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Unformatted text preview: ME 114 Homework Assignment 3 Solutions Problem 1
Problem 224 . Ti=1100°C To“ =30°C h = 10 W/m2 "C L = 0.3 m ' m, 'T. .. Assumptions: 1. Steady state. 2. One dimensional temperature distribution. 3. Constant properties. . Solution: I ,
From Table SE, for fine clay brick k 1 W/m °C '
9: Ti .'l‘m 1100 _— 30
‘T = if = m—
r + r T + 1— .  ‘ . e i ' HT. I r ' T”
q;= 2575 WIm2 . , , ‘A. MW TED WW (1:: h (Tm _‘ T") = 2675"? 10.): (TB; ElO) Ir ‘ T r 'I —
‘T = 297.5%: . H v ' 13* 11A ‘50 Comment: The outer surface temperature of the furnace is too high for operating personnel. It
should be brought below = 60°C by adding insulation, ‘ Problem 2 _ Problem 222 ‘
T‘ = 30°C Tm = 20°C L'= 1 cm Width = 20 cm Height ; go cm
347 St. Steel
Assumptions: 1. Steady state ' 2. Constant properties
3. Onedimensional temperature distribution
4. Heat transfer takes place only from the outer surfaces exposed to air. All other surfaees are perfectly insulated.
Solution: From Table 1, for Type 347 stainless steel k =‘ 14.2 W/mK ' Thermal circuit making use of the symmetry of the arrangement To. . pt T T51“ 30 20 _+ 1_“ 0.01 1 RA hA —14.2 x 0—,; {"072 + 'h—x 0.2? 0.2 n = 20.3 W/mz,°C Problem 3
Problem 225 ri = 0.1 m rD = 0.108 m k = 0.05 W/m K Ti=50°C T =10°c. ~110=7wimz°c hi = 20 W/mz, °c
Assumptiens: 1. Steady state.
2. One dimensional temperature diSnibution. 3. Constant properties. Solution:
Ti  Tm
Q: = ha Ai (T3 ' T3) = ln(ro/ri_) 1
’“2 z: k L‘ + —h A
' O O
Dividing by L, . I . .
q ‘ T T 4 .  i c  __*_.::I_ T '
f—hizmi (T3 T9 "ln(rolri) 1 g Ti To TN '
21:1: +hznr 1 ‘
° ° ‘ ___ _~ luau/I‘D i
20x2nx0.1(Tz50)
_ so  (10)
‘ 1n(0.108/o.1)+ 1
2 Tax 0.05 7 x2nx0.108
T = 50.5 “c '
8 Comment: To prevent condensations of acids when burning wood, temperature of the gases should be above the dew point temperature. Wot \el/W Al 348 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For speciﬁed top and bottom surface temperatures, the rate of heat transfer along the cylinders and
the temperature drop at the interface are to be determined. Assumptions 1 Steady operating conditions exist. 2
Heat transfer is onedimensional in the axial direction Interface since the lateral surfaces of both cylinders are well— insulated. 3 Thermal conductivities are constant. W W Properties The thermal conductivity of aluminum bars
is given to be k = 176 W/m°C. The contact
conductance at the interface of aluminumaluminum plates for the case of ground surfaces and of 20 atrn z
2 MPa pressure is h0 = 11,400 W/m2°C (Table 32). Analysis (a) The thermal resistance network in this
case consists of two conduction resistance and the
contact resistance, and they are determined to be 1 l m R = __ = ————_ = 0.0447 °C/W
mm hcAc (11,400 W/m2 °C)[7r(0.05 m)2 /4]
Rpm = L 0'13 m = 0.4341 °C/W E _ (176 W/m  °C)[7r(0.05 1102/4]
Then the rate of heat transfer is determined to be 0. _ AT _ AT (150—20)°c 12total 12contact + 2Rbar = (0.0447 + 2 x 0.4341) °C/W Therefore, the rate of heat transfer through the bars is 142.4 W.
(b) The temperature drop at the interface is determined to be Armterface = QRcontact = (142.4 W)(0.0447°C/W) = 6.4°C = 142.4 W 354 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is
to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is onedimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is
disregarded. Properties The thermal conductivities are given to be k = 0.72 W/m°C for bricks, k = 0.22 W/m°C for
plaster layers, and k= 0.026 W/mDC for the rigid foam. Analysis We consider 1 m deep and 0.33 m high portion of wall which is representative of the entire wall.
The thermal resistance network and individual resistances are R3
Tw1—/\/V\/VV\/—/\/\/\/———/W\/——’\/V\/—Tw2 RI. =RCW1 =_1_=———7——1——7_=0.303°C/W
’ h1A (10W/m~°C)(0.33x1m)
R1=Rfaam ='L_= “0.02m 7 =2.33 oC/VV
kA (0.026W/m°C)(0.33><1m‘)
R2 = R5 =RWW =_L_=———M__7_=O.27S°C/W
side IcA (0.22W/m°C)(0.33><lm“)
R3:R5=Rplaster: L = "—2 =54550C/W
center hoA (0.22W/m4°C)(0.015x1m)
R4 =Rbrick =‘L—=—_9.—18—m——7‘=0833 OC/W
kA (0.72W/m°C)(0.30><1m‘)
R0 =Rcon112 = 1 = ——"—‘_‘l 7 =0.152°C/W
' th (20W/m°C)(o.33x1m)
1 1 + 1 + 1 = 1 + 1 +;——>Rmid=0.81°C/W Rmm = 723— 72: E 54.55 0.333 54.55
R,.,,,,, = R, +121 +2122 +Rm,.d +12, = 0.303 +2.33+2(0.275)+0.8l+ 0.152 = 4.145 °C/W The steady rate of heat transfer through the wall per 0.33 m2 is
0 : T... — .2 = [(22—(—4)]°C = ~ —— 6.27 W
Rm, 4.145°C/W Then steady rate of heat transfer through the entire wall becomes QM, = (6.27 mm = 456 w 0.33 m ...
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 Spring '08
 Okamoto
 Heat Transfer

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