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Chapter 1: Problem 20

A wall has inner and outer surface temperatures of 16 and \(6^{\circ} \mathrm{C}\), respectively. The interior and exterior air temperatures are 20 and \(5^{\circ} \mathrm{C}\), respectively. The inner and outer convection heat transfer coefficients are 5 and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

Short Answer

Expert verified
In conclusion, the heat flux from the interior air to the wall and from the wall to the exterior air are both \(20\mathrm{~W} / \mathrm{m}^{2}\). The wall is under steady-state conditions as the heat flux entering the inner surface equals the heat flux leaving the outer surface.

Step by step solution

01

Identify the heat transfer equations for calculation

We can define both the heat flux equations corresponding to inner and outer surfaces as follows: For inner surface (interior air to wall): \(q_{i} = h_{i} \cdot A \cdot (T_{a} - T_{w, i})\) For outer surface (wall to exterior air): \(q_{o} = h_{o} \cdot A \cdot (T_{w, o} - T_{out})\) where q is the heat flux, h is the convection heat transfer coefficient, A is the area of the wall, T is the temperature, and the subscripts indicate inner (i) and outer (o) surface, air (a), wall (w), and outdoor (out) conditions.
02

Calculate the heat flux from the interior air to the wall

Using the inner heat transfer equation, plug in the known values for the inner convection heat transfer coefficient (\(h_{i} = 5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), interior air temperature (\(T_{a} = 20^{\circ}\mathrm{C}\)), and inner wall temperature (\(T_{w, i} = 16^{\circ}\mathrm{C}\)): \(q_{i} = 5\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (20^{\circ}\mathrm{C} - 16^{\circ}\mathrm{C})\) \(q_{i} = 20\mathrm{~W} / \mathrm{m}^{2}\)
03

Calculate the heat flux from the wall to the exterior air

Using the outer heat transfer equation, plug in the known values for the outer convection heat transfer coefficient (\(h_{o} = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), exterior air temperature (\(T_{out} = 5^{\circ}\mathrm{C}\)), and outer wall temperature (\(T_{w, o} = 6^{\circ}\mathrm{C}\)): \(q_{o} = 20\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (6^{\circ}\mathrm{C} - 5^{\circ}\mathrm{C})\) \(q_{o} = 20\mathrm{~W} / \mathrm{m}^{2}\)
04

Check if the wall is under steady-state conditions

A wall is under steady-state conditions when the heat flux entering the inner surface equals the heat flux leaving the outer surface. Compare the heat flux calculated in step 2 and 3: \(q_{i} = q_{o}\) \(20\mathrm{~W} / \mathrm{m}^{2} = 20\mathrm{~W} / \mathrm{m}^{2}\) Since the heat fluxes are equal, the wall is under steady-state conditions. In conclusion, the heat flux from the interior air to the wall and from the wall to the exterior air are both \(20\mathrm{~W} / \mathrm{m}^{2}\). The wall is under steady-state conditions as the heat flux entering the inner surface equals the heat flux leaving the outer surface.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer between surfaces and fluids is crucial for controlling temperatures. It occurs when heat moves because of fluid motion over a surface. The rate of heat transfer depends on the convection heat transfer coefficient (\( h \)), which varies based on factors like fluid speed and properties.

In our scenario, the wall's surfaces are interacting with air on both sides. The **interior air** moving over the inner surface has a convection coefficient of \(5 \, \text{W/m}^2 \cdot \text{K}\). The **exterior air** has a higher coefficient of \(20 \, \text{W/m}^2 \cdot \text{K}\). A higher coefficient on the outside indicates more efficient heat removal.

Using these coefficients, we can calculate how much heat flows from the air to the wall and vice versa using the formula:
  • For the inner surface: \( q_i = h_i \cdot A \cdot (T_a - T_{w, i})\)
  • For the outer surface: \( q_o = h_o \cdot A \cdot (T_{w, o} - T_{out})\)
This tells us the heat per unit area moving due to the temperature differences.
Steady-State Conditions
Steady-state conditions occur when the heat entering and leaving a system is balanced, showing no net change over time. In other words, the temperatures remain constant.

For the wall in this exercise, at **steady-state**:
  • The heat flow from the interior air to the wall equals the heat flow from the wall to the exterior air.
In mathematical terms: \( q_i = q_o \).

The calculations showed that the heat flux is \(20 \, \text{W/m}^2\) on both sides, confirming steady-state conditions. This means the amount of heat entering the wall matches the amount being transferred out, ensuring no accumulation of heat within the wall itself.
Heat Flux Calculation
Heat flux quantifies the rate of heat transfer through a surface area. It is critically important for designing systems that manage heat. The formula for calculating heat flux in convection is:
  • \( q = h \cdot (T_1 - T_2)\)
This represents how heat flows between two points, with units typically in \( \text{W/m}^2\).

For this wall:
  • The **interior to wall** heat flux was calculated as \( q_i = 20 \, \text{W/m}^2\).
  • The **wall to exterior** heat flux was also \( q_o = 20 \, \text{W/m}^2\).
These consistent values assure us that heat movement through the wall is stable鈥攁nd efficient. Understanding heat flux allows engineers and architects to ensure structures can maintain desired temperatures without excess energy input.

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Most popular questions from this chapter

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?

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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