Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 55

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)

Short Answer

Expert verified
Answer: The average heat transfer coefficient during the cooling process is approximately 1.5 Btu/h·ft²·°F.

Step by step solution

01

Calculate the Biot number (Bi)

To begin, we need to calculate the Biot number (Bi), which is defined as the ratio of the internal resistance to heat conduction to the surface heat transfer resistance. The formula for the Biot number is given by: Bi = hLc/k, where h is the convective heat transfer coefficient, Lc is the characteristic length, and k is the thermal conductivity. In this case, the characteristic length can be taken as half of the slab thickness, since it is a plane wall. Therefore, Lc = 6 inches / 2 = 3 inches = 0.25 ft. Since we are asked to find the average heat transfer coefficient, we will express the Biot number in terms of the desired h, as follows: Bi = h(0.25)/0.26.
02

Calculate the Fourier number (Fo)

Next, we need to calculate the Fourier number (Fo), which represents the ratio of the diffusive heat transfer to the convective heat transfer in a solid. The Fourier number is given by the formula: Fo = αt/Lc², where α is the thermal diffusivity, t is the cooling time, and Lc is the characteristic length, which we already found in Step 1. In this case, α = 1.4 x 10^-6 ft²/s, and t = 12 h = 43200 s. Therefore, the Fourier number is: Fo = (1.4 x 10^-6)(43200)/(0.25)² = 0.099648.
03

Determine the temperature ratio

Now, we shall calculate the temperature ratio, which is needed to find the temperature changes from initial to final time. The ratio of change in temperature to the initial difference between the slab and air temperatures is given by: (T2-T1)/(Ti-Ta) = (36-50)/(50-23). Therefore, the temperature ratio is: -14/27 ≈ -0.5185.
04

Use the Heisler charts

As stated in the problem, we are required to use the Heisler charts to solve for the heat transfer coefficient. We will use the Heisler chart for a plane wall, which requires the Fourier number (Fo) on the horizontal axis, and the Biot number (Bi) on the vertical axis. First, find the value of the temperature ratio (-0.5185) on the curve corresponding to Fo = 0.099648. Next, read the corresponding Biot number value from the vertical axis, Bi ≈ 0.960.
05

Calculate the average heat transfer coefficient (h)

Recall that the Biot number is given by Bi = hLc/k, thus to find h, we can rearrange the formula as follows: h = Bi*k/Lc. Plugging in our values, we get: h = 0.960 * 0.26 / 0.25 ≈ 1.5 Btu/h·ft²·°F. Thus, the average heat transfer coefficient during the cooling process is approximately 1.5 Btu/h·ft²·°F.

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 stainless steel slab \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and a copper slab \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Using \(\mathrm{EES}\) (or other) software, investigate the effect of time on the temperatures of both materials at the depth of \(5 \mathrm{~cm}\) from the surface. By varying the time from 1 to \(80 \mathrm{~s}\) after the slabs have received the energy pulse, plot the temperatures at \(5 \mathrm{~cm}\) from the surface as a function of time.

Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known.

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties \(\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), \(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and \(\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}\). Assuming the tip of the thermometer is at one- third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and \((c)\) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

A \(10-\mathrm{cm}\) thick aluminum plate \(\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(903 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=97.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) is being heated in liquid with temperature of \(500^{\circ} \mathrm{C}\). The aluminum plate has a uniform initial temperature of \(25^{\circ} \mathrm{C}\). If the surface temperature of the aluminum plate is approximately the liquid temperature, determine the temperature at the center plane of the aluminum plate after 15 seconds of heating. Solve this problem using analytical one- term approximation method (not the Heisler charts).

How does \((a)\) the air motion and (b) the relative humidity of the environment affect the growth of microorganisms in foods?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Electrostatics

Read Explanation

Medical Physics

Read Explanation

Thermodynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Engineering Physics

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