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Unformatted text preview: PROBLEM " 44;} KNOWN: Dimensions of a thermopane window. Room and ambient air conditions. FIND: (:1) Heat loss through window, (In) Effect of variation in outside convection coefficient for double
and triple pane construction. SCHEMATIC (Double Pane): _
Glass 1..L+L+L..1  Window. 0.8 m x 0.5 m ‘5 : =o.oo
111 111 L 7'“
T i=2000 » ‘ L'O=10°C
hT'=1ow1m2 K ho=30 W’mz'K
Tomi Too,0
>
.1. .L. _L._ 1... 1_ q
MA I g.41 ltaA 19.4 h A ASSUMPTIONS: (1) Steadystate conditions. (2) Onedimensional heat transfer. (3') Consrant
properties. (4) Negligible radiation effects. (5) Air between glass is stagnant. PROPERTIES: Tobie A3, Glass (300 K): kg = 1.4 W/mrK; Table A4, Air (T i 278 K): k1. = 0.0245
WlmK. ANALYSIS: (a) From the thermal circuit, the heat loss is _ Tad—Tame
1 1 L L L 1
——+—+—+—+—
A hi kg kn kg ha
20°C—(—10°C)
q" 1 1 + 0.00m g 0.007111 1. 0.007m + 1
0.4m1 tow/nFK 1.4W/mK'0.0245W/rnKT14W/rnK BOW/mzK
30"c 30°C
=mz_=29.4w <
q (0.25+0.0125+0.715+0.0125+0.03125)K/W 1.021K/W . (b) For the triple pane window. the additional pane and airspace increase the total resistance from 1.02] ‘
KfW to 1.749 KIW, thereby reducing the heat loss from 29.4 to 17.2 W. The effect of lag, on the heat loss
is plotted as follows. Heal loss, qlW] .___i_ l 1 » — —— —a—«———L. .__.'_......_. e—.r:— «ma—l—n {1 _...._'_,.... .,M_.c1__' . l i  Outside ccnvactlon coefﬁcient. nelernﬂan Doubl pane
F9“ Triple pane Continued... PROBLEM 3.10 lCont.) Changes in ht. inﬂuence the heat loss at small values of hm for which the outside convection resistance is
not negligible relative to the total resistance. However. the resistance becomes negligible with increasing
ha, particularly for the triple pane window, and changes in ht. have little effect on the heat loss. COMMENTS: The largest contribution to the thermal resistance is due to conduction across the
enclosed air. Note that this air could be in motion due to free convection currents. If the corresponding . . . . " . .
convecnon coeff1c1ent exceeded 3.: W/m’K. the thermal re5tstance would be less than that predicted by
assuming conduction across stagnant air. ' ' ‘ PROBLEM as; KNOWN; Dimensions and materials asswiated with a composite wall (2.5m x 6.5111; 10 studs
each 2.5m high). ' ‘ ‘ FIND: Wall thermal resistance. SCHEMATIC: Heats: . Hardwood siding (A)
I _ , Emma
Instr/after: . ‘40 717777
Glass Fiber; 1  HagdwooLd (B) Imam:
paper faced (D) ~ .  .7 . a. c ASSUMPTIONS: (1) SteadyState conditions, (2) Temperature of composite ‘depends only on
x (surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resisrance. PROPERTIES: Table A3 (T: 300K): Hardwood siding. kA =0.094 W/mK; Hardwood,
k3 = 0.16 W/InK; Gypsum, kg = 0.17 W/mK; Insulation (glass ﬁber paper faced, .28 kg/mB),
kD = 0.038 W/mK. ' ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a
single unit (enclosed by dashed lines) of the wall is L [k A ' ‘ a a B LC/kCAC
LA/k‘A“ A //< A ‘ o n o ,  _ _ 0.008111
(LA/kAAA) ‘ (1094 W/mK (0.65mx25m) 0.13m ’
A =___________________= .12 ‘ ’
(LB/RB B) 0.16 W/mKC0.04mx2.Sm) 8 SW 0.13m
0.038 W/m‘K (0.61mx2.5m) 0.012m .
aC/kCAC) _ 0.17 WImK(0.65m><2.5m) "0343“; l ' The equivalent resistance of the core is qu =(1/a13 + mam—1 = (1/8125 + 1/2243)‘I = 1.758 W
and the total unit resistance is Rm“ = RA +Reel +RC =1.854 KfW.
With 10 such units in parallel, the total wall resistance is = 0.0524 K/W (Lg/snap) = = 2.243 m , Rm=(10><l/le)"I = 0.1854 K/w .  <1 COMMENTS: If surfaces parallel to the heat ﬂow direction are assumed adiabatic the thermal
circuit and the value of Rmt will differ. ' PROBLEM 3 KNOWN: Height and inside diameter of hot water tank. Thickness and thermal conductivity
of insulation. Temperamm and convection coefﬁcient of ambient air. Unit energy costs and storage temperature. FIND: Daily energy cost. SCHEMATIC:
T .. M d R if
.0  can conv L=2m 7 7;;=5'5°C ‘
EJO C, AL Ure‘f‘hane insulaﬁon {frgbaﬂ‘omsidesl
[1:10W/m3K ‘ _,+=o,o4m,k=o.ozaw/mK, CﬁlS/kWh 96 ASSUMPTIONSh (l) Steadystate conditions, (2) Constant k. (3) Onedimensional conduction
through insulation (x or r). 0
ANALYSIS: The daily energy cost is
ID = (q, + (1,, + (11,)(2‘5‘t10'3 kWIWm h/d
with '
Tm — T. q: Remd'l'Remv. It follows that, for the Side Wall: Rm _ Inst/n) _ 1n(0.44/O.4D)
d ' 2m: ‘ 21t(2m)0.026W/r§1'1{ Rm“, = (h2m,L *1 = (210410 W/mgKx0.44m><2n1)“ = 0.018 KM
q, = (55 — 10)°C/0.310 KJW = 145.3 w. = 0.292 W Top and Bottom: 0 t 0.04 m
Rm = —— = ————— = . z
'1 kA (0.026 WlmK)It(0.8 uni/4 3 06 W Rm = [WIDE/41“ = [min WImltqos inf/414‘ = 0.20 WW
0: = qt = (55 10)°C/3.26 KJW = 13.3 wa. Hence, D = (145.3 + 2x13.8)Wx0.lS/kthlO‘3 kaWx24 h/d = 62.20%! 4 COMMENTS: Note that the area based on the inside diameter is appropriate for evaluating the
top and bottom losses. If the side insulation is approximated as a plane wall, the conduction resistance is (t/kA) = [0.04 III/(0.025 W/mK)7t(0.82 m)2m] = 0.299 W, and the heat loss ((1,
= 142.1 W) is slightly nnderpredicted. 325 _ Six identical power transistors are attached on a copper plate. For a maximum case temperature of
75°C, the maximum powar dissipation and the temperature jump at the interface are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be approximated as being one
dimensional, although it is recognized that heat conduction in some parts of the plate will be two
dimensional since the plate area is much larger than the base area of the transistor. But the large thermal
conductivity of copper will minimize this effect. 3 All the heat generated at the junction is dissipated
through the back surface of the plate since the transistors are covered by a thick plexiglass layer. 4 Thermal
conductivities are constant. Properties The thermal conductivity of copper is given to be k = 336 W/m“C. The contact conductance at the interface of copperaluminum plates for the case of 1.171.4 um roughness and 10 MPa pressure is he =
49,000 W/mz“C (Table 3—2). Analysis The contact area between the case and the plate is given to be 9 01312, and the plate area for each
transistor is 100 cm3. The thermal resistance network of this problem consists of three resistances in series
(contact, plate, and convection) which are determined to be I 1 a = — = ———————_ = 0.0227 Dcrw
“m” hurt. (49,000 was: °C)(9 ><10"4 m2)
9
Rplale = i = ' "'_O'l—"“"“01 m ., = 0.0031 c"C/VV
M (386 Who  rJ<:)(0.01 m“)
a = L _ 1 = 3.333 “CIW conVectiuﬂ hu'A _ (30 was2 °C)(0.01m1)
The total thermal resistance is then Riotal = Rcontact + Rplme + Roonvectinn
= 0.0227 + 0.0031+ 3.333 = 3.359 °CIW Note that the thermal resistance of copper plate is
very small and can be ignored all together. Then
the rate of heat transfer is determined to be .7 5
AT _(75 3)C=15.5W " am, _ 3.359°cxw Therefore, the power transistor should not be operated at power levels greater than 15.5 W if the case
temperature is not to exceed 75“C. The temperature jump at the interface is determined from
ATimerrm = QRcuntﬂct = (15.5 W)(0.0227 “C/W) = 035°C which is not very large. Therefore, even if we eliminate the thermal contact resistance at the interface
completely, we will lower the operating temperature of the transistor in this case by less than 1°C. PROPRIETARY MATERIAL. ID 2007 The McGrawHill Companies, inc. Limited distribution permitted only to teachers and
educators for course preparation. if you are a student using this Manual, you are using it without permission. I ‘ A ﬁn is attached to a surface. The percent error in the rate of heat transfer from the fin when the
mﬁmtely long ﬁn assumption is used instead of the adiabatic ﬁn tip assumption is to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction
only (normal to the plate). 3 The heat transfer coefﬁcient is constant and uniform over the entire ﬁn surface. 4 The thermal properties of the ﬁns are constant. 5 The heat transfer coefﬁcient accounts for the
effect of radiation from the ﬁns. Properties The thermal conductivity of the aluminum ﬁn is given to be k = 237 Wlm“C. Analysis The expressions for the heat transfer from a ﬁn under
inﬁnitely long ﬁn and adiabatic ﬁn tip assumptions are Qlongﬁn = Vim“: (Th —Tac)
out. = Jane. (Tb r..)muh(mn The percent error in using long ﬁn assumption can be expressed as %Errcr=———_—_....________.__.____1
Q ins. up 1/ hpkAc (TI, — IL.) tan11(m[.) tanh(mL)
where m = (so = (12 WIm.°c)2r(o.004 m) = 7.1 16 mi
~ 104.. (237 W/m.°C)7r(D.OO4 m)2 [4
Substituting,
1 1 %Error=———1= , ~—1=0.635=63.5°/
tanh(mL) tanhl(7.116 m*)(o.1o m)l ” This result shows that using inﬁnitely long ﬁn assumption may yield results grossly in enor. 372 \0 i A commercially available heat sinlc is to be selected to keep the case temperature of a transistor
W below 90°C in an environment at 20°C. Assumptions 1 Steady operating conditions exist. 2 The
transistor case is isothermal at 90°C. 3 The contact
resistance between the transistor and the heat sink is negligible.
Analysis The thermal resistance between the transistor
attached to the sink and the ambient air is determined to be . AT T — Ta (90— 20)°C ,,
0 = —""_—5" Reuseambient = m = —— = Renaeambient Q The thermal resistance of the heat sink must be below 1.75°C/W. Table 36 retreats that H36071 in vertical
position, H85030 and H561 15 in both horizontal and vertical position can be selected. ...
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 Rhee,Jinny

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