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Chapter 1: Problem 11

The heat flux that is applied to one face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The opposite face is exposed to air at temperature \(30^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the wall exposed to air is measured and found to be \(50^{\circ} \mathrm{C}\). Do steady-state conditions exist? If not, is the temperature of the wall increasing or decreasing with time?

Short Answer

Expert verified
Steady-state conditions do not exist in this case since the heat transfer from the wall to the air (\(q^{\prime\prime}_{out} = 400 \, \mathrm{W/m^2}\)) is greater than the heat flux entering the wall (\(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\)). Since the wall is losing heat faster than it is gaining it, the temperature of the wall is decreasing with time.

Step by step solution

01

Calculate the heat transfer from the wall to the air

Using the convection heat transfer formula, we can calculate the heat transfer rate between the wall and the air: \[q' = hA(T_s - T_\infty)\], where: - \(q'\) is the heat transfer per unit area, - \(h\) is the convection heat transfer coefficient, - \(A\) is the area of the surface, - \(T_s\) is the surface temperature, - \(T_\infty\) is the ambient air temperature. In this case, we are interested in calculating the heat transfer per unit area (\(q^{\prime\prime}\)). So we can write the formula: \[q^{\prime\prime} = h(T_s - T_\infty)\]
02

Plug in the given values

We are given: - \(h = 20 \, \mathrm{W/m^2 \cdot K}\) - \(T_s = 50^{\circ}\mathrm{C}\) - \(T_\infty = 30^{\circ}\mathrm{C}\) - \(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\) (the given heat flux entering the wall) Plugging these values in the formula, we get: \[q^{\prime\prime}_{out} = 20(50 - 30)\]
03

Calculate the heat transfer per unit area

We calculate the heat transfer per unit area (the heat transfer rate between the wall and the air) using the formula derived in step 1: \[q^{\prime\prime}_{out} = 20(50 - 30) = 20 \times 20 = 400 \, \mathrm{W/m^2}\]
04

Compare the heat transfer into and out of the wall

We now compare the heat transfer into and out of the wall: - \(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\) - \(q^{\prime\prime}_{out} = 400 \, \mathrm{W/m^2}\) Since \(q^{\prime\prime}_{out} > q^{\prime\prime}_{in}\), the heat transfer away from the wall to the air is greater than the heat flux entering the wall; thus, steady-state conditions do not exist.
05

Determine if the temperature of the wall is increasing or decreasing with time

Since the heat transfer from the wall to the air is greater than the heat flux entering the wall, the wall is losing heat faster than it is gaining it. Therefore, the temperature of the wall is decreasing with time.

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Key Concepts

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

Steady-State Conditions
In the context of heat transfer, steady-state conditions occur when the temperature profile within a material does not change over time. This means that the amount of heat entering the system is equal to the amount of heat leaving it, resulting in no net change in energy. Think about it like a perfectly tuned seesaw; both sides are balanced, and there is no movement in either direction.

For the exercise given, we need to verify whether the heat entering the wall equals the heat leaving it. The heat flux into the wall is specified as 20 W/m虏, and via calculation, it鈥檚 found that the heat flux out of the wall is 400 W/m虏. Clearly, the heat leaving the wall is much greater than the heat entering it.

This imbalance implies that the conditions are not steady-state. Instead, the system is in a transient state, meaning the temperature within the wall will be changing over time as it seeks equilibrium.
Heat Flux
Heat flux is a measure of the rate of heat energy transfer per unit area. It essentially describes how much heat flows through a surface in a specific amount of time. The units for heat flux are typically watts per square meter (W/m虏), which indicates power per area.

In our exercise, there are two key heat flux values: the input heat flux and the output heat flux.
  • The **input heat flux** is given as 20 W/m虏, representing the heat applied to the wall.
  • The **output heat flux** calculated as 400 W/m虏 illustrates how much heat is being transferred from the wall to the surrounding air.
    This calculation uses the formula: \(q'' = h(T_s - T_\infty)\) where \(h\) is the convection heat transfer coefficient, \(T_s\) is the wall temperature, and \(T_\infty\) is the air temperature.
The significant difference between these two values highlights that more heat is leaving the wall than entering it. This differential triggers changes in the wall's temperature over time.
Temperature Gradient
A temperature gradient describes the rate at which temperature changes over a certain distance. In heat transfer problems, understanding the temperature gradient is crucial as it drives the flow of heat from warmer regions to cooler ones.

Temperature gradients are steep when the difference in temperature over distance is large, leading to faster heat transfer. Conversely, a gentle gradient indicates slower heat transfer.
  • In this problem, the temperature on the surface of the wall is 50掳C, while the air temperature is only 30掳C. This 20掳C difference across the wall-air boundary creates a significant temperature gradient, propelling a strong heat flow from the wall to the cooler air.
This gradient is what ultimately results in the calculated heat flux of 400 W/m虏 leaving the wall. Understanding how the temperature changes across surfaces helps predict how rapidly energy balance will be achieved or why non-steady-state conditions might occur, as it does in this scenario.

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Most popular questions from this chapter

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that must be applied to the top and side walls to ensure a heat load of less than \(500 \mathrm{~W}\), when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

The 5 -mm-thick bottom of a \(200-\mathrm{mm}\)-diameter pan may be made from aluminum \((k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) or copper \((k=390 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). When used to boil water, the surface of the bottom exposed to the water is nominally at \(110^{\circ} \mathrm{C}\). If heat is transferred from the stove to the pan at a rate of \(600 \mathrm{~W}\), what is the temperature of the surface in contact with the stove for each of the two materials?

A small sphere of reference-grade iron with a specific heat of \(447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass of \(0.515 \mathrm{~kg}\) is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to \(14^{\circ} \mathrm{C}\) in \(6.35 \mathrm{~s}\). The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of \(1.263 \mathrm{~kg}\). If the same observed temperature change occurs in \(4.59 \mathrm{~s}\), what is the specific heat of the unknown material?

Chips of width \(L=15 \mathrm{~mm}\) on a side are mounted to a substrate that is installed in an enclosure whose walls and air are maintained at a temperature of \(T_{\text {sur }}=25^{\circ} \mathrm{C}\). The chips have an emissivity of \(\varepsilon=0.60\) and a maximum allowable temperature of \(T_{s}=85^{\circ} \mathrm{C}\). (a) If heat is rejected from the chips by radiation and natural convection, what is the maximum operating power of each chip? The convection coefficient depends on the chip-to-air temperature difference and may be approximated as \(h=C\left(T_{s}-T_{\infty}\right)^{1 / 4}\), where \(C=4.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{5 / 4}\). (b) If a fan is used to maintain airflow through the enclosure and heat transfer is by forced convection, with \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum operating power?
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