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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

Following the hot vacuum forming of a paper-pulp mixture, the product, an egg carton, is transported on a conveyor for \(18 \mathrm{~s}\) toward the entrance of a gas-fired oven where it is dried to a desired final water content. Very little water evaporates during the travel time. So, to increase the productivity of the line, it is proposed that a bank of infrared radiation heaters, which provide a uniform radiant flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), be installed over the conveyor. The carton has an exposed area of \(0.0625 \mathrm{~m}^{2}\) and a mass of \(0.220 \mathrm{~kg}, 75 \%\) of which is water after the forming process. The chief engineer of your plant will approve the purchase of the heaters if they can reduce the water content by \(10 \%\) of the total mass. Would you recommend the purchase? Assume the heat of vaporization of water is \(h_{f g}=2400 \mathrm{~kJ} / \mathrm{kg}\).

For a boiling process such as shown in Figure \(1.5 c\), the ambient temperature \(T_{\infty}\) in Newton's law of cooling is replaced by the saturation temperature of the fluid \(T_{\text {sat }}\). Consider a situation where the heat flux from the hot plate is \(q^{\prime \prime}=20 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). If the fluid is water at atmospheric pressure and the convection heat transfer coefficient is \(h_{w}=20 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the upper surface temperature of the plate, \(T_{s, w^{\circ}}\). In an effort to minimize the surface temperature, a technician proposes replacing the water with a dielectric fluid whose saturation temperature is \(T_{\text {sat,d }}=52^{\circ} \mathrm{C}\). If the heat transfer coefficient associated with the dielectric fluid is \(h_{d}=3 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), will the technician's plan work?

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?

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

The concrete slab of a basement is \(11 \mathrm{~m}\) long, \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02 / \mathrm{MJ}\), what is the daily cost of the heat loss?
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