HWS_CH8_EGN3358_Sleiti - PR M- KEQWN: The heat flux in a...

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Unformatted text preview: PR M- KEQWN: The heat flux in a stainless steel (A181 304) pan containing boiling water may be as high as 5.0 x 105 W/m2 and the temperature at the water/pan interface is 120°C. FIND ; The temperature gradient in the pan at the water/pan interface. CHEMATIC: W One-dimensional heat conduction. BMW From table A-14 for stainless steel (M81 304) at 393.15 K: k stainless steel: 16.48 WIm-"C W ci'=-kg-1]I=5x105[§]=—16.48[fi5]% 5 105 1V— a g—E: = -30.34 x 103 <=<=<=<=<=<= MMENT DIS CU ION: The wmperature gradient is influenced by the surface temperature in the stainless Steel since the themal conductivity is temperature dependenL It the thermal conductivity is evaluate at 400 K, k = 16.6 WIm-"C and BTIBT] = {50.12 x 103 °Clm. This Difference would be negligible for most engineering calculations. PR BLEM - E EEQWN : The heat flux on the 30° diagonal surface of the Bakelite wedge is 680 Btufhr-ft2 in the normal direction. FIND; Determine the heat flux and temperature gradient in the x and y directions. SQHEMATIQ; ASS LIMPTIQNS: The heat flux is constant Over the entire wedged surface. PR! IPERTIES ; From table 8-5 for Bakelite at 80°F: k Bakelite = 0.8089 Btu/hr-ft»°F 9 = 30“ ANALYSIS: o, the angle between nETA and x axis. is 90° + 6 = 120° q"; = on (cos ¢) = 680 cos 120" = - 340 '='<=== hr-ft hr-ft q} = q'}, (sin 4») = 680 sin 120 ° = 588.9 =<==<= hr-ft br-ft C] y = 'kBakelite 5; and q x = ‘kBakelite a—x - 340 a ‘31; = M = 420.3 [—F] =‘=‘=‘= 3X13 akelile -03039 ft - hr-ftv'F 588.9 ,, Bl = hr t't2 ___ _ 728IO[ F] <=<=<=<= aYBakettte 413039 ft hr-ft-‘F C MMENT DISCU I N: The heat flux and the resulting thermal gradient can be broken down into its x and y vector components with textbook trigonometry. PRQBLEM 8-2 KNQWN: (a) k] = 14 W/m-“C T1 = 100°C Lb) Q” = 0.86 kW/m2 Ta, = 20°C [(2 = 0.5 W/m’C T2 = 30°C ' k1 = 15 WIm-"C h = 10 W/msz k2 = 60 W/m-°C IND“ Draw the temperature profiles in the two composite section shown. Calculate the total rate of heat transfer per unit area for case (a) SQHEMATIQ; lcm ASS UMPTIQNS; The thermal properties are constant and the tem is steady state. The two-dimensional conducfion from the ends is negligible, i.e. the materials are treated as infinite slabs. P ER I ' Stated above in Known section. QNALYSIs. a) q" ___ _kl (T1 1312) = _k2 (Tl-1:210) __w__ (100 - T1—2)['C] _ _ty__ (T1~2 - 30) ['C] 14 [m-'C] 0.01 [m] - 0'5 [mt] 0.03 [m] Tm = 99.17 ['C] q" = 1152. cm: b) q" = h (Tsurface ‘ a.) Tsurl'ace = 106cc :4: = _k 3_T= _k (Tsurrace - T1-2 ) ___ _15 (106 - Tt-z )i'C] 3x 3‘1 m-‘C 0.02 [In] 0.02 [m] o a T1_2=86 th]. ___-_— +106 c =107.15 c (L: J gem: T24 =860[Mg- —°—-—"‘— +107.15°c =107.29°c m 603%] an: TH —86({l‘i]- E%':l + 107.29°c = 108.44°C :c=<= QQMMENTS DISC Q SSIQN ; Thermal contact resistances between the materials has been neglected. PR BLE -11 KN WN: A Plain carbon steel pipe (5.25 cm inside diameter and 6.03 cm outside diameter) is covered with 2 cm thick six-ply asbestos corrugated paper. The temperature of the steam inside is 150°C and that of the surrounding air is 25°C. h SW = 1500 Wlmz-“C h ah- : 5 Wlm2-°C l Estimate the outside surface temperature of the insulation and the rate of heat transfer per meter of pipe length. ' SQflEMATIC: 2 cm thick. 0 sixvply asbestos T” = 25 C r corrugated paper h m 3-9 mm thick pipe 5. . hls‘ t" l l I ASS flMPTIONS ; The thermal properties are constant. The heat conduction from the ends is negligible, i.e. the materials are treated as infinite cylinders. The heat l transfer is one dimensional and steady state. EBQPERTIES: h swam =1500 Wlm2-°C hat.- =5 Wlm2-°C from problem statement. k carbon steel = 56.7 WIm-"C at 400 K from Table A-l4 k asbest con-ugalcd “Um-“C, K, A‘15.4. A N A LYS IS : cm; (Tsteam - T..) 1.; z Rl-L fi . - w I I z RrL = ] | + 1n(routpipe’rinpiE) + ln(routineulJlfroutpiE) I . 2 . 1; hip: kinsulation Ilit-ipiptrl'lsteam I'outiusul.'hair z RpL =_L { 1 +ln(3.015/2.625)+1n(5.015/3.015) 1 }= 1.592 I ‘~ 2- n 0026254500 56.7 0-035 0.05.015-5 1: , - _ (150 - 25) _ E ¢=¢=¢=c=¢=¢: . .. gash—1592 -7s.523[m] (Ii/L = 2"it'l'tiutinsulfl‘air(_Toutside surface temp. ' Too) = 2'75'0»050 1 5'S(T0l115ldt: surface temp. ' 25) 5- Toutside surface temp. = 74.84%: ¢=<=<=<:¢=¢= W KNQWN; A plain carbon steel pipe with an inside diameter of 9 cm and a thickness of 0,5 cm carries hot water which is at a temperature of 150°C. The pipe is insulated with 5 cm thick 85% magnesia. The inside heat transfer coefficient is 7100 Wlm2-'C. The convection heat transfer coefficient on the outside of the insulation is 57 Wlm2-'C. Scale builds up on .the inside of the pipe yielding a fouling factor of 0.0005 m2-'CIW. I D ' Determine: (a) The overall heat transfer coefficient based on the outside surface - area of the insulation, in watts per meter squared-degree (b) The rate of heat transfer lost from the pipe per meter length if the outside air temperature is 20°C SQHEMAIIE; Tw= 150°C hw==7100 WImZBC rw= 0.0005 m2-'C/W W EBQPERTIES; Thermal conductivity of pipe and insulation; Pipe; kp = 55.83 WIm-“C @423 K Table (A-l4) Insulation; 1:: = 0.055 WIm-“C @365 K Table (A-15.4) ANALYSIS; 2 R (a) ‘ hw-rt-dyL marl. amp-L 2-rr-ki-L ho-n-dg-L 2R‘=___l____+_0.9_005_+ 1n 10/9 + ln(20l10 +____1____ 7100-rr-(0.09}l «(0.09M 2-n-55.33-l 2-n~0.055-l 57-1r-(0.20)—l 2 R‘ = {4.981 x 10" + 1.768 x 10" + 3.004 x 10" + 2.006 + 2.792 x 10'2} 2 R.=2.036[°%] ; Uo-A3=EL{ ; Uo=—-1——— where A3=1t-d3-(l) l U =0.782[ 2v ] ¢=¢=<=¢== m -'C 0 = ____l___ n-(O.2)-(2.036) (b) (1/1. = UJ'A3'AT =(0.782)-(1t)-(0.2)-(150 - 20): 63.87 since L = 1 meter ¢=¢=<=¢=<= PRQBLEM B-lfi KN WN: An electrical heating element is shrunk in a hollow cylinder of amorphous carbon, k = 1.6 WIm-“C, Fig. P8-16. The outside of the carbon is in contact with 20 °C air. The convection heat transfer coefficient is 40 Wlm2-°C. FIND: Determine the maximum electrical heating (Watts per meter of cylinder length) which can be applied if the maximum allowable temperature of the carbon is 200°C. The resistance to the transfer of heat within the heating element may be neglected. SQHEMATIC: ASSUMPTIONS; PR! QPE RTIES: Carbon kc = 1.6 W/m-"C from problem statement. ANALYSIS; T. - To. in [dgidfl 1 . Where 2 R. — zfl'kc‘L + hum-dTL . LetL— 1 meter _ In (0.02/0.01) _ _ Z Rt _——————2'n.“ ASH + ———1————40_M0'02H _ 0.0689 + 0.3979 _ 0.4668 w 3 = ' = 385.6 Wlm (since L = 1 meter) <=<=<=<=<=<=<= PR BLE -1 KN! [WE ; The ceiling of a house is composed of 1.5 cm plaster board. From a typical winter day, the r00m temperature is 21"C and the air temperature in the attic is -10”C. IND ' Determine the amount of heat saved if the attic floor is covered with 1.5 cm plywood _ and the 9 cm space between this floor and plaster board is filled with a glass fiber blanket with a density of 16 kg/m3, Fig. P8—19. The convection film convection film coefficient are both 40 W/m2-'C. . SCHEMATIQ: See below: ASSUMPIIQES; BRQPERTIES: Thermal Properties: Plaster board kpa = 0.17 W/m-° Table (A~15.2) C @ 300 K Plywood kpy = 0.12 WIm-“C @ 300 K Table (A-lS.2) Fiberglass kf = 0.046 W/m-"C @ 300 K Table (A—15.3) - .___...__ ._ |__ _p__.-—-— ANALY§I§: (a) Plaster board alone; Tn: 40°C h...= 40 Win? .°C - A" = 1-5 cm BL in EL 1 pa 0 Ti = 21°C h; = 40 W/m2-°C _._A_T__ =_(2_1;(;1_0))__= 31 _ = 11 = .\_V_ i M _1_ L+Q.D_Li+_t_ 0.025 +0.033+0.025 0.138 224‘6Ln2] ‘=‘=‘=‘= hi +kpa+hm 40 0.17 40 i: A (b) For insulated ceiling, 1;. = -10"C 11.; 40 WIm2 -°C mm w -'-'V'-‘<V='>_ ‘t, :::-:;'3;-g¢-3_-0W;W9m+ it: ééitéliié'iiiaflon 2:1: . § asseé‘éséfigflwg o kpa kf kpg It???" “No.63:- -'- .. (AT J_+A_xt+é£2+é£1+J_ ho kpa ho ‘ .0. l J_+Q.QL5_+.Q.Q9_+Q.D_L1+_L 2.219 ' m2 40 0.17 0.046 0.12 40 Savings A—q = 224.63 - 13.97 = 210.7 [31] =<=<=<=¢=<= A m2 ERQ LEM 3-23 KJQW_N; Hot water at 98°C flows through commercial bronze tube having an inside diameter of 2 cm. The tube is extruded and has the cross sectional profile shown in Figure P 8-23. The outside diameter of the finned tube is 4.8 cm and the fins are 1 cm long and 2 mm thick. The convection heat transfer coefficient on the water side of the tube is 1200 W/m2 °C. The finned tube is surrounded by 15 “C air and the convection heat transfer coefficient is 5 W/m2 “C. ' E IND ; Determine the rate of heat transfer per meter of tube length. SQEEMATI; 2; ASS UMPTIQNS: The thermal properties are constant and the problem is steady state. EBQPERTIES: k bronze = 52 WIm-"C @ 300 K, from table A-14. A NALYS IS ; The problem is very similar to the previous two problems and example 8-4 in the text. V For 1 meter axial length; L; Ab = it'd — 12-t = “It-0.028 - 120.002 = 0.064 m2 _ ln (rout/fin) _ln (2.8/2.0) _ _3 _ l _ I _ E Rp—z'fl‘kbronZe‘L _ 2.1r-52-l _l’03X10 w R” Thump!) ‘5.0_064 '3-125 [w] A. = L . thickness = 10.002: 0.002 [ml] P = 2 (L) = 2.0 [m] = '—= —-—"—=9.806 I.c=L+AC/P=0.01+—0308-2-[m]=0,011[m] m VioAc .V 520.002 R = _ __1_ .._= =0.760 ‘ Ntlh-P-k-Ac tanh [xi-LC 121520520002 tanh (9.80.011) _ Rb-Rf _ 3.1250760 _ . R“? ‘Rb + R, ‘ 3.125 + 0.760 “0'6” [ CM] =_1__ = 3 = n 2 R MAin + RP +Req momma-02 + 0.00103 + 0.611 0.625 [ CIW] =Imter'_Tair. —m = 132.3[31] c=¢=¢=¢= m .g _ “L 2R 0.625['CIW] h I . l i PRQB LEM 3-39 I KNQ E N ; The thermocouple shown in the sketch, Fig. P8—30, is to be used as the sensing _ _ element in a temperature control unit The control unit is set to take corrective action if the fluid temperature is equal to or greater than 150°C. The normal operating fluid temperature is 100 “C. A malfunction in the system results in an instantaneous increase in the fluid temperature to 200 °C. The thennocouple diameter is 0.5 mm. F D ' How long will it take the thermocouple to sense that the control unit must take i corrective action if the convective heat transfer coefficient is 500 W/m2-°C? SQHE M ATIQ ; a d = 0.5 mm ‘ > i To = a 1;: 200°C 2 i h = 500 W/m -°C 5 ASSQMPTIQNS; 1 i ERQPE RTIE§: The average thermocouple properties from the problem statement are: k = 23 W/m'"C p = 8920 kglm3 c = 334 Ifkg-“C / r g N Y I - " 1 Check Bi number: 3‘ ‘ k ‘ L‘- A J 4 n r3 r Q (0.0005) _ I =——-—-—=—= = =8.33X10 m I 3 (4 m2) 3 6 6 S[ 1 1' _ -5 Bi=w=1311x103 .l 23 Since Bi < 0.10 use the Lumped-Capacity method. T‘ To. -h»t To - 11,, ‘ exp [p-c-Lc J ; and To = 100°C; T,” = 200°C; T = 150°C j M _ ex _____'10_9L___ — - - ! _ P —exp[ 1.752 t] - 100 — 200 [(8920}(384)-(8.33 x 10-5) '. 0.5 = exp [-0.693]: 1.752-t = 0.693 or t = 0.396 sec. <=<=<=<=<=<= PRQB LEM 8-45 K EQWE: Most foods are blanched in a water bath before they can be processed for canning. In the blanching process the minimum temperature of the food must be high encugh to destroy the undesirable enzymes. The specific produce of concern is mushrooms which have an average spherical shape of 2 cm in diameter. The thermodynamic propenies of the mushrooms are approximately the same as those for water and the convection heat transfer coefficient is 1400 WImZ-‘C. EIND ; If the minimum allowable temperature is 75°C. determine the minimum amount of time that the mushrooms must remain in the water bath. The initial temperature of the mushroom entering the water is 10°C and the bath temperature is 95°C. HEMATIC: / Evaluate thermal Tof 10F . T—%C properties at: u T": 75.C (75+95)/2 = 85 C h = 1400 Wimzrc Diameter = 2 cm ASS UMPTIQNS; Treat the mushroom as a sphere with R = 1 cm (0.01 m). Assume center must reach 75°C. PRQPERTIES: From Table A-9 @ 85 "C p = 968 kg/m3; c = 4.201 x 103 JIkg-‘C; k = 0.670 WIm-“C ANALYSIS; a = JL = _.Mla_._ = 1.655 x 10-1 [W2] P'c 968-4201 x 103 J . . -7. Bi :3: 1400 (0'01 = 20.8 ; Fo =H=M1= 1.655x 10-1: 1: 0.673 R2 (0.01)2 Use Figure 8-26: Biz-Fo;100 ; (20.8)2-(1.655x 10-3-t)=100 t: 139.6 seconds ¢=¢=¢=¢=¢=¢=¢=¢= ...
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HWS_CH8_EGN3358_Sleiti - PR M- KEQWN: The heat flux in a...

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