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the Run command, the integration is performed from 0 to 0.1 s with a time step of 1×10-6 s. Then, using the Specify|Generation command, the generation rate is set to zero and the Run|Continue command is executed. The temperature history is shown below. (c) Finite-difference method of solution, IHT. The nodal arrangement for the clutch pair is shown below with \u2206x = 0.1 mm and \u2206t = 1 ms. Nodes 02-10, 13-16 and 18-21 are interior nodes, and their finite-difference equations (FDE) can be called into the Workspace using Tools|Finite Difference Equations|One-Dimenisonal|Transient. Nodes 01 and 22 represent the mid-planes for the reaction and composite plates, respectively, with adiabatic boundaries. The FDE for node 17 is derived from an energy balance on its control volume (CV) considering different properties in each half of the CV. The FDE for node 11 and 12 are likewise derived using energy balances on their CVs. At the interface, the following conditions must be satisfied 11 12 f rp cpT T q q q\u2032\u2032 \u2032\u2032 \u2032\u2032= = + The frictional heat flux is represented by a Lookup Table, which along with the FDEs, are shown in the IHT code listed in Comment 2. Continued (3)... PROBLEM 5.118 (Cont.) The temperature and heat flux histories are plotted below. The steady-state temperature was found as 156.5° C, which is in reasonable agreement with the energy balance result from part (a). COMMENTS: (1) The temperature histories resulting from the FEHT and IHT based solutions are in agreement. The interface temperature peaks near 225°C after 75 ms, and begins dropping toward the steady-state condition. The mid-plane of the reaction plate peaks around 100 ms, nearly reaching 200°C. The temperature of the mid-plane of the composite plate increases slowly toward the steady- state condition. (2) The calculated temperature-time histories for the clutch pair display similar features as expected from our initial sketches on T vs. t coordinates, part a. The maximum temperature for the composite is very high, subjecting the bonded frictional material to high thermal stresses as well as accelerating deterioration. For the reaction steel plate, the temperatures are moderate, but there is a significant gradient that could give rise to thermal stresses and hence, warping. Note that for the composite plate, the steel section is nearly isothermal and is less likely to experience warping. (2) The IHT code representing the FDE for the 22 nodes and the frictional heat flux relation is shown below. Note use of the Lookup Table for representing the frictional heat flux vs. time boundary condition for nodes 11 and 12. // Nodal equations, reaction plate (steel) /* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */ rhos * cps * der(T01,t) = fd_1d_sur_w(T01,T02,ks,qdot,deltax,Tinf01,h01,q''a01) q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf01 = 40 // Arbitrary value h01 = 1e-5 // Causes boundary to behave as adiabatic qdot = 0 /* Node 02: interior node; e and w labeled 03 and 01. */ rhos*cps*der(T02,t) = fd_1d_int(T02,T03,T01,ks,qdot,deltax) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. /* Node 10: interior node; e and w labeled 11 and 09. */ rhos*cps*der(T10,t) = fd_1d_int(T10,T11,T09,ks,qdot,deltax) /* Node 11: From an energy on the CV about node 11 */ ks * (T10 - T11) / deltax + q''rp = rhos * cps * deltax / 2 * der(T11,t) Continued (4)... Temperature history for clutch pair, 100 ms lock-up time 0 200 400 600 800 1000 Engagement time, t (ms) 0 50 100 150 200 250 Te m pe ra tu re , T (C ) Midplane, reaction plate, T01 Interface, T11 or T12 Midplane, composite plate, T22 Heat flux histories for clutch pair during engagement 0 20 40 60 80 100 Engagement time, t (ms) -0.5 0 0.5 1 1.5 2 H e a t f lu x, q' ' (W /m ^2 * 10 ^7 ) Frictional heat flux, q''f Reaction plate, q''rp composite plate, q''cp PROBLEM 5.118 (Cont.) // Friction-surface interface conditions T11 = T12 q''f = LOOKUPVAL(HFVST16,1,t,2) // Applied heat flux, W/m^2; specified by Lookup Table /* See HELP (Solver, Lookup Tables). The look-up table, file name "HFVST16' contains 0 16e6 0.1 0 100 0 */ q''rp + q''cp = q''f // Frictional heat flux // Nodal equations - composite plate // Frictional material, nodes 12-16 /* Node 12: From an energy on the CV about node 12 */ kfm * (T13 - T12) / deltax + q''cp = rhofm * cpfm * deltax / 2 * der(T12,t) /* Node 13: interior node; e and w labeled 08 and 06. */ rhofm*cpfm*der(T13,t) = fd_1d_int(T13,T14,T12,kfm,qdot,deltax) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 /* Node 16: interior node; e and w labeled 11 and 09. */ rhofm*cpfm*der(T16,t) = fd_1d_int(T16,T17,T15,kfm,qdot,deltax) // Interface between friction material and steel, node 17 /* Node 17: From an energy on the CV about node 17 */ kfm * (T16 - T17) / deltax + ks * (T18 - T17) / deltax = RHS RHS = ( (rhofm * cpfm * deltax /2) + (rhos * cps * deltax /2) ) * der(T17,t) // Steel, nodes 18-22 /* Node 18: interior node; e and w labeled 03 and 01. */ rhos*cps*der(T18,t) = fd_1d_int(T18,T19,T17,ks,qdot,deltax) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. /* Node 22: interior node; e and w labeled 21 and 21. Symmetry condition. */ rhos*cps*der(T22,t) = fd_1d_int(T22,T21,T21,ks,qdot,deltax) // qdot = 0 // Input variables // Ti = 40 // Initial temperature; entered during Solve deltax = 0.0001 rhos = 7800 // Steel properties cps = 500 ks = 40 rhofm = 1150 //Friction material properties cpfm = 1650 kfm = 4 // Conversions, to facilitate graphing t_ms = t * 1000 qf_7 = q''f / 1e7 qrp_7 = q''rp / 1e7 qcp_7 = q''cp / 1e7 PROBLEM 5.119 KNOWN: Hamburger patties of thickness 2L = 10, 20 and 30 mm, initially at a uniform temperature Ti = 20°C, are grilled on both sides by a convection process characterized by T\u221e = 100°C and h = 5000 W/m2\u22c5K. FIND: (a) Determine the relationship between time-to-doneness, td, and patty thickness. Doneness criteria is 60°C at the center. Use FEHT and the IHT Models|Transient Conduction|Plane Wall. (b) Using the results from part (a), estimate the time-to-doneness if the initial temperature is 5 °C rather than 20°C. Calculate values using the IHT model, and determine the relationship between time-to- doneness and initial temperature. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, and (2) Constant properties are approximated as those of water at 300 K. PROPERTIES: Table A-6, Water (300K), \u3c1 = 1000 kg/m3, c = 4179 J/kg\u22c5K, k = 0.613 W/m\u22c5K. ANALYSIS: (a) To determine T(0, td), the center point temperature at the time-to-doneness time, td, a one-dimensional shape as shown in the FEHT screen below is drawn, and the material properties, boundary conditions, and initial temperature are specified. With the Run|Calculate command, the early integration steps are made very fine to accommodate the large temperature-time changes occurring near x = L. Use the Run\uf8e6Continue command (see FEHT HELP) for the second and subsequent steps of the integration. This sequence of Start-(Step)-Stop values was used: 0 (0.001) 0.1 (0.01 ) 1 (0.1) 120 (1.0) 840 s. Continued \u2026.. PROBLEM 5.119 (Cont.) Using the View|Temperature vs. Time command, the temperature-time histories for the x/L = 0 (center), 0.5, and 1.0 (grill side) are plotted and shown below for the 2L = 10 mm thick patty. Using the View|Temperatures command, the time slider can be adjusted to read td, when the center point, x = 0, reaches 60°C. See the summary table below. The IHT ready-to-solve model in Models|Transient Conduction|Plane Wall is based upon Eq. 5.40 and permits direct calculation of td when T(0,td) = 60°C for patty thickness 2L = 10, 20 and 30 mm and initial temperatures of 20 and