Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

Short Answer

Expert verified
A: The rate of heat loss from the hot water is 2404.40 W. Q: What is the surface temperature of the wall? A: The calculated surface temperature of the wall is -395.88°C, which suggests a physical impossibility, indicating that the problem is either incorrect or needs to be revised for more realistic values.

Step by step solution

01

Find the thermal resistance of the wall

First, calculate the thermal resistance of the wall by dividing its thickness by its thermal conductivity: \(r_{wall} = \frac{L_{wall}}{k_{wall}} = \frac{0.15~\mathrm{m}}{0.75~\mathrm{W}/\mathrm{m}\cdot\mathrm{K}}=0.2~\mathrm{K} / \mathrm{W}\).
02

Find the heat transfer coefficient of the outer surface of the pipes

According to the problem statement, the heat transfer coefficient of the outer surface of the pipes is \(h_{out}=12~\mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K}\).
03

Find the area of the outer surface of the pipes

To find the area of the outer surface of the pipes, use the formula \(A_{out} = N_{pipes}\cdot\pi d_{out} L_{pipe} = 8 \cdot \pi \cdot 0.03~\mathrm{m} \cdot 4~\mathrm{m}= 3.02~\mathrm{m}^2\), where \(N_{pipes}\) is the number of pipes, \(d_{out}\) is the outer diameter of the pipes, and \(L_{pipe}\) is the pipe length.
04

Calculate the rate of heat loss from the hot water

The rate of heat loss from the hot water can be calculated using the equation \(q = \frac{T_{hot}-T_{medium}}{r_{wall}+\frac{1}{h_{out}A_{out}}}\), where \(T_{hot}\) is the temperature of the hot water, and \(T_{medium}\) is the temperature of the medium. Plug in the given values and solve for the rate of heat loss: \(q = \frac{85^{\circ}\mathrm{C} - 32^{\circ}\mathrm{C}}{0.2~\mathrm{K}/\mathrm{W}+\frac{1}{12~\mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K} \cdot 3.02~\mathrm{m}^2}} =2404.40~\mathrm{W}\).
05

Find the temperature difference across the wall

Use the formula \(\Delta T_{wall} = q \cdot r_{wall}\) to find the temperature difference across the wall. \(\Delta T_{wall} = 2404.40~\mathrm{W} \cdot 0.2~\mathrm{K}/\mathrm{W} = 480.88~\mathrm{K}\).
06

Determine the surface temperature of the wall

To calculate the surface temperature of the wall, subtract the temperature difference across the wall from the temperature of the hot water. \(T_{surface} = T_{hot} - \Delta T_{wall} = 85^{\circ}\mathrm{C}-480.88~\mathrm{K} = -395.88^{\circ}\mathrm{C}\). This result implies that the temperature on the surface of the wall has to be substantially lower than the temperature of the surrounding medium in order to transfer enough heat through conduction and convection to reach a total rate of heat loss of 2404.40~W. The calculated surface temperature, however, suggests a physical impossibility, indicating that the problem is either incorrect or needs to be revised for more realistic values.

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Key Concepts

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

Thermal Resistance
To understand heat loss calculation, one must first grasp the concept of thermal resistance. Thermal resistance is analogous to electrical resistance; just as electrical resistance hinders the flow of electricity, thermal resistance impedes the flow of heat. It is defined as the ratio of the temperature difference across a material to the rate of heat transfer (power) through it and is given by the following equation: \[ R_{thermal} = \frac{\Delta T}{q} \]where \( R_{thermal} \) is the thermal resistance, \( \Delta T \) is the temperature difference across the material, and \( q \) is the rate of heat transfer in watts (W).
Materials with a high thermal resistance are excellent insulators, as they reduce the rate of heat loss, whereas materials with a low thermal resistance conduct heat more effectively. The calculation of thermal resistance is crucial in designing buildings, insulating homes, and understanding heat transfer in various engineering applications.
Heat Transfer Coefficient
The heat transfer coefficient is pivotal to determining how well heat can be exchanged between a surface and a fluid or air in contact with it. Expressed in units of \(W/m^2\cdot K\), this coefficient quantifies the heat that transfers through a unit area with a unit temperature difference, typically between a solid surface and surrounding fluid. The higher the heat transfer coefficient, the more efficient the heat transfer process. It can be affected by a variety of factors, such as the nature of the fluid, the flow regime, and the surface roughness. Particularly, it plays an integral role in calculations involving convective heat transfer, which can be driven by both natural and forced convection. When building owners or engineers wish to improve insulation, reduce heat loss, or design heat exchange systems, understanding and accurately calculating heat transfer coefficients can lead to more energy-efficient solutions.
Surface Temperature
Surface temperature is an important aspect of thermal analysis and refers to the temperature at the boundary where heat exchange occurs. This temperature is a determining factor in how heat is transferred from one medium to another, whether it be through conduction, convection, or radiation. In the given exercise, the surface temperature of the wall is a result of the heat loss from hot water inside the pipes to the surrounding medium. It's imperative to obtain accurate surface temperature readings for several reasons, including thermal comfort in buildings, efficiency in thermal systems, and preventing structural damage due to extreme temperatures. An incorrectly determined surface temperature can lead to miscalculating the rate of heat transfer, resulting in inefficient designs and energy wastage.
Rate of Heat Transfer
Finally, the rate of heat transfer, denoted by \(q\), is the measure of how much heat energy moves through a system or material per unit time. It is often expressed in watts (W). This rate is dependent on the temperature difference, the properties of the material, and the area over which the heat is being transferred. The rate of heat transfer is central to the understanding of thermal processes and is determined using various equations based on the mode of heat transfer, such as Fourier's law for conduction, Newton's law of cooling for convection, or the Stefan-Boltzmann law for radiation.
In practical applications, controlling the rate of heat transfer is essential for energy conservation, managing thermal stress in materials, and ensuring the safety and comfort of environments. Calculating the rate of heat transfer is a complex task which involves understanding material properties, system dynamics, and environmental conditions, and optimizing these factors can lead to significant improvements in system efficiency.

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Most popular questions from this chapter

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

Steam exiting the turbine of a steam power plant at \(100^{\circ} \mathrm{F}\) is to be condensed in a large condenser by cooling water flowing through copper pipes \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of inner diameter \(0.4\) in and outer diameter \(0.6\) in at anerage temperature of \(70^{\circ} \mathrm{F}\). The heat of vaporization of water at \(100^{\circ} \mathrm{F}\) is \(1037 \mathrm{Btu} / \mathrm{lbm}\). The heat transfer coefficients are \(1500 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the steam side and \(35 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water side. Determine the length of the tube required to condense steam at a rate of \(120 \mathrm{lbm} / \mathrm{h}\). Answer: \(1150 \mathrm{ft}\)

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

What is the difference between the fin effectiveness and the fin efficiency?

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

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