Heat Transfer Problem Set 4 PDF

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Heat Transfer: A Practical Approach - Yunus A Cengel Fall 2003, Assignment 4 Wednesday, September 24, 2003 Chapter 4, Problem 19. During a picnic on a hot summer day, all the cold drinks disappeared quickly, and the only available drinks were those at the ambient temperature of 80F. In an effort to cool a 12fluid-oz drink in a can, which is 5 in. high and has a diameter of 2.5 in., a person grabs the can and starts shaking it in the iced water of the chest at 32F. The temperature of the drink can be assumed to be uniform at all times, and the heat transfer coefficient between the iced water and the aluminum can is 30 Btu/h  ft2  F. Using the properties of water for the drink, estimate how long it will take for the canned drink to cool to 45F?

Figure P4.19 Chapter 4, Solution 19 A person shakes a can of drink in a iced water to cool it. The cooling time of the drink is to be determined. Assumptions 1 The can containing the drink is cylindrical in shape with a radius of r0 = 1.25 in. 2 The Water thermal properties of the cola are taken to be the same 32F as those of water. 3 Thermal properties of the cola are Cola constant at room temperature. 4 The heat transfer 75F coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the cola is stirred constantly, so that its temperature remains uniform at all times. Properties The density and specific heat of water at room temperature are  = 62.22 lbm/ft3, and Cp = 0.999 Btu/lbm.F (Table A-9E). Analysis Application of lumped system analysis in this case gives Lc =

r o 2 L  (1.25 / 12 ft) 2 (5 / 12 ft) V = = = 0.04167 ft As 2ro L + 2ro 2 2 (1.25 / 12 ft)(5/12 ft) + 2 (1.25 / 12 ft) 2

b=

hAs

C pV

=

h 30 Btu/h.ft 2 .F = = 11.583 h -1 = 0.00322 s -1 C p L c (62.22 lbm/ft 3 )(0.999 Btu/lbm. F)(0.04167 ft)

-1 T ( t ) − T 45 − 32 = e − bt   = e −( 0.00322 s ) t   t = 406 s Ti − T 80 − 32

Therefore, it will take 7 minutes and 46 seconds to cool the canned drink to 45F.

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Heat Transfer: A Practical Approach - Yunus A Cengel Fall 2003, Assignment 4 Wednesday, September 24, 2003 Chapter 4, Problem 23.

Carbon steel balls (r = 7833 kg/m3, k = 54 W/m  C, Cp = 0.465 kJ/kg  C, and  = 1.474  10-6 m2/s) 8 mm in diameter are annealed by heating them first to 900C in a furnace and then allowing them to cool slowly to 100C in ambient air at 35C. If the average heat transfer coefficient is 75 W/m2  C, determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air. Figure P4.23 Chapter 4, Solution 23 A number of carbon steel balls are to be annealed by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The time of annealing and the total rate of heat transfer from the balls to the ambient air are to be determined. Assumptions 1 The balls are spherical in shape with a radius of r0 = 4 mm. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 54 W/m.C,  = 7833 kg/m3, and Cp = 0.465 kJ/kg.C. Analysis The characteristic length of the balls and the Biot number are 3

Lc =

V  D / 6 D 0.008 m = = = = 0.0013 m 6 6 As D 2

Furnace

(75 W/m 2 . C)(0.0013 m ) = = 0.0018  0.1 Bi = k (54 W/m. C ) hL c

Steel balls 900C

Air, 35C

Therefore, the lumped system analysis is applicable. Then the time for the annealing process is determined to be b=

hAs

C pV

=

h 75 W/m 2 . C -1 = 0. 01584 s = 3 C p L c (7833 kg/m )(465 J/kg.C)(0.0013 m)

-1 T ( t ) − T 100 − 35 = e −bt   = e −( 0.01584 s )t   t = 163 s = 2.7 min Ti − T 900 − 35

The amount of heat transfer from a single ball is m = V = 

 D3

= (7833 kg / m 3 )

 (0.008 m) 3

= 0.0021 kg 6 6 Q = mCp [Tf − Ti ] = (0.0021 kg )(465 J / kg. C )(900 − 100) C = 781 J = 0.781 kJ (per ball)

Then the total rate of heat transfer from the balls to the ambient air becomes Q& = n& ball Q = (2500 balls/h)  (0.781 kJ/ball ) = 1,953 kJ/h = 543 W

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Heat Transfer: A Practical Approach - Yunus A Cengel Fall 2003, Assignment 4 Wednesday, September 24, 2003 Chapter 4, Problem 34. An ordinary egg can be approximated as a 5.5-cmdiameter sphere whose properties are roughly k = 0.6 -6 2 W/m  C and  = 0.14  10 m /s. The egg is initially at a uniform temperature of 8C and is dropped into boiling water at 97C. Taking the convection heat transfer coefficient to be h = 1400 W/m2  C, determine how long it will take for the center of the egg to reach 70C. Figure P4.34 Chapter 4, Solution 34

An egg is dropped into boiling water. The cooking time of the egg is to be determined.  Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is  > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and diffusivity of the eggs are given to be k = 0.6 W/m.C and  = 0.1410-6 m2/s. Analysis The Biot number for this process is Bi =

2 hro (1400 W / m . C)(0.0275 m) = = 64.2 k ( 0.6 W / m. C)

Water 97C

The constants 1 and A1 corresponding to this Biot number are, from Table 4-1, by linear interpolation 1 = 3.0877 and A1 = 1.9969

Egg Ti =

Then the Fourier number becomes

0, sph =

2 70 − 97 T0 − T − 2 − = A1 e 1    = (3.0897) e (1.9969 )     = 0.1977  0.2 8 − 97 Ti − T

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the time required for the temperature of the center of the egg to reach 70C is determined to be

t=

 ro 2 (0.198)(0.0275 m) 2 = = 1067 s = 17.8 min  (0.14 10− 6 m 2 /s)

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