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
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Chapter 18, Problem 88P
To determine

The center temperature after 10, 20 and 60 min for each geometry.

Expert Solution & Answer
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Explanation of Solution

Calculation:

For cubic block:

The Biot number corresponding to h=40W/m2°C is,

  Bi=hLk=(40W/m2°C)(0.025m)2.5W/m2=0.400

Hence the constants λ1and A1 are 0.5932 and 1.0580 respectively.

The Biot number corresponding to h=80W/m2°C is,

  Bi=hLk=(80W/m2°C)(0.025m)2.5W/m2=0.800

Hence the constants λ1and A1 are 0.7910 and 1.1016 respectively.

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

  θ(0,0,0,t)block =[θ(0,t)wall ]2[θ(0,t)wall ]T(0,0,0,t)TTiT=(A1eλ12τ)2(A1eλ12τ)T(0,0,0,t)50020500=[(1.0580)e(0.5932)2τ]2[(1.1016)e(0.7910)2τ]

After 10 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(10×60 s)(0.025 m)2=1.104>0.2

The center temperature is,

  T(0,0,0,10)50020500=[(1.0580)e(0.5932)2(1.104)]2[(1.1016)e(0.7910)2(1.104)]T(0,0,0,10)=364°C

After 20 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(20×60 s)(0.025 m)2=2.208>0.2

The center temperature is,

  T(0,0,0,20)50020500=[(1.0580)e(0.5932)2(2.208)]2[(1.1016)e(0.7910)2(2.208)]T(0,0,0,20)=469°C

After 60 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(60×60 s)(0.025 m)2=6.624>0.2

The center temperature is,

  T(0,0,0,60)50020500=[(1.0580)e(0.5932)2(6.624)]2[(1.1016)e(0.7910)2(6.624)]T(0,0,0,60)=500°C

Thus, the center temperature after 10, 20 and 60 min for cubic block is 364°C, 469°C and 500°C respectively.

For cylinder:

The Biot number is,

  Bi=hrok=(40W/m2°C)(0.025m)2.5W/m2=0.400

Hence the constants λ1and A1 are 0.8516 and 1.0931 respectively.

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

  θ(0,0,t)block =[θ(0,t)wall ][θ(0,t)cyl ]T(0,0,t)TTiT=(A1eλ12τ)wall(A1eλ12τ)T(0,0,t)50020500=[(1.1016)e(0.7910)2τ][(1.0931)e(0.8516)2τ]

After 10 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(10×60 s)(0.025 m)2=1.104>0.2

The center temperature is,

  T(0,0,t)50020500=[(1.1016)e(0.7910)2(1.104)][(1.0931)e(0.8516)2(1.104)]T(0,0,10)=370°C

After 20 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(20×60 s)(0.025 m)2=2.208>0.2

The center temperature is,

  T(0,0,20)50020500=[(1.1016)e(0.7910)2(2.208)][(1.0931)e(0.8516)2(2.208)]T(0,0,20)=471°C

After 60 min:

The Fourier number is,

  τ=αtL2=(1.15×106m2/s)(60×60 s)(0.025 m)2=6.624>0.2

The center temperature is,

  T(0,0,60)50020500=[(1.1016)e(0.7910)2(6.624)][(1.0931)e(0.8516)2(6.624)]T(0,0,60)=500°C

Thus, the center temperature after 10, 20 and 60 min for cylinder is 370°C, 471°C and 500°C respectively.

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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
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