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Chapter 4: Problem 75

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=\) \(0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) initially at a uniform temperature of \(15^{\circ} \mathrm{C}\) are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be \(0.95 \mathrm{~g} / \mathrm{cm}^{3}\) and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and \(45 \mathrm{~min}\). Also, determine if any part of the chicken will freeze during this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Short Answer

Expert verified
Answer: To determine whether any part of the chicken is freezing during the cooling process, follow the steps mentioned in the solution, and calculate the surface and center temperature of the chicken. Compare these temperatures with the freezing point of chicken. If any temperature is less than or equal to the freezing point, then that part will undergo freezing during the cooling process.

Step by step solution

01

Find the properties of the chicken

The first step is to find other properties of the chicken using the given mass, density, and shape. We will start by calculating the volume of the chicken. The chicken is treated as a sphere and its mass (\(m\)) is given. We can calculate the volume of the chicken (\(V\)) using the density (\(\rho\)) formula: $$V = \frac{m}{\rho}$$ Once we have the volume, we can calculate the radius (\(r\)) of the sphere using the volume formula: $$V = \frac{4}{3}\pi r^3$$
02

Calculate Biot number and Fourier number

We need to find the Biot number (Bi) which gives an idea of how heat transfer resistance is distributed between the body and the surroundings. It can be calculated using the formula: $$Bi = \frac{h \cdot r}{k}$$ Where \(h\) is the heat transfer coefficient, \(r\) is the radius, and \(k\) is the thermal conductivity. Next, we need to find the Fourier number (Fo) which represents the ratio of heat conduction within the body to heat storage. It is defined as: $$Fo = \frac{\alpha \cdot t}{r^2}$$ Where \(\alpha\) is the thermal diffusivity, and \(t\) is the time in seconds.
03

Calculate temperature distribution

As the problem requires an analytical one-term approximation method solution instead of using Heisler charts, we can find the dimensionless temperature distribution inside the chicken using the following equation: $$\frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{2}{\sqrt{\pi}}\frac{1}{Bi}(e^{-\beta^2Fo} - \beta\sqrt{Fo}erfc(\beta\sqrt{Fo}))$$ where \(T_{\infty}\) is the brine temperature, \(T_0\) is the initial temperature of chicken, \(\beta\) is the dimensionless parameter \(\frac{\sqrt{3Bi}}{2}\), and \(erfc\) is the complementary error function. First, we need to find \(\beta\) in terms of the Biot number. Then plug it in the temperature distribution equation to find out the temperature distribution inside the chicken.
04

Calculate surface and center temperature

Now that we have found the temperature distribution, we can calculate the surface and center temperature of chicken at the given time. The surface temperature can be found at the value of \(r=0\), and the center temperature will be found at the value of \(r=R\) (radius of chicken sphere). Plug in proper values in the temperature distribution equation for surface and center temperature.
05

Check if any part of the chicken is freezing

To determine if any part of the chicken is freezing, we need to compare the temperatures we calculated in step 4 with the freezing point of chicken. If any temperature is less than or equal to the freezing point, then that part will undergo freezing during the cooling process.

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Most popular questions from this chapter

How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

We often cut a watermelon in half and put it into the freezer to cool it quickly. But usually we forget to check on it and end up having a watermelon with a frozen layer on the top. To avoid this potential problem a person wants to set the timer such that it will go off when the temperature of the exposed surface of the watermelon drops to \(3^{\circ} \mathrm{C}\). Consider a 25 -cm- diameter spherical watermelon that is cut into two equal parts and put into a freezer at \(-12^{\circ} \mathrm{C}\). Initially, the entire watermelon is at a uniform temperature of \(25^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the watermelon to have the properties of water, determine how long it will take for the center of the exposed cut surfaces of the watermelon to drop to \(3^{\circ} \mathrm{C}\).

Polyvinylchloride automotive body panels \((k=\) \(\left.0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, \rho=1714 \mathrm{~kg} / \mathrm{m}^{3}\right), 3-\mathrm{mm}\) thick, emerge from an injection molder at \(120^{\circ} \mathrm{C}\). They need to be cooled to \(40^{\circ} \mathrm{C}\) by exposing both sides of the panels to \(20^{\circ} \mathrm{C}\) air before they can be handled. If the convective heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation is not considered, the time that the panels must be exposed to air before they can be handled is (a) \(1.6 \mathrm{~min}\) (b) \(2.4 \mathrm{~min}\) (c) \(2.8 \mathrm{~min}\) (d) \(3.5 \mathrm{~min}\) (e) \(4.2 \mathrm{~min}\)

A large cast iron container \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(1.70 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) with 5 -cm- thick walls is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) and is filled with ice at \(0^{\circ} \mathrm{C}\). Now the outer surfaces of the container are exposed to hot water at \(60^{\circ} \mathrm{C}\) with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer to the ice through a \(1.2-\mathrm{m}\)-wide and \(2-\mathrm{m}\)-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to \(0.1^{\circ} \mathrm{C}\).

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.
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