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Chapter 3: Problem 87

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

Short Answer

Expert verified
Question: Determine the rate of heat loss from hot water flowing through a pipe and the average velocity of the water using the given heat transfer coefficients, dimensions, and thermal conductivity of the pipe. Answer: To find the rate of heat loss and average velocity of water flowing through the pipe, perform the following steps: 1. Calculate the heat loss through conduction by applying the equation for steady-state, one-dimensional heat conduction and accounting for the pipe's dimensions. 2. Determine heat loss through convection from the outer surface of the pipe to the cold air in the basement using the given heat transfer coefficient. 3. By conservation of energy, find the total heat loss rate by equating the heat loss through conduction and the heat loss through convection. 4. Use an energy balance equation, the given temperature change, and the total heat loss rate to solve for the average velocity of water in the pipe.

Step by step solution

01

Determine heat loss through conduction

First, we need to find the heat loss through conduction in the pipe wall. To do this, we can apply the equation for steady-state, one-dimensional heat conduction: $$q_{cond} = kA\frac{T_i-T_o}{\Delta x}$$ where \(q_{cond}\) is the heat loss through conduction, \(k\) is the thermal conductivity of the pipe, \(A\) is the area of the pipe wall, \(T_i\) and \(T_o\) are the temperatures at the inner and outer surfaces of the pipe, and \(\Delta x\) is the thickness of the pipe. To find the thickness and the surface area, we'll use the given diameters. Pipe thickness (\(\Delta{x}\)) can be found by taking the difference between the outer and inner radii: $$\Delta x = \frac{D_o - D_i}{2}$$ Furthermore, we can find the surface area of the pipe wall by the equation: $$A = \pi(D_o + D_i)L$$ where \(L\) is the length of the pipe. Now we can substitute the known values and solve for \(q_{cond}\).
02

Calculate heat loss through convection

Next, we need to determine the heat loss through convection from the outer surface of the pipe to the cold air in the basement. We can do this using the equation for convection: $$q_{conv} = h_{conv}A(T_o - T_\infty)$$ where \(q_{conv}\) is the heat loss through convection, \(h_{conv}\) is the heat transfer coefficient at the outer surface of the pipe, \(T_\infty\) is the temperature of the basement air, and \(A\) is the surface area of the pipe wall (\(A = \pi D_o L\)). Now we can substitute the known values and solve for \(q_{conv}\).
03

Find the total heat loss

By the conservation of energy, the heat loss through the pipe wall by conduction must be equal to the heat loss by convection; thus, the total heat loss rate is: $$q_{total} = q_{cond} = q_{conv}$$
04

Calculate the average velocity of the water

Now, to find the average velocity of the water, we can apply the energy balance to the hot water. The energy input to the water should be equal to the energy taken out by the heat loss through the pipe wall. For this, we can use the equation: $$m\dot{V}c_p(T_{in} - T_{out}) = q_{total}$$ where \(m\dot{V}\) is the mass flow rate of the water, \(c_p\) is the specific heat capacity of the water, \(T_{in}\) and \(T_{out}\) are the inlet and outlet temperatures of the water, and \(q_{total}\) is the total heat loss rate. We are given the temperature drop as \(3^{\circ}C\). Thus, the temperature change, \((T_{in} - T_{out}) = 3^{\circ}C\). By rearranging the equation, we can find the average velocity of the water: $$\bar{v} = \frac{q_{total}}{\rho A_c c_p (T_{in} - T_{out})}$$ where \(\bar{v}\) is the average velocity, \(\rho\) is the density of the water, and \(A_c\) is the cross-sectional area of the pipe (\(A_c = \pi (\frac{D_i}{2})^2\)). Now we can substitute the known values and solve for the average velocity of the water in the pipe.

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Key Concepts

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

Conduction Heat Transfer
In the context of the exercise, heat loss through conduction in pipes occurs when heat energy travels from the hot water inside the pipe to the colder exterior environment. Conduction is a mode of heat transfer that involves direct molecular interactions and the passage of heat through a material without the movement of the material itself.

For our cast iron pipe, the thermal energy moves from the hot water at its inner surface to the cooler outer surface. The rate at which this energy is transferred depends on several factors: the material's thermal conductivity (represented by the symbol k), the temperature difference between the inner and outer walls (Ti and To, respectively), the thickness of the pipe wall (Δx), and the surface area through which the heat is being transferred (A).

The formula qcond = kA(Ti - To)/Δx quantifies how much heat is lost by conduction. The thermal conductivity k is a measure of the material's ability to conduct heat; a higher k means more efficient heat transfer. The exercise gives us all the needed values to calculate the heat loss due to conduction through the pipe wall.
Convective Heat Transfer
Once the heat reaches the outer surface of the pipe, it is then transferred to the surrounding air via convective heat transfer. Convection involves the transfer of heat by the physical movement of a fluid (which can be liquid or gas) over a surface.

In this scenario, the pipe's outer surface is exposed to the basement air, initiating convective heat transfer. The efficiency of this transfer is quantified by the heat transfer coefficient, denoted as hconv, which in our exercise is 15 W/m2·K. The greater this coefficient, the more efficient the convection process.

The convection heat loss can be calculated using the equation qconv = hconvA(To - T∞), where T∞ represents the air temperature outside the pipe. This tells us how much heat the pipe loses to the surrounding air due to convection.
Thermal Conductivity
Thermal conductivity is a fundamental property of materials that indicates the ease with which heat can pass through them. Represented by the symbol k, it is a measure of the material's ability to conduct thermal energy. In our exercise, the cast iron pipe has a thermal conductivity of 52 W/m·K.

Different materials have different thermal conductivities. Metals, for example, typically have high thermal conductivities, which makes them good conductors of heat. Insulators, such as rubber or wood, have low thermal conductivities, and thus are poor heat conductors. The value for thermal conductivity is crucial in determining the rate of heat loss through a material, and it plays a direct role in the equations we use to calculate both conduction and convection heat loss.
Heat Transfer Coefficient
The heat transfer coefficient (;emh;) is a value that represents the convective heat transfer capability between a solid surface and a fluid. It's an indicator of how well the fluid absorbs heat from the surface or releases heat to the surface.

In the given exercise, the heat transfer coefficient for the cold air in the basement is 15 W/m2·K, whereas the inner surface of the pipe has a value of 120 W/m2·K due to the flowing hot water. The inner heat transfer coefficient is typically higher because liquids, especially when they are forced to move as in this scenario, tend to have higher convective heat exchange rates compared with gases.

In our calculations, the coefficient helps us understand the rate at which the heat energy is being transferred from the pipe's surface to the air or vice versa. It's a vital part of the convective heat loss equation and can dramatically impact thermal management in systems such as heating, ventilation, and air conditioning.

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Most popular questions from this chapter

Explain how the thermal contact resistance can be minimized.

Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made of five layers of \(0.1-\mathrm{mm}\)-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(1.5\)-mm- thick air space \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the layers. Assuming the inner surface temperature of the jacket to be \(28^{\circ} \mathrm{C}\) and the surface area to be \(1.25 \mathrm{~m}^{2}\), determine the rate of heat loss through the jacket when the temperature of the outdoors is \(0^{\circ} \mathrm{C}\) and the heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What would your response be if the jacket is made of a single layer of \(0.5-\mathrm{mm}\)-thick synthetic fabric? What should be the thickness of a wool fabric ( \(k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) if the person is to achieve the same level of thermal comfort wearing a thick wool coat instead of a five-layer ski jacket?

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.

What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not?

In an experiment to measure convection heat transfer coefficients, a very thin metal foil of very low emissivity (e.g., highly polished copper) is attached on the surface of a slab of material with very low thermal conductivity. The other surface of the metal foil is exposed to convection heat transfer by flowing fluid over the foil surface. This setup diminishes heat conduction through the slab and radiation on the metal foil surface, while heat convection plays the prominent role. The slab on which the metal foil is attached to has a thickness of \(25 \mathrm{~mm}\) and a thermal conductivity of \(0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). In a condition where the surrounding room temperature is \(20^{\circ} \mathrm{C}\), the metal foil is heated electrically with a uniform heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). If the bottom surface of the slab is \(20^{\circ} \mathrm{C}\) and the metal foil has an emissivity of \(0.02\), determine \((a)\) the convection heat transfer coefficient if air is flowing over the metal foil and the surface temperature of the foil is \(150^{\circ} \mathrm{C}\); and \((b)\) the convection heat transfer coefficient if water is flowing over the metal foil and the surface temperature of the foil is \(30^{\circ} \mathrm{C}\).
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