(a) Explanation Given data: The diameter of the cylindrical canister is D = 75 mm . The length of cylindrical canister is L = 120 mm . The operating temperature is T f c = 800 °C . The convective heat transfer coefficient is h = 12 W / m 2 ⋅ K . The surface temperature is T s u r = 300 K . The ambient temperature is T ∞ = 298 K . Formula used: The expression for the temperature of canister surface is given as, T s = ( T b 4 ε s ) 1 4 Here, ε s is the emissivity of canister surface, σ is the Stefan Boltzmann’s Constant, and T b is the surface temperature. The expression for the radiation heat transfer coefficient is given as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) Here, h r is the radiation heat transfer coefficient. The expression for the heat loss from the cylindrical wall is given as, q = q ∞ + q s u r T f c [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] Here, k is the thermal conductivity, L is the length of the cylindrical canister, r 1 is the internal diameter of cylinder, t is the thickness of cylinder, h is the convective heat transfer coefficient, and T ∞ is the ambient temperature. Calculation: Calculations for the Case i: The outer surface is covered with the thin layer of dirt and insulating material is Calcium silicate with ε s = 0.90 and k = 0.09 W / m ⋅ K . The surface temperature can be calculated as, T s = ( T b 4 ε s ) 1 4 T s = ( ( 304 K ) 4 0.90 ) 1 4 T s = 312.1 K The radiation heat transfer coefficient can be calculated as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) h r = ( 0.9 ) × ( 5.67 × 10 − 8 ) × ( 312.1 K + 300 K ) × [ ( 312.1 K ) 2 + ( 300 ) 2 ] h r = 5.853 W / m 2 ⋅ K The required thickness can be calculated as, T f c − T s [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] { ( 800 °C + 273 ) K [ ln ( ( 37.5 mm × 1 m 10 3 mm ) + t ( 37.5 mm × 1 m 10 3 mm ) ) 2 π × ( 0.09 W / m ⋅ K ) × ( 120 mm × 1 m 10 3 mm ) ] = 312.1 K − 298 K [ 1 ( 12 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] + 312.1 K − 300 K [ 1 ( 5.853 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] } 51.633 ln ( 37.5 × 10 − 3 m + t 37.5 × 10 − 3 m ) = 180.968 × ( 37.5 × 10 − 3 m + t ) t = 0.1436 m Calculations for the Case ii: The outer surface is covered with the thin layer of dirt and insulating material is Calcium silicate with ε s = 0.08 and k = 0.09 W / m ⋅ K . The surface temperature can be calculated as, T s = ( T b 4 ε s ) 1 4 T s = ( ( 304 K ) 4 0.08 ) 1 4 T s = 571.61 K The radiation heat transfer coefficient can be calculated as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) h r = ( 0.08 ) × ( 5.67 × 10 − 8 W / m 2 ⋅ K 4 ) × ( 571.61 K + 300 K ) × [ ( 571.61 K ) 2 + ( 300 ) 2 ] h r = 1.647 W / m 2 ⋅ K The required thickness can be calculated as, T f c − T s [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] { ( 800 °C + 273 ) K − 571.16 K [ ln ( ( 37.5 mm × 1 m 10 3 mm ) + t ( 37.5 mm × 1 m 10 3 mm ) ) 2 π × ( 0.09 W / m ⋅ K ) × ( 120 mm × 1 m 10 3 mm ) ] = 571.61 K − 298 K [ 1 ( 12 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] + 571.61 K − 300 K [ 1 ( 1.647 W / m 2 ⋅ K ) 2 π ( ( 37

6th Edition

ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Videos

Question

Chapter 3, Problem 3.23P

(a)

To determine

The thickness of insulation in case i.

The thickness of insulation in case ii.

The thickness of insulation in case iii.

The thickness of insulation in case iv.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3, Problem 3.22P

Chapter 3, Problem 3.24P

Students have asked these similar questions

Question 4: A fire in a 4 m x 5 m compartment can be representative of a "medium" f-fire. This compartment has wood (with a moisture content of 12%) as fuel. The peak heat release rate to be achieved is 10 MW. Assuming the calorific value of dry wood is 16.7 MJ/kg, the fuel load (on floor area) is 30 kg/m2. Find the duration of the steady-state burning phase and the total duration of the fire buning due to dry wood.

A boiler delivers steam at 100 bar and 500 C. the feed water inlet temperature is 160 C. the steam is produced at the rate of 100 tonnes/h and the boiler efficiency is 88%. Estimate the fuel burning rate in kg/h if the calorific value of the fuel is 21MJ/kg

Help!

Chapter 3 Solutions

Introduction to Heat Transfer

Ch. 3 - Consider the plane wall of Figure 3.1, separating...Ch. 3 - A new building to be located in a cold climate is...Ch. 3 - The rear window of an automobile is defogged by...Ch. 3 - The rear window of an automobile is defogged by...Ch. 3 - A dormitory at a large university, built 50 years...Ch. 3 - In a manufacturing process, a transparent film is...Ch. 3 - Prob. 3.7P Ch. 3 - A t=10-mm-thick horizontal layer of water has a...Ch. 3 - Prob. 3.9P Ch. 3 - The wind chill, which is experienced on a cold,...

Ch. 3 - Prob. 3.11P Ch. 3 - A thermopane window consists of two pieces of...Ch. 3 - A house has a composite wall of wood, fiberglass...Ch. 3 - Prob. 3.14P Ch. 3 - Prob. 3.15P Ch. 3 - Work Problem 3.15 assuming surfaces parallel to...Ch. 3 - Consider the oven of Problem 1.54. The walls of...Ch. 3 - The composite wall of an oven consists of three...Ch. 3 - The wall of a drying oven is constructed by...Ch. 3 - The t=4-mm-thick glass windows of an...Ch. 3 - Prob. 3.21P Ch. 3 - In the design of buildings, energy conservation...Ch. 3 - Prob. 3.23P Ch. 3 - Prob. 3.24P Ch. 3 - Prob. 3.25P Ch. 3 - A composite wall separates combustion gases at...Ch. 3 - Prob. 3.27P Ch. 3 - Prob. 3.28P Ch. 3 - Prob. 3.29P Ch. 3 - The performance of gas turbine engines may...Ch. 3 - A commercial grade cubical freezer, 3 m on a...Ch. 3 - Prob. 3.32P Ch. 3 - Prob. 3.33P Ch. 3 - Prob. 3.34P Ch. 3 - A batt of glass fiber insulation is of density...Ch. 3 - Air usually constitutes up to half of the volume...Ch. 3 - Prob. 3.37P Ch. 3 - Prob. 3.38P Ch. 3 - The diagram shows a conical section fabricatedfrom...Ch. 3 - Prob. 3.40P Ch. 3 - From Figure 2.5 it is evident that, over a wide...Ch. 3 - Consider a tube wall of inner and outer radii ri...Ch. 3 - Prob. 3.43P Ch. 3 - Prob. 3.44P Ch. 3 - Prob. 3.45P Ch. 3 - Prob. 3.46P Ch. 3 - To maximize production and minimize pumping...Ch. 3 - A thin electrical heater is wrapped around the...Ch. 3 - Prob. 3.50P Ch. 3 - Prob. 3.51P Ch. 3 - Prob. 3.52P Ch. 3 - A wire of diameter D=2mm and uniform temperatureT...Ch. 3 - Prob. 3.54P Ch. 3 - Electric current flows through a long rod...Ch. 3 - Prob. 3.56P Ch. 3 - A long, highly polished aluminum rod of diameter...Ch. 3 - Prob. 3.58P Ch. 3 - Prob. 3.59P Ch. 3 - Prob. 3.60P Ch. 3 - Prob. 3.61P Ch. 3 - Prob. 3.62P Ch. 3 - Consider the series solution, Equation 5.42, for...Ch. 3 - Prob. 3.64P Ch. 3 - Copper-coated, epoxy-filled fiberglass circuit...Ch. 3 - Prob. 3.66P Ch. 3 - A constant-property, one-dimensional Plane slab of...Ch. 3 - Referring to the semiconductor processing tool of...Ch. 3 - Prob. 3.69P Ch. 3 - Prob. 3.70P Ch. 3 - Prob. 3.71P Ch. 3 - The 150-mm-thick wall of a gas-fired furnace is...Ch. 3 - Steel is sequentially heated and cooled (annealed)...Ch. 3 - Prob. 3.74P Ch. 3 - Prob. 3.75P Ch. 3 - Prob. 3.76P Ch. 3 - Prob. 3.77P Ch. 3 - Prob. 3.78P Ch. 3 - The strength and stability of tires may be...Ch. 3 - Prob. 3.80P Ch. 3 - Prob. 3.81P Ch. 3 - A long rod of 60-mm diameter and thermophysical...Ch. 3 - A long cylinder of 30-min diameter, initially at a...Ch. 3 - Work Problem 5.47 for a cylinder of radius r0 and...Ch. 3 - Prob. 3.85P Ch. 3 - Prob. 3.86P Ch. 3 - Prob. 3.87P Ch. 3 - Prob. 3.88P Ch. 3 - Prob. 3.89P Ch. 3 - Prob. 3.90P Ch. 3 - Prob. 3.91P Ch. 3 - Prob. 3.92P Ch. 3 - In Section 5.2 we noted that the value of the Biot...Ch. 3 - Prob. 3.94P Ch. 3 - Prob. 3.95P Ch. 3 - Prob. 3.96P Ch. 3 - Prob. 3.97P Ch. 3 - Prob. 3.98P Ch. 3 - Work Problem 5.47 for the case of a sphere of...Ch. 3 - Prob. 3.100P Ch. 3 - Prob. 3.101P Ch. 3 - Prob. 3.102P Ch. 3 - Prob. 3.103P Ch. 3 - Consider the plane wall of thickness 2L, the...Ch. 3 - Problem 4.9 addressed radioactive wastes stored...Ch. 3 - Prob. 3.106P Ch. 3 - Prob. 3.107P Ch. 3 - Prob. 3.108P Ch. 3 - Prob. 3.109P Ch. 3 - Prob. 3.110P Ch. 3 - A one-dimensional slab of thickness 2L is...Ch. 3 - Prob. 3.112P Ch. 3 - Prob. 3.113P Ch. 3 - Prob. 3.114P Ch. 3 - Prob. 3.115P Ch. 3 - Derive the transient, two-dimensional...Ch. 3 - Prob. 3.117P Ch. 3 - Prob. 3.118P Ch. 3 - Prob. 3.119P Ch. 3 - Prob. 3.120P Ch. 3 - Prob. 3.121P Ch. 3 - Prob. 3.122P Ch. 3 - Consider two plates, A and B, that are each...Ch. 3 - Consider the fuel element of Example 5.11, which...Ch. 3 - Prob. 3.125P Ch. 3 - Prob. 3.126P Ch. 3 - Prob. 3.127P Ch. 3 - Prob. 3.128P Ch. 3 - Prob. 3.129P Ch. 3 - Consider the thick slab of copper in Example 5.12,...Ch. 3 - In Section 5.5, the one-term approximation to the...Ch. 3 - Thermal energy storage systems commonly involve a...Ch. 3 - Prob. 3.133P Ch. 3 - Prob. 3.134P Ch. 3 - Prob. 3.135P Ch. 3 - A tantalum rod of diameter 3 mm and length 120 mm...Ch. 3 - A support rod k=15W/mK,=4.0106m2/s of diameter...Ch. 3 - Prob. 3.138P Ch. 3 - Prob. 3.139P Ch. 3 - A thin circular disk is subjected to induction...Ch. 3 - An electrical cable, experiencing uniform...Ch. 3 - Prob. 3.142P Ch. 3 - Prob. 3.145P Ch. 3 - Consider the fuel element of Example 5.11, which...Ch. 3 - Prob. 3.147P Ch. 3 - Prob. 3.148P Ch. 3 - Prob. 3.149P Ch. 3 - Prob. 3.150P Ch. 3 - In a manufacturing process, stainless steel...Ch. 3 - Prob. 3.153P Ch. 3 - Carbon steel (AISI 1010) shafts of 0.1-m diameter...Ch. 3 - A thermal energy storage unit consists of a large...Ch. 3 - Small spherical particles of diameter D=50m...Ch. 3 - A spherical vessel used as a reactor for producing...Ch. 3 - Batch processes are often used in chemical and...Ch. 3 - Consider a thin electrical heater attached to a...Ch. 3 - An electronic device, such as a power transistor...Ch. 3 - Prob. 3.161P Ch. 3 - In a material processing experiment conducted...Ch. 3 - Prob. 3.165P Ch. 3 - Prob. 3.166P Ch. 3 - Prob. 3.167P Ch. 3 - Prob. 3.168P Ch. 3 - Prob. 3.173P Ch. 3 - Prob. 3.174P Ch. 3 - Prob. 3.175P Ch. 3 - Prob. 3.176P Ch. 3 - Prob. 3.177P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The thickness of insulation in case i. The thickness of insulation in case ii. The thickness of insulation in case iii. The thickness of insulation in case iv.

Solution Summary: The author explains the thickness of insulation in case i, and the convective heat transfer coefficient for the cylindrical canister.

Videos

The thickness of insulation in case i. The thickness of insulation in case ii. The thickness of insulation in case iii. The thickness of insulation in case iv.

Want to see the full answer?

Chapter 3 Solutions

The thickness of insulation in case i.

The thickness of insulation in case ii.

The thickness of insulation in case iii.

The thickness of insulation in case iv.