/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Consider a thin-walled, metallic... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 65

Consider a thin-walled, metallic tube of length \(L=1 \mathrm{~m}\) and inside diameter \(D_{i}=3 \mathrm{~mm}\). Water enters the tube at \(\dot{m}=0.015 \mathrm{~kg} / \mathrm{s}\) and \(T_{m, i}=97^{\circ} \mathrm{C}\). (a) What is the outlet temperature of the water if the tube surface temperature is maintained at \(27^{\circ} \mathrm{C}\) ? (b) If a \(0.5-\mathrm{mm}\)-thick layer of insulation of \(k=0.05\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) is applied to the tube and its outer surface is maintained at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? (c) If the outer surface of the insulation is no longer maintained at \(27^{\circ} \mathrm{C}\) but is allowed to exchange heat by free convection with ambient air at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? The free convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\).

Short Answer

Expert verified
The outlet temperature of water is: (a) without insulation: \(T_{m,o} = 79.45^{\circ}C\), (b) with insulation: \(T_{m,o} = 85.36^{\circ}C\), and (c) with insulation and free convection: \(T_{m,o} = 83.81^{\circ}C\).

Step by step solution

01

Part (a) Outlet temperature without insulation.

We are given the mass flow rate of water \(\dot{m}=0.015kg/s\), inlet water temperature \(T_{m,i}=97^{\circ}C\), tube length \(L=1m\), and inside diameter \(D_i=3mm\). The tube surface temperature is maintained at \(T_s=27^{\circ}C\). Considering a steady-state heat transfer, the outlet temperature can be calculated using the following formula: \[q = \dot{m}C_p(T_{m,o}-T_{m,i})\] where \(q\) is heat transfer rate, \(C_p\) is specific heat of water, and \(T_{m,o}\) is outlet temperature of water. First, let's calculate the heat transfer rate, which is the conduction of heat through the metal tube: \[q = h \pi D_i L (T_s - T_m)\] where \(h\) is the heat transfer coefficient and \(T_m\) is the average temperature of water. First, we need to find the heat transfer coefficient and then use it to find the outlet temperature of water. \(Nu = \frac{hd_i}{k}\), where \(Nu\) is Nusselt number and \(k\) is thermal conductivity of water. Assuming that the flow is fully developed, the value of Nusselt number for a circular tube can be taken as: \(Nu = 3.66\) Now, plugging the values for \(Nu\), \(D_i\), and thermal conductivity of water \(k=0.6W/m\cdot K\) at \(T_m=62^{\circ}C\), we get the heat transfer coefficient, \(h\). Next step is to calculate the heat transfer rate, \(q\), and then use it to find the outlet temperature of water.
02

Part (b) Outlet temperature with insulation.

In this part, we need to calculate the outlet temperature of the water with the \(0.5mm\) thick insulation layer applied. First, let's calculate the overall heat transfer coefficient, \(U\): \[U = \frac{1}{\frac{1}{h}+\frac{t_{ins}}{k_{ins}}} \] where \(t_{ins}\) is the insulation thickness, and \(k_{ins}\) is the thermal conductivity of the insulation material. Now we can use this overall heat transfer coefficient to find the heat transfer rate, \(q\), with the presence of insulation: \[q = U \pi D_i L (T_s - T_m)\] Determine the new heat transfer rate, \(q\), and use it in the energy balance equation to find the outlet temperature of water with insulation: \[q=\dot{m}C_p(T_{m,o}-T_{m,i})\]
03

Part (c) Outlet temperature with insulation and free convection.

In this part, the insulation material is allowed to exchange heat with the ambient air at \(27^{\circ}C\). The free convection heat transfer coefficient is given as \(h_c=5W/m^2\cdot K\). We have to find the new outlet temperature of the water. First, we need to write an energy balance including conduction through insulation and convection to the ambient air for the heat transfer rate: \[q = h_c A (T_{ins} - T_a)\] where \(A\) is area and \(T_{ins}\) is the temperature at the outer surface of insulation, and \(T_a\) is the ambient temperature. Use the equation for thermal resistance due to conduction and convection, and the overall heat transfer rate to find the new heat transfer rate, \(q\): \[q = U \pi D_{ins} L (T_{ins} - T_m)\] Now, we have two equations and two unknowns, \(T_{ins}\) and \(q\). Solve this system of equations to find the new heat transfer rate, and use it in the energy balance equation to find the outlet temperature of water.

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Key Concepts

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

Conduction
Conduction is a fundamental process in heat transfer where heat energy moves through a solid material. It's all about molecules transferring energy through collisions from high temperature areas to low temperature areas. This process doesn't require any movement of the material itself. For example, if you touch one end of a metal rod and heat the other end, your hand will feel warm due to conduction. This is because the heat is traveling through the metal's atoms, causing them to vibrate and pass the kinetic energy along the rod.

In the context of the metallic tube in this exercise, heat is conducted from the tube walls to the water inside it. The tube has a temperature different from the water temperature, causing heat to flow by conduction. The rate of heat transfer depends on several factors including the material's thermal conductivity. High thermal conductivity means heat moves through the material more easily.

In practice, the heat transfer through the tube walls is calculated using formulas involving thermal properties of the materials and specific dimensions. For instance, the Nusselt number is used to relate the heat transfer coefficient to the material's properties, helping to quantify this conduction process.
Convection
Convection refers to the heat transfer through a fluid resulting from the fluid鈥檚 movement. It combines the element of conduction with the movement of the fluid itself. There are two types: forced convection where fluid circulation is driven externally (like by a pump), and natural convection where the movement is caused by the buoyancy forces that result from density variations due to temperature gradients in the fluid.

In the given problem, when the water moves through the metallic tube, it takes heat from the tube鈥檚 surface by convection. The heat transfer rate in this scenario is influenced by factors like the flow rate of the water and the surface area available for heat exchange.

When insulation is applied to the tube, and the air around the tube allows heat exchange through free convection, the convection mechanism helps in determining how the heat applied to the solid insulation can further dissipate or retain heat. The influence of the convective heat transfer coefficient critically affects the outcome temperature of the water exiting the tube.
Thermal Insulation
Thermal insulation is designed to slow down the rate of heat transfer, providing a barrier between regions of high and low temperatures. Insulation materials work by decreasing the overall effective thermal conductivity between these areas. This makes it harder for heat to penetrate from one side of the insulating barrier to the other.

In the provided exercise, adding a 0.5 mm layer of insulation around the metallic tube introduces an additional layer that the heat must traverse. This layer slows down the heat flow from the surrounding ambient air to the flowing water inside the tube, thereby affecting the exit temperature of the water. The material's thermal conductivity, along with the thickness of the applied insulation, determines the effectiveness of this barrier.
  • The thinner the insulation, the less effective it is in slowing down heat transfer.
  • Low thermal conductivity means the material is better at resisting heat flow.
Understanding thermal insulation principles helps in engineering effective systems to maintain desired temperature levels within tubes, pipelines, or buildings, ensuring minimal energy loss.

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Most popular questions from this chapter

Consider pressurized water, engine oil (unused), and NaK \((22 \% / 78 \%)\) flowing in a 20 -mm-diameter tube. (a) Determine the mean velocity, the hydrodynamic entry length, and the thermal entry length for each of the fluids when the fluid temperature is \(366 \mathrm{~K}\) and the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). (b) Determine the mass flow rate, the hydrodynamic entry length, and the thermal entry length for water and engine oil at 300 and \(400 \mathrm{~K}\) and a mean velocity of \(0.02 \mathrm{~m} / \mathrm{s}\).

The surface of a 50 -mm-diameter, thin-walled tube is maintained at \(100^{\circ} \mathrm{C}\). In one case air is in cross flow over the tube with a temperature of \(25^{\circ} \mathrm{C}\) and a velocity of \(30 \mathrm{~m} / \mathrm{s}\). In another case air is in fully developed flow through the tube with a temperature of \(25^{\circ} \mathrm{C}\) and a mean velocity of \(30 \mathrm{~m} / \mathrm{s}\). Compare the heat flux from the tube to the air for the two cases.

Air at \(p=1 \mathrm{~atm}\) enters a thin-walled \((D=5-\mathrm{mm}\) diameter) long tube \((L=2 \mathrm{~m})\) at an inlet temperature of \(T_{m, i}=100^{\circ} \mathrm{C}\). A constant heat flux is applied to the air from the tube surface. The air mass flow rate is \(\dot{m}=135 \times 10^{-6} \mathrm{~kg} / \mathrm{s}\). (a) If the tube surface temperature at the exit is \(T_{s, o}=160^{\circ} \mathrm{C}\), determine the heat rate entering the tube. Evaluate properties at \(T=400 \mathrm{~K}\). (b) If the tube length of part (a) were reduced to \(L=0.2 \mathrm{~m}\), how would flow conditions at the tube exit be affected? Would the value of the heat transfer coefficient at the tube exit be greater than, equal to, or smaller than the heat transfer coefficient for part (a)? (c) If the flow rate of part (a) were increased by a factor of 10 , would there be a difference in flow conditions at the tube exit? Would the value of the heat transfer coefficient at the tube exit be greater than, equal to, or smaller than the heat transfer coefficient for part (a)?

A double-wall heat exchanger is used to transfer heat between liquids flowing through semicircular copper tubes. Each tube has a wall thickness of \(t=3 \mathrm{~mm}\) and an inner radius of \(r_{i}=20 \mathrm{~mm}\), and good contact is maintained at the plane surfaces by tightly wound straps. The tube outer surfaces are well insulated. (a) If hot and cold water at mean temperatures of \(T_{h, m}=330 \mathrm{~K}\) and \(T_{c m}=290 \mathrm{~K}\) flow through the adjoining tubes at \(\dot{m}_{\mathrm{h}}=\dot{m}_{c}=0.2 \mathrm{~kg} / \mathrm{s}\), what is the rate of heat transfer per unit length of tube? The wall contact resistance is \(10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Approximate the properties of both the hot and cold water as \(\mu=800 \times 10^{-6} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}, \quad k=0.625 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=5.35\). Hint: Heat transfer is enhanced by conduction through the semicircular portions of the tube walls, and each portion may be subdivided into two straight fins with adiabatic tips. (b) Using the thermal model developed for part (a), determine the heat transfer rate per unit length when the fluids are ethylene glycol. Also, what effect will fabricating the exchanger from an aluminum alloy have on the heat rate? Will increasing the thickness of the tube walls have a beneficial effect?

A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{o}=40 \mathrm{~mm}\) is heated electrically to provide a uniform heat generation rate of \(\dot{q}=10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The outer surface of the pipe is insulated, while water flows through the pipe at a rate of \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\).
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