/*! 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 9 In order to prevent fogging, the... [FREE SOLUTION] | 91Ó°ÊÓ

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In order to prevent fogging, the \(3 \mathrm{~mm}\)-thick rear window of an automobile has a transparent film electrical heater bonded to the inside of the glass. During a test, \(200 \mathrm{~W}\) are dissipated in a \(0.567 \mathrm{~m}^{2}\) area of window when the inside and outside air temperatures are \(22^{\circ} \mathrm{C}\) and \(1^{\circ} \mathrm{C}\), and the inside and outside heat transfer coefficients are \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(22 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. Determine the temperature of the inside surface of the window.

Short Answer

Expert verified
The inside surface temperature of the window is approximately \(61.15 \, ^\circ\mathrm{C}\).

Step by step solution

01

Identify Known Values

First, identify the values provided in the problem:- Power dissipated: \( P = 200 \, \mathrm{W} \)- Area: \( A = 0.567 \, \mathrm{m}^2 \)- Inside air temperature: \( T_{\text{in}} = 22 \, ^\circ\mathrm{C} \)- Outside air temperature: \( T_{\text{out}} = 1 \, ^\circ\mathrm{C} \)- Inside heat transfer coefficient: \( h_{\text{in}} = 9 \, \mathrm{W}/\mathrm{m}^2 \mathrm{K} \)- Outside heat transfer coefficient: \( h_{\text{out}} = 22 \, \mathrm{W}/\mathrm{m}^2 \mathrm{K} \)
02

Surface Energy Balance Equation

Apply an energy balance on the inside surface of the window. The energy balance equation is:\[ q - h_{\text{in}} A (T_{\text{si}} - T_{\text{in}}) - h_{\text{out}} A (T_{\text{so}} - T_{\text{out}}) = 0 \]where - \( q = \frac{P}{A} \) is the heat flux,- \( T_{\text{si}} \) is the temperature of the inside surface,- \( T_{\text{so}} \) is the temperature of the outside surface (unknown).Simplifying for the heat flux, we have:\[ q = h_{\text{in}} (T_{\text{si}} - T_{\text{in}}) + h_{\text{out}} (T_{\text{so}} - T_{\text{out}}) \]
03

Calculate Heat Flux

Calculate the heat flux using the given power and area.\[ q = \frac{P}{A} = \frac{200 \, \mathrm{W}}{0.567 \, \mathrm{m}^2} \approx 352.38 \, \mathrm{W}/\mathrm{m}^2 \]
04

Solve for Inside Surface Temperature

We rearrange the energy balance equation to solve for \( T_{\text{si}} \).Since the heat transfer through convection across the outside surface is unknown, consider simplifying the problem by assuming steady state conditions and a negligible temperature drop across the window, leading to:\[ T_{\text{so}} \approx T_{\text{si}} \]Thus, solve\[ q = h_{\text{in}} (T_{\text{si}} - T_{\text{in}}) + h_{\text{out}} (T_{\text{so}} - T_{\text{out}}) \approx h_{\text{in}} (T_{\text{si}} - T_{\text{in}}) \]Plug in the known values to solve for \( T_{\text{si}} \):\[ 352.38 = 9 (T_{\text{si}} - 22) \]\[ T_{\text{si}} - 22 = \frac{352.38}{9} \approx 39.15 \]\[ T_{\text{si}} = 39.15 + 22 \approx 61.15 \, ^\circ\mathrm{C} \]
05

Verify and Conclude

Double check assumptions and calculations for consistency. Given the values and assumptions, the calculation provides an estimate of the surface temperature of the inside of the window. The assumption regarding negligible conduction through the window is a simplification but illustrates the main effect on the surface temperature.

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Key Concepts

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

Surface Energy Balance
In the world of heat transfer, the concept of energy balance is essential for understanding how heat is managed in systems. The surface energy balance takes into account the various forms of energy entering and leaving a system, ensuring that energy conservation principles are met.
For a problem like preventing fogging on an automobile window, the surface energy balance involves the calculation of the heat that flows through the window, noting how much energy is dissipated by the heater, and how much is transferred to the surrounding air.
  • The input power, which is the energy provided by the heater, is part of the equation.
  • Heat is transferred both into the interior of the car and out into the environment, each path subjected to different coefficients of heat transfer.
The primary equation used to set up the energy balance around the window is: \[ q - h_{\text{in}} A (T_{\text{si}} - T_{\text{in}}) - h_{\text{out}} A (T_{\text{so}} - T_{\text{out}}) = 0 \] where the different terms represent the heat flux, the heat transfer coefficients, and the temperatures on either side of the window. Understanding the surface energy balance is crucial in solving such problems, as it ensures every form of energy that crosses the boundary is accounted for.
Heat Flux Calculation
Heat flux represents the rate of heat energy transfer through a given surface, typically measured in watts per square meter (W/m²). Calculating heat flux is integral in thermal analysis to ensure the effectiveness of heating or cooling strategies in a system.
In the automotive window problem, the heat flux is calculated to determine how much heat per unit area is being transferred through the window. The formula used is quite straightforward: \[ q = \frac{P}{A} \] where \( P \) stands for the power dissipated, and \( A \) is the area through which that power is being distributed.
This simple division gives the density of heat transfer across the surface. The resulting value of 352.38 W/m² indicates how much effort is needed from the heater to maintain a comfortable temperature balance, preventing window fogging. Thus, calculating the heat flux helps engineers design heating solutions that are both efficient and effective for maintaining visibility in vehicles.
Convection Heat Transfer
Convection is a mode of heat transfer where heat is transported by the movement of fluids. In contexts like an automobile window, convection occurs between the window surface and the air inside and outside the vehicle.
Understanding convection heat transfer involves looking at how fluid flow, which can be air or liquid, contributes to the heat exchange. It's characterized by the heat transfer coefficient, denoted as \( h \), which in this problem has different values for the inside and outside air.
  • Inside: \( h_{\text{in}} = 9 \text{ W/m}^2 \text{K} \)
  • Outside: \( h_{\text{out}} = 22 \text{ W/m}^2 \text{K} \)
These coefficients are determined by factors like fluid velocity, viscosity, and temperature difference at the surface.
The formula to find how much heat is transferred through convection can be seen in the energy balance equation, where the term \( h \times A \times (T_{\text{surface}} - T_{\text{fluid}}) \) tells us how convection affects the temperature difference across the surface.
In practical applications, engineers use these coefficients to design systems that control temperatures effectively, ensuring reliable performance under varying environmental conditions.

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Most popular questions from this chapter

A \(5 \mathrm{~mm} \times 2 \mathrm{~mm} \times 1 \mathrm{~mm}\)-thick semiconductor laser is mounted on a \(1 \mathrm{~cm}\) cube copper heat sink and enclosed in a Dewar flask. The laser dissipates \(2 \mathrm{~W}\), and a cryogenic refrigeration system maintains the copper block at a nearly uniform temperature of \(90 \mathrm{~K}\). Estimate the top surface temperature of the laser chip for the following models of the dissipation process: (i) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer underneath the top surface of the laser. (ii) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer at the midplane of the chip. (iii) The energy is dissipated uniformly through the chip. Take \(k=170 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the chip, and neglect parasitic heat gains from the Dewar flask.

A test technique for measuring the thermal conductivity of copper-nickel alloys is based on the measurement of the tip temperature of pin fins made from the alloys. The standard fin dimensions are a diameter of \(5 \mathrm{~mm}\) and a length of \(20 \mathrm{~cm}\). The test fin and a reference brass fin \((k=111 \mathrm{~W} / \mathrm{m} \mathrm{K})\) are mounted on a copper base plate in a wind tunnel. The test data include \(T_{B}=100^{\circ} \mathrm{C}, T_{e}=20^{\circ} \mathrm{C}\), and tip temperatures of \(64.2^{\circ} \mathrm{C}\) and \(49.7^{\circ} \mathrm{C}\) for the brass and test alloy fins, respectively. (i) Determine the conductivity of the test alloy. (ii) If the conductivity should be known to \(\pm 1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), how accurately should the tip temperatures be measured?

An air-cooled R-22 condenser has 10 mm-O.D., 9 mm-I.D. aluminum tubes with \(50 \mathrm{~mm}-\mathrm{O} . \mathrm{D} .\) annular rectangular aluminum fins of thickness \(0.2 \mathrm{~mm}\) at a pitch of \(2 \mathrm{~mm}\). The heat transfer coefficient inside the tubes is \(800 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and on the fins it is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the \(\mathrm{R}-22\) is at \(320 \mathrm{~K}\) and the air is at \(300 \mathrm{~K}\), calculate the heat transfer per unit length of tube. Take \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

A test rig for the measurement of interfacial conductance is used to determine the effect of surface anodization treatment on the interfacial conductance for aluminum-aluminum contact. The specimens themselves form the heat flux meters because they are each fitted with a pair of thermocouples as shown. The heat flux is determined from the measured temperature gradient and a known thermal conductivity of the aluminum of \(185 \mathrm{~W} / \mathrm{m} \mathrm{K}\) as $$ q_{1}=\frac{k\left(T_{1}-T_{2}\right)}{L_{1}} ; \quad q_{2}=\frac{k\left(T_{3}-T_{4}\right)}{L_{2}} ; \quad q=\frac{1}{2}\left(q_{1}+q_{2}\right) $$ and the interfacial temperatures obtained by linear extrapolation. $$ \begin{gathered} T_{5}=\frac{1}{2}\left(T_{1}+T_{2}\right)-\left[\frac{L_{3}+\frac{L_{1}}{2}}{L_{1}}\right]\left(T_{1}-T_{2}\right) ; \\ T_{6}=\frac{1}{2}\left(T_{3}+T_{4}\right)+\left[\frac{L_{4}+\frac{L_{2}}{2}}{L_{2}}\right]\left(T_{3}-T_{4}\right) \end{gathered} $$ The interfacial conductance is then obtained as $$ h_{i}=\frac{q}{\left(T_{5}-T_{6}\right)} $$ The main possible sources of error in the values of \(h_{i}\) so determined are due to uncertainty in temperature measurement and thermocouple locations. Since only temperature differences are involved, the absolute uncertainty in the individual temperature measurements is not of concern; rather it is the relative uncertainties. Previous calibrations of similar type and grade thermocouples indicate that the relative uncertainties are \(\pm 0.2^{\circ} \mathrm{C}\). Also, the techinician who drilled the thermocouple holes and installed the thermocouples estimates an uncertainty of \(\pm 0.5 \mathrm{~mm}\) in the thermocouple junction locations. At a particular pressure, the temperatures recorded are \(T_{1}=338.7 \mathrm{~K}, T_{2}=328.7 \mathrm{~K}, T_{3}=305.3 \mathrm{~K}\), \(T_{4}=295.0 \mathrm{~K}\). (i) Assuming that the uncertainties in temperature measurement and thermocouple location can be treated as random errors, estimate the uncertainty in the interfacial conductance. (ii) In reality, the uncertainties in temperature and location are bias errors (for example, the location of a thermocouple does not vary from test to test). Thus it is more appropriate to determine bounds on the possible error in \(h_{i}\), by considering best and worst cases. Determine these bounds.

A gas turbine rotor has 54 AISI 302 stainless steel blades of dimensions \(L=6 \mathrm{~cm}\), \(A_{c}=4 \times 10^{-4} \mathrm{~m}^{2}\), and \(\mathscr{P}=0.1 \mathrm{~m}\). When the gas stream is at \(900^{\circ} \mathrm{C}\), the temperature at the root of the blades is measured to be \(500^{\circ} \mathrm{C}\). If the convective heat transfer coefficient is estimated to be \(440 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), calculate the heat load on the rotor internal cooling system.

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