/*! 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 55 Layers of 6-in-thick meat slabs ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

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Key Concepts

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

Biot Number
The Biot number (Bi) is a crucial concept in heat transfer. It helps us understand the relationship between internal and surface heat transfer resistance in solid objects. Simply put, it compares how well heat moves within the solid to how easily it leaves the surface.
The formula for Biot number is:
  • \( \text{Bi} = \frac{hL_c}{k} \)
Here:
  • \( h \) is the convective heat transfer coefficient, which measures how well heat is removed from the surface by the surrounding fluid.
  • \( L_c \) is the characteristic length, typically half the thickness for a slab or an appropriate dimension for other shapes.
  • \( k \) is the thermal conductivity, defining how well the material conducts heat internally.
When the Biot number is low (much less than one), it indicates that the surface heat resistance is greater than the internal resistance. This typically means the object will heat or cool uniformly. Conversely, a high Biot number suggests that temperature differences arise within the solid, as internal heat transfer resistance is significant.
Fourier Number
The Fourier number (Fo) is instrumental in transient heat conduction analysis. It provides insight into the efficiency of heat diffusion within a material during a period.
Mathematically, it is represented as:
  • \( \text{Fo} = \frac{\alpha t}{L_c^2} \)
Here:
  • \( \alpha \) is the thermal diffusivity, indicating how fast heat conducts relative to how fast it stores energy.
  • \( t \) is the time over which the process occurs.
  • \( L_c \) again is the characteristic length, which affects how far the heat can travel.
The Fourier number illustrates whether sufficient time has passed for the heat to diffuse through an object. A higher Fourier number shows that the heat has had time to spread within the object, leading to a more even temperature distribution.
Heisler Charts
Heisler Charts are a specialized tool used in the analysis of transient heat conduction problems for one-dimensional objects such as slabs, cylinders, and spheres subjected to convective boundary conditions. They are particularly helpful when simple analytical solutions are not feasible.
In practice, the Heisler Charts allow you to determine temperature distribution and heat transfer properties using normalized Fourier and Biot numbers. These charts plot the dimensionless temperature against the Fourier number for several Biot numbers. Through this visualization, you can match a known progression of temperatures with the corresponding Buot number.
In the problem at hand, Heisler Charts were used to estimate the heat transfer coefficient by finding the intersection of the Fourier and Biot number line for a specific temperature ratio. This intersection allowed for reading of the Biot number, which was then used to solve for the average heat transfer coefficient, linking the theoretical analysis directly to practical application.

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Most popular questions from this chapter

Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time?

How can the contamination of foods with microorganisms be prevented or minimized? How can the growth of microorganisms in foods be retarded? How can the microorganisms in foods be destroyed?

Oranges of \(2.5\)-in-diameter \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(78^{\circ} \mathrm{F}\) are to be cooled by refrigerated air at \(25^{\circ} \mathrm{F}\) flowing at a velocity of \(1 \mathrm{ft} / \mathrm{s}\). The average heat transfer coefficient between the oranges and the air is experimentally determined to be \(4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine how long it will take for the center temperature of the oranges to drop to \(40^{\circ} \mathrm{F}\). Also, determine if any part of the oranges will freeze during this process.

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).

Long cylindrical AISI stainless steel rods \((k=\) \(7.74 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\) and \(\left.\alpha=0.135 \mathrm{ft}^{2} / \mathrm{h}\right)\) of 4 -in-diameter are heat treated by drawing them at a velocity of \(7 \mathrm{ft} / \mathrm{min}\) through a 21 -ft-long oven maintained at \(1700^{\circ} \mathrm{F}\). The heat transfer coefficient in the oven is \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the rods enter the oven at \(70^{\circ} \mathrm{F}\), determine their centerline temperature when they leave. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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