Heat Transfer. Mechanical Engineering

March 14 Homework Solutions 4-38 Reconsider problem 4-37 using EES (or other software) investigate the effect of the final center temperature of the egg on the time it will take for the center to reach this temperature. Let the temperature vary from 50oC to 95oC. Plot the time versus the temperature and discuss the results. Problem 4-37, assigned on last week’s homework is copied here: An ordinary egg can be approximated as a 5.5-cm-diameter sphere whose properties are roughly k = 0.6 W/m2oC and α = 0.14x10-6 m2/s. The egg is initially at a uniform temperature of 8oC and is dropped into boiling water at 97oC. Taking the convection heat transfer coefficient to be h = 1400 W/m2·oC, determine how long it will take for the center of the egg to reach 70oC. Last week’s solutions showed how the problem can be solved using the charts. To compute several times, we will use the approximate solution approach. At the end we will check to make sure that the dimensionless time, τ = αt/ro2 is always greater than 0.2. In order to use the approximate solution we have to know the values of A1 and λ1 corresponding to the Biot number computed below, where the sphere radius = (5.5cm)/2 = 0.0275 m. Bisphere =  m⋅o C hro 1400 W = 2 o (0.0275 m ) k m ⋅ C  0.6 W   = 64.17  Interpolating in Table 4-2 for the sphere we find that A1 = 3.0863 and λ1 = 1.9969 for Bi = 64.17. We use equation (4-28) on page 230 of the text for the center temperature of the sphere. Θ0 = 2 T0 − T∞ = A1e −λ1τ Ti − T∞ ⇒ τ= αt r02 =−  Θ0  r02  Θ 0     ln ln t ⇒ = − λ12  A1  αλ12  A1  1 We want to set this equation so that we can solve for t as a function of T0. We can substitute the known data to get a working equation as follows. 2  1  Θ0  T0 − 97 o C  ( 0.0275 m )  =− t = − 2 ln ln 3.0863 8 o C − 97 o C  αλ1  A1  0.14 x10 −6 m 2 2  1.9969 s  274.68 o C  1 min  274.68 o C  1 min  274.68 o C      = (1354.6 s ) ln ln = (22.577 min )ln = (1354.6 s ) o 60 s  T0 − 97 o C  60 s  T0 − 97 o C  T C 97 −   0 r02 ( ) The final equation gives the time in minutes when the desired center temperature is entered in degrees Celsius. This formula is evaluated for temperatures between 50oC and 95oC and the results are plotted using an Excel spreadsheet. The results are shown below. The table shows that all values of t are greater than 0.2 an ex post facto demonstration that the use of the oneterm approximate solution is valid. As expected the time increases as the desired temperature increases, however the computed times seem quite long for cooking an egg. The assumptions used in this problem are undoubtedly invalid. Perhaps the property data for the egg or the heat transfer coefficient are incorrect. 50 55 60 65 70 75 80 85 90 39.9 42.4 45.3 48.5 52.4 57.0 62.8 70.7 82.9 τ 0.4427 0.4709 0.5027 0.5391 0.5817 0.6331 0.6978 0.7851 0.9203 Time Required to Heat Egg 90 80 . t (min) 70 60 Time (min) T0 (oC) 50 40 30 20 10 0 50 55 60 65 70 75 80 85 90 95 Center Temperature (oC) 4-43E Long cylindrical AISI stainless steel rods (k = 7.74 Btu/h·ft·oF and α = 0.135 ft2/h) of 4-in diameter are heat treated by drawing them at a velocity of 7 ft/min through a 21-ft-long oven maintained at 1700oF. The heat transfer coefficient in the oven is 20 Btu/h·ft2·oF. If the rods enter at 70oF, determine their centerline temperature when they leave. Assuming that the heat transfer is in the radial direction only, we can use equation (4.27) on page 230 to find the centerline temperature. First we find the Fourier number to make sure that we can use the approximate solution. Since the rods proceed at a velocity of 7 ft/min through the 21 ft oven, their transit time is 3 min = 0.05 h. With this time, the data given for α and the radius = 2 in, the Fourier number, τ, can be found. 0.135 ft 2 (0.05 h ) αt h = 0.243 τ= 2 = 2 r0  1 ft  (2 in )  12 in   Next we have to find the Biot number so that we can find the correct values of A1 and λ1. Bi = hro 20 Btu 1 ft  h ⋅ ft ⋅o F   = 0.4307  ( ) 2 = in 12 in  7.74 Btu  k h ⋅ ft 2 ⋅o F For this value of Biot number, we find the value of A1 = 1.0996 and λ1 = 0.8790 from the data for the infinite cylinder in Table 4-2. Now we can apply the approximate solution in equation (4-27) Θ0 = 2 2 T0 − T∞ = A1e −λ1τ = 1.0995e −(0.8790 ) (0.243 ) = 0.911 Ti − T∞ We can then find the centerline temperature. Θ0 = T0 − T∞ = 0.911 ⇒ T0 = T∞ + (Ti − T∞ )(0.911) = 1700 o F + 70 o F − 1700 o F (0.911) Ti − T∞ ( ) T0 = 215oF 4-45 A long cylindrical wood log (k = 0.17 W/m·oC and α = 1.28x10-7 m2/s) is 10 cm in diameter and is initially at a uniform temperature of 15oC. It is exposed to hot gases at 550oC in a fireplace with a heat transfer coefficient of 13.6 W/m2·oC on the surface. If the ignition temperature of the wood is 420oC, determine how long it will be before the log ignites. Assuming that the heat transfer is in the radial direction only, we can use equation (4.24) on page 230 to find the surface temperature. Because we are trying to find the time, we cannot start by finding the Fourier number to make sure that we can use the approximate solution. However, we can find this as part of the solution and then make sure that our use of the approximate solution is justified. We first find the Biot number to determine the correct values of A1 and λ1. Bicylinder  m⋅o C hro 13.6 W = = 2 o (0.05 m ) k m ⋅C  0.17 W   = 4.00  For this value of Biot number, we find the value of A1 = 1.4698 and λ1 = 1.9081 from the data for the infinite cylinder in Table 4-2. Now we can apply the approximate solution in equation (4-24) where we find the value of the Bessel function by interpolation in Table 4-3 on page 231. (This interpolation gives almost the same result as using the besselj(x,n) function in Excel since there is only a small change from the tabulated values due to the interpolation.) For the surface of the cylinder we set r/r0 = 1. Θ= 2 2 2  r T (r , t ) − T∞ = A1e − λ1τ J 0  λ1  = 1.4698e − (1.9081) τ J 0 (1.9081(1)) = 1.4698e − (1.9081) τ (0.27711) Ti − T∞  r0  Solving this equation for the Fourier number, τ, when T(r = r0) = 420oC gives. 2 T (r , t ) − T∞ 420o C − 550o C = = 0.243 = 1.4698e −(1.9081) τ (0.27711) o o Ti − T∞ 15 C − 550 C τ =−   1 0.243 ln  = 0.142 2 (1.9081) 1.4698(0.27711) This means that all our calculations have been in vain! The chart for finding temperatures at locations other than the center of the body is based on the same approximate solution process that we tried to use here. Since we do have a Fourier number, we can see what time it is equivalent to, even though we know that this will not be an accurate result. t=− r02τ (0.05 m ) (0.142) = 2770 s α 0.128 x10 −6 m 2 s 2 Converting to minutes gives t = 46.2 min . Solving the exact differential equation for this case (which we have not covered in class) gives a solution of τ = 0.152; so applying the approximate solution at this point causes an error of about 7%. 4-72 A thick wood slab (k = 0.17 W/m·oC and α = 1.28x10-7 m2/s) that is initially at a uniform temperature of 25oC. It is exposed to hot gases at 550oC for a period of 5 minutes. The heat transfer coefficient between the gases and the wood slab is 35 W/m2·oC on the surface. If the ignition temperature of the wood is 450oC, determine if the wood will ignite. We are not given the dimension of the wood slab, but we are told that it is thick. Let’s assume that we may model the surface of this “thick” slab as a semi-infinite surface. For such a surface the following equation describes the temperature at any time and x location when the wood is introduced into a convection environment at t = 0. T − Ti  x = erfc T∞ − Ti  4αt  hx h 2αt    k 2    k + −e   x h αt erfc + k  4αt     In this case we want to find the surface temperature so we have x = 0, and the problem is simplified to  h 2 αt   k 2    T − Ti = erfc(0) − e T∞ − Ti  h 2 αt   k 2    h αt    = 1 − e erfc   k   h αt   erfc   k  In the above equation we use the value of erfc(0) = 1. The terms in this equation are evaluated below. h αt 35 W m⋅o C 1.28 x10−7 m 2 (5 min ) 60 s = 1.276 = 2o k m ⋅ C 0.17 W s min h 2αt = 1.2762 = 1.628 2 k With these values and Table 4-4 to find erfc(1.276), we can now find the dimensionless temperature. T − Ti = 1 − e1.628erfc(1.276 ) = 1 − (5.094 )(0.0712 ) = 0.637 T∞ − Ti We can now find the surface temperature at the end of 5 minutes. T − Ti T − 25o C = = 0.637 T = 25o C + 525o C (0.637 ) = 360oC o o T∞ − Ti 550 C − 25 C ( ) This is less than the ignition temperature so the wood will not ignite . 4-83 A 5-cm-high rectangular ice block (k = 2.22 W/m·oC and α = 0.124 x 10-7 m2/s) initially at -20oC is placed on table on its square base 4 cm by 4 cm in size in a room at 18oC. The heat transfer coefficient on the exposed surfaces of the ice block is 12 W/m2·oC. Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear? We can find the solution to this problem as the solution to three separate one-dimensional infinite slabs. The first slab, which is parallel to the table, will have a half length, L = 5 cm, the height of the block. This is valid because we are told to assume that there is no heat flow through the table. A condition of no heat flow is the same as a condition of zero temperature gradient, which is the condition at the centerline of our infinite slab. The other two solutions will be for the sides of the ice block. Both of these solutions will be for the same dimensions – a slab with a half-length of 2 cm. For the three-dimensional solution, the dimensionless temperature at any point (x,y,z) in the slab, at any time t, is given by the product of three one-dimensional solutions. Θ( x, y, z , t ) = Θ1 ( x, t )Θ1 ( y, t )Θ1 ( z , t ) We will assume that the Fourier number, τ, is greater than 0.2 so that we can use the approximate solution and check this assumption after we solve for τ. If we denote the coordinate distance as xi and the half-length as Li for each of the coordinate directions, the one-dimensional solution for an infinite slab is. 2 Θ1 (xi , t ) = A1e − λ1τ cos λ1ξ = A1e − λ21αt L2i  x  cos λ1 i   Li  We expect that the first drops of liquid to form will be at the upper corners of the ice block where the exposure to air is the greatest and the overall distance from the center of the cold ice is the greatest. For this corner, the dimensionless coordinates, xi/Li, in the above equation, are 1 for all three directions. We first have to find the Biot number to determine the values of A1 and λ1. In general we would have three such calculations for a three-dimensional problem, but here we have only two calculations since two of the dimensions are the same. For the vertical distance of 5 cm = 0.05 m, the Biot number is. Bislab =  m⋅o C hL 12 W = 2 o (0.05 m )  2.22 W k m ⋅ C    = 0.2703   Interpolating in Table 4-2 we find that A1 = 1.0408 and λ1 = 0.4951 for this Biot number. Both of the horizontal solutions will have the same Biot number for their half thickness of 2 cm = 0.02 m. Bislab =  m⋅o C hL 12 W = 2 o (0.02 m )  2.22 W k m ⋅ C    = 0.1091   Interpolating in Table 4-2 we find that A1 = 1.0713 and λ1 = 0.3208 for this Biot number. Since two of our one-dimensional solutions are the same, we can write our three-dimensional product solution as follows. Θ( x, y, z , t ) = Θ12 ( x, t )Θ1 ( z , t ) The value of τ = αt/Li2, will be different for each dimension because the length parameter will be different for each dimension. However, the value of time, t, will be the same. So we have to write the equation in terms of the time, t. Substituting the one dimensional solutions, with the appropriate values of A1, λ1, α, and the length parameter for each dimension and setting the dimensionless distance in the cosine term to 1 for each solution gives. 2 T − T∞ Ti − T∞ (0.3208 )2 0.124 x10 − 7 m 2 t   −   0 C − 18 C s ( 0.02 m )2 cos(0.3208) • = = 0.4737 = 1.0713e o o − 20 C − 18 C     2 −7 2 (0.4951) 0.124 x10 m t   −   s ( 0.02 m )2 cos(0.4951) 1.0408e     o o We can not solve this equation explicitly for t, but we can solve it by an iteration algorithm or using a calculator or a program such as Matlab or Excel. Doing this we find that t = 77,500 s = 21.5 h. We have to check the value of t to make sure that it is greater than 0.2. We only have to check the value for a length parameter of 0.05 m. If t > 2 for this length, then it will also be > 0.2 for the length parameter of 0.02 m. τ= αt 0.124 x10 −7 m 2 (77,500 s ) = = 0.384 > 0.2 s L2 (0.05 m )2 So the criterion that t > 0.2 is satisfied and our approach is justified; the answer is t = 21.5 h . 4-86 Consider a cubic block whose sides are 5 cm long and a cylindrical block whose height and diameter are also 5 cm. Both blocks are initially at 20oC and are made of granite (k = 2.5 W/m·oC and α = 1.15x10-6 m2/s). Now both blocks are exposed to hot gases at 500oC in a furnace on all of their surfaces with a heat transfer coefficient of 40 W/m2·oC. Determine the center temperature of each geometry after 10, 20, and 60 min. For the cube we have three product solutions, but they are all identical because the lengths are identical and the location at which we want to evaluate the temperature is identical for each coordinate. Thus we can write Θ( x , y , z , t ) = T − T∞ = Θ1 ( x, t )Θ1 ( y, t )Θ1 ( z , t ) = [Θ1 ( xi , t )]3 Ti − T∞ For the minimum time of 10 min = 600 s, we can find the Fourier number, τ, = αt/L2, where L is the half thickness. For this problem, L = (5 cm)/2 = 2.5 cm = 0.025 m, and the Fourier number is τ= αt 1.15 x10 −6 m 2 (600 s ) = =1.104 > 0.2 s L2 (0.025 m )2 Since τ > 0.2, we can apply the approximate solution Θ1 ( xi , t ) = A1e − λ21τ cos λ1ξ = A1e − λ21αt L2i  x  cos λ1 i   Li  We can find the values of A1 and λ1 from the Biot number. Bislab =  m⋅o C hL 40 W = 2 o (0.025 m )  2.5 W k m ⋅ C    = 0.4   For Bi = 0.4 we find that A1 = 1.0580 and λ1 = 0.5932 from Table 4-2. We can now compute the center temperature of the cube where x = y = z = 0. (Recall that cos(0) = 1.) 3  − λ21αt   − λ21αt  2 2   x   T − T∞  Θ( x , y , z , t ) = =  A1e Li cos λ1 i  =  A1e Li  Ti − T∞  Li         3 The argument of the exponential is λ21αt − 2 L = (0.5932 )2 1.15 x10 −6 m 2 (600 s ) = 0.3885 s (0.025 m )2 Substituting this value and the value for A1 into our equation for Θ gives the temperature as 3  − λ21αt  2  3  T = T∞ + (Ti − T∞ ) A1e Li  = 500o C + 25o C − 500o C 1.050e − 0.3885 = 323o C     ( )[ ] At t = 20 minutes = 1200 s and 60 minutes = 3600 s, -λ12αt/L2 = 0.7770 and 2.331, respectively. Applying the same equation gives T = 445oC and 500oC at these times. The cylinder can be solved as the product of the solution for the infinite cylinder and that of the infinite slab. Since the cylinder radius is the same as the half width of the cube solved above, the Fourier number will be the same and all times will have a Fourier number greater than 0.2 allowing the use of the approximate solution. The temperature will be the product of two solutions. Θ( x , r , t ) = 2   x  T − T∞ = Θ slab (r , t )Θcyl (r , t ) =  A1e − λ1τ cos λ1 i  Ti − T∞  Li    − λ2 τ  r   A1e 1 J 0  λ1   r0  c s The subscripts s for slab and c for cylinder indicate that the values of A1 and λ1 will be different for the different solutions. The Biot numbers will be the same for both solutions because the halflength of the cylinder height is the same as its radius. Bislab =  m⋅o C hL hr 40 W = Bicyl = o = 2 o (0.025 m )  2.5 W k k m ⋅ C    = 0.4   For Bi = 0.4 Table 4-2 shows that that A1 = 1.0580 and λ1 = 0.5932 for the slab and A1 = 1.0931 and λ1 = 0.8516 for the cylinder. We can now compute the center temperature of the cylinder where x = r = 0. (Note that cos(0) = J0(0) = 1.) 2 2 2  0  T − T∞  − λ21τ  0   cos λ1   A1e − λ1τ J 0  λ1  =  A1e − λ1τ   A1e − λ1τ  =  A1e  s   c Ti − T∞   L  s   r0  c  Substituting numerical values for t = 10 min = 600 s, where we have previously computed τ = 1.104 gives. 2 2 T − T∞  − λ21τ   − λ21τ  A1e = 1.0580e − (0.5932 ) (1.104 )  1.0931e − (0.8516 ) (1.104 )  = 0.352 = A1e     s   c  Ti − T∞  We can then find the temperature. ( )[ ] 3 T = T∞ + (Ti − T∞ )Θ = 500o C + 25o C − 500o C 1.050e − 0.3885 = 331o C At t = 20 minutes = 1200 s and 60 minutes = 3600 s, τ = 2.208 and 6.624, respectively. Applying the same equations give T = 449oC and 500oC at these times. The answers to this problem are summarized in the table below. Geometry Cube Cylinder Temperatures at times shown below 10 min 20 min 60 min 323oC 445oC 500oC o o 331 C 449 C 500oC At the large values of Fourier number in this problem, t = 1.1, 2.2, and 6.6, the temperature approaches and then equals the value of T∞.