/*! 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 19 Hot water at \(85^{\circ} \mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 19

Hot water at \(85^{\circ} \mathrm{C}\) flws through a thin-walled copper tube of \(30-\mathrm{mm}\) diameter. The tube is enclosed by an eccentric cylindrical shell that is maintained at \(35^{\circ} \mathrm{C}\) and has a diameter of \(120 \mathrm{~mm}\). The eccentricity, defined as the separation between the centers of the tube and shell, is \(20 \mathrm{~mm}\). The space between the tube and shell is filled with an insulating material having a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Calculate the heat loss per unit length of the tube, and compare the result with the heat loss for a concentric arrangement.

Short Answer

Expert verified
The heat loss per unit length for the eccentric arrangement is approximately 11.72 W/m, while for the concentric arrangement, it is approximately 5.22 W/m. The heat transfer rate is higher in the eccentric arrangement due to the smaller distance between the copper tube and the outer shell compared to the concentric arrangement.

Step by step solution

01

Determine the temperature difference

The temperature difference (螖T) between the hot water in the tube and the outer shell can be calculated by simply subtracting the temperature of the outer shell from the temperature of the hot water: \(\Delta T = T_1 - T_2 = 85^{\circ} \mathrm{C} - 35^{\circ} \mathrm{C} = 50^{\circ} \mathrm{C}\)
02

Calculate the surface area of the copper tube per unit length

To calculate the heat loss per unit length of the tube, we first need to determine the surface area of the copper tube per unit length. The surface area can be calculated using the formula: \(A = \pi D\) where D is the diameter of the copper tube. In this case, the diameter given is 30 mm: \(A = \pi (30 \times 10^{-3} \mathrm{m}) = 0.0942 \mathrm{m^2/m}\)
03

Calculate the distance (L) between the tube and the point of heat loss at the shell

Next, we need to determine the distance between the copper tube wall and the point where the heat loss reaches the outer shell. In this eccentric arrangement, the given eccentricity is 20 mm: \(L = 20 \times 10^{-3} \mathrm{m}\)
04

Calculate the heat loss per unit length of the tube

Now we have all the values to calculate the heat loss per unit length of the tube using the formula mentioned earlier: \(Q = \frac{kA\Delta T}{L} = \frac{0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 0.0942 \mathrm{m^2/m} \cdot 50^{\circ} \mathrm{C}}{20 \times 10^{-3} \mathrm{m}} = 11.72 \mathrm{~W} / \mathrm{m}\) So, the heat loss per unit length of the tube is approximately 11.72 W/m.
05

Calculate heat loss per unit length for a concentric arrangement

For a concentric arrangement, the distance between the tube and the cylindrical shell would be equal to the tube's radius, given by: \(L_{conc} = \frac{D_{shell}}{2} - \frac{D_{tube}}{2}\) \(L_{conc} = \frac{120 \times 10^{-3} \mathrm{m}}{2} - \frac{30 \times 10^{-3} \mathrm{m}}{2} = 45 \times 10^{-3} \mathrm{m}\) Using the same formula for heat loss, we can calculate the heat loss per unit length of the tube for a concentric arrangement: \(Q_{conc} = \frac{kA\Delta T}{L_{conc}} = \frac{0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 0.0942 \mathrm{m^2/m} \cdot 50^{\circ} \mathrm{C}}{45 \times 10^{-3} \mathrm{m}} = 5.22 \mathrm{~W} / \mathrm{m}\) Now we have the heat loss for both eccentric and concentric arrangements. The heat loss in the eccentric arrangement is approximately 11.72 W/m, and in the concentric arrangement, it is approximately 5.22 W/m. The heat transfer rate is higher in the eccentric arrangement due to the smaller distance between the copper tube and the outer shell compared to the concentric arrangement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental concept in the study of heat transfer. It represents a material's ability to conduct heat.
The value of thermal conductivity is denoted by the symbol \( k \) and is expressed in watts per meter-kelvin (\( W / m \cdot K \)).
In the problem, we considered an insulating material with a thermal conductivity of \( 0.05 \, W / m \cdot K \). When we calculate heat loss, thermal conductivity comes into play as it helps determine how easily heat moves through the material between two different temperature points.
A material with a high thermal conductivity allows heat to pass through it more readily, whereas one with a low thermal conductivity resists heat flow.Thus, to find the rate of heat transfer across the insulating layer, we use the formula:
  • \(Q = \frac{k \, A \, \Delta T}{L} \)
where:
  • \( Q \) is the heat transfer per unit length,
  • \( k \) is the thermal conductivity,
  • \( A \) is the surface area through which heat is transferred,
  • \( \Delta T \) is the temperature difference, and
  • \( L \) is the distance heat travels through.
By understanding thermal conductivity, we can better understand how different materials affect the rate and efficiency of heat transfer.
Eccentric Cylindrical Shell
An eccentric cylindrical shell involves a configuration where the cylinder inside (in this case, the copper tube) is not centered with the outer cylindrical shell. The eccentricity is the distance between the center of the inner tube and the center of the outer shell, which is crucial in determining the efficiency of heat transfer.
In this exercise, it was given as 20 mm. This offset causes a non-uniform gap in the insulating material between the tube and the outer shell. This variation in spacing affects how heat transfers through the insulation because of the variability in the path heat must take to travel from one surface to the other.
  • Eccentric arrangements can lead to higher rates of heat transfer as the shorter path in some sections can facilitate faster heat movement.
  • The eccentricity can drastically influence thermal performance compared to perfectly centered arrangements.
Understanding the eccentricity helps in accurately calculating the heat loss as it requires customized approaches to handle the non-uniform characteristics of such arrangements.
Concentric Arrangement
In a concentric arrangement, the inner cylinder is perfectly centered within the outer shell. This symmetry leads to a uniform layer of insulation between the two surfaces. This uniform gap, calculated as 45 mm for the concentric shell arrangement in the problem, can influence heat transfer efficiency in such systems. The even distance in a concentric configuration means the path for heat movement is consistent and predictable. This impacts the calculation because
  • All sections of the insulating material contribute equally to resisting heat flow.
  • This often results in lower heat loss compared to eccentric configurations due to the greater average distance heat must travel.
Therefore, understanding concentric arrangements allows for more straightforward calculations as opposed to systems with variable gaps, like eccentric setups, and is vital when aiming for efficient thermal management within such structures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A long constantan wire of \(1-\mathrm{mm}\) diameter is butt welded to the surface of a large copper block, forming a thermocouple junction. The wire behaves as a fin, permitting heat to flow from the surface, thereby depressing the sensing junction temperature \(T_{j}\) below that of the block \(T_{\sigma}\). Copper block, \(T_{o}\) (a) If the wire is in air at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the measurement error \(\left(T_{j}-T_{o}\right)\) for the thermocouple when the block is at \(125^{\circ} \mathrm{C}\). (b) For convection coefficients of 5,10 , and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the measurement error as a function of the thermal conductivity of the block material over the range 15 to \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Under what circumstances is it advantageous to use smaller diameter wire?

Small-diameter electrical heating elements dissipating \(50 \mathrm{~W} / \mathrm{m}\) (length normal to the sketch) are used to heat a ceramic plate of thermal conductivity \(2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The upper surface of the plate is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a convection coeftient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\Delta x=6 \mathrm{~mm}\) and \(\Delta y=2 \mathrm{~mm}\), obtain the temperature distribution within the plate. (b) Using the calculated nodal temperatures, sketch four isotherms to illustrate the temperature distribution in the plate. (c) Calculate the heat loss by convection from the plate to the fluid. Compare this value with the element dissipation rate. (d) What advantage, if any, is there in not making \(\Delta x=\Delta y\) for this situation? (e) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the temperature field within the plate and the rate of heat transfer from the plate. Under no circumstances may the temperature at any location in the plate exceed \(400^{\circ} \mathrm{C}\). Would this limit be exceeded if the airfw were terminated and heat transfer to the air were by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ?

An igloo is built in the shape of a hemisphere, with an inner radius of \(1.8 \mathrm{~m}\) and walls of compacted snow that are \(0.5 \mathrm{~m}\) thick. On the inside of the igloo, the surface heat transfer coefficient is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); on the outside, under normal wind conditions, it is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal conductivity of compacted snow is \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the ice cap on which the igloo sits is \(-20^{\circ} \mathrm{C}\) and has the same thermal conductivity as the compacted snow. (a) Assuming that the occupants' body heat provides a continuous source of \(320 \mathrm{~W}\) within the igloo, calculate the inside air temperature when the outside air temperature is \(T_{\infty}=-40^{\circ} \mathrm{C}\). Be sure to consider heat losses through the floor of the igloo. (b) Using the thermal circuit of part (a), perform a parameter sensitivity analysis to determine which variables have a significant effect on the inside air temperature. For instance, for very high wind conditions, the outside convection coefficient could double or even triple. Does it make sense to construct the igloo with walls half or twice as thick?

A tube of diameter \(50 \mathrm{~mm}\) having a surface temperature of \(85^{\circ} \mathrm{C}\) is embedded in the center plane of a concrete slab \(0.1 \mathrm{~m}\) thick with upper and lower surfaces at \(20^{\circ} \mathrm{C}\). Using the appropriate tabulated relation for this configuration, find the shape factor. Determine the heat transfer rate per unit length of the tube.

The top surface of a plate, including its grooves, is maintained at a uniform temperature of \(T_{1}=200^{\circ} \mathrm{C}\). The lower surface is at \(T_{2}=20^{\circ} \mathrm{C}\), the thermal conductivity is \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the groove spacing is \(0.16 \mathrm{~m}\). (a) Using a finite-difference method with a mesh size of \(\Delta x=\Delta y=40 \mathrm{~mm}\), calculate the unknown nodal temperatures and the heat transfer rate per width of groove spacing \((w)\) and per unit length normal to the page. (b) With a mesh size of \(\Delta x=\Delta y=10 \mathrm{~mm}\), repeat the foregoing calculations, determining the temperature field and the heat rate. Also, consider conditions for which the bottom surface is not at a uniform temperature \(T_{2}\) but is exposed to a fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\). With \(\Delta x=\Delta y=10 \mathrm{~mm}\), determine the temperature field and heat rate for values of \(h=5,200\), and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), as well as for \(h \rightarrow \infty\).
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation

Astrophysics

Read Explanation

Oscillations

Read Explanation

Linear Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.