Sample problems MAE 655 PDF

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Problem #1 In a production facility, 3-cm-thick large brass plates (k  110 W/m·C,   8530 kg/m3, cp  380 J/kg·C, and   33.9  106 m2/s) that are initially at a uniform temperature of 25 C are heated by passing them through an oven maintained at 700 C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h  80 W/m2·C, determine the surface temperature of the plates when they come out of the oven. Is the lumped system analysis applicable for this problem? Why or why not? If yes, determine the surface temperature of the plates when they come out of the oven by using lumped system analysis and compare results. Assumptions 1. 1-D. 2. The thermal properties of the plate are constant. 3. The heat transfer coefficient is constant and uniform over the entire surface. The Biot number for this process is Bi 

hL (80 W/m 2 .C)(0.015 m)   0.0109 k (110 W/m.C)

The constants 1 and A1 corresponding to this Biot number are, from Table 4-2, 1  0.1035 and A1  1.0018

Plates 25C

The Fourier number is  

t 2

L



(33.9  10  6 m 2 /s)(10 min  60 s/min) (0.015 m) 2

 90.4  0.2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the plates becomes  ( L, t) wall 

2 2 T ( x, t )  T  A1e  1  cos( 1L / L)  (1.0018)e (0.1035 ) (90.4) cos(0.1035)  0.378 Ti  T

T ( L, t)  700  0.378  T (L , t )  445 C 25  700

This problem can be solved easily using the lumped system analysis since Bi < 0.1, and thus the lumped system analysis is applicable. It gives  b

110 W/m  C k k  3.245  10 6 W  s/m 3  C   cp   -6 2 c  p  33.9 10 m / s hA hA   Vc p  (LA )c

T (t )  T   e bt Ti  T 

 p



80 W/m 2  C h h  0.001644 s -1   Lc p L (k /  ) (0.015 m)(3.245 106 W  s/m3  C)

bt T( t)  T  ( Ti  T ) e   700 C  (25 - 700 C) e  (0.001644 s

-1

)( 600 s)

 448 C

which is almost identical to the result obtained above. Problem #2 A 65 kg beef carcass (k  0.5 W/m·C and   0.15106 m2/s) initially at a uniform temperature of 35 C is to be cooled by refrigerated air at10 C flowing at a velocity of 1 m/s. The average heat transfer coefficient between the carcass and the air is 20 W/m2·C. Treating the carcass as a

cylinder of diameter 25 cm and height 1.5 m and disregarding heat transfer from the base and top surfaces, determine how long it will take for the center temperature of the carcass to drop to 5 C. Also, determine if any part of the carcass will freeze during this process. Assumptions 1. The beef carcass can be approximated as a cylinder with insulated top and base surfaces having a radius of ro = 12 cm and a height of H = 1.4 m. 2. 1-D. 3. The thermal properties of the carcass are constant. 4. The heat transfer coefficient is constant and uniform over the entire surface. First we find the Biot number: Air hro ( 20 W/m 2 . C) (0.125 m ) -10 C  5 Bi  1 m/s 0.5 W/m. C k Beef From Table 4-2 we read, for a cylinder, 1 = 1.9898 and A1 = 35  C 1.5029. Substituting these values into the one-term solution gives 2 2 T T 5  ( 10)  0  0   A1e 1    1.5029e (1.9898 )    = 0.38 Ti  T 35  ( 10) which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes t r 2 (0.38)(0.125 m) 2  39,622 s  11 h   2  t o   ro 0.15  10 -6 m 2 / s The lowest temperature during cooling will occur on the surface (r/ro = 1), and is determined to be 2 T T T (r )  T  T( r)  T   0 J 0 ( 1r / ro ) = o  J 0 ( 1ro / ro )  A1 e 1 J 0(  1 r / ro )  o Ti  T T i T Ti  T 

Substituting,

T (r0 )  ( 10)  5  ( 10)     J 0 (1 )  0.33 0.23  0.076   T (ro ) = -6.6 C 35  ( 10)  35  ( 10) 

which is below the freezing temperature. Therefore, the outer part of the beef carcass will freeze during this cooling process. Problem #3 Citrus fruits are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy them. Consider a 10-cm-diameter orange that is initially at 15 C. A cold front moves in one night, and the ambient temperature suddenly drops to 5 C, with a heat transfer coefficient of 15 W/m2·C. Using the properties of water for the orange and assuming the ambient conditions to remain constant for 5 h before the cold front moves out, determine if any part of the orange will freeze that night. Assumptions 1. The orange is spherical in shape with a diameter of 10 cm. 2. 1-D. 3. The thermal properties of the orange are constant.

4. The heat transfer coefficient is constant and uniform over the entire surface. The properties of the orange are approximated by those of water at the average temperature of about 5 C, k = 0.571 W/m.C and   k / c p 0.571 /(999.9 4205)  0.136 106 m2 /s (Table A-9). The Biot number is

hro (15 W/m 2 . C)(0.05 m)   1.313 (0.571 W/m. C) k The constants 1 and A1 corresponding to this Biot number are, from Table 4-2, 1  1.72 and A1  1.33 Bi 

The Fourier number is t (0.136  10 6 m 2 /s)(5 h  3600 s/h)   2   0.98  0.2 (0.05 m) 2 ro

Air T = -5 C

Orange Ti = 15 C

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the oranges becomes 2 sin(  r / r ) T( ro, t)  T 2 sin(1.72 rad) o 1 o  A1e 1   (1.33) e(1.72 ) (0.98 )  0.042  (ro , t ) sph  1 ro / ro Ti  T 1.72 T (ro , t )  ( 5)  0.042   T (ro , t )  - 4.2 C 15  ( 5) which is less than 0C. Therefore, the oranges will freeze. Problem #4 A person puts a few apples into the freezer at 15 C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20 C, and the heat transfer coefficient on the surfaces is 8 W/m2·C. Treating the apples as 9-cm-diameter spheres and taking their properties to be   840 kg/m3, cp  3.81 kJ/kg·C, k  0.418 W/m·C, and   1.3  107 m2/s, determine the center and surface temperatures of the apples in 1 hour. Also, determine the amount of heat transfer from each apple. Assumptions 1. The apples are spherical in shape with a diameter of 9 cm. 2. 1-D. 3. The thermal properties of the apples are constant. 4. The heat transfer coefficient is constant and uniform over the entire surface. 5. Assume  > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Air The Biot number is (8 W/m 2 . C)(0.045 m)   0.861 Bi  k (0.418 W/m.C)

T = -15C

hro

The constants 1 and A1 corresponding to this Biot number are, from Table 4-2, 1  1.476 and A1  1.2390

Apple Ti = 20C

The Fourier number is  

t



ro2

(1.3 107 m 2 /s)(1 h  3600 s/h) (0.045 m) 2

 0.231  0.2

Then the temperature at the center of the apples becomes  0,sph 

2 2 T  ( 15) T0  T  A1 e 1    0  (1.239)e (1.476 ) (0.231 )  0.749  T0  11.2C 20  ( 15) Ti  T

The temperature at the surface of the apples is  ( r o, t) sph 

T (ro , t )  T

 A1 e 1  2

2 sin( 1 ro / ro ) sin(1.476 rad )  (1.239) e (1.476 ) (0 .231 )  0.505 1.476  1 ro / ro

T i T  T (r o , t )  ( 15)  0.505   T (ro , t )  2.7C 20  ( 15)

The maximum possible heat transfer is 4 3  4  ro  (840 kg/m 3 )   (0.045 m) 3.   0.3206 kg 3  3  mc p (T i  T )  (0.3206 kg )(3.81 kJ/kg. C)20  ( 15)  C  42.75 kJ

m  V   Q max

Then the actual amount of heat transfer becomes sin( 1 )   1 cos( 1 ) sin(1.476 rad)  (1.476) cos(1.476 rad) Q  1  3 o ,sph  1  3(0.749)  0.402 3 Q max (1.476) 3 1 Q  0. 402 Qmax  (0.402)( 42.75 kJ)  17.2 kJ

Problem #5 White potatoes (k  0.50 W/m·C and   0.13106 m2/s) that are initially at a uniform temperature of 25 C and have an average diameter of 6 cm are to be cooled by refrigerated air at 2 C flowing at a velocity of 4 m/s. The average heat transfer coefficient between the potatoes and the air is experimentally determined to be 19 W/m2·C. Determine how long it will take for the center temperature of the potatoes to drop to 6 C. Also, determine if any part of the potatoes will experience chilling injury during this process. Consider the potatoes are spherical in shape with a diameter of 6 cm. Assumptions 1. The potatoes are spherical in shape with a radius of r0 = 3 cm. 2. 1-D. 3. The thermal properties of the potato are constant. 4. The heat transfer coefficient is constant and uniform over the entire surface. 5. Assume  > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Air First we find the Biot number: Bi 

hro (19 W/m 2 .C)(0.03 m)  1.14  0.5 W/m.C k

From Table 4-2 we read, for a sphere, 1 = 1.635 and A1 = 1.302. Substituting these values into the one-term solution gives

2C 4 m/s Potato Ti = 25C

0 

2 2 T 0 T 6 2  A1e  1   1.302e  (1.635 )    = 0.753 25  2 Ti  T

which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes  

t ro2

  t 

ro2 (0.753)(0.03 m) 2   5213 s  1.45 h  0.13 10 -6 m 2 / s

The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be 2 sin( 1ro / ro ) To  T sin( 1ro / ro ) sin(  1r / ro ) T ( ro )  T T ( r) T  0  A1 e1  = 1 r / ro 1ro /ro 1ro / ro Ti  T Ti  T  Ti  T 

Substituting,

T( ro )  2  6  2  sin(1.635 rad)    T (ro ) = 4.44 C  25  2 1.635  25  2 

which is above the temperature range of 3 to 4 C for chilling injury for potatoes. Therefore, no part of the potatoes will experience chilling injury during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows:  0.50W/m. o C k   0.877  2 o Bi hro (19 W/m . C)(0.03m) t   = 2  0.75 6 2 T 0  T ro    0.174  Ti  T  25  2  1



Therefore,

t

(Fig. 4 - 17a)

(0.75)(0.03)2  ro2   5192 s  1.44 h  0.13 10 6 m2 / s

The surface temperature is determined from 1 k    0.877 Bi hro  T ( r)  T  0.6  r  T o  T 1  ro

(Fig. 4  17b)

which gives Tsurface  T  0.6(To  T )  2  0.6(6  2)  4.4C Problem #6 In areas where the air temperature remains below 0 C for prolonged periods of time, the freezing of water in underground pipes is a major concern. Fortunately, the soil remains relatively warm during those periods, and it takes weeks for the subfreezing temperatures to reach the water mains in the ground. Thus, the soil effectively serves as an insulation to protect the water from the freezing atmospheric temperatures in winter. The ground at a particular location is covered with snow pack at 8 C for a continuous period of 60 days, and the average soil properties at that location are k  0.35 W/m·C and   0.15  806 m2/s. Assuming an initial uniform temperature of 8 C for the ground, determine the minimum burial depth to prevent the water pipes from freezing. Assumptions 1. The soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant.

The length of time the snow pack stays on the ground is Ts =-8C

t  (60 days)(24 hr/days)(3600 s/hr)  5.184 10 6 s

The surface is kept at -8 C at all times. The depth at which freezing at 0 C occurs can be determined from the analytical solution,  x  T( x, t)  Ti   erfc    Ts Ti  t 

Soil Ti = 8C Water pipe

  0 8 x    erfc   6 2 6 88  2 (0.15 10 m /s)(5.184 10 s)    x   0.5  erfc   1.7636 

Then from Table 4-4 we get

x  0.4796   x  0.846 m 1.7636

Problem #7 A hot dog can be considered to be a 12-cm-long cylinder whose diameter is 2 cm and whose properties are   980 kg/m3, cp  3.9 kJ/kg·C, k  0.76 W/m·C, and   2  107 m2/s. A hot dog initially at 5 C is dropped into boiling water at 100 C. The heat transfer coefficient at the surface of the hot dog is estimated to be 600 W/m2·C. If the hot dog is considered cooked when its center temperature reaches 80 C, determine how long it will take to cook it in the boiling water. Assumptions 1. Heat conduction in the hot dog is 2-D, and thus the temperature varies in both the axial x- and the radial r- directions. 2. The thermal properties of the hot dog are constant. 3. 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). This hot dog can physically be formed by the intersection of an infinite plane wall of thickness 2L = 12 cm, and a long cylinder of radius ro = D/2 = 1 cm. The Biot numbers and corresponding constants are first determined to be Bi 

hL (600 W/m 2 .C)(0.06 m)   47.37   1  1.5380 and A1  1.2726 k (0.76 W/m.C)

Bi 

hro (600 W/m 2 .C)(0.01 m)  1  2.1249 and A1 1.5514   7.895  k (0.76 W/m.C)

Noting that   t / L2 and assuming  > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable, the product solution for this problem can be written as

 2  2  (0,0, t ) block   (0, t ) wall  (0, t ) cyl   A1e 1   A1 e 1  







 80 100  (2  10 7 ) t   (1.2726) exp  (1.5380) 2  5 100  ( 0.06) 2   

Water 100C

  (2  10 7 )t   (1.5514) exp  (2.1249) 2   0.2105 (0.01) 2   

2 cm Hot dog Ti = 5C

which gives t  244 s = 4.1 min

Therefore, it will take about 4.1 min for the hot dog to cook. Note that  cyl 

t 2



(2 107 m2 /s)(244 s)

ro

(0.01 m) 2

 0.49  0.2

and thus the assumption  > 0.2 for the applicability of the one-term approximate solution is verified. Problem #8 Consider a 8-m-high, 8-m-long, and 0.22-m-thick wall whose representative cross section is as given in the figure. The thermal conductivities of various materials used, in W/m·C, are kA  kF  5, kB  10, kC  20, kD  15, and kE  40. The left and right surfaces of the wall are maintained at uniform temperatures of 300 C and 100 C, respectively. Assuming heat transfer through the wall to be one-dimensional, determine (a) the rate of heat transfer through the wall; (b) the temperature at the point where the sections B, D, and E meet; and (c) the temperature drop across the section F. Disregard any contact resistances at the interfaces.

Assumptions 1 Steady-State. 2 1-D. 3 k = constant. 4 Thermal contact resistances at the interfaces are disregarded. (a) The representative surface area is A  0.12 1  0.12 m 2 . The thermal resistance network and the individual thermal resistances are R2 R1

R3

T1 R4

R5

R7 T2

R6

0.01 m  L R1  RA      0.017  C/W 2  kA  A (5 W/m   C) (0.12 m ) 0.05 m  L R2  R4  RC      0.063 C/W 2  kA C ( 20 W/m   C) (0.04 m ) 0.05 m  L R3  RB      0.125 C/W 2  kA  B (10 W/m  C) (0.04 m )

0.1 m  L R5  RD      0.111 C/W o 2  kA  D (15 W/m  C) (0.06 m ) 0.1 m  L R6  RE      0.042 o C/W 2 kA   ( 40 W/m C) ( 0 . 06 m )  E  L  0.06 m R7  RF      0.1 C/W 2  kA F (5 W/m   C) (0.12 m ) 1 1 1 1 1 1 1         Rmid ,1  0.025  C/W R mid ,1 R 2 R 3 R 4 0.063 0.125 0.063

1 R mid ,2



1 1 1 1      R mid ,2  0.03 C/W R 5 R 6 0.111 0.042

R total  R 1  R mid ,1  R mid ,2  R 7  0.017  0.025  0.03  0.1  0.172 C/W T  T2 (300  100) C Q  1   1163 W (for a 0.12 m 1 m section) Rtotal 0.172 C/W

Then steady rate of heat transfer through entire wall becomes (8 m) (8 m)  6.2  10 5 W Qtotal  (1163 W) 2 0.12 m (b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is Rtotal  R1  Rmid ,1  0.017  0.025  0.042 C/W Then the temperature at the point where the sections B, D, and E meet becomes TT   T  T1  Q Rtotal  300 C  (1163 W)(0.042 C/W)  251 C Q  1 Rtotal (c) The temperature drop across the section F can be determined from T  T  Q R F  (1163 W)(0.1 C/W)  116 C Q  RF Problem #9 Steam at 320 C flows in a stainless steel pipe (k  15 W/m·C) whose inner and outer diameters are 5 cm and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k  0.038 W/m·C). Heat is lost to the surroundings at 5 C by convection, with a convection heat transfer coefficient of 15 W/m2·C. Taking the heat transfer coefficient inside the pipe to be 80

W/m2·C, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation. Assumptions 1 Steady-State. 2 1-D. 3 k = constant. 4 Thermal contact resistances at the interfaces are disregarded. The inner and the outer surface areas of the insulated pipe per unit length are Ai  Di L   (0.05 m) (1 m)  0.157 m 2 Ao   Do L   (0.055 0.06 m) (1 m)  0.361 m 2

The individual thermal resistances are R1

Ri

R2

Ro T2

T1

1 1   0.08 C/W 2 hi Ai (80 W/m  C)(0.157 m 2 ) ln( r2 / r1 ) ln( 2.75 / 2.5) R1  R pipe    0.00101 C/W 2 k1 L 2 (15 W/m  C)(1 m) Ri 

R2  Rinsulation  Ro 

1 ho Ao



ln( r3 / r2 ) 2 k 2 L



ln(5.75 / 2. 75)  3.089 C/W 2 (0.038 W/m  C)(1 m)

1 (15 W/m2   C)(0.361 m 2 )

 0.1847  C/W

Rtotal  Ri  R1  R2  Ro  0.08  0.00101  3.089  0.1847  3.355 C/W

Then the steady rate of heat loss from the steam per m. pipe length becomes T  T 2 (320  5)C Q  1  93.9 W  3.355 C/W Rtotal

The temperature drops across the pipe and the insulation are Tpipe  Q Rpipe  (93.9 W)(0.0010 1 C/W)  0.095 C  Tinsulation  Q Rinsulation  (93.9 W)(3.089  C/W)  290 C

Problem #10 A 50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open space at 15 C. The average temperature of the outer surface of the pipe is measured to be 150 C. If the combined heat transfer coefficient on the outer surface of the pipe is 20 W/m 2·C, determine (a) the rate of heat loss from the steam pipe; (b) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 % and the price of natural gas is $0.52/therm (1 therm  105,500 kJ); and (c) the thickness of fiberglass insulation (k  0.035 W/m·C) needed in order to save 90 % of the heat lost. Assume the pipe temperature to remain constant at 150 C.

Assumptions 1 Steady-State. 2 1-D. 3 k = constant. 4 The thermal contact resistance at the interface is negligible. 5 The pipe temperature remains constant at about 150C with or without insulation. 6 The combined heat transfer coe...