in assessing the thermal performance of the reactor and is a key aspect in thermal management strategies to ensure safe and effective nuclear reactor operations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Typically, air is heated in a hair dryer by blowing it across a coiled wire through which an electric current is passed. Thermal energy is generated by electric resistance heating within the wire and is transferred by convection from the surface of the wire to the air. Consider conditions for which the wire is initially at room temperature, \(T_{i}\), and resistance heating is concurrently initiated with airflow at \(t=0\). (a) For a wire radius \(r_{o}\), an air temperature \(T_{\infty}\), and a convection coefficient \(h\), write the form of the heat equation and the boundary/initial conditions that govern the transient thermal response, \(T(r, t)\), of the wire. (b) If the length and radius of the wire are \(500 \mathrm{~mm}\) and \(1 \mathrm{~mm}\), respectively, what is the volumetric rate of thermal energy generation for a power consumption of \(P_{\text {elec }}=500 \mathrm{~W}\) ? What is the convection heat flux under steady-state conditions? (c) On \(T-r\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (d) On \(q_{r}^{\prime \prime}-t\) coordinates, sketch the variation of the heat flux with time for locations at \(r=0\) and \(r=r_{o^{*}}\).

At a given instant of time, the temperature distribution within an infinite homogeneous body is given by the function $$ T(x, y, z)=x^{2}-2 y^{2}+z^{2}-x y+2 y z $$ Assuming constant properties and no internal heat generation, determine the regions where the temperature changes with time.

Consider a small but known volume of metal that has a large thermal conductivity. (a) Since the thermal conductivity is large, spatial temperature gradients that develop within the metal in response to mild heating are small. Neglecting spatial temperature gradients, derive a differential equation that could be solved for the temperature of the metal versus time \(T(t)\) if the metal is subjected to a fixed surface heat rate \(q\) supplied by an electric heater. (b) A student proposes to identify the unknown metal by comparing measured and predicted thermal responses. Once a match is made, relevant thermophysical properties might be determined, and, in turn, the metal may be identified by comparison to published property data. Will this approach work? Consider aluminum, gold, and silver as the candidate metals.

A spherical particle of radius \(r_{1}\) experiences uniform thermal generation at a rate of \(\dot{q}\). The particle is encapsulated by a spherical shell of outside radius \(r_{2}\) that is cooled by ambient air. The thermal conductivities of the particle and shell are \(k_{1}\) and \(k_{2}\), respectively, where \(k_{1}=2 k_{2}\). (a) By applying the conservation of energy principle to spherical control volume \(A\), which is placed at an arbitrary location within the sphere, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(0 \leq r \leq r_{1}\). (b) By applying the conservation of energy principle to spherical control volume \(\mathrm{B}\), which is placed at an arbitrary location within the spherical shell, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(r_{1} \leq r \leq r_{2}\). (c) On \(T-r\) coordinates, sketch the temperature distribution over the range \(0 \leq r \leq r_{2}\).

Consider a one-dimensional plane wall with constant properties and uniform internal generation \(\dot{q}\). The left face is insulated, and the right face is held at a uniform temperature. (a) Using the appropriate form of the heat equation, derive an expression for the \(x\)-dependence of the steady-state heat flux \(q^{\prime \prime}(x)\). (b) Using a finite volume spanning the range \(0 \leq\) \(x \leq \xi\), derive an expression for \(q^{\prime \prime}(\xi)\) and compare the expression to your result for part (a).
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Radiation

Read Explanation

Space Physics

Read Explanation

Magnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.