Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 58

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

Short Answer

Expert verified
#tag_title# Answer #tag_content# To solve the problem, follow these steps: 1. Calculate the Biot number: \(\text{Bi} = \frac{h L_c}{k} = \frac{128 \cdot 6.25 \times 10^{-3}}{80.2} \approx 0.01\) Since Bi << 1, the lumped capacitance method can be used. 2. Calculate the surface area and volume of the rod: \(A_p = \pi DL = \pi (25 \times 10^{-3})(0.6) \approx 0.0471 \ \mathrm{m^2}\) \(V = \frac{\pi D^2 L}{4} = \frac{\pi (25 \times 10^{-3})^2 (0.6)}{4} \approx 7.33 \times 10^{-5} \ \mathrm{m^3}\) 3. Calculate the time required for the rod temperature to reach \(200^{\circ}\mathrm{C}\): \(150 = 650e^{-\frac{128 \cdot 0.0471}{7874 \cdot 7.33 \times 10^{-5} \cdot 444}t}\) After solving for t, we get: \(t \approx 626.9 \ \mathrm{s}\) The time required for the rod surface temperature to cool down to \(200^{\circ}\mathrm{C}\) is approximately 626.9 seconds.

Step by step solution

01

Calculate the Biot Number

To calculate the Biot number, we use the relation Bi = hL_c/k, where h is the convection heat transfer coefficient, L_c is the characteristic length, and k is the thermal conductivity of the material. For a cylinder (the rod), the characteristic length can be considered as L_c = D/4, where D is the diameter. \(h = 128 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) \(k = 80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(D = 25 \times 10^{-3}\ \mathrm{m}\) \(L_c = \frac{D}{4} = 6.25 \times 10^{-3}\ \mathrm{m}\) \(\text{Bi} = \frac{h L_c}{k} = \frac{128 \cdot 6.25 \times 10^{-3}}{80.2}\) Calculate the value of the Biot number using the given values.
02

Apply the lumped capacitance method

Now we need to apply the lumped capacitance method to determine the temperature of the rod as a function of time. The temperature equation for the lumped capacitance method is: \(T(t) - T_\infty = (T_i - T_\infty)e^{-\frac{hA_p}{\rho V c_p}t}\) where: \(T(t)\) is the rod temperature at time \(t\) \(T_\infty\) is the temperature of the water bath (50°C) \(T_i\) is the initial temperature of the rod (700°C) \(A_p\) is the area of the rod surface \(V\) is the volume of the rod \(\rho\), \(c_p\), and \(k\) are the density, specific heat, and thermal conductivity of the iron, respectively. To find the time required for the rod surface temperature to cool to \(200^{\circ} \mathrm{C}\), we will use the above equation and solve for \(t\) when \(T(t) = 200^{\circ}\mathrm{C}\). First, calculate the surface area and volume of the rod.
03

Calculate the surface area and volume of the rod

The surface area (\(A_p\)) of a cylinder can be found using the formula: \(A_p = 2\pi rL = \pi DL\) Similarly, the volume (\(V\)) can be found using the formula: \(V = \pi r^2 L = \frac{\pi D^2 L}{4}\) Replace all the known values, and calculate the surface area and volume of the rod.
04

Calculate the time required for the rod temperature to reach \(200^{\circ}\mathrm{C}\)

Now, substitute the known values in the temperature equation derived using the lumped capacitance method and solve for \(t\) when \(T(t) = 200^{\circ} \mathrm{C}\). \(200 - 50 = (700 - 50)e^{-\frac{128A_p}{\rho V c_p}t}\) Simplify the equation and solve for \(t\). The result will be the time required for the rod surface temperature to cool to \(200^{\circ} \mathrm{C}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 5-cm-high rectangular ice block \((k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(-20^{\circ} \mathrm{C}\) is placed on a table on its square base \(4 \mathrm{~cm} \times 4 \mathrm{~cm}\) in size in a room at \(18^{\circ} \mathrm{C}\). The heat transfer coefficient on the exposed surfaces of the ice block is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear?

It is claimed that beef can be stored for up to two years at \(-23^{\circ} \mathrm{C}\) but no more than one year at \(-12^{\circ} \mathrm{C}\). Is this claim reasonable? Explain.

Consider a short cylinder whose top and bottom surfaces are insulated. The cylinder is initially at a uniform temperature \(T_{i}\) and is subjected to convection from its side surface to a medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). Is the heat transfer in this short cylinder one- or twodimensional? Explain.

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) is reentering the atmosphere. The shell satellite is made of stainless steel with properties of \(\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after 5 minutes of reentry. Assume heat transfer occurs only on the satellite shell.

Thick slabs of stainless steel \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and copper \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) are subjected to uniform heat flux of \(8 \mathrm{~kW} / \mathrm{m}^{2}\) at the surface. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Determine the temperatures of both slabs, at \(1 \mathrm{~cm}\) from the surface, after \(60 \mathrm{~s}\) of exposure to the heat flux.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Fields in Physics

Read Explanation

Force

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Solid State Physics

Read Explanation

Electrostatics

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.

Sign-up for free