/*! 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 29 A disk-shaped turbine rotor is h... [FREE SOLUTION] | 91Ó°ÊÓ

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A disk-shaped turbine rotor is heat-treated by quenching in water at \(p=1 \mathrm{~atm}\). Initially, the rotor is at a uniform temperature of \(T_{i}=1100^{\circ} \mathrm{C}\) and the water is at its boiling point as the rotor is lowered into the quenching bath by a harness. (a) Assuming lumped-capacitance behavior and constant properties for the rotor, carefully plot the rotor temperature versus time, pointing out important features of your \(T(t)\) curve. The rotor is in Orientation A. (b) If the rotor is reoriented so that its large surfaces are horizontal (Orientation B), would the rotor temperature decrease more rapidly or less rapidly relative to Orientation A?

Short Answer

Expert verified
##Short answer## The temperature of the disk-shaped turbine rotor as a function of time can be obtained using the lumped capacitance model and Newton's Law of Cooling, resulting in the equation: \[T(t) = T_\infty + (T_i - T_\infty) e^{-h_A t / mC_p}\] The temperature of the rotor decreases rapidly at first and slows down as it approaches the boiling point of water. If the rotor's large surfaces are horizontal (Orientation B), heat transfer will be enhanced due to an increase in the area available for convective heat transfer, resulting in a more rapid decrease in temperature compared to Orientation A.

Step by step solution

01

Understanding Lumped Capacitance Model

Lumped capacitance model is a simplification used in transient conduction when the heat conduction within the body is much faster than the heat convection from the surface to the surrounding. The model assumes uniform temperature throughout the body. Newton's Law of Cooling can be used to analyze the temperature change with respect to time, as the internal heat transfer is negligible compared to the external heat transfer. ##Step 2: Apply Newton's Law of Cooling##
02

Apply Newton's Law of Cooling

Newton's Law of Cooling states that the rate of temperature change of an object is proportional to the temperature difference between the object and the surrounding medium: \[\frac{dT}{dt} = -h_A \frac{(T-T_\infty)}{mC_p}\] where: - \(T\) is the temperature of the object (ºC) - \(T_\infty\) is the temperature of the surrounding medium (ºC) - \(h\) is the heat transfer coefficient (\(\mathrm{W/m^2K}\)) - \(A\) is the surface area of the object (\(\mathrm{m^2}\)) - \(m\) is the mass of the object (kg) - \(C_p\) is the specific heat of the object (\(\mathrm{J/kgK}\)) ##Step 3: Solve the Differential Equation for Temperature##
03

Solve the Differential Equation for Temperature

Integrate both sides of the differential equation with respect to time to get the expression for temperature as a function of time: \[T(t) = T_\infty + (T_i - T_\infty) e^{-h_A t / mC_p}\] where: - \(T_i\) is the initial temperature of the object. - \(t\) is the time (s) Notice that this equation represents an exponential decay, which means that the temperature of the rotor will decrease rapidly at first and then slower as it approaches the temperature of the surrounding water. Now we can carefully sketch the temperature versus time curve, ensuring that important features are pointed out, such as the initial temperature drop and the curve reaching the water boiling point as the time increases.

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Key Concepts

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

Newton's Law of Cooling
Newton's Law of Cooling is a powerful tool used to describe how the temperature of an object changes over time when exposed to a different surrounding temperature. It simplifies the process by assuming that the rate of heat loss of a body is directly proportional to the difference between the body's temperature and its environment. This means, simply put, that the larger the initial temperature difference, the faster the object will cool down.
However, this rate decreases progressively as the temperature difference reduces. In the context of the lumped capacitance model, we use the formula:
  • \(\frac{dT}{dt} = -h_A \frac{(T-T_\infty)}{mC_p}\)
  • \(T\) is the object temperature.
  • \(T_\infty\) is the ambient temperature.
  • \(h\) is the heat transfer coefficient.
  • \(A\) is the surface area.
  • \(m\) is the mass.
  • \(C_p\) is the specific heat capacity.
This model shows us that as time goes on, the temperature approaches that of the surroundings, following an exponential decay.
Transient Conduction
Transient conduction is a temporary state of heat transfer where the temperature within a material changes with time until a steady state is achieved. This is in contrast to steady-state conduction, where temperature remains constant over time. In the exercise, when the turbine rotor is introduced to the quenching water, it initially experiences transient conduction.
During this phase, the heat is conducted from the hotter inner parts towards the cooler outer surfaces while also transferring to the water. The assumption here is that the internal conduction is much faster than external convection, making the lumped capacitance model applicable.
This means that all parts of the rotor are approximately at the same temperature at any moment in time. This uniformity simplifies the analysis greatly, allowing us to focus on the heat transfer dynamics between the rotor and the water.
Heat Transfer Coefficient
The heat transfer coefficient \(h\) plays a crucial role in quantifying how readily heat is transferred between different substances. It is defined as the amount of heat that passes through a unit area of a surface per unit time per degree difference in temperature between the surface and the surrounding fluid.
In essence, the larger the heat transfer coefficient, the more readily and quickly heat will move from the rotor to the water. Various factors influence \(h\), including:
  • The nature of the surface (smooth or rough surface).
  • The type of fluid (is it water, air, oil, etc.).
  • The movement of the fluid (is it still or flowing).
In this exercise, the rotor in water scenario suggests a relatively high heat transfer coefficient due to the moving water facilitating heat dispersion.
Exponential Decay in Heat Transfer
Exponential decay is a key concept in the lumped capacitance model and is described by the equation:\[T(t) = T_\infty + (T_i - T_\infty) e^{-h_A t / mC_p}\]This equation highlights how temperature changes over time as the hot rotor cools to ambient temperature. At first, the rotor temperature drops quickly because of a large difference between its initial temperature and the water's temperature.
As time passes, this temperature difference becomes smaller, causing the rate of temperature decrease to slow down. This is a classic example of exponential decay: fast changes initially, then gradually slowing.

Understanding exponential decay is crucial for predicting how fast an object will change temperature in given conditions, which in engineering, implies being able to control process durations, energy consumption, and material properties.

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Most popular questions from this chapter

Advances in very large scale integration (VLSI) of electronic devices on a chip are often restricted by the ability to cool the chip. For mainframe computers, an array of several hundred chips, each of area \(25 \mathrm{~mm}^{2}\), may be mounted on a ceramic substrate. A method of cooling the array is by immersion in a low boiling point fluid such as refrigerant R-134a. At 1 atm and \(247 \mathrm{~K}\), properties of the saturated liquid are \(\mu=1.46 \times 10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), \(c_{p}=1551 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\operatorname{Pr}=3.2\). Assume values of \(C_{s, f}=0.004\) and \(n=1.7\). (a) Estimate the power dissipated by a single chip if it is operating at \(50 \%\) of the critical heat flux. What is the corresponding value of the chip temperature? (b) Compute and plot the chip temperature as a function of surface heat flux for \(0.25 \leq q_{s}^{\prime \prime} / q_{\max }^{\prime \prime} \leq 0.90\).

A silicon chip of thickness \(L=2.5 \mathrm{~mm}\) and thermal conductivity \(k_{s}=135 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is cooled by boiling a saturated fluorocarbon liquid \(\left(T_{\text {sat }}=57^{\circ} \mathrm{C}\right)\) on its surface. The electronic circuits on the bottom of the chip produce a uniform heat flux of \(q_{o}^{\prime \prime}=5 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2}\), while the sides of the chip are perfectly insulated. Properties of the saturated fluorocarbon are \(c_{p, l}=\) \(1100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, h_{f g}=84,400 \mathrm{~J} / \mathrm{kg}, \rho_{l}=1619.2 \mathrm{~kg} / \mathrm{m}^{3}\), \(\rho_{v}=13.4 \mathrm{~kg} / \mathrm{m}^{3}, \sigma=8.1 \times 10^{-3} \mathrm{~N} / \mathrm{m}, \mu_{1}=440 \times 10^{-6}\) \(\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\), and \(P r_{I}=9.01\). In addition, the nucleate boiling constants are \(C_{s, f}=0.005\) and \(n=1.7\). (a) What is the steady-state temperature \(T_{o}\) at the bottom of the chip? If, during testing of the chip, \(q_{o}^{\prime \prime}\) is increased to \(90 \%\) of the critical heat flux, what is the new steady-state value of \(T_{o}\) ? (b) Compute and plot the chip surface temperatures (top and bottom) as a function of heat flux for \(0.20 \leq q_{o}^{\prime \prime} / q_{\max }^{\prime \prime} \leq 0.90\). If the maximum allowable chip temperature is \(80^{\circ} \mathrm{C}\), what is the maximum allowable value of \(q_{e}^{\text {"? }}\) ?

Consider refrigerant R-134a flowing in a smooth, horizontal, 10-mm-inner- diameter tube of wall thickness \(2 \mathrm{~mm}\). The refrigerant is at a saturation temperature of \(15^{\circ} \mathrm{C}\) (for which \(\rho_{v \text { sat }}=23.75 \mathrm{~kg} / \mathrm{m}^{3}\) ) and flows at a rate of \(0.01 \mathrm{~kg} / \mathrm{s}\). Determine the maximum wall temperature associated with a heat flux of \(10^{5} \mathrm{~W} / \mathrm{m}^{2}\) at the inner wall at a location \(0.4 \mathrm{~m}\) downstream from the onset of boiling for tubes fabricated of (a) pure copper and (b) AISI 316 stainless steel.

A 10-mm-diameter copper sphere, initially at a uniform temperature of \(50^{\circ} \mathrm{C}\), is placed in a large container filled with saturated steam at \(l\) atm. Using the lumped capacitance method, estimate the time required for the sphere to reach an equilibrium condition. How much condensate \((\mathrm{kg})\) was formed during this period?

Saturated steam at \(0.1\) bar condenses with a convection coefficient of \(6800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside of a brass tube having inner and outer diameters of \(16.5\) and \(19 \mathrm{~mm}\), respectively. The convection coefficient for water flowing inside the tube is \(5200 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Estimate the steam condensation rate per unit length of the tube when the mean water temperature is \(30^{\circ} \mathrm{C}\).

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