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giving the following example, find the temperature distribution inside the apple after 1.05 hours, depending on the ditance from the center.

giving the following example, find the temperature distribution inside the apple after 1.05 hours, depending on the ditance from the center.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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giving the following example, find the temperature distribution inside the apple after 1.05 hours, depending on the ditance from the center.

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Before apples are loaded into a cool store, it is wished to chill them to a central temperature of 5°C so as to avoid problems of putting warm apples with the colder ones in storage. The apples, initially at 25°C, are considered to be spheres of 7 cm diameter and the chilling is to be carried out using air at -1°C and at a velocity which provides a surface heat-transfer coefficient of 30 J m2 s-1 °C-1. The physical properties of the apples are K = 0.5 J m¹ s1 °C-1, p = 930 kg m-³, c = 3.6 kJ kg-1 °C-1. Calculate the time necessary to chill the apples so that their centres reach 5°C. This is an example of unsteady-state cooling and can be solved by application of Fig. 5.3, Bi = hsr/k = (30 x 0.035)/0.5 = 2.1 1/Bi = 0.48 (T-To)/(T1-To) = [5-(-1)]/[25- (-1)] = 0.23 and so, reading from Fig.5.3 Fo= 0.46 = kt/pcr² t=Fo pcr² /k so t = [0.46x 3600 x 930 x (0.035)²1/0.5 = 3773 s = 1.05 h
Transcribed Image Text:Before apples are loaded into a cool store, it is wished to chill them to a central temperature of 5°C so as to avoid problems of putting warm apples with the colder ones in storage. The apples, initially at 25°C, are considered to be spheres of 7 cm diameter and the chilling is to be carried out using air at -1°C and at a velocity which provides a surface heat-transfer coefficient of 30 J m2 s-1 °C-1. The physical properties of the apples are K = 0.5 J m¹ s1 °C-1, p = 930 kg m-³, c = 3.6 kJ kg-1 °C-1. Calculate the time necessary to chill the apples so that their centres reach 5°C. This is an example of unsteady-state cooling and can be solved by application of Fig. 5.3, Bi = hsr/k = (30 x 0.035)/0.5 = 2.1 1/Bi = 0.48 (T-To)/(T1-To) = [5-(-1)]/[25- (-1)] = 0.23 and so, reading from Fig.5.3 Fo= 0.46 = kt/pcr² t=Fo pcr² /k so t = [0.46x 3600 x 930 x (0.035)²1/0.5 = 3773 s = 1.05 h
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