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Chapter 8: Problem 74

Water at \(290 \mathrm{~K}\) and \(0.2 \mathrm{~kg} / \mathrm{s}\) flows through a Teflon tube \((k=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner and outer radii equal to 10 and \(13 \mathrm{~mm}\), respectively. A thin electrical heating tape wrapped around the outer surface of the tube delivers a uniform surface heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\), while a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained on the outer surface of the tape by ambient air at \(300 \mathrm{~K}\). What is the fraction of the power dissipated by the tape, which is transferred to the water? What is the outer surface temperature of the Teflon tube?

Short Answer

Expert verified
The fraction of power transferred to the water is given by \(f=\frac{R_h}{R_c+R_h}\), and the outer surface temperature of the Teflon tube is given by \(T_{outer}=T_{ambient}+\Delta T\), where \(\Delta T=\frac{Q_{in}}{R_t}\) and \(R_t=R_c+R_h\).

Step by step solution

01

Calculate the power dissipated by the heating tape

First, let's determine the power delivered by the electrical heating tape. The surface area of the outer Teflon tube is given by: \(A_s=2\pi R_o L\), where \(R_o\) is the outer radius of the tube and \(L\) is the length. Therefore, the power dissipated by the heating tape is given by: \(Q_{in}=q_sA_s\), where \(q_s\) is the surface heat flux.
02

Calculate the thermal resistance of the Teflon tube

The thermal resistance of the Teflon tube can be found using the following formula for conduction through a cylinder: \[R_c=\frac{\ln\frac{R_o}{R_i}}{2\pi kL}\], where \(R_i\) is the inner radius of the tube, and \(k\) is the thermal conductivity of Teflon.
03

Calculate the thermal resistance of the convection in the ambient air

The thermal resistance of the convection in the ambient air can be found using the following formula: \[R_{h}=\frac{1}{hA_c}\] where \(h\) is the convection coefficient and \(A_c\) is the outer contact area of the heating tape, which is equal to the surface area of the outer Teflon tube (\(A_s\)).
04

Calculate the total thermal resistance

The total thermal resistance is the sum of the thermal resistance of the Teflon tube and ambient air: \(R_t=R_c+R_{h}\).
05

Calculate the fraction of power transferred to the water

The fraction of power transferred to the water \(f\) can be calculated as follows: \[f=\frac{R_h}{R_c+R_h}\].
06

Calculate the temperature difference between the tube and the ambient air

The temperature difference between the outer surface of the Teflon tube and the ambient air can be found using the formula: \[\Delta T=\frac{Q_{in}}{R_t}\].
07

Calculate the outer surface temperature of the Teflon tube

Finally, we can calculate the outer surface temperature of the Teflon tube by adding the temperature difference to the ambient air temperature: \[T_{outer}=T_{ambient}+\Delta T\].

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Key Concepts

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

Conduction in Cylindrical Coordinates
Understanding heat transfer through cylindrical objects, such as tubes, is vital for various engineering applications, including the scenario where water is flowing through a Teflon tube with electrical heating tape. When analyzing conduction in cylindrical coordinates, we account for the radial temperature distribution within the material, which can change along the length and thickness of the tube. To calculate the heat conduction through the cylindrical wall, we must consider the cylinder's thermal conductivity, inner and outer radii, and length. The formula that encapsulates this is ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline. ewline
Thermal Resistance
Thermal resistance is a measure of a material's ability to resist heat flow. It is analogous to electrical resistance in circuits and is inversely related to thermal conductivity. Higher thermal resistance means that the material is less effective at conducting heat. In the context of the Teflon tube wrapped with heating tape, we analyze the thermal resistance of both the Teflon material and the convection process occurring in the ambient air. The overall heat transfer from the heating tape to the water will depend on the sum of these thermal resistances. Understanding the formula to calculate it, which involves natural logarithms and known physical properties, helps in determining how efficient the system is at transferring heat.
Convection Heat Transfer
Convection heat transfer is the process of heat transfer between a solid surface and the fluid moving over it. It involves energy transfer due to bulk movement and molecular motion of the fluid. In our exercise, convection heat transfer occurs on the outer surface of the electrical heating tape, with air moving over it. The efficiency of this heat transfer process is represented by the convection heat transfer coefficient (ewline ewline ewline ewline ewline ewline ewline ewline ewline). A higher value indicates more effective heat transfer from the surface to the air. It's crucial to calculate the thermal resistance associated with convection to evaluate how much heat is dissipated to the surroundings compared to that transferred to the water.
Electrical Heating Tape
Electrical heating tape is a tool used to apply heat to surfaces, such as pipes or tubes, by converting electrical energy into thermal energy. The tape delivers a uniform surface heat flux, described in watts per square meter. For our problem, we quantify the power dissipated by the heating tape based on this heat flux and the outer surface area of the Teflon tube. This power input is crucial for determining the portion of heat that actually transfers to the water inside the tube versus the heat lost to the ambient environment.
Surface Heat Flux
Surface heat flux is the rate of heat energy transfer per unit area, generally expressed in watts per square meter. It is an indicator of how much heat is being delivered or removed from a surface - in this case, the heat provided by electrical heating tape to the Teflon tube. By calculating the product of the given surface heat flux and the surface area of the Teflon tube, we can determine how much power the heating tape is dissipating. This allows us to assess the efficiency of the heating process and the subsequent temperature changes on the tube's surface.
Temperature Gradient
A temperature gradient is the rate of temperature change in a particular direction. It is essential in the study of heat transfer, as it drives heat exchange through conduction. In our scenario, the temperature gradient exists between the heated outer surface of the Teflon tube and the cooler water inside. The strength of the temperature gradient influences how quickly heat transfers from one location to another. A steep gradient implies rapid heat transfer, while a shallow gradient indicates slower transfer. We see its direct effect in calculating the temperature difference across the thermal resistance, which in turn gives us the outer surface temperature of the tube when combined with the ambient temperature.

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Most popular questions from this chapter

A cold plate is an active cooling device that is attached to a heat-generating system in order to dissipate the heat while maintaining the system at an acceptable temperature. It is typically fabricated from a material of high thermal conductivity, \(k_{\text {cp, }}\), within which channels are machined and a coolant is passed. Consider a copper cold plate of height \(H\) and width \(W\) on a side, within which water passes through square channels of width \(w=h\). The transverse spacing between channels \(\delta\) is twice the spacing between the sidewall of an outer channel and the sidewall of the cold plate. Consider conditions for which equivalent heat-generating systems are attached to the top and bottom of the cold plate, maintaining the corresponding surfaces at the same temperature \(T_{s}\). The mean velocity and inlet temperature of the coolant are \(u_{m}\) and \(T_{m i}\), respectively. (a) Assuming fully developed turbulent flow throughout each channel, obtain a system of equations that may be used to evaluate the total rate of heat transfer to the cold plate, \(q\), and the outlet temperature of the water, \(T_{m, o}\), in terms of the specified parameters. (b) Consider a cold plate of width \(W=100 \mathrm{~mm}\) and height \(H=10 \mathrm{~mm}\), with 10 square channels of width \(w=6 \mathrm{~mm}\) and a spacing of \(\delta=4 \mathrm{~mm}\) between channels. Water enters the channels at a temperature of \(T_{m, i}=300 \mathrm{~K}\) and a velocity of \(u_{m}=2 \mathrm{~m} / \mathrm{s}\). If the top and bottom cold plate surfaces are at \(T_{s}=360 \mathrm{~K}\), what is the outlet water temperature and the total rate of heat transfer to the cold plate? The thermal conductivity of the copper is \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while average properties of the water may be taken to be \(\rho=984 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=489 \times\) \(10^{-6} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=3.15\). Is this a good cold plate design? How could its performance be improved?

A hot fluid passes through a thin-walled tube of \(10-\mathrm{mm}\) diameter and 1-m length, and a coolant at \(T_{\infty}=25^{\circ} \mathrm{C}\) is in cross flow over the tube. When the flow rate is \(\dot{m}=18 \mathrm{~kg} / \mathrm{h}\) and the inlet temperature is \(T_{m, i}=85^{\circ} \mathrm{C}\), the outlet temperature is \(T_{m \rho}=78^{\circ} \mathrm{C}\). Assuming fully developed flow and thermal conditions in the tube, determine the outlet temperature, \(T_{m, a}\) if the flow rate is increased by a factor of 2 . That is, \(\dot{m}=36 \mathrm{~kg} / \mathrm{h}\), with all other conditions the same. The thermophysical properties of the hot fluid are \(\rho=\) \(1079 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2637 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=0.0034 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), and \(k=0.261 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider pressurized liquid water flowing at \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\) in a circular tube of diameter \(D=0.1 \mathrm{~m}\) and length \(L=6 \mathrm{~m}\). (a) If the water enters at \(T_{m, i}=500 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=510 \mathrm{~K}\), determine the water outlet temperature \(T_{\text {m,o. }}\). (b) If the water enters at \(T_{m, i}=300 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=310 \mathrm{~K}\), determine the water outlet temperature \(T_{\text {m, } \sigma}\). (c) If the water enters at \(T_{m, i}=300 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=647 \mathrm{~K}\), discuss whether the flow is laminar or turbulent.

8.105 The mold used in an injection molding process consists of a top half and a bottom half. Each half is \(60 \mathrm{~mm} \times 60 \mathrm{~mm} \times 20 \mathrm{~mm}\) and is constructed of metal \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=450 \mathrm{~J} / \mathrm{kg}+\mathrm{K}\right)\). The cold mold \(\left(100^{\circ} \mathrm{C}\right)\) is to be heated to \(200^{\circ} \mathrm{C}\) with pressurized water (available at \(275^{\circ} \mathrm{C}\) and a total flow rate of \(0.02 \mathrm{~kg} / \mathrm{s}\) ) prior to injecting the thermoplastic material. The injection takes only a fraction of a second, and the hot mold \(\left(200^{\circ} \mathrm{C}\right)\) is subsequently cooled with cold water (available at \(25^{\circ} \mathrm{C}\) and a total flow rate of \(0.02 \mathrm{~kg} / \mathrm{s})\) prior to ejecting the molded part. After part ejection, which also takes a fraction of a second, the process is repeated. (a) In conventional mold design, straight cooling (heating) passages are bored through the mold in a location where the passages will not interfere with the molded part. Determine the initial heating rate and the initial cooling rate of the mold when five 5 -mm-diameter, 60-mm-long passages are bored in each half of the mold (10 passages total). The velocity distribution of the water is fully developed at the entrance of each passage in the hot (or cold) mold. (b) New additive manufacturing processes, known as selective freeform fabrication, or \(S F F\), are used to construct molds that are configured with conformal cooling passages. Consider the same mold as before, but now a 5 -mm- diameter, coiled, conformal cooling passage is designed within each half of the SFF-manufactured mold. Each of the two coiled passages has \(N=2\) turns. The coiled passage does not interfere with the molded part. The conformal channels have a coil diameter \(C=50 \mathrm{~mm}\). The total water flow remains the same as in part (a) \((0.01 \mathrm{~kg} / \mathrm{s}\) per coil). Determine the initial heating rate and the initial cooling rate of the mold. (c) Compare the surface areas of the conventional and conformal cooling passages. Compare the rate at which the mold temperature changes for molds configured with the conventional and conformal heating and cooling passages. Which cooling passage, conventional or conformal, will enable production of more parts per day? Neglect the presence of the thermoplastic material.

Water at \(300 \mathrm{~K}\) and a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) enters a black, thin-walled tube, which passes through a large furnace whose walls and air are at a temperature of \(700 \mathrm{~K}\). The diameter and length of the tube are \(0.25 \mathrm{~m}\) and \(8 \mathrm{~m}\), respectively. Convection coefficients associated with water flow through the tube and airflow over the tube are \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Write an expression for the linearized radiation coefficient corresponding to radiation exchange between the outer surface of the pipe and the furnace walls. Explain how to calculate this coefficient if the surface temperature of the tube is represented by the arithmetic mean of its inlet and outlet values. (b) Determine the outlet temperature of the water, \(T_{m, o^{\circ}}\)
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