ࡱ> S` 0bbjbj 8 : : : $^ ***P+ ",^ V,/L///4N<5$`5 UUUUUUU$yXhZ2UM: 64^4662U //VCCC6 l/: / UoX C6 UCCM ,: N/, 0׮ed*<NRV0VN<[;?[0N[: Nt505"C55t5t5t52U2UAt5t5t5V6666^ ^ ^ "^ ^ ^ "^ ^ ^   INCLUDEPICTURE "http://www.csun.edu/~pubrels/NewLogoSm.JPG" \* MERGEFORMATINET  College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 375 Heat Transfer INCLUDETEXT "C:\\Documents and Settings\\Larry Caretto\\My Documents\\Courses\\ME 375\\include.doc" firstPage \* MERGEFORMAT Spring 2007 Number 17629 Instructor: Larry Caretto 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 weeks 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 a = 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/m2oC, 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, t = at/ro2 is always greater than 0.2. In order to use the approximate solution we have to know the values of A1 and l1 corresponding to the Biot number computed below, where the sphere radius = (5.5cm)/2 = 0.0275 m.  EMBED Equation.3  Interpolating in Table 4-2 for the sphere we find that A1 = 3.0863 and l1 = 1.9969 for Bi = 64.17. We use equation (4-28) on page 230 of the text for the center temperature of the sphere.  EMBED Equation.3  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.  EMBED Equation.3  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 one-term 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. T0 (oC)t (min)t5039.90.44275542.40.47096045.30.50276548.50.53917052.40.58177557.00.63318062.80.69788570.70.78519082.90.92034-43E Long cylindrical AISI stainless steel rods (k = 7.74 Btu/hftoF and a = 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/hft2oF. 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 a and the radius = 2 in, the Fourier number, t, can be found.  EMBED Equation.3  Next we have to find the Biot number so that we can find the correct values of A1 and l1.  EMBED Equation.3  For this value of Biot number, we find the value of A1 = 1.0996 and l1 = 0.8790 from the data for the infinite cylinder in Table 4-2. Now we can apply the approximate solution in equation (4-27)  EMBED Equation.3  We can then find the centerline temperature.  EMBED Equation.3  T0 = 215oF 4-45 A long cylindrical wood log (k = 0.17 W/moC and a = 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/m2oC 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 l1.  EMBED Equation.3  For this value of Biot number, we find the value of A1 = 1.4698 and l1 = 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.  EMBED Equation.3  Solving this equation for the Fourier number, t, when T(r = r0) = 420oC gives.  EMBED Equation.3   EMBED Equation.3  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.  EMBED Equation.3  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 t = 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/moC and a = 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/m2oC 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. Lets 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.  EMBED Equation.3  In this case we want to find the surface temperature so we have x = 0, and the problem is simplified to  EMBED Equation.3  In the above equation we use the value of erfc(0) = 1. The terms in this equation are evaluated below.  EMBED Equation.3  With these values and Table 4-4 to find erfc(1.276), we can now find the dimensionless temperature.  EMBED Equation.3  We can now find the surface temperature at the end of 5 minutes.  EMBED Equation.3  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/moC  and a = 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/m2oC. 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.  EMBED Equation.3  We will assume that the Fourier number, t, is greater than 0.2 so that we can use the approximate solution and check this assumption after we solve for t. 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.  EMBED Equation.3  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 l1. 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.  EMBED Equation.3  Interpolating in Table 4-2 we find that A1 = 1.0408 and l1 = 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.  EMBED Equation.3  Interpolating in Table 4-2 we find that A1 = 1.0713 and l1 = 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.  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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, l1, a, and the length parameter for each dimension and setting the dimensionless distance in the cosine term to 1 for each solution gives.  EMBED Equation.3  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.  EMBED Equation.3  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/moC and a = 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/m2oC. 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  EMBED Equation.3  For the minimum time of 10 min = 600 s, we can find the Fourier number, t, = at/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  EMBED Equation.3  Since t > 0.2, we can apply the approximate solution  EMBED Equation.3  We can find the values of A1 and l1 from the Biot number.  EMBED Equation.3  For Bi = 0.4 we find that A1 = 1.0580 and l1 = 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.)  EMBED Equation.3  The argument of the exponential is  EMBED Equation.3  Substituting this value and the value for A1 into our equation for Q gives the temperature as  EMBED Equation.3  At t = 20 minutes = 1200 s and 60 minutes = 3600 s, -l12at/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.  EMBED Equation.3  The subscripts s for slab and c for cylinder indicate that the values of A1 and l1 will be different for the different solutions. The Biot numbers will be the same for both solutions because the half-length of the cylinder height is the same as its radius.  EMBED Equation.3  For Bi = 0.4 Table 4-2 shows that that A1 = 1.0580 and l1 = 0.5932 for the slab and A1 = 1.0931 and l1 = 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.)  EMBED Equation.3  Substituting numerical values for t = 10 min = 600 s, where we have previously computed t = 1.104 gives.  EMBED Equation.3  We can then find the temperature.  EMBED Equation.3  At t = 20 minutes = 1200 s and 60 minutes = 3600 s, t = 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. GeometryTemperatures at times shown below10 min20 min60 minCube323oC445oC500oCCylinder331oC449oC500oCAt 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(.     Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail:  HYPERLINK mailto:lcaretto@csun.edu lcaretto@csun.edu 8348 Fax: 818.677.7062  REF date \h  \* MERGEFORMAT March 14 homework solutions  INCLUDETEXT "C:\\Documents and Settings\\Larry Caretto\\My Documents\\Courses\\ME 375\\include.doc" header \* MERGEFORMAT ME 375, L. S. 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"Gd|tn#@q0`\mxڥMhAlfmB-Z I)~ @O$jLh17*E^A"E"?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~T Root Entry F,hd@Data 5WordDocumentObjectPool);ed,hd_1234874982 F;ed;edOle CompObjfObjInfo  !$'()*+,-.1456789:;<=>?@ABCDEHKLMNORUVWXYZ[\]^_`abcdefghijklmopqrsvyz{|} FMicrosoft Equation 3.0 DS Equation Equation.39qs"<  .  @&h & MathType-```v `%OlePres000 Equation Native P_1234870006 F;ed;edOle "SymbolX! xww 0wf-2 ](ySymbol Axww 0wf-2 i)y``@Times New Romanxww 0wf-2 m172  .72 642 642 a.42 04 2 02752 .22 02 2 f 1400 Times New Romanxww 0wf-2 E 24SymbolX! xww 0wf-2 a=42 Tp42 p42 Jp42 p42 T42 42 J42 42 42  42 >=42 =4Times New Romanxww 0wf-2 W42 C42 m42 Dm42  C42  m42  W42 jk42 hr2 FBi Times New Romanxww 0wf-2 #Qoi2 E oi2 $oi 2 sphere & "Systemf !-NANI4  Bi sphere =hr o k=1400Wm 2 ", o C0.0275m()m", o C0.6W()=64.17 FMicrosoft Equation 3.0 DS Equation Equation.39q0   0 =T 0 "T " T i "CompObj #fObjInfo%Equation Native &_1234875356F;ed;edT " =A 1 e " 12  !=tr 02 ="1 12 ln 0 A 1 ()!t="r 02  12 ln 0 A 1 ()Ole /CompObj0fObjInfo2Equation Native 3 FMicrosoft Equation 3.0 DS Equation Equation.39q‡  t="r 02  12 ln 0 A 1 ()="0.0275m() 2 0.14x10 "6 m 2 s1.9969 2 ()ln13.0863T 0 "97 o C8 o C"97 o C()=1354.6s()1min60sln274.68 o CT 0 "97 o C()=1354.6s()1min60sln274.68 o CT 0 "97 o C()=22.577min()ln274.68 o CT 0 "97 o C() FMicrosoft Equation 3.0 DS Eq_1234877131F;ed;edOle FCompObjGfObjInfoIuation Equation.39q)@  =tr 02 =0.135ft 2 h0.05h()2in()1ft12in[] 2 =0.243Equation Native JE_1234876984F;ed;edOle PCompObjQf FMicrosoft Equation 3.0 DS Equation Equation.39q##<   .  &h & MathType-`/`9``u ObjInfoSOlePres000TZEquation Native ns_1234877679$!F;ed;edSymbolN" xww 0wI f-2  (ySymbol! xww 0wI f-2 i)y````5Times New Romanxww 0wI f- 2 43072 7.32 w032 742 .42 742 122 S122 " 222 20 Times New Romanxww 0wI f-2 B 20SymbolN" xww 0wI f-2 P=02 Te02 e02 Je02 e02 T&02 &02 J&02 &02 02 +02 Z 02 02 =02 =0Times New Romanxww 0wI f- 2 Btu72 Ft2 ft2 7ht2 in2 ft2 +in2 S Fn2 ft2 ht 2  Btu72 kt2 Ihr2 FBi Times New Romanxww 0wI f-2 #Ooi2 B oi2 &oi & "SystemI f !-NANIW  Bi=hr o k=20Btuh"ft 2 ", o F2in()1ft12inh"ft", o F7.74Btu()=0.4307 FMicrosoft Equation 3.0 DS Equation Equation.39qOle tCompObj "ufObjInfo#wEquation Native xuY   0 =T 0 "T " T i "T " =A 1 e " 12  =1.0995e "0.8790() 2 0.243() =0.911 FMicrosoft Equation 3.0 DS Eq_1234877783&F;ed;edOle ~CompObj%'fObjInfo(uation Equation.39q¾  0 =T 0 "T " T i "T " =0.911!T 0 =T " +T i "T " ()0.911()=1700 o F+70 oEquation Native _1234878586[+F;ed;edOle CompObj*-f F"1700 o F()0.911() FMicrosoft Equation 3.0 DS Equation Equation.39q <  .  &ObjInfoOlePres000,.Equation Native P_12349842721F;ed;ed`h & MathType-`>`H` `,Symbol xww 0w f-2 a(ySymbolX! xww 0w f-2 )y` `~Times New Romanxww 0w f-2 002 .02 402 >172 .72 072 052 .52 052  652 S .52  13 Times New Romanxww 0w f-2 BQ 23Symbol xww 0w f-2 =32 T32 32 J32 32 TE32 E32 JE32 E32 32  32 =32  =3Times New Romanxww 0w f-2 W32 C32 m32 m32  C32 ! m32  W32 k32 Xhr2 FBi Times New Romanxww 0w f-2 #.oi2 BN oi2 &oi 2 cylinder & "System f !-NANI4  Bi cylinder =hr o k=13.6Wm 2 ", o C0.05m()m", o C0.17W()=4.00 FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObj02fObjInfo3Equation Native f&J\ =T(r,t)"T " T i "T " =A 1 e " 12  J 0  1 rr 0 ()=1.4698e "1.9081() 2  J 0 1.90811()()=1.4698e "1.9081() 2  0.27711() FMicrosoft Equation 3.0 DS Equation Equation.39q›, T(r,t)_12348795656F;ededOle CompObj57fObjInfo8Equation Native _12348848284>;FedledOle CompObj:<f"T " T i "T " =420 o C"550 o C15 o C"550 o C=0.243=1.4698e "1.9081() 2  0.27711() FMicrosoft Equation 3.0 DS Equation Equation.39q(  ="11.9081() 2 ln0.2431.46980.27711()[]=0.142    $6 !"#%)&'(*+,-./0312457;89:<=?>B@ACDEFGHIJKLMQNOPRUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ObjInfo=Equation Native _1234885053@FledledOle CompObj?AfObjInfoBEquation Native '_12349526359OEFledled FMicrosoft Equation 3.0 DS Equation Equation.39q    t="r 02 0.05m() 2 0.142()0.128x10 "6 m 2 s=2770sOle CompObjDGfObjInfoOlePres000FH FMicrosoft Equation 3.0 DS Equation Equation.39q# S .  @ & $ & MathType-@PF* *[  -2[  - 0 0 0M  m -j5-F*5-25|-000'G-8-@>Symbol ;xww 0wf-2 ny2 4ny2 ny2 >ny2 y2 4y2 y2 >y2 `+y2 `{ -y2 t y2  y2 ~ y2 t<y2 <y2 ~<y2 `=y2 -y2 r-y Symbol axww 0wf-2 H]y2 ]y2 ]y2 *]y2 HGy2 Gy2 Gy2 *Gy2 +y2 yTimes New Romanxww 0wf-2 &ky2 rty2 rhy2 ty2 rFxy 2 `erfc2 `er2  tr2 rl xr 2 `erfc2 Tr2 BTr2 rTr2 rTr Times New Romanxww 0wf-2 kr2 Ltr2 Lyhr2 kr2 Lhx2 ix2 ixSymbol =xww 0wf-2 rax2 ax2  ax Symbol cxww 0wf-2 LRaxTimes New Romanxww 0wf-2 4x2  4x`Times New Romanxww 0wf-2 D_2x2 2x & "Systemf !-NANIEquation Native _1234952764KFledledOle CompObjJMf    !"#$%&'()*+,-./012356789:;>ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abdefghijmpqrstuvwxyz{|}~rp T"T i T " "T i =erfcx 4t  ()"e hxk+h 2 tk 2 () erfcx 4t  +h t  k() FMicrosoft Equation 3.0 DS Equation Equation.39q`(   .  $&`$$ & MathType-@SSymbol xww 0wE!fX-2 ^D(yObjInfo OlePres000LN  Equation Native 4_1234953028IUQFledledSymbolq xww 0wE!fX-2 ^ )y-y E- -X-`?_--!P!-P!!-!!!~#+ #Symbol xww 0wE!fX-2 #y2 4#y2 #y2 >#y2 ey2 4ey2 ey2 >ey2 `l-y2 `}=y2 y2 4y2 y2 >y2 &y2 4&y2 &y2 >&y2 `- -y2 `=y2 -y2 r-y Symbolq xww 0wE!fX-2 Hy2 y2 y2 *y2 H8y2 8y2 8y2 *8y2 Hmy2 my2 my2 *my2 H y2  y2  y2 * y2 yTimes New Romanxww 0wE!fX-2 !ky2 r"ty2 rE hy 2 ` erfc2 `er2 Ekr2 rtr2 rhr 2 `erfc2 `G er 2 `erfc2 Tr2 BTr2 rTr2 rTr Times New Romanxww 0wE!fX-2 $kr2 L0tr2 Lhr2  kr2 L tr2 L hr2 ir2 irSymbol xww 0wE!fX-2 r!ar2 rar Symbolq xww 0wE!fX-2 Lar2 Lb ar`Times New Romanxww 0wE!fX-2 D2r2 N2r2 Do 2r2  2rTimes New Romanxww 0wE!fX-2 `1r2 `0r & "SystemXE!fX !-NANI³l T"T i T " "T i =erfc0()"e h 2 tk 2 () erfch t  k()=1"e h 2 tk 2 () erfch t  k() FMicrosoft Equation 3.0 DS Equation Equation.39qOle <CompObjPS=fObjInfo?OlePres000RT@6<T * .  1&1u & MathType-4~e -e-@T' g  !Symbol xww 0w f}-2 -=52 '=52 =52 2 52 52 =5 Symbol xww 0w f}-2 F-5Times New Romanxww 0w f}-2 /%k52 &t52 $h52 hs52 s52 m52 x52 U W52  C52 $ m52 C52 nm52 WW52 k52 t52 Zh5 Times New Romanxww 0w f}-2 F o52 \yo5Symbol @xww 0w f}-2 %a52  a5 & "System} f} !-NANI  h t  k=35Wm 2 ", o Cm", o C0.17W 1.28x10 "7 m 2 s5min()Equation Native c_1234953226WFledledOle kCompObjVYlf60smin  =1.276h 2 tk 2 =1.276 2 =1.628 FMicrosoft Equation 3.0 DS Equation Equation.39q+& u ObjInfonOlePres000XZo,Equation Native ?_1234953374Cf]Fledled.  @"&`" & MathType-@RSymbol xww 0w f9-2 ^ (ySymbol xww 0w f9-2 ^{)ySymbol xww 0w f9-2 ^@(ySymbol xww 0w f9-2 ^)ySymbol xww 0w f9-2 ^{(ySymbol xww 0w f9-2 ^)yTimes New Romanxww 0w f9- 2 ` 637e2 `.32 `03 2 ` 07122 `.72 `07 2 `09422 `l.92 `592 `=19 2 `/27622 ` .72 ` 172 `17 Times New Romanxww 0w f9- 2  62822 .22 f12Symbol ߴxww 0w f9-2 `=22 `'-22 `7=22 `-22 `=22 -22 r-2 Symbol xww 0w f9-2 2Times New Romanxww 0w f9- 2 `d erfc2 `er2 Tr2 BTr2 rTr2 rTr Times New Romanxww 0w f9-2 ir2 ir & "System9 f9 !-NANI#`4= T"T i T " "T i =1"e 1.628 erfc1.276()=1"5.094()0.0712()=0.637 FMicrosoft Equation 3.0 DS Equation Equation.39q. J .  @*&*% & MathTypeOle CompObj\_fObjInfoOlePres000^`-@@@S@@kSymbol hxww 0wa f-2 (ySymbol;" nxww 0wa f-2  )ySymbol ixww 0wa f-2 q (ySymbol;" oxww 0wa f-2 U$)yTimes New Romanxww 0wa f-2 (Cy2 Cy2 [Cy2 Ty2 D Cy2 Cy2  Cy2 ZTy2 Ty2 BTy2 Ty2 Ty Times New Romanxww 0wa f-2 {(oy2 xoy2 oy2  oy2 Goy2 K oy2 (iy2 iyTimes New Romanxww 0wa f- 2 5&360e 2 "637e2 !.32  03 2 7525e2 E25 2 637e2 .32 032 C 25 2 550e2  25Symbol;" qxww 0wa f-2 %=52 +52 =52 =52 # -52 -52 =52 -52 -5 Symbol lxww 0wa f-2 (5 & "Systema f !-NANIs4= T"T i T " "T i =T"25 o C550, o C"25, o C=0.637T=25 o C+525, o C()0.637()=360, o CEquation Native _1234960579cFledledOle CompObjbdf FMicrosoft Equation 3.0 DS Equation Equation.39q& x,y,z,t()= 1 x,t() 1 y,t() 1 z,t()ObjInfoeEquation Native _1234974726whFledledOle  FMicrosoft Equation 3.0 DS Equation Equation.39q&i%  1 x i ,t()=A 1 e " 12  cos 1 =A 1 e " 12CompObjgifObjInfojEquation Native _1234960974aqmFledled tL i2 cos 1 x i L i () FMicrosoft Equation 3.0 DS Equation Equation.39q   Ole CompObjlofObjInfoOlePres000np.  &@ & MathType-. Symbol$ ww 0wJf-2  (ySymbol ww 0wJf-2 Z)y}Times New Romanww 0wJf- 2 Y27032 .72 9072 222 W.22 222 c052 .52 C 052 12Times New Romanww 0wJf-2  22Symbol$ ww 0wJf-2 =22 !22 !22 !22 !22 22 22 22 22 922 0z 22 ^=22 =2Times New Romanww 0wJf-2 jW22 5C22 +m22 4m22 0v C22 0m22  W22 k22 HhL2 FBiTimes New Romanww 0wJf-2 Foi2  oi 2 @slab & "SystemJf !-NANI&$ Bi slabEquation Native 2_1234961154sFledledOle CompObjruf =hLk=12Wm 2 ", o C0.05m()m", o C2.22W()=0.2703 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoOlePres000tvEquation Native 2_1234961457kyFledled     !"#$%&),-./036789:;>ABCDGJKLMNORUVWXYZ[\]^_`abcdefghijkmnopsvwxyz{|}   .  &@ & MathType-. Symbol ww 0wJf-2  (ySymbol ww 0wJf-2 `)yTimes New Romanww 0wJf- 2 _10912 .02 ?002 222 ].22 222 c022 .22 C 022 12Times New Romanww 0wJf-2  22Symbol ww 0wJf-2 =22 '22 '22 '22 '22 22 22 22 22 ?22 0z 22 ^=22 =2Times New Romanww 0wJf-2 pW22 ;C22 1m22 :m22 0v C22 0m22  W22 k22 HhL2 FBiTimes New Romanww 0wJf-2 Foi2  oi 2 @slab & "SystemJf !-NANI&xt; Bi slab =hLk=12Wm 2 ", o C0.02m()m", o C2.22W()=0.1091 FMicrosoft Equation 3.0 DS Equation Equation.39qOle  CompObjxzfObjInfo{Equation Native &Û x,y,z,t()= 12 x,t() 1 z,t() FMicrosoft Equation 3.0 DS Equation Equation.39q_1234972970~FledledOle CompObj}fObjInfoEquation Native _1234973144|FledledOle 'CompObj(f&À T"T " T i "T "   =0 o C"18 o C"20 o C"18 o C=0.4737=1.0713e "0.3208() 2 0.02m() 2 0.124x10 "7 m 2 st cos0.3208()[] 2 " 1.0408e "0.4951() 2 0.02m() 2 0.124x10 "7 m 2 st cos0.4951()[] FMicrosoft Equation 3.0 DS Equation Equation.39q&'8<  =tL 2 =0.124x10 "7 m 2 s77,500s()0.05m() 2 =0.384>0.2ObjInfo*Equation Native +C_1234974944FledledOle 1 FMicrosoft Equation 3.0 DS Equation Equation.39q&mh x,y,z,t()=T"T " T i "T " = 1 xCompObj2fObjInfo4Equation Native 5_1234973541Fledled,t() 1 y,t() 1 z,t()= 1 x i ,t()[] 3 FMicrosoft Equation 3.0 DS Equation Equation.39qOle <CompObj=fObjInfo?Equation Native @7&   =tL 2 =1.15x10 "6 m 2 s600s()0.025m() 2 =1.104>0.2 FMicrosoft Equation 3.0 DS Equation Equation.39q_1234973709FledledOle ECompObjFfObjInfoH&i@  1 x i ,t()=A 1 e " 12  cos 1 =A 1 e " 12 tL i2 cos 1 x i L i ()Equation Native I_1234973905FledledOle PCompObjQf FMicrosoft Equation 3.0 DS Equation Equation.39q   .  @& & MathType-. ObjInfoSOlePres000TEquation Native l&_1234984228/FledledSymbol ww 0wf-2  (ySymbol ww 0wf-2 )y=Times New Romanww 0wf-2 S4y2 .y2 30y2 w5y2 .y2 W2y 2 c025e2 .22 C 022  40Times New Romanww 0wf-2  20Symbol ww 0wf-2  =02 02 02 02 02 x02 x02 x02 x02 02 0z 02 ^=02 =0Times New Romanww 0wf-2 dW02 C02 m02 m02 0v C02 0m02  W02 k02 HhL2 FBiTimes New Romanww 0wf-2 Foi2  oi 2 @slab & "Systemf !-NANI& $ Bi slab =hLk=40Wm 2 ", o C0.025m()m", o C2.5W()=0.4 FMicrosoft Equation 3.0 DS Equation Equation.39q& x,y,z,t()=T"T " T i "T " =A 1 eOle qCompObjrfObjInfotEquation Native u " 12 tL i2 cos 1 x i L i ()[] 3 =A 1 e " 12 tL i2 [] 3 FMicrosoft Equation 3.0 DS Eq%"< m au=BJ5QcL8gke{ @Ow?=G_0ۀӋl}Iocє%1f5WU;i}~02̟6^г([ߝ؏W=~Wsj: \w.xlhWjLa2U822AL<Hn[I\E2aF+=zq VZ*ɔeɹ&7g Z3akE8ZR=Ȳ7Oi'k+~N$u/N$zɟ/&ؓp:I TY &ݺ> >+Ec}^nǦw|-o ~䋊n~?G l?*}3g62A#rPTx [dI⸅ O ǩ2q !r~4`$2)KV JYw͆,Q'e$e=-Rv'Zų& fLֻNvmGKnZ!tFp9g17O!筴_l9fphs~/xi<(BWRntrͷÔ38\?O_?BWBzw'7W-W2u?9TpġB##i>\/w='^ʭ7| "qAO'sX<>5O8<.2y8lǹθ]p/>!x苪AQݏ}qVRQ_V{qO@WK?cSs},D_}fRB밂בqzQꃻ/{;)E=YI&oEGl69U>^By*-~>.e6O 3 t^MO4i~M} *X"iy{iO^pcK.v<2z~PPzn>3 >x&Dd h$b  c $A? ?3"`?2&# M7 `!&# @: XMZ(txڥmL[U^ZZ>*QBiK63"QE3I"FpQkb,Y4[Bc8ls1G$Fb7"S=s{JCJy{Ϲr $lFHFe2"#Q#A&GHo2YK\&ӆJ6_Ҏ1!4j1 ꍽE褨 D?mh|qPGyъmX)J[?SXZ=ON\X־NS ϞWW"Z-e:H\ cuA^_Λ\LZ3oBemw,͒F]RWgAu]K]*e5,8X 'k&RW;VR׌Uj\'PIg8Žw̐#*tDy#cՈQ)o뼦,`nSs~-F~>9RNQN;J9G9FۗA?rr¯ GT ? 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