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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

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

Hot water is flowing at an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) through a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\), respectively. The pipe passes through a \(15-\mathrm{m}\)-long section of a basement whose temperature is \(15^{\circ} \mathrm{C}\). If the temperature of the water drops from \(70^{\circ} \mathrm{C}\) to \(67^{\circ} \mathrm{C}\) as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. Answer: \(272.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Hot water \(\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through a 200-m-long PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\), entering at \(40^{\circ} \mathrm{C}\). If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(39^{\circ} \mathrm{C}\) (b) \(38^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(36^{\circ} \mathrm{C}\) (e) \(35^{\circ} \mathrm{C}\)

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Consider two walls, \(A\) and \(B\), with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is \(k_{A} / k_{B}=4\) and the ratio of the wall thicknesses is \(L_{A} / L_{B}=2\). The ratio of heat transfer rates through the walls \(\dot{Q}_{A} / \dot{Q}_{B}\) is (a) \(0.5\) (b) 1 (c) \(2 \quad(d) 4\) (e) 8
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