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4 4e o sur c E E 0 xI P x T T 0. A \u3c1 \u3b5\u3c3 \u2212 = \u2206 \u2212 \u2206 \u2212 = \ufffd \ufffd Substituting numerical values, find To = 2003K. It is unlikely that the present rod would ever reach this steady-state, maximum temperature. That is, the effect of conduction along the rod will cause the center temperature to be less than this value. PROBLEM 5.115 KNOWN: Support rod spanning a channel whose walls are maintained at Tb = 300 K. Suddenly the rod is exposed to cross flow of hot gases with T \u221e = 600 K and h = 75 W/m2\u22c5K. After the rod reaches steady-state conditions, the hot gas flow is terminated and the rod cools by free convection and radiation exchange with surroundings. FIND: (a) Compute and plot the midspan temperature as a function of elapsed heating time; compare the steady-state temperature distribution with results from an analytical model of the rod and (b) Compute the midspan temperature as a function of elapsed cooling time and determine the time required for the rod to reach the safe-to-touch temperature of 315 K. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Constant properties, (3) During heating process, uniform convection coefficient over rod, (4) During cooling process, free convection coefficient is of the form h = C\u2206Tn where C = 4.4 W/m2\u22c5K1.188 and n = 0.188, and (5) During cooling process, surroundings are large with respect to the rod. ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtained using the IHT Finite-Difference Equation, One-Dimensional, Transient Extended Surfaces Tool. The temperature-time history for the midspan position T10 is shown in the plot below. The steady-state temperature distribution for the rod can be determined from Eq. 3.75, Case B, Table 3.4. This case is treated in the IHT Extended Surfaces Model, Temperature Distribution and Heat Rate, Rectangular Pin Fin, for the adiabatic tip condition. The following table compares the steady-state temperature distributions for the numerical and analytical methods. Method Temperatures (K) vs. Position x (mm) 0 10 20 30 40 50 Analytical 300 386.1 443.4 479.5 499.4 505.8 Numerical 300 386.0 443.2 479.3 499.2 505.6 The comparison is excellent indicating that the nodal mesh is sufficiently fine to obtain precise results. 0 100 200 300 400 500 600 Elapsed heating time, t (s) 300 400 500 600 M id sp an te m pe ra tu re , T 10 (K ) Continued... PROBLEM 5.115 (Cont.) (b) The same finite-difference approach can be used to model the cooling process. In using the IHT tool, the following procedure was used: (1) Set up the FDEs with the convection coefficient expressed as hm = hfc,m + hr,m, the sum of the free convection and linearized radiation coefficients based upon nodal temperature Tm. ( )pfc,m mh C T T\u221e= \u2212 ( ) ( )2p p 2r,m m sur m surh T T T T\u3b5\u3c3 \uf8eb \uf8f6= + +\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 (2) For the initial solve, set hfc,m = hr,m = 5 W/m2\u22c5K and solve, (3) Using the solved results as the Initial Guesses for the next solve, allow hfc,m and hr,m to be unknowns. The temperature-time history for the midspan during the cooling process is shown in the plot below. The time to reach the safe-to-touch temperature, T Kp10 315= , is t = 550 s < 0 200 400 600 800 1000 Elapsed cooling time, t (s) 300 400 500 600 M id sp an te m pe ra tu re , T 10 (K ) PROBLEM 5.116 KNOWN: Thin metallic foil of thickness, w, whose edges are thermally coupled to a sink at temperature, Tsink, initially at a uniform temperature Ti = Tsink, is suddenly exposed on the top surface to an ion beam heat flux, sq\u2032\u2032 , and experiences radiation exchange with the vacuum enclosure walls at Tsur. Consider also the situation when the foil is operating under steady-state conditions when suddenly the ion beam is deactivated. FIND: (a) Compute and plot the midspan temperature-time history during the heating process; determine the elapsed time that this point on the foil reaches a temperature within 1 K of the steady-state value, and (b) Compute and plot the midspan temperature-time history during the cooling process from steady-state operation; determine the elapsed time that this point on the foil reaches the safe-to-touch temperature of 315 K. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient conduction in the foil, (2) Constant properties, (3) Upper and lower surfaces of foil experience radiation exchange with the large surroundings, (4) Ion beam incident on upper surface only, (4) Foil is of unit width normal to the page. ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtained using the IHT Finite-Difference Equation, One-Dimensional, Transient, Extended Surfaces Tool. In formulating the energy-balance functions, the following steps were taken: (1) the FDE function coefficient h must be identified for each node, e.g., h1 and (2) coefficient can be represented by the linearized radiation coefficient, e.g., ( )( )2 21 1 sur 1 surh T T T T\u3b5\u3c3= + + , (3) set a oq q 2\u2032\u2032 \u2032\u2032= since the ion beam is incident on only the top surface of the foil, and (4) when solving, the initial condition corresponds to Ti = 300 K for each node. The temperature-time history of the midspan position is shown below. The time to reach within 1 K of the steady-state temperature (374.1 K) is ( )10 h hT t 373K t 136s= = < (b) The same IHT workspace may be used to obtain the temperature-time history for the cooling process by taking these steps: (1) set sq\u2032\u2032 = 0, (2) specify the initial conditions as the steady-state temperature (K) distribution tabulated below, T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 374.1 374.0 373.5 372.5 370.9 368.2 363.7 356.6 345.3 327.4 (3) when performing the integration of the independent time variable, set the start value as 200 s and (4) save the results for the heating process in Data Set A. The temperature-time history for the heating and cooling processes can be made using Data Browser results from the Working and A Data Sets. The time required for the midspan to reach the safe-to-touch temperature is ( )10 c cT t 315K t 73s= = < Continued... PROBLEM 5.116 (Cont.) 0 100 200 300 400 500 Heating or cooling time, t (s) 300 320 340 360 380 400 M id ps an te m pe ra tu re , T 1 (K ) Heating process Cooling process COMMENTS: The IHT workspace using the Finite-Difference Equations Tool to determine the temperature-time distributions is shown below. Some of the lines of code were omitted to save space on the page. // Finite Difference Equations Tool: One-Dimensional, Transient, Extended Surface /* Node 1: extended surface interior node; transient conditions; e and w labeled 2 and 2. */ rho * cp * der(T1,t) = fd_1d_xsur_i(T1,T2,T2,k,qdot,Ac,P,deltax,Tinf, h1,q''a) q''a1 = q''s / 2 // Applied heat flux, W/m^2; on the upper surface only h1 = eps * sigma * (T1 + Tsur) * (T1^2 + Tsur^2) sigma = 5.67e-8 // Boltzmann constant, W/m^2.K^4 /* Node 2: extended surface interior node; transient conditions; e and w labeled 3 and 1. */ rho * cp * der(T2,t) = fd_1d_xsur_i(T2,T3,T1,k,qdot,Ac,P,deltax,Tinf, h2,q''a2) q''a2 = 0 // Applied heat flux, W/m^2; zero flux shown h2 = eps * sigma * (T2+ Tsur) * (T2^2 + Tsur^2) ....... ....... /* Node 10: extended surface interior node; transient conditions; e and w labeled sk and 9. */ rho * cp * der(T10,t) = fd_1d_xsur_i(T10,Tsk,T9,k,qdot,Ac,P,deltax,Tinf, h10,q''a) q''a10 = 0 // Applied heat flux, W/m^2; zero flux shown h10 = eps * sigma * (T10 + Tsur) * (T10^2 + Tsur^2) // Assigned variables deltax = L / 10 // Spatial increment, m Ac = w * 1 // Cross-sectional area, m^2 P = 2 * 1 // Perimeter, m L = 0.150 // Overall length, m w = 0.00025 // Foil thickness, m eps = 0.45 // Foil emissivity