PROBLEM 5.1 NOWN: Electrical heater attached to backside of plate while front surface is exposed to convection process (T,h); initially plate is at a uniform temperature of the ambient air and suddenly heater power is switched on providing a constant q o. FIND: (a) Sketch temperature distribution, T(x,t), (b) Sketch the heat flux at the outer surface, q L,t as a function of time. x ASSUMPTIONS: (1) One-dimensional conduction, () Constant properties, (3) Negligible heat loss from heater through insulation. ANALYSIS: (a) The temperature distributions for four time conditions including the initial distribution, T(x,0), and the steady-state distribution, T(x, ), are as shown above. Note that the temperature gradient at x = 0, -dt/dx) x=0, for t > 0 will be a constant since the flux, q x ( 0, ) is a constant. Noting that T o = T(0, ), the steady-state temperature distribution will be linear such that To T( L, ) q = k = h T( L, ) T. o L q x L,t = k dt/dx. From the x=l temperature distribution, we can construct the heat flux-time plot. (b) The heat flux at the front surface, x = L, is given by COMMENTS: At early times, the temperature and heat flux at x = L will not change from their initial values. Hence, we show a zero slope for q x ( L,t) at early times. Eventually, the value of q x ( L,t) will reach the steady-state value which is q o.
PROBLEM 5.9 NOWN: Diameter and radial temperature of AISI 1010 carbon steel shaft. Convection coefficient and temperature of furnace gases. FIND: me required for shaft centerline to reach a prescribed temperature. ASSUMPTIONS: (1) One-dimensional, radial conduction, () Constant properties. PROPERTIES: AISI 1010 carbon steel, Table A.1 ( T = 550 ): ρ = 783 kg / m 3, k = 51. W/m, c = 541 J/kg, α = 1.1 10-5 m /s. ANALYSIS: The Biot number is hr o / 100 W/m ( 0.05 m/) Bi= = = 0.0488. k 51. W/m Hence, the lumped capacitance method can be applied. From Equation 5.6, T T has 4h = exp t = exp t T ρvc ρcd 800 100 4 100 W/m ln 0.811 t 300 100 = = 783 kg/m 3 541 J/kg 0.1 m t= 859 s. COMMENTS: To check the validity of the foregoing result, use the one-term approximation to the series solution. From Equation 5.49c, ( ς ) To T 400 = = 0.444= C 1 exp 1 Fo T 900 For Bi = hr o /k = 0.0976, Table 5.1 yields ς 1 = 0.436 and C 1 = 1.04. Hence 5 ( 0.436) ( 1. 10 m /s) t= 915 s. ( 0.05 m) t = ln 0.434 = 0.835 The results agree to within 6%. The lumped capacitance method underestimates the actual time, since the response at the centerline lags that at any other location in the shaft.
PROBLEM 5.15 NOWN: Thickness and properties of furnace wall. Thermal resistance of film on surface of wall exposed to furnace gases. Initial wall temperature. FIND: (a) me required for surface of wall to reach a prescribed temperature, (b) Corresponding value of film surface temperature. ASSUMPTIONS: (1) Constant properties, () Negligible film thermal capacitance, (3) Negligible radiation. PROPERTIES: Carbon steel (given): ρ = 7850 kg/m 3, c = 430 J/kg, k = 60 W/m. ANALYSIS: The overall coefficient for heat transfer from the surface of the steel to the gas is 1 1 1 1 1 U ( Rtot) Rf 10 = = + m /W 0 W/m. h = + = 5 W/m Hence, UL 0 W/m 0.01 m Bi = = = 0.0033 k 60 W/m and the lumped capacitance method can be used. (a) It follows that T T = exp T t/ τt = exp t/rc = exp Ut/ ρlc 3 ρlc T T 7850 kg/m ( 0.01 m) 430 J/kg 100 1300 t ln = = ln U T i T 0 W/m 300 1300 t= 3886s= 1.08h. (b) Performing an energy balance at the outer surface (s,o), h T Ts,o = Ts,o T s,i /R f ht - + T s,i/r f 5 W/m 1300 + 100 /10 m /W Ts,o = = h+ 1/R 5+ 100 W/m ( f ) Ts,o = 10. COMMENTS: The film increases τ t by increasing R t but not C t.
PROBLEM 5.49 NOWN: A long cylinder, initially at a uniform temperature, is suddenly quenched in a large oil bath. FIND: (a) me required for the surface to reach 500, (b) Effect of convection coefficient on surface temperature history. ASSUMPTIONS: (1) One-dimensional radial conduction, () Constant properties, (3) Fo > 0.. ANALYSIS: (a) Check first whether lumped capacitance method is applicable. For h = 50 W/m, hlc h ( ro ) 50 W m ( 0.015m / ) Bic = = = = 0.1. k k 1.7 W m Since Bi c > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder, ( r, Fo) C1exp ( 1 Fo) J0( 1r ) θ * * = ζ ζ * (1) Solving for Fo and setting r * = 1, find 1 θ * Fo = ln ζ CJ 10( ζ1) 1 where * T( r o, to) T ( 500 350) θ = ( 1, Fo) = = = 0.31. T ( 1000 350) From Table 5.1, with Bi = 0.441, find ζ 1 = 0.888 rad and C 1 = 1.1019. From Table B.4, find J 0 (ζ 1 ) = 0.811. Substituting numerical values into Eq. (), 1 Fo = ln[ 0.31 1.1019 0.811] = 1.7. ( 0.888) From the definition of the Fourier number, Fo = r o ρc t = Fo = Fo ro α k α t r o, and α = k/ρc, t 1.7( 0.015m) 400kg m 3 1600J kg 1.7 W m 145s = =. (b) Using the IHT Transient Conduction Model for a Cylinder, the following surface temperature histories were obtained. Continued...
PROBLEM 5.49 (Cont.) 1000 Surface temperature, T() 900 800 700 600 500 400 300 0 50 100 150 00 50 300 me, t(s) h = 50 W/m^. h = 50 W/m^. Increasing the convection coefficient by a factor of 5 has a significant effect on the surface temperature, greatly accelerating its approach to the oil temperature. However, even with h = 50 W/m, Bi = 1.1 and the convection resistance remains significant. Hence, in the interest of accelerated cooling, additional benefit could be achieved by further increasing the value of h. COMMENTS: For Part (a), note that, since Fo = 1.7 > 0., the approximate series solution is appropriate.
PROBLEM 5.51 NOWN: Sapphire rod, initially at a uniform temperature of 800 is suddenly cooled by a convection process; after 35s, the rod is wrapped in insulation. FIND: Temperature rod reaches after a long time following the insulation wrap. ASSUMPTIONS: (1) One-dimensional radial conduction, () Constant properties, (3) No heat losses from the rod when insulation is applied. PROPERTIES: Table A-, Aluminum oxide, sapphire (550): ρ = 3970 kg/m 3, c = 1068 J/kg, k =.3 W/m, α = 5.59 10-5 m /s. ANALYSIS: First calculate the Biot number with L c = r o /, h L c h ( r o /) 1600 W/m ( 0.00 m/) Bi = = = = 0.7. k k.3 W/m Since Bi > 0.1, the rod cannot be approximated as a lumped capacitance system. The temperature distribution during the cooling process, 0 t 35s, and for the time following the application of insulation, t > 35s, will appear as Eventually (t ), the temperature of the rod will be uniform at T ( ). To find T, write the conservation of energy requirement for the rod on a time interval basis, Ein Eout = E Efinal E initial. Using the nomenclature of Section 5.5.3 and basing energy relative to T, the energy balance becomes Q = ρ cv( T( ) T ) Qo where Q o = ρcv(t i - T ). Dividing through by Q o and solving for T ( ), find T( ) = T + ( T T )( 1 Q/Q ). i o From the Groeber chart, Figure D.6, with hr o 1600 W/m 0.00m Bi= = = 1.43 k.3 W/m ( α o ) ( ) Bi Fo = Bi t/r = 1.43 5.59 10-6 m /s 35s/ 0.00m = 0.95. find Q/Q o 0.57. Hence, T 300 ( 800 300) ( 1-0.57) 515. = + = COMMENTS: From use of Figures D.4 and D.5, find T(0,35s) = 55 and T(r o,35s) = 43.
PROBLEM 5.83 NOWN: Initial temperature of copper and glass plates. Initial temperature and properties of finger. FIND: Whether copper or glass feels cooler to touch. ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, () Constant properties, (3) Negligible contact resistance. PROPERTIES: Skin (given): ρ = 1000 kg/m 3, c = 4180 J/kg, k = 0.65 W/m ; Table A-1 (T = 300), Copper: ρ = 8933 kg/m 3, c = 385 J/kg, k = 401 W/m ; Table A-3 (T = 300), Glass: ρ = 500 kg/m 3, c = 750 J/kg, k = 1.4 W/m. ANALYSIS: Which material feels cooler depends upon the contact temperature T s given by Equation 5.63. For the three materials of interest, 1/ 1/ k c = 0.65 1000 4180 = 1,616 J/m s 1/ 1/ 1/ k c = 401 8933 385 = 37,137 J/m s 1/ 1/ 1/ k c = 1.4 500 750 = 1,60 J/m s 1/. glass ( ρ ) skin ( ρ ) cu ( ρ ) k c 1/ 1/ >> cu k c glass, the copper will feel much cooler to the touch. From Equation Since ( ρ ) ( ρ ) 5.63, 1/ 1/ ( k ρ c) A TA,i+ ( k ρ c) B TB,i Ts = 1/ 1/ ( k ρ c) + A ( k ρ c) B Tscu + 1,616 310 37,137 300 = = 300.4 1,616+ 37,137 + 1,616 310 1,60 300 Tsglass = = 305.0. 1,616+ 1,60 COMMENTS: The extent to which a material s temperature is affected by a change in its thermal environment is inversely proportional to (kρc) 1/. Large k implies an ability to spread the effect by conduction; large ρc implies a large capacity for thermal energy storage.