PROBLEM 5.5 KNOWN: Diameter and radial temperature of AISI 00 carbon steel shaft. Convection coefficient and temperature of furnace gases. FIND: me required for shaft centerline to reach a prescribed temperature. ASSUMPTIONS: () One-dimensional, radial conduction, () Constant properties. PROPERTIES: AISI 00 carbon steel, Table A. T 550 K : 783 kg / m 3, k = 5. W/mK, c = 54 J/kgK, =.0-5 m /s. ANALYSIS: The Biot number is hr o / 00 W/m K 0.05 m/ Bi 0.0488. k 5. W/m K Hence, the lumped capacitance method can be applied. From Equation 5.6, T T has 4h exp t exp t T Vc cd 800 00 400 W/m K ln 0.8 t 300 00 783 kg/m 3 54 J/kg K 0. m t 859 s. COMMENTS: To check the validity of the foregoing result, use the one-term approximation to the series solution. From Equation 5.5c, To T 400 0.444 C exp Fo T 900 For Bi = hr o /k = 0.0976, Table 5. yields = 0.436 and C =.04. Hence 0.436 5. 0 m / s t 95 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 5.3 KNOWN: Thickness and properties of strip steel heated in an annealing process. Furnace operating conditions. FIND: (a) me required to heat the strip from 300 to 600 C. Required furnace length for prescribed strip velocity (V = 0.5 m/s), (b) Effect of wall temperature on strip speed, temperature history, and radiation coefficient. ASSUMPTIONS: () Constant properties, () Negligible temperature gradients in transverse direction across strip, (c) Negligible effect of strip conduction in longitudinal direction. PROPERTIES: Steel: ρ = 7900 kg/m 3, c p = 640 J/kg K, k = 30 W/m K, ε= 0.7. ANALYSIS: (a) Considering a fixed (control) mass of the moving strip, its temperature variation with time may be obtained from an energy balance which equates the change in energy storage to heat transfer by convection and radiation. If the surface area associated with one side of the control mass is designated as A s, A s,c = A s,r = A s and V = δa s in Equation 5.5, which reduces to dt ρδ c = h T T + T T dt or, introducing the radiation coefficient from Equations.8 and.9 and integrating, t T f f = h( T T ) hr( T Tsur) dt ρc o + 4 4 ( ) εσ ( sur ) ( δ ) Using the IHT Lumped Capacitance Model to integrate numerically with T i = 573 K, we find that T f = 873 K corresponds to t f 09s in which case, the required furnace length is L = Vtf 0.5m s 09s 05m (b) For T w = 3 K and 73 K, the numerical integration yields t f 0s and 6s respectively. Hence, for L = 05 m, V = L/t f yields V ( Tw = 3K) =.03m s V ( Tw = 73K) =.69 m s Continued...
PROBLEM 5.3 (Cont.) which correspond to increased process rates of 06% and 38%, respectively. Clearly, productivity can be enhanced by increasing the furnace environmental temperature, albeit at the expense of increasing energy utilization and operating costs. If the annealing process extends from 5 C (98 K) to 600 C (873 K), numerical integration yields the following results for the prescribed furnace temperatures. 600 00 Temperature, T(C) 500 400 300 00 00 Radiation coefficient, hr(w/m^.k) 50 00 0 0 50 00 50 00 50 300 50 0 50 00 50 00 50 300 Annealing time, t(s) Annealing time, t(s) Tsur = nf = 000 C Tsur = nf = 850 C Tsur = nf = 700 C Tsur = nf = 000 C Tsur = nf = 850 C Tsur = nf = 700 C As expected, the heating rate and time, respectively, increase and decrease significantly with increasing T w. Although the radiation heat transfer rate decreases with increasing time, the coefficient h r increases with t as the strip temperature approaches T w. COMMENTS: To check the validity of the lumped capacitance approach, we calculate the Biot number based on a maximum cumulative coefficient of (h + h r ) 300 W/m K. It follows that Bi = (h + h r )(δ/)/k = 0.06 and the assumption is valid.
PROBLEM 5.57 KNOWN: Initial temperature, thickness and thermal diffusivity of glass plate. Prescribed surface temperature. FIND: (a) me to achieve 50% reduction in midplane temperature, (b) Maximum temperature gradient at that time. ASSUMPTIONS: () One-dimensional conduction, () Constant properties. ANALYSIS: Prescribed surface temperature is analogous to h and T = T s. Hence, Bi =. Assume validity of one-term approximation to series solution for T (x,t). (a) At the midplane, Hence To T s o 0.50 Cexp Fo Ts tan Bi /. 4sin 4 C.73 sin ln o / C Fo 0.379 FoL 0.3790.0 m t 63 s. 60 7 m / s (b) With Cexp Fo cosx T T T T T i s i s C exp Fo sinx x L x L 300 C T/ x T/ x 0.5.360 4 C/m. max x 0.0 m COMMENTS: Validity of one-term approximation is confirmed by Fo > 0..
PROBLEM 5.66 KNOWN: Long plastic rod of diameter D heated uniformly in an oven to T i and then allowed to convectively cool in ambient air (T, h) for a 3 minute period. Minimum temperature of rod should not be less than 00C and the maximum-minimum temperature within the rod should not exceed 0C. FIND: Initial uniform temperature T i to which rod should be heated. Whether the 0C internal temperature difference is exceeded. ASSUMPTIONS: () One-dimensional radial conduction, () Constant properties, (3) Uniform and constant convection coefficients. PROPERTIES: Plastic rod (given): k = 0.3 W/mK, c p = 040 kj/m 3 K. ANALYSIS: For the worst case condition, the rod cools for 3 minutes and its outer surface is at least 00C in order that the subsequent pressing operation will be satisfactory. Hence, hr o 8 W/m K 0.05 m Bi 0.40 k 0.3 W/m K t k t 0.3 W/m K 360s Fo 0.308. r cp 3 3 o ro 0400 J/m K 0.05 m Using Eq. 5.5a and 0.856 rad and C =.093 from Table 5., T r o, t T C J0 ro exp Fo. T from Table B.4, With r o, J0 Jo 0.856 0.863, giving 00 5.093 0.863exp 0.856 0.308 T i 54 C. 5 At this time (3 minutes) what is the difference between the center and surface temperatures of the rod? From Eq. 5.5b, T r o, t T 00 5 J0 r o 0.863 o T 0,t T T 0,t 5 which gives T(0,t) = 37C. Hence, T T 0,80s T r o,80s 37 00 C 37 C. Hence, the desired max-min temperature difference sought (0C) is not achieved. COMMENTS: T could be reduced by decreasing the cooling rate; however, h can not be made much smaller. Two solutions are (a) increase ambient air temperature and (b) non-uniformly heat rod in oven by controlling its residence time.
PROBLEM 5.7 KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs at locations with T > 000 K. FIND: me required to harden outer layer of mm. ASSUMPTIONS: () One-dimensional radial conduction, () Constant properties, (3) Fo 0.. ANALYSIS: Since any location within the ball whose temperature exceeds 000 K will be hardened, the problem is to find the time when the location r = 9 mm reaches 000 K. Then a mm outer layer will be hardened. Begin by finding the Biot number. h r o 5000 W/m K 0.00 m/ Bi.00. k 50 W/m K Using the one-term approximate solution for a sphere, find Fo ln / C sin r. r From Table 5. with Bi =.00, for the sphere find 5708. rad and C =.73. With r* = r/r o = (9 mm/0 mm) = 0.9, substitute numerical values. 000 300 K Fo ln /.73 sin.5708 0.9 rad 0.44..5708 300 300 K.5708 0.9 From the definition of the Fourier number with = k/c, ro c 0.00 m kg J t Fo Fo ro 0.44 7800 500 / 50 W/m K 3.4 s. k m 3 kg K COMMENTS: () Note the very short time required to harden the ball. At this time it can be easily shown the center temperature is T(0, 3.4 s) = 87 K.
PROBLEM 5.85 KNOWN: Asphalt pavement, initially at 50C, is suddenly exposed to a rainstorm reducing the surface temperature to 0C. FIND: Total amount of energy removed (J/m ) from the pavement for a 30 minute period. ASSUMPTIONS: () Asphalt pavement can be treated as a semi-infinite solid, () Effect of rainstorm is to suddenly reduce the surface temperature to 0C and is maintained at that level for the period of interest. PROPERTIES: Table A-3, Asphalt (300K): = 5 kg/m 3, c = 90 J/kgK, k = 0.06 W/mK. ANALYSIS: This solution corresponds to Case, Figure 5.7, and the surface heat flux is given by Eq. 5.6 as / q s t k Ts T i / t () The energy into the pavement over a period of time is the integral of the surface heat flux expressed as t Q q 0 s t dt. () Note that q s t is into the solid and, hence, Q represents energy into the solid. Substituting Eq. () for q s t into Eq. () and integrating find / t -/ k Ts Q k / Ts T i / t dt t. (3) 0 / Substituting numerical values into Eq. (3) with k 0.06 W/m K 3.80 8 m / s c 5 kg/m 3 90 J/kg K find that for the 30 minute period, 0.06 W/m K 0 50K Q 3060s / 4.990 5 J/m. -8 / 3.80 m / s COMMENTS: Note that the sign for Q is negative implying that energy is removed from the solid.