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Find the center temperature of the cylindrical block after 19.1 min.

Find the center temperature of the cylindrical block after 19.1 min.

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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Consider a cubic block whose sides are 5 cm long and a cylindrical block whose height and diameter are also 5 cm. Both blocks are initially at 20 °C and are made of granite (k = 2.5 W/m·°C and α = 1.15 × 10−6 m2/s). Now both blocks are exposed to hot gases at 500°C in a furnace on all of their surfaces with a heat transfer coefficient of 40 W/m2·°C.

Find the center temperature of the cylindrical block after 19.1 min.

The center temperature of the cylindrical block is

### Heat Transfer: Conduction and Geometry In this section, we explore how the geometry of an object affects the conduction of heat. Consider two objects with different shapes but similar dimensions: a cube and a cylinder. #### Cube - **Dimensions:** Each side of the cube measures 5 cm. - **Initial Temperature (Ti):** The cube starts at an initial temperature of \( 20^\circ C \). - **Surrounding Environment:** The cube is exposed to hot gases at \( 500^\circ C \). #### Cylinder - **Dimensions:** - **Height:** 5 cm - **Diameter:** 5 cm, making the radius 2.5 cm - **Initial Temperature (Ti):** The cylinder also starts at an initial temperature of \( 20^\circ C \). - **Surrounding Environment:** The cylinder is exposed to the same hot gases at \( 500^\circ C \). These diagrams serve to illustrate how the surface area and volume for different shapes can influence the rate at which they heat up when exposed to the same environment. To further understand this, consider the following details: 1. **Surface Area to Volume Ratio:** - **Cube:** - Surface Area = 6 * (side length)^2 = 6 * (5 cm)^2 = 150 cm² - Volume = (side length)^3 = (5 cm)^3 = 125 cm³ - Surface Area to Volume Ratio = 150 cm² / 125 cm³ = 1.2 cm⁻¹ - **Cylinder:** - Surface Area = 2π(radius)^2 + 2π(radius)(height) = 2π(2.5 cm)^2 + 2π(2.5 cm)(5 cm) ≈ 157.08 cm² - Volume = π(radius)^2(height) = π(2.5 cm)^2(5 cm) ≈ 98.17 cm³ - Surface Area to Volume Ratio ≈ 157.08 cm² / 98.17 cm³ ≈ 1.6 cm⁻¹ 2. **Heat Exposure:** - Heat transfer will be different in each case due to varying surface area-to-volume ratios. Objects with higher surface area-to-volume ratios will typically heat up or cool down more rapidly than those with lower ratios. By comparing the
Transcribed Image Text:### Heat Transfer: Conduction and Geometry In this section, we explore how the geometry of an object affects the conduction of heat. Consider two objects with different shapes but similar dimensions: a cube and a cylinder. #### Cube - **Dimensions:** Each side of the cube measures 5 cm. - **Initial Temperature (Ti):** The cube starts at an initial temperature of \( 20^\circ C \). - **Surrounding Environment:** The cube is exposed to hot gases at \( 500^\circ C \). #### Cylinder - **Dimensions:** - **Height:** 5 cm - **Diameter:** 5 cm, making the radius 2.5 cm - **Initial Temperature (Ti):** The cylinder also starts at an initial temperature of \( 20^\circ C \). - **Surrounding Environment:** The cylinder is exposed to the same hot gases at \( 500^\circ C \). These diagrams serve to illustrate how the surface area and volume for different shapes can influence the rate at which they heat up when exposed to the same environment. To further understand this, consider the following details: 1. **Surface Area to Volume Ratio:** - **Cube:** - Surface Area = 6 * (side length)^2 = 6 * (5 cm)^2 = 150 cm² - Volume = (side length)^3 = (5 cm)^3 = 125 cm³ - Surface Area to Volume Ratio = 150 cm² / 125 cm³ = 1.2 cm⁻¹ - **Cylinder:** - Surface Area = 2π(radius)^2 + 2π(radius)(height) = 2π(2.5 cm)^2 + 2π(2.5 cm)(5 cm) ≈ 157.08 cm² - Volume = π(radius)^2(height) = π(2.5 cm)^2(5 cm) ≈ 98.17 cm³ - Surface Area to Volume Ratio ≈ 157.08 cm² / 98.17 cm³ ≈ 1.6 cm⁻¹ 2. **Heat Exposure:** - Heat transfer will be different in each case due to varying surface area-to-volume ratios. Objects with higher surface area-to-volume ratios will typically heat up or cool down more rapidly than those with lower ratios. By comparing the
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