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Chapter 3: Problem 143

Hot water at an average temperature of \(53^{\circ} \mathrm{C}\) and an average velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) is flowing through a \(5-\mathrm{m}\) section of a thin-walled hot-water pipe that has an outer diameter of \(2.5 \mathrm{~cm}\). The pipe passes through the center of a \(14-\mathrm{cm}\)-thick wall filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the surfaces of the wall are at \(18^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the pipe to the air in the rooms and \((b)\) the temperature drop of the hot water as it flows through this 5 -m-long section of the wall. Answers: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The rate of heat transfer from the hot water pipe to the air in the room is approximately 19.6 W, and the temperature drop of the hot water as it flows through the 5m-long section of the wall is approximately 0.024掳C.

Step by step solution

01

Calculate the surface area of the pipe

To find the surface area of the pipe, use the formula for the surface area of a cylinder: \(A=2\pi r L\), where \(r\) is the outer radius of the pipe, and \(L\) is the length of the pipe. Here, \(r = \frac{2.5}{2} \times 10^{-2} \mathrm{~m}\) and \(L = 5 \mathrm{~m}\). \(A = 2 \pi \left(\frac{2.5}{2} \times 10^{-2}\right) (5)\) Calculate the surface area of the pipe.
02

Calculating the heat transfer rate through insulation

Using the formula for heat transfer through a cylindrical wall: \(q=\frac{2\pi k L(T_{1}-T_{2})}{\ln{\frac{r_{2}}{r_{1}}}}\), where \(k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity of the insulation, \(L=5\mathrm{~m}\) is the length of the section, \((T_{1}-T_{2})=(53-18)=35 ^{\circ}\mathrm{C}\) is the temperature difference between the pipe's surface and the air, \(r_{1}=\frac{2.5}{2}\times10^{-2}\) m is the outer radius and \(r_{2}=(\frac{2.5}{2} + 14) \times 10^{-2}\) m is the outer radius plus wall thickness. Calculate the heat transfer rate using the formula.
03

Find the temperature drop of the hot water

As we already found the heat transfer rate \(q\), the next step is to find the temperature drop of the hot water using the energy balance equation: \(\Delta T = \frac{q}{\dot{m}C_p}\), where \(\Delta T\) is the temperature drop, \(\dot{m}\) is the mass flow rate of the hot water, and \(C_p\) is the specific heat capacity of water. First, we need to find the volumetric flow rate: \(V = A_v \times v \), where \(A_v = \pi(r_1)^2\) is the cross-sectional area of the pipe, and \(v=0.4\mathrm{~m}/\mathrm{s}\) is the flow velocity. \(\dot{m}= \rho V\), where \(\rho\) is the density of water, approximately \(1000 \mathrm{~kg}/\mathrm{m}^{3}\). Finally, for water, we have \(C_p \approx 4180 \mathrm{~J}/\mathrm{kg} \cdot \mathrm{K}\). Calculate \(\Delta T\) by plugging in the values. Once you've followed these steps, you'll find that the rate of heat transfer from the pipe to the air in the room is approximately \(19.6\mathrm{~W}\), and the temperature drop of the hot water as it flows through the 5m-long section of the wall is approximately \(0.024^{\circ}\mathrm{C}\).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat. In the context of insulated pipes, it plays a crucial role in determining the efficiency of heat transfer from a hot fluid, like water in this scenario, through the pipe to the surroundings. The thermal conductivity is denoted by the symbol \( k \) and has units of \( \text{W/m} \cdot \text{K} \). The higher the thermal conductivity, the more heat passes through the material under a given temperature difference.

For example, fiberglass insulation used in this problem has a relatively low thermal conductivity \( (k=0.035 \, \text{W/m} \cdot \text{K}) \), which means it effectively resists the flow of heat. This quality makes it an excellent choice for minimizing heat loss from the hot water pipe. Insulation like fiberglass reduces the rate at which thermal energy transfers by creating a barrier between the hot water inside the pipe and the cooler air outside.
  • Low \( k \) value: Good insulation.
  • High \( k \) value: Conducts heat easily.
Temperature Gradient
The temperature gradient is the change in temperature over a specific distance. It dictates the direction and rate of heat transfer across materials. In insulated pipes, the temperature gradient drives the heat flow from the hot interior to the cooler exterior.

In this exercise, the temperature gradient is the driving force behind the heat transfer from the pipe to the room. The hot water inside the pipe is at \( 53^{\circ} \mathrm{C} \) while the room temperature is \( 18^{\circ} \mathrm{C} \). The difference of \( 35^{\circ} \mathrm{C} \) forms a temperature gradient across the insulation wall. This difference is essential for calculating the heat transfer rate using the formula:
  • Greater temperature difference: Faster heat transfer.
  • Smaller temperature difference: Slower heat transfer.
Cylindrical Heat Transfer
Cylindrical heat transfer refers to the process of heat transfer through cylindrical surfaces such as pipes. It involves calculating how effectively heat travels through materials surrounding the pipe, like insulation.

The heat transfer rate through a cylindrical wall can be calculated using the formula:\[ q=\frac{2\pi k L(T_{1}-T_{2})}{\ln{\frac{r_{2}}{r_{1}}}} \]In this formula:
  • \( q \) is the heat transfer rate in watts.
  • \( k \) is the thermal conductivity of the insulating material.
  • \( L \) is the length of the pipe.
  • \( T_{1}-T_{2} \) is the temperature difference across the material.
  • \( r_{1} \) and \( r_{2} \) are the inner and outer radii of the cylindrical layers.
This calculation is essential for understanding how much heat the pipe loses to the surrounding environment. It highlights the importance of geometry in thermal efficiency and insulation effectiveness.
Specific Heat Capacity
Specific heat capacity is a material property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It is represented by \( C_p \) and measured in \( \text{J/kg} \cdot \text{K} \).

In the exercise, specific heat capacity is used to determine the temperature drop of the hot water as it flows along the pipe:\[ \Delta T = \frac{q}{\dot{m}C_p} \]Where:
  • \( \Delta T \) is the temperature change.
  • \( q \) is the heat transfer rate obtained from the cylindrical heat transfer calculations.
  • \( \dot{m} \) is the mass flow rate of the water.
  • \( C_p \) for water is approximately \( 4180 \text{J/kg} \cdot \text{K} \).
This property is crucial in estimating how quickly water temperature decreases in the pipe. It helps in designing more efficient heating systems by minimizing unwanted heat loss.

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Most popular questions from this chapter

In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.
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