/*! 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 23 A transmission case measures \(W... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 23

A transmission case measures \(W=0.30 \mathrm{~m}\) on a side and receives a power input of \(P_{i}=150 \mathrm{hp}\) from the engine. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The required heater power to operate the dryer at an air temperature of \(50^{\circ}\mathrm{C}\) is approximately \(111862.5\text{W}\).

Step by step solution

01

Convert the power input from hp to Watts

First, we need to convert the given power input \(P_i\) from horsepower (hp) to Watts (W) to match the units of other given values: \[P_i = 150hp \cdot \frac{746W}{1hp} = 111900W\]
02

Calculate the exposed switch surface area in m虏

Next, we'll convert the exposed switch surface area from \(30~\text{mm}^2\) to \(\text{m}^2\): \[A = 30~\text{mm}^2 \cdot \left(\frac{1~\text{m}}{1000~\text{mm}}\right)^2 = 3.0 \times 10^{-5}~\mathrm{m}^2\]
03

Calculate the convective heat transfer rate

We're given the convection heat transfer coefficient, \(h=25\mathrm{~W} / \mathrm{m}^{2}\mathrm{K}\), the air temperature \(T_{\infty}=50^{\circ}\mathrm{C}\), and the switch temperature \(T_s=70^{\circ}\mathrm{C}\). Now, we can calculate the convective heat transfer rate, \(q\): \[q=hA(T_s-T_{\infty}) = 25\frac{\mathrm{W}}{\mathrm{m}^2 \mathrm{K}} \cdot 3.0 \times 10^{-5}\mathrm{m}^2 \cdot (70-50)\mathrm{K} = 37.5\mathrm{W}\]
04

Calculate the required heater power

Finally, we can find the desired heater power (\(P_e\)) by subtracting the convective heat transfer rate (\(q\)) from the power input (\(P_i\)): \[P_e=P_i-q = 111900\mathrm{W} - 37.5\mathrm{W} =111862.5\mathrm{W}\] The required heater power to operate the dryer at an air temperature of \(50^{\circ}\mathrm{C}\) is approximately \(111862.5\text{W}\).

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Key Concepts

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

Heat Transfer Coefficient
Understanding the heat transfer coefficient is essential in solving problems related to convective heat transfer. It is a property of the particular system and is denoted by the symbol 'h'. This coefficient quantifies how well heat is transferred from a surface to a fluid or vice versa. Its units are typically expressed in watts per square meter per Kelvin \( \frac{W}{m^2 \cdot K} \) and it's influenced by several factors including the fluid's velocity, its properties, and the nature of the surface.

In the given exercise, the convective heat transfer coefficient between the air and the switch surface is provided as \(25~\frac{W}{m^2 \cdot K} \). Knowing this, we can determine the rate of heat transfer from the switch to its surroundings, which in turn allows us to calculate the required power that the heater must provide. Keep in mind that higher heat transfer coefficients indicate more efficient heat transfer, which is desirable in many engineering applications.
Convection
Convection is one of the three modes of heat transfer, the others being conduction and radiation. When we talk about convective heat transfer, we refer to the process of heat moving with a fluid (which could be a gas or a liquid). This can occur naturally, driven by temperature differences within the fluid, or can be forced by an external source like a pump or a fan.

In the context of the exercise, we deal with forced convection as there's a flow of air facilitated by the dryer mechanism. To calculate the convective heat loss \( q \) from the switch to the dryer air, we use the formula \( q = hA(T_s - T_{\infty}) \). This equation takes into account the heat transfer coefficient (h), the surface area (A), and the temperature difference between the switch \( (T_s) \) and the ambient air \( (T_{\infty}) \). It's fundamental to understand how convection works to determine the thermal management within systems like this dryer.
Power Conversion
Power conversion involves changing one form of power to another. In terms of thermal systems, it often refers to converting electrical power to heat. Most appliances with heating elements, like dryers or heaters, function through power conversion.

In our textbook problem, we convert electrical power (measured in watts) into thermal energy, which is necessary to maintain the switch's temperature against the convective cooling effect of the air passing over it. The exercise first requires converting the power input from horsepower to watts, a common task in power conversion problems since power can be expressed in various units. Once we know the heat being lost to convection, we then determine the required electrical power the heater needs to provide to maintain the desired temperature. Understanding power conversion is critical for designers to ensure that devices function efficiently under different operating conditions.

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Most popular questions from this chapter

The heat flux that is applied to the left face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The wall is of thickness \(L=10\) \(\mathrm{mm}\) and of thermal conductivity \(k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the surface temperatures of the wall are measured to be \(50^{\circ} \mathrm{C}\) on the left side and \(30^{\circ} \mathrm{C}\) on the right side, do steady-state conditions exist?

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

Electronic power devices are mounted to a heat sink having an exposed surface area of \(0.045 \mathrm{~m}^{2}\) and an emissivity of \(0.80\). When the devices dissipate a total power of \(20 \mathrm{~W}\) and the air and surroundings are at \(27^{\circ} \mathrm{C}\), the average sink temperature is \(42^{\circ} \mathrm{C}\). What average temperature will the heat sink reach when the devices dissipate \(30 \mathrm{~W}\) for the same environmental condition?

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~mm}\) and emissivity \(\varepsilon=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the temperature range \(40 \leq T \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by considering values of \(0.20\) and \(0.30\).

1.42 One method for growing thin silicon sheets for photovoltaic solar panels is to pass two thin strings of high melting temperature material upward through a bath of molten silicon. The silicon solidifies on the strings near the surface of the molten pool, and the solid silicon sheet is pulled slowly upward out of the pool. The silicon is replenished by supplying the molten pool with solid silicon powder. Consider a silicon sheet that is \(W_{\mathrm{si}}=85 \mathrm{~mm}\) wide and \(t_{\mathrm{si}}=150 \mu \mathrm{m}\) thick that is pulled at a velocity of \(V_{\mathrm{si}}=20 \mathrm{~mm} / \mathrm{min}\). The silicon is melted by supplying electric power to the cylindrical growth chamber of height \(H=350 \mathrm{~mm}\) and diameter \(D=300 \mathrm{~mm}\). The exposed surfaces of the growth chamber are at \(T_{s}=\) \(320 \mathrm{~K}\), the corresponding convection coefficient at the exposed surface is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surface is characterized by an emissivity of \(\varepsilon_{s}=0.9\). The solid silicon powder is at \(T_{\mathrm{s}, i}=298 \mathrm{~K}\), and the solid silicon sheet exits the chamber at \(T_{\text {si, } o}=420 \mathrm{~K}\). Both the surroundings and ambient temperatures are \(T_{\infty}=T_{\text {sur }}=298 \mathrm{~K}\). (a) Determine the electric power, \(P_{\text {elec }}\), needed to operate the system at steady state. (b) If the photovoltaic panel absorbs a time-averaged solar flux of \(q_{\text {sol }}^{\prime \prime}=180 \mathrm{~W} / \mathrm{m}^{2}\) and the panel has a conversion efficiency (the ratio of solar power absorbed to electric power produced) of \(\eta=0.20\), how long must the solar panel be operated to produce enough electric energy to offset the electric energy that was consumed in its manufacture?
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