chap3assign38-41

chap3assign38-41 - PROBLEM 3.38 KNOWN: Inner and outer...

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Unformatted text preview: PROBLEM 3.38 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner and outer wall temperatures. Temperature of fluid adjoining outer wall. FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefficient on total heater power and heat rates to outer fluid and inner surface. SCHEMATIC: T0 = 25 0C Electrical heater, (1' W 2 ro=75mm 10éhé1000W/m «K T. : no: ‘1 0 OC ri=25 mm ’ 1éké200 Wlm-K OéR'tcéDJ m-K/W 0 5. 7e qf M R't,c (1l21t rah) ‘70 275k q" ASSUMPTIONS: (1) Steady—state conditions, (2) One—dimensional conduction, (3} Constant properties, (4) Negligible temperature drop across heater, (5) Negligible radiation. ANALYSIS: Applying an energy balance to a control surface about the heater, d=d+% q, T0 —T1 + T0 —Tw 111(r0/r1)+ R, (l/erroh) Zak t’c Selecting nominal values of k = 10 W/m-K, R'LC = 0.01 m-KIW and h = 100 W/mZ-K, the following parametric variations are obtained 350° ' —r l "Ilfl—F"‘fg'—1 3000 3000 ~ 7- "I"? v \ "7 2500 2500 ‘ r - P; E ' , “ x a ‘“ “g 2000 / r ., . - .< g 0“— +>~fi v 7, 10., u: . ‘ . . . " Tl; ‘ In: 1500 I e’ . E I 1000 r , _ . _ I ‘ 500 7 - ‘- - , v1 ’ " / fAFl-r ‘ p it 0 -—‘- , —y— "r— a z i l , r‘r—l 0 50 100 150 200 D 0.02 0.04 0.06 0.08 0.1 Thermal conductivity, lefm K) Contact resistance, RlclmK/W) -E— q; —El— q. fl" 0 H3" 0 , E, qo e9" qo Continued... PROBLEM 3.38 (Cont) 20000 i I 16000 ’E‘ - 12000 E 2 S a 3000 a; I 4000 o i t t 0 200 400 600 800 Convection coefficient, htW/m"2.K) —FJ— qi ’5 q + W For a prescribed value of h, (12) is fixed, while qfl , and hence q' , increase and decrease, respectively, with increasing k and R'LC. These trends are attributable to the effects of k and R'LC on the total (conduction plus contact) resistance separating the heater from the inner surface. For fixed k and RLC , qi is fixed, while q'0 , and hence q' , increase with increasing h due to a reduction in the convection resistance. COMMENTS: For the prescribed nominal values of k, R'LC and h, the electric power requirement is q' z 2377 W/rn. To maintain the prescribed heater temperature, q' would increase with any changes which reduce the conduction, contact and/‘or convection resistances. PROBLEM 3.41 KNOWN: Thin electrical heater fitted between two concentric cylinders, the outer surface of which experiences convection. FIND: (a) Electrical power required to maintain outer surface at a specified temperature, (b) Temperature at the center SCHEMATIC: AJkA=O.I5fl%/(- r, :20?” m 73in electrical hea‘fcr '3 = 4 0m 7" (r: ) Tear-I 0C a B,k3=1.5fl-,fl_/K h=505W/m2'K ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Heater element has negligible thickness, (4) Negligible contact resistance between cylinders and heater, (5) Constant properties, (6) No generation. ANALYSIS: (3) Perform an energy balance on the composite system to determine the power required to maintain T(r2) = TS = 5°C. Ein _ Ebut + Egen : Est +qelec ‘qconv : 0- Using Newton’s law of cooling, qelec : qconv = h‘ 27: r2 (Ts ‘Too) qelec=50 EV x27z(0.040m) [5—(~15):|0C:251W/m. < m -K (b) From a control volume about Cylinder A, we recognize that the cylinder must be isothermal, that is, T(0) = Tm). Represent Cylinder B by a thermal circuit: T0,) 7; W0 qr:T(rl),_Ts R 3 RB For the cylinder, from Eq. 3.28, Rig :lnr2/r1/27r kB giving T(r1)=TS +q'R'B : 5”C+253.l—“—]——hlfiO/L— = 235°C m 27r><l.5 W/m-K Hence, T(0) = T(r1) : 235°C. < Note that k A has no influence on the temperature T(0). ...
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chap3assign38-41 - PROBLEM 3.38 KNOWN: Inner and outer...

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