Tutorial 4Question 1 4-23 A brick of 203 x 102 x 57 mm in dimension is being burned in a kiln to ll00 °C, and then allowed to cool in a room with ambient air temperature of 30°C and convection heat transfer coefficient of 5 W/m2-K. If the brick has properties of ρ= 1920 kg/m3 , cp = 790 J/kg- K, and k= 0.90 W/m-K, determine the time required to cool the brick to a temperature difference of 5 oC from the ambient air temperature. Question 2 4-37 An experiment is to be conducted to determine heat transfer coefficient on the surfaces of tomatoes that are placed in cold water at 7°C. The tomatoes (k =0.59 W/m-K, α = 0.141 x 10-6 m2/s, ρ = 999 kg/m3, cp = 3.99 kJ/kg-K) with an initial uniform temperature of 30°C are spherical in shape with a diameter of 8 cm. After a period of 2 hours, the temperatures at the center and the surface of the tomatoes are measured to be l0.0°C and 7.l°C, respectively. Using analytical one-term approximation method (not the Heisler charts), determine the heat transfer coefficient and the amount of heat transfer during this period if there are eight such tomatoes in water. Question 3 4-39 A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°C and convection heat transfer coefficient of 220 W/m2-K. The 10-cm thick brass plate (ρ = 8530 kg/m3, cp = 380 J/kg.K, k= 110 W/m-K, and α =33.9 10-6 m2/s) has a uniform initial temperature of 650°C, and the bottom surface of the plate is insulated. Determine the temperature at the center plane of the brass plate after 3 minutes of cooling. 91 x10-6 m2/s). and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers 60°C.45 W/m. and (c) the amount of heat transferred to the rib. and α =0. ρ = 1200 kg/m3. . 3-cm-thick large brass plates (k = 110 W/m. determine the surface temperature of the plates when they come out of the oven. Taking the convection heat transfer coefficient to be h= 80 W/m2-K. Determine (a) the heat transfer coefficient at the surface of the rib: (b) the temperature of the outer surface of the rib when it is done. cp = 4.2-kg rib initially at 4. Question 5 4-49 ln Betty Crocker’s Cookbook. ρ = 8530 kg/m3.9 x10-6 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 l0 min. (d) Using the values obtained.K.K. It is recommended that a meat thermometer be used to monitor the cooking. cp = 380 J/kg-K.5°C “rare” in an oven maintained at 163°C. The rib can be treated as a homogeneous spherical object with the properties (k = 0.1 J/kg-K.Question 4 4-42 ln a production facility. and α =33. it is stated that it takes 2 h 45 min to roast 21 3. Do you agree with this recommendation‘? . Compare your result to the listed value of 3 h 20 min. it is recommended that the rib be taken out of the oven when the thermometer registers about 4°C below the indicated value because the rib will continue cooking even after it is taken out of the oven.predict how long it will take to roast this rib to “medium” level. which occurs when the innermost temperature of the rib reaches 71°C. If the roast rib is to be set on the counter for about l5 min before it is sliced. 057 ) m 2 hLc (5 W/m 2 K)(0.C.2.14110-6 m2/s. Assumptions 1 Thermal properties are constant. the lumped system analysis is applicable.522 10 4 s 7 hours b Ti T 2. Properties The properties of the tomatoes are given to be k = 0.C. 4 The heat transfer coefficient is constant and uniform over the entire surface. 4-40 Tomatoes are placed into cold water to cool them.0861 0.102 ) 2(0.01549 m As 2(0. Properties The properties of the brick are given as = 1920 kg/m3. Assumptions 1 The tomatoes are spherical in shape.90 W/m ∙ K.203 0. The ratio of the dimensionless temperatures at the Ti = 30C surface and center of the tomatoes are . it takes days to cool bricks coming out of kilns.59 W/m.128 10 4 s -1 c pV ρc p Lc (1920 kg/m 3 )( 790 J/kg K)( 0. Therefore one-term solution is applicable.01549 m) T (t ) T e bt Ti T or 1 T (t ) T 1 5 t ln ln 2. 3 The thermal properties of the tomatoes are constant.635 ro2 (0.Solutions The time required to cool a brick from 1100°C to a temperature difference of 5°C from the ambient air temperature is to be determined. 2 Heat conduction in the tomatoes is one-dimensional because of symmetry about the midpoint.057 ) 2(0. since they are being burned and cooled in bulk.90 W/m K Since Bi < 0.1 k 0. Analysis For a brick. 3 Heat transfer by radiation is negligible. = 999 kg/m3 and cp = 3. Analysis The Fourier number is t (0. cp = 790 J/kg ∙ K. = 0.99 kJ/kg. 5 The Fourier number is > 0. Then the time required to cool a brick from 1100°C to a temperature difference of 5°C from the ambient air temperature is hAs h 5 W/m 2 K b 2.128 10 s4 -1 1100 30 Discussion In practice.04 m) 2 7C Tomato which is greater than 0.057 ) m 3 Lc 0. 2 Convection heat transfer coefficient is uniform.01549 m) Bi 0.203 0.1. and k = 0.102 0. the characteristic length and the Biot number are V (0.102 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).141 10 6 m 2 /s)(2 3600 s) Water 0. The heat transfer coefficient and the amount of heat transfer are to be determined.203 0. 1 hr kBi (0.0311 The Fourier number is t (33 .0401 ) 1 3 0 1 3 0.10 m) Bi 0. C)(31 .sph T0 T T0 T A1e 12 1 Ti T Substituting.sph T T T T 1 sin(1 ) i s 0.9 10 6 m 2 /s)(3 60 s) 2 0.0401 10 7 1 From Table 4-2.4328 and A1 1.143 kg Qmax mc p [Ti T ] (2. 7.1 7 sin(1 ) 1 3.6102 0.08 m) 3 / 6] 2.99 kJ/kg. Assumptions 1 Heat conduction is one-dimensional. and = 33.2 k 110 W/m K From Table 4-2.04 m) The maximum amount of heat transfer is m 8V 8D 3 / 6 8(999 kg/m 3 )[ (0. the corresponding Biot number and the heat transfer coefficient are Bi 31 .5) after 3 minutes of cooling is .0401 ) cos(3.143 kg )(3.9565 Q T T 30 7 max cyl i 1 3 (3.10 m) 2 The temperature at the center plane of the plate (x/L = 0. 2 Thermal properties are constant.9 10−6 m2/s.2 L (0. cp = 380 J/kg ∙ K. 4 Heat transfer by radiation is negligible.9565 Qmax Q 0.1) Bi o h 459 W/m 2 . 3 Convection heat transfer coefficient is uniform. Analysis The Biot number for this process is hL (220 W/m 2 K)(0.0401 ) (3. k = 110 W/m ∙ K. the corresponding constants 1 and A1 are 1 0. C)(30 7)C 196.C k ro (0.6 kJ Then the actual amount of heat transfer becomes Q T T sin 1 1 cos 1 10 7 sin(3.6 kJ ) 188 kJ 4-42 The temperature at the center plane of a brass plate after 3 minutes of cooling by impinging air jet is to be determined.59 W/m.0401 ) 3 Q 0. Ts T sin(1 ) A1e 1 2 s.9565 (196 . Properties The properties of the brass plate are given as = 8530 kg/m3. 1035) (90. Properties The properties of brass at room temperature are given to be k = 110 W/m. t ) 700 0.700 C) e (0. 4-45 Large brass plates are heated in an oven.1035 and A1 1. C) The constants 1 and A1 corresponding to this Biot number Plates are. 1 0. t ) T A1e 1 cos(1 L / L) (1. .4328 )( 0.1035 ) 0.6102) 585 C Discussion The insulated bottom surface of the brass plate is treated as a thermally symmetric boundary.4) cos(0. the one-term approximate solution (or the transient temperature charts) is applicable.015 m) Bi 0.05 m.378 T ( L.001644s -1 )( 600 s) 448 C Ti T which is almost identical to the result obtained above.5) 15 C 2 ( 0.1. = 33.4 0. The surface temperature of the plates leaving the oven is to be determined.4328) cos(0. from Table 4-2.C.0109 k (110 W/m. It gives k k 110 W/m C c p 3.015 m) 2 Therefore.245 10 6 W s/m3 C c p 33.C)(0.0018 25C The Fourier number is t (33 . and thus the lumped system analysis is applicable.9 10 6 m 2 /s)(10 min 60 s/min) 90 .015 m)( 3. t ) (Ti T ) A1e 1 cos(1 x / L) T 2 T (0. Then the temperature at the surface of the plates becomes T ( x.9 10 -6 m 2 / s hA hA h h 80 W/m 2 C b 0.2 L2 (0.378 2 2 ( L. Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.910-6 m2/s Analysis The Biot number for this process is hL (80 W/m 2 .245 10 6 W s/m3 C) T (t ) T e bt T (t ) T (Ti T )e bt 700 C (25 . t ) 445C 25 700 Discussion This problem can be solved easily using the lumped system analysis since Bi < 0. t ) T A1e 1 cos(1 x / L) 2 wall Ti T T ( x. 4 The heat transfer coefficient is constant and uniform over the entire surface.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). T ( x. 3 The thermal properties of the plate are constant. 5 The Fourier number is > 0.0311 )e (0. t ) wall Ti T T ( L. 180 s) (650 C 15 C)(1.0018 )e (0.001644 s -1 Vc p ( LA)c p Lc p L(k / ) (0. 3 The thermal properties of the rib are constant.91 10 7 m 2 /s)(2 3600 + 45 60)s 0. Therefore. which corresponds to 1 3.1 kJ/kg.65 A1e 1 (0.45 W/m. = 1200 kg/m3.2 kg m V V 0.9110-7 m2/s.1217) 2 2 0.45 W/m.0372 and A1 1.5 163 It is determined from Table 4-2 by trial and error that this equation is satisfied when Bi = 30. (b) The temperature at the surface of the rib is T (ro . the temperature of the outer surface of the rib and the amount of heat transfer when it is rare done are to be determined.9898 )e (3.2.9898 . Assumptions 1 The rib is a homogeneous spherical object. Analysis (a) The radius of the roast is determined to be m 3.08603 m) This value seems to be larger than expected for problems of this kind. and = 0.4-52 A rib is roasted in an oven.08603 m 3 4 4 Oven 4. 5 The Fourier number is > 0. t ) 159. t ) sph Ti T 1 ro / ro 3.1217 ro2 (0.08603 m) 2 which is somewhat below the value of 0.1217) 2 2 (ro . t ) 163 0. The heat transfer coefficient at the surface of the rib. sph Ti T 4.002667 m 3 ) Rib V ro3 ro 3 0. with the understanding that the error involved will be a little more than 2 percent.0372 rad ) A1e 1 (1.5C 4.0372 T (ro . Then the one-term solution can be written in the form T0 T 60 163 A1e 1 0. the one-term approximate solution (or the transient temperature charts) can still be used.002667 m 3 1200 kg/m 3 4 3V 3 3(0. 4 The heat transfer coefficient is constant and uniform over the entire surface. 2 Heat conduction in the rib is one-dimensional because of symmetry about the midpoint.0222 T (ro .2. This is probably due to the Fourier number being less than 0. Properties The properties of the rib are given to be k = 0.5C The Fourier number is 163C t (0. t ) T sin(1 ro / ro ) sin(3.0372) (0.C.C.9 W/m 2 . The time it will take to roast this rib to medium level is also to be determined.2 so that the one- term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).C k ro (0. Then the heat transfer coefficient can be determined from hro kBi (0.5 163 (c) The maximum possible heat transfer is . cp = 4. C)(30 ) Bi h 156. 0372 ) 1 3 o. sph 1 3(0.783 Qmax 1 3 (3.0372 ) (3.783 Qmax (0.866 s 181 min 3 hr (0. sph 0.C)(163 4.9898 )e (3.08603 m) 2 t 10. Therefore.2 and thus the error in the one-term approximation. The recommendation is logical.1 kJ/kg. The difference between the two results is due to the Fourier number being less than 0.0372) 2 2 0. .0372 ) cos(3. Qmax mc p (T Ti ) (3.2 kg)(4. there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference.0372 ) 3 Q 0.91 10 7 m 2 /s) This result is close to the listed value of 3 hours and 20 minutes.1336 Ti T 4.5 163 ro2 (0.1336 )( 0. Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven.783 )( 2080 kJ) 1629 kJ (d) The cooking time for medium-done rib is determined to be T0 T 71 163 A1e 1 (1.5)C 2080 kJ Then the actual amount of heat transfer becomes Q sin(1 ) 1 cos(1 ) sin(3.65) 0.