Exam1 - ME 364 Exam#1 in-class portion Name Lecture time(circle one 11:00 or 2:25 Note there are 4 multiple choice questions worth 40 pts follmzved

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Unformatted text preview: ME 364 Exam #1, in-class portion Name Lecture time (circle one): 11:00 or 2:25 Note # there are 4 multiple choice questions worth 40 pts follmzved by a problem that is worth 60 pts. There is an extra credit question at the end that is worth 10 pts. You only have 50 minutes — make sure you budget your time appropriately. Multiple choice questions £40 pts) 1. (10 pts) Two solid rods are made of the same material with the same thermal conductivity. They are insulated on their outer radius and have the same temperatures on either end, as shown. The diameter of rod A is twice that of rod B and the length of rod A is half that of rod B. direction of heat flow \X‘x‘x‘x‘xm‘x\h\\\\\\\\\\\\\\\\\ "~ Circle the letter of the correct statement about the thermal resistance to conduction associated with the flow of heat through these rods (along their centerline). a. The thermal resistances ofthe two rods are equal (R3 = RA). b. The thermal resistance ofrod B is 2 times that of rod A (R3 = 2 RA). c. The thermal resistance of rod B is 4 times that of rod A (RB : 4 RA). ; The thermal resistance of rod B is 8 times that of rod A (RB : 8 RA). tie/.9 The thermal resistance of rod B is equal to the thermal resistance of rod A squared (R3 = RE). The thermal resistance to conduction is Rm 2 i : L 2 k A k 71' D Therefore. the ratio ot‘the thermal resistance of rod 8 to rod A is; RB _ LB MD: _ (2124) Di 2 RA_k7rD§ LA _ L [D 2 J : 8 and the correct answer is (d) A A Z. (10 pts) The steady-state temperature distribution within a plane wall with a c0nstant thermal conductivity is shown below. T(X) x=0 x=L Circle the letter of the correct statement (there is only one). a. The left hand side ofthe wall is adiabatic. b. There is no internal generation of energy within the wall. 0. At x30 (the left side of the wall) heat is being transferred out of the wall (1. At x=L (the right side of the wall) heat is being transferred into the wall. (a) the temperature gradient at the left hand side of the wall is not zero — it cannot be adiabatic and (a) is not correct (b) there must be internai generation to account for the nonlinear temperature distribution and so (b) is not correct @t x:0 the temperature gradient is positive, therefore the heat transfer rate must he in the negative x-direction or out ofthe wall. Therefore (0) is the correct answer. (d) at XZL, the temperature gradient is negative, therefore the heat transfer rate must he in the positive x—direction or out ofthe wali and (d) is not correct (10 pts) The composite wall is composed of two layers. Layer A has a thermal conductivity that is higher than layer B. The thermal conductivity in both layers is constant. For steady state conduction of heat through the wall from left to right, with no internal generation of energy, circle the letter of the temperature distribution that is possible. layer A has a higher thermal conductivity than layer B direction of heat _—. transfer AB 'AB m A B (a) (b) (c) (e) The temperature gradient must be iinear in both layers if there is no generation — therefore (a) cannot be correct. The temperature must decrease from Eeft to right given the direction of heat transfer and therefore (b) and (c) cannot be correct. The temperature gradient within layer A. the high conductivity layer. must be less than the ' temperature gradient in layer B. Therefore, (d) must be the correct answer. A cylindrical heater element is encased in a layer of ceramic, The outer surface of the ceramic both radiates and convects to its surroundings. surface of ceramic experiences convection and radiation with surroundings R = ‘10 W conv R =100 KAN rad heat is generated by cylindrical heater interface between the heater and the ceramic - heater is surrounded by ceramic R = 1.0 KAN cond,cer The thermal resistances have been calculated for you; the conductive resistance of the ceramic is 1.0 K/W, the resistance due to convection is It} K/W, and the resistance due to radiation is 100 K/W. Circle the quantity below that has the largest effect on the temperature at the interface between the cylindrical heater and the ceramic. a. The emissivity of ceramic surface. The heat transfer coefficient at the surface orthe ceramic. c. The conductivity of the ceramic material, C]. The conductivity of the heater material. The conductivity ofthe heater material does not play any role with regard to the temperature at the interface — therefore (d) is not the correct answer. The resistance network representing this process is conduction through the ceramic (Rmdm) in series with two resistors in paratlel that represent convection and radiation (Rwy and Rm). The parallel combination of convection and radiation has a value of about 9 WW which is much. iarger than the conduction resistance — as a result, the conduction resistance does not have a significant impact on the problem and we can eliminate (c) as an answer (recail i in a series combination of resistances the biggest one dominates the answer). So now the question is whether convection or radiation is most important. In a parallel combination of resistances, the smaliest resistor dominates {none of the heat wants to go through the big one), The convection resistance is much iess than the radiation resistance. Therefore. the heat transfer coefficient at the surface of the ceramic dominates the problem and (b) is the correct answer. Problem 160 ptsl You have designed a wall for a freezer. A cross-section of your freezer wall is shown below. The wall separates the freezer air (at Y}: 40°C) from air within the room (at T r = 20°C). The heat transfer coefficient between the freezer air and the inner wall of the freezer (hf) is 10 W/mZ-K and the heat transfer coefficient between the room air and the outer wall of the freezer (P1,) is also 10 W/mZ-K. The wall is composed of a 1.0 cm thick layer of fiberglass blanket sandwiched between two 5.0 mm sheets of stainless steel. The thermal conductivity of fiberglass (kb) and stainless steel (kw) are 0.06 W/m-K and 15 W/m-K, respectively. Assume that the area ofthe wall is 1 1112. tb=1.0cm tw=5.0mm—.‘ Fig la—tfl=5.0mm room air at Tr = 200 room air to wall heat transfer coefficient is hr: 10 W/mZ-K freezer air at T, = 40°C freezer air to wall heat transfer ‘ coefficient is hf = 10 W/mZ—K stainless steel outer wall kW 2 15 W/m-K stainless steel inner wall kw = t5 W/m—K fiberglass blanket kb: 0.06 W/m-K a.) (10 pts) Draw a resistance network to illustrate this problem. Be sure to label the resistances in your network so that it is clear what each is meant to represent. Calculate the value of each of the resistances in your network (K/W). The resistance network is shown below: R R R R conv,r cond,w cond. b condw Rco nv,f Moving from left to right, the resistors are (assuming a .1 m2 area): M 2 convection with the room air Rm", r = ~1— = m K 2 2 0.15 ' hr A 10 W lrn W t. “0.005ran K conduction through the outer stainless wall Rma, w = 2 : 0.0003— ‘ k A 15W 1 m W w t!) 0.01 m mK K conduction through the fiberglass blanket Rmd h = ——-—— = 2 = 0.17— ’ kh A 0.06W 1 m W conduction through the outer stainless wall Rmd W m r‘” z 0005 m m K 2 a 00003 K ’ kw A 15 W 1 in W 2 convection with the freezer air R. 1 m K = 0.115w ZZZ: 10 w 11m2 W .) (5 pts) Determine the rate of heat transfer from the room air to the freezer (W). The heat transfer rate for l m2 is given by: . 3)- Tl—Tf 20—(—10)Kw Ki: (1 = —'¢— = — z 82 w RCGWJ‘ + Rcondm‘ + Ra)me + Rcond,w + Roomuf K .) (10 pts) Your boss wants to make a more energy efficient freezer by reducing the rate of heat transfer to the freezer. He suggests that you double the thickness of the stainless steel wall panels from 5.0 mm to 10.0 mm. Is this a good idea? Justify your answer briefly. No i the resistance of the wall panels is negligible. in a series resistance network the small resistors don’t matter and, changing the thickness of the wall panel will not affect the heat transfer rate to the freezer. .) (10 pts) What design change to your wall would you suggest in order to improve the energy efficiency ofthe freezer. The biggest resistor in the network is related to conduction through the fibergiass blanket. To make a real difference in the heat transfer 1 would suggest making this blanket thicker. ,) (15 pts) One of your design requirements is that no condensation must form on the external surface of your freezer wall, even if the relative humidity in the room reaches 75%. This implies that the extemal surface of the freezer wall must be greater than 15°C, as shown below. Does your freezer wall satisfy this requirement? Calculate the external surface temperature (°C). freezer air room air you have to avoid any / condensation on this surface, therefore, this surface temperature must be at least 15°C We can calculate the surface temperature, TS, which is l beied on the resistance network from (a), below. Tr Ts Tf 7! m!” WWW—WWW «3 > K, a '\ Rconv,r Rcondw Rcondb RCDndM Rconv,f -1} E ’IN "I .’ «j o 82W 0.1K (2/ a w . . . 7; : TF .3430 = 20 C——— =11.8 C so the answer 13 that there Will be condensanon OIII'J‘ W ) (10 pts) Suggest a design change that could prevent condensation. What would be the impact of your proposed design change on the energy consumption of the freezer (i.e., would it increase, stay the same, or decrease) and why? there are several correct answers to this problem; a couple of these are listed below. You could install a heater somewhere near the outer wall to keep the surface temperature at 15 deg. C, this could cause the freezer energy consumption to increase because the surface temperature, TM has gone up driving more heat into the freezer which must be removed by the refrigeration system. You could increase the thickness ofthe fiberglass blanket m this would tend to increase the amount of resistance between 2’; and I} and therefore drive T5 closer to Tr. The energy consumption would decrease in this case as the amount of heat driven into the freezer is reduced. Extra Credit Problem [worth 10 extra points! Consider a two—part composite wall. . . .o e N. | o t t d 6' Q t t I $0. vat-ares o o o 00$ ho ' u 0' +39 .4 " ..,, . o .. “9:. .,....... .. - - .. «3 my convection to air, '~::::::::,:g§§::+:::¢:. . h, T v ++u '. ,o o. 3: . 2:5 . . . .. ... nu... on... :53. 0.0.304}. .339; ’ '0 o e‘ .. "$.43: ..¢3‘$.€ ... nu... . .... ..¢.. §~o¢~0¢§+++g “W... 926. Avatéafififififiéw. ‘ a‘.‘ *.°.:.°.*:.:.:.:.:.°f,;2:. ‘3' 45¢. .wad‘s-é‘ .‘. we: «we. .‘.‘. - m3. ....... .3... “‘5... r we». or ..§+.+.H.'.§+.+.+ ' v1.4" .. o . . o . Was-mam a” 5.; o o f a . $3? a £0330 4 1p 1‘: ‘n Each part (A and B) has some volumetric heat generation (4) and thermal conductivity (k) that is spatially uniform within that part. However, the-thermal conductivities and volumetric heat generation rates are such that the following is true: 2 ¢ i k MaterialA k MaterialB The heat that is generated is removed by convection from the two outer surfaces of the wall; the heat transfer coetficient and ambient temperature are the same on both sides. The temperature gradient at the interface between the two materials (F0) is the same in both materials. In other words, all _'£ 055 MateriaIA,@x:0 dx MaterialB,@ x=0 There is no contact resistance at the interface. Circle the single, correct answer below: a) This is not possible kwEnergy generated in A is not transferred to B c) At the interface, the temperature profile is discontinuous id) The temperature gradients in the wall at the convection boundary are different in parts A and B. This is a hard problem — i’in not sure I got it right. My solution follows: The governing equation for a plane wail is: dZT ‘ . . . . . . . . dxz = *2 integrating twice leads to an expression for the temperature distribution in parts A and B ‘l 2 w 2 a w [4(x)=— E x +Clx+C2 and T3(x):i E x +C3x+C4 A B where (,7; through (74 are constants of integration that must be obtained from the boundary conditions. Extra Credit Problem (worth 10 extra points) Consider a two—part composite wall. convection to air, h! Tair convection to air, h, T , arr o .50 0:4: o‘noo tat-w. :gwfiay +33 mi aka .‘W‘ “new ., .. m... . «a was 4:. ~ vi + . 2% a? o :0 +4 ... 03:. + 1-»;51 o o .0 4 .. Fig To?» 0 o 9,... 35' ,. . ‘3'?" . . .ofo:o o‘. ,. wage ‘0'... Each part (A and B) has some volumetric heat generation (cf) and thermal conductivity (k) that is spatially uniform within that part. However, therthermal conductivities and volumetric heat generation rates are such that the following is true: 2 i i k Material/1 k MaterialB The heat that is generated is removed by convection from the two outer surfaces of the wall; the heat transfer coefficient and ambient temperature are the same on both sides. The temperature gradient at the interface between the two materials (x=0) is the same in both materials. In other words, dT dT dx MaterialA,@x:0 dx MaterialB,@ x=0 There is no contact resistance at the interface. Circle the single, correct answer below: a) This is not possible iJEnergy generated in A is not transferred to B 0c) At the interface, the temperature profile is discontinuous The temperature gradients in the wall at the convection boundary are different in parts A and B. This is a hard problem i i’m not sure I got it right. My solution follows: The governing equation for a plane wail is: all?" ' . . . . . . . . Ex? = *1 integrating twrce leads to an expressron for the temperature distribution in parts A and B q 2 v 2 ‘4 ‘1 T4(x)=— E x +C1x+C2 and T3(x)=i E x +(ssx-vaA A 8 where C, throogb C4 are constants of integration that must be obtained from the boundary conditions. At the interface, x = 0 we have an energy baiance: dT dT ' x z 0 = ' x 2 0 which im lies that: —k — = ‘k _ qA( ) gB( ) p (1i dxjm‘o 5[ dx 19,va dx dx only the volumetric generation is different for the two parts. Plugging the temperature distribution formulae into this boundary condition we get: C1 = C3 Given the statement in the problem that = we know that k4 = in; which implies that Aixifl B,X=0 The other boundary condition at the interface is that the temperature distribution at the interface cannot be discontinuous so TA (x = 0) 2 TB (x : 0) which leads to C2 = C4 The temperature distributions are therefore: TA(x)=—[%] x2+Clx+C2 and TB(x)=—{%] x2+Cix+C2 A B ‘ where C, and C; would come from boundary conditions at the convection interfaces. At this point iets look at our choices: a. This is not possible WI don‘t see why it wouldn’t be possible. Nothing we’ve found so far seems physically impossible. I don‘t think (a) is correct. b. Energy generated in A is not transferred to B — I can't determine whether this is true or not w my guess is that the material with the higher heat generation rate wilt tend to transfer heat to the material with the tower heat generation. lfA has a higher heat generation then this will not be true, ldonit think (b) is correct. c. At the interface= the temperature profile is discontinuous. No — the temperature profile is continuous; the first derivative of the temperature profile is aiso continuous. The governing equation shows that the second derivative of the temperature profile is not continuous but that is not the definition ofdiscontinuous * I think (c) is incorrect. d, The temperature gradients in the wait at the convection boundary are different in parts A and B. By process of elimination (d) must be correct. Hm ever, this also makes sense 7 the probicni is not symmetric if the generation rates are different. More heat must be convected from one side of the composite than from the other: given that the thermal conductivity and heat transfer coefficients are the same in both parts the difference in the heat transfer rate must manifest itself as a difference in the temperature gradients. (d) is correct ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at Wisconsin.

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Exam1 - ME 364 Exam#1 in-class portion Name Lecture time(circle one 11:00 or 2:25 Note there are 4 multiple choice questions worth 40 pts follmzved

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