Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 18, Problem 91P

(a)

To determine

The center temperature of the cylinder.

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The Biot number for the plane wall is,

  Bi=hLk=(40W/m2°C)(0.075 m)110W/m2=0.02727

Hence the constants λ1and A1 are 0.1620 and 1.0045 respectively.

The Biot number for the long cylinder is,

  Bi=hrok=(40W/m2°C)(0.04 m)110W/m2=0.01455

Hence the constants λ1and A1 are 0.1677 and 1.0036 respectively.

The center temperature is determined using the product solution method as follows:

  θ(0,0,t)short cyl =[θ(0,t)wall ][θ(0,t)cyl ]T(0,0,t)TTiT=(A1eλ12τ)(A1eλ12τ)T(0,0,t)2015020=[(1.0045)e(0.1620)2τ][(1.0036)e(0.1677)2τ]

The Fourier number for the plane wall is,

  τ=αtL2=(3.39×105m2/s)(15×60 s)(0.075m)2=5.434>0.2

The Fourier number for the cylinder is,

  τ=αtro2=(3.39×105m2/s)(15×60 s)(0.04m)2=19.07>0.2

The center temperature is,

  T(0,0,15)2015020=[(1.0045)e(0.1620)2(5.424)][(1.0036)e(0.1677)2(19.07)]T(0,0,15)=86.5°C

Thus, the center temperature of the cylinder is 86.5°C.

(b)

To determine

The center temperature of the top surface of the cylinder.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The center temperature of the top surface of the cylinder is determined using the product solution method as follows:

  θ(L,0,t)short cyl =[θ(L,t)wall ][θ(0,t)cyl ]T(L,0,t)TTiT=[(A1eλ12τ)cos(λ1LL)](A1eλ12τ)T(L,0,t)2015020=[(1.0045)e(0.1620)2τcos(0.162)][(1.0036)e(0.1677)2τ]T(L,0,15)2015020=[(1.0045)e(0.1620)2(5.424)cos(0.162)][(1.0036)e(0.1677)2(19.07)]T(L,0,15)=85.6°C

Thus, the center temperature at the top surface of the cylinder is 85.6°C.

(c)

To determine

The total heat transfer from the cylinder after 15 min from the start of cooling.

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The maximum heat that can be transferred from the cylinder is,

  Qmax=mcp(TiT)=ρVcp(TiT)=(8530kg/m3)[π4(0.08 m)2(0.15 m)](0.389kJ/kg°C)(15020)°C=325.2 kJ

The dimensionless heat transfer ratio for the plane wall is,

  (QQmax)wall=1θo,wallsin(λ1)λ1=1[(1.0045)e(0.1620)2(5.424)]sin(0.1620)0.1620=0.1326

The dimensionless heat transfer ratio for the cylinder is,

  (QQmax)cyl=12θo,cylJ1(λ1)λ1=12[(1.0036)e(0.1677)2(19.07)]0.083480.1677=0.4156

The heat transfer rate for the short cylinder is,

  (QQmax)short cylinder =(QQmax)planewall +(QQmax)long cylinder [1(QQmax)planewall ]=0.1326+(0.4156)(10.1326)=0.4931

Calculate the total heat transfer from the short cylinder.

  Q=0.4931Qmax=0.4931(325.2 kJ)=160 kJ

Thus, the total heat transfer from the cylinder after 15 min from the start of cooling is 160 kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 18 Solutions

Fundamentals of Thermal-Fluid Sciences

Ch. 18 - Prob. 11PCh. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - Prob. 14PCh. 18 - Consider a 1000-W iron whose base plate is made of...Ch. 18 - Metal plates (k = 180 W/m·K, ρ = 2800 kg/m3, and...Ch. 18 - A 5-mm-thick stainless steel strip (k = 21 W/m·K,...Ch. 18 - A long copper rod of diameter 2.0 cm is initially...Ch. 18 - Prob. 21PCh. 18 - Steel rods (ρ = 7832 kg/m3, cp = 434 J/kg·K, and k...Ch. 18 - Prob. 23PCh. 18 - The temperature of a gas stream is to be measured...Ch. 18 - Prob. 25PCh. 18 - A thermocouple, with a spherical junction diameter...Ch. 18 - Prob. 27PCh. 18 - Prob. 28PCh. 18 - Carbon steel balls (ρ = 7833 kg/m3, k = 54 W/m·K,...Ch. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Prob. 33PCh. 18 - Prob. 34PCh. 18 - Prob. 35PCh. 18 - Prob. 36PCh. 18 - Prob. 37PCh. 18 - Prob. 38PCh. 18 - Prob. 39PCh. 18 - A body at an initial temperature of Ti is brought...Ch. 18 - In a meat processing plant, 2-cm-thick steaks (k =...Ch. 18 - Prob. 42PCh. 18 - Prob. 43PCh. 18 - Prob. 45PCh. 18 - Prob. 46PCh. 18 - Prob. 47PCh. 18 - Prob. 48PCh. 18 - Prob. 49PCh. 18 - A long iron rod (ρ = 7870 kg/m3, cp = 447 J/kg·K,...Ch. 18 - Prob. 51PCh. 18 - A long 35-cm-diameter cylindrical shaft made of...Ch. 18 - Prob. 54PCh. 18 - Prob. 55PCh. 18 - Prob. 56PCh. 18 - A father and son conducted the following simple...Ch. 18 - Prob. 58PCh. 18 - Prob. 59PCh. 18 - Citrus fruits are very susceptible to cold...Ch. 18 - Prob. 61PCh. 18 - Prob. 63PCh. 18 - Prob. 64PCh. 18 - Prob. 65PCh. 18 - White potatoes (k = 0.50 W/m·K and α = 0.13 × 10−6...Ch. 18 - Prob. 67PCh. 18 - Prob. 68PCh. 18 - Prob. 69PCh. 18 - Consider a hot semi-infinite solid at an initial...Ch. 18 - Prob. 71PCh. 18 - Prob. 72PCh. 18 - Prob. 73PCh. 18 - Prob. 74PCh. 18 - Prob. 75PCh. 18 - Prob. 76PCh. 18 - Prob. 77PCh. 18 - Prob. 78PCh. 18 - Prob. 79PCh. 18 - Prob. 81PCh. 18 - Prob. 82PCh. 18 - Prob. 83PCh. 18 - Prob. 84PCh. 18 - Prob. 85PCh. 18 - Prob. 86PCh. 18 - Prob. 88PCh. 18 - Prob. 89PCh. 18 - A 2-cm-high cylindrical ice block (k = 2.22 W/m·K...Ch. 18 - Prob. 91PCh. 18 - Prob. 93PCh. 18 - Prob. 94RQCh. 18 - Large steel plates 1.0-cm in thickness are...Ch. 18 - Prob. 96RQCh. 18 - Prob. 97RQCh. 18 - Prob. 98RQCh. 18 - Prob. 99RQCh. 18 - Prob. 100RQCh. 18 - Prob. 101RQCh. 18 - Prob. 102RQCh. 18 - The water main in the cities must be placed at...Ch. 18 - Prob. 104RQCh. 18 - Prob. 105RQCh. 18 - Prob. 106RQCh. 18 - Prob. 107RQ
Knowledge Booster
Background pattern image
Mechanical Engineering
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.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Understanding Conduction and the Heat Equation; Author: The Efficient Engineer;https://www.youtube.com/watch?v=6jQsLAqrZGQ;License: Standard youtube license