This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ME 364 Exam #1, inclass 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 steadystate 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 xdirection 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 crosssection 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/mZK and the heat
transfer coefﬁcient between the room air and the outer wall of the freezer (P1,) is also 10 W/mZK. 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/mK
and 15 W/mK, 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/mZK 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/mK
stainless steel inner wall
kw = t5 W/m—K fiberglass blanket
kb: 0.06 W/mK 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
$0. vatares o
o
o 00$
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‘sé‘
.‘. we: «we. .‘.‘. 
m3. ....... .3... “‘5...
r we». or
..§+.+.H.'.§+.+.+
' v1.4"
.. o
. .
o
. Wasmam
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, thethermal 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 tatw.
: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 +(ssxvaA 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 ...
View
Full Document
 Fall '08
 Rothamer
 Heat, Heat Transfer, energy consumption, DI, rod B

Click to edit the document details