sm3-003.pdfsm3-007.pdfsm3-010.pdfsm3-030.pdfsm3-052.pdfsm3-067.pdfHW4bsolution.pdfPROBLEM 3.3 KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window. FIND: (a) Inner and outer window surface temperatures, Ts,i and Ts,o, and (b) Ts,i and Ts,o as a function of the outside air temperature T∞,o and for selected values of outer convection coefficient, ho. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation effects, (4) Constant properties. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12, (),i ,o22oi40 C 10 CTTq1L1 1 0.004m 1hkh 1.4WmK65W m K 30W m K∞∞−−−′′==++ + +⋅⋅⋅oo ()2250 Cq 969W m0.0154 0.0029 0.0333 m K W′′==++ ⋅o. Hence, with ()i,i ,oqhT T∞∞′′=−, the inner surface temperature is 2s,i ,i2iq969WmTT 40C 7.7Ch30W m K∞′′=−= − =⋅oo < Similarly for the outer surface temperature with ()os,o ,oqhT T∞′′=− find 2s,o ,o2oq969WmTT 10C 4.9Ch65W m K∞′′=+=−+ =⋅oo < (b) Using the same analysis, Ts,i and Ts,o have been computed and plotted as a function of the outside air temperature, T∞,o, for outer convection coefficients of ho = 2, 65, and 100 W/m2⋅K. As expected, Ts,i and Ts,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m2⋅K, Ts,i - Ts,o, is too small to show on the plot. Continued …PROBLEM 3.3 (Cont.) -30 -25 -20 -15 -10 -5 0Outside air temperature, Tinfo (C)-30-20-10010203040Surface temperatures, Tsi or Tso (C)Tsi; ho = 100 W/m^2.KTso; ho = 100 W/m^2.KTsi; ho = 65 W/m^2.KTso; ho = 65 W/m^2.KTsi or Tso; ho = 2 W/m^.K COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values of Ts,i and Ts,o could be increased by increasing the value of hi. (2) The IHT Thermal Resistance Network Model was used to create a model of the window and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinfo // Outside air temperature, C //q1 = // Heat rate, W T2 = Tso // Outer surface temperature, C q2 = 0 // Heat rate, W; node 2, no external heat source T3 = Tsi // Inner surface temperature, C q3 = 0 // Heat rate, W; node 2, no external heat source T4 = Tinfi // Inside air temperature, C //q4 = // Heat rate, W // Thermal Resistances: R21 = 1 / ( ho * As ) // Convection thermal resistance, K/W; outer surface R32 = L / ( k * As ) // Conduction thermal resistance, K/W; glass R43 = 1 / ( hi * As ) // Convection thermal resistance, K/W; inner surface // Other Assigned Variables: Tinfo = -10 // Outside air temperature, C ho = 65 // Convection coefficient, W/m^2.K; outer surface L = 0.004 // Thickness, m; glass k = 1.4 // Thermal conductivity, W/m.K; glass Tinfi = 40 // Inside air temperature, C hi = 30 // Convection coefficient, W/m^2.K; inner surface As = 1 // Cross-sectional area, m^2; unit areaPROBLEM 3.7 KNOWN: Thicknesses and thermal conductivities of refrigerator wall materials. Inner and outer air temperatures and convection coefficients. FIND: Heat gain per surface area. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Negligible contact resistance, (4) Negligible radiation, (5) Constant properties. ANALYSIS: From the thermal circuit, the heat gain per unit surface area is ()()()()(),o ,iippiipp oTTq1/hL/k L/kL/k1/h∞∞−′′=++++ ()()()( )225 4 Cq2 1/5W / m K 2 0.003m/60W / m K 0.050m / 0.046W / m K−°′′=⋅+ ⋅+ ⋅ ()2221 Cq 14.1 W / m0.4 0.0001 1.087 m K / W°′′==++ ⋅ < COMMENTS: Although the contribution of the panels to the total thermal resistance is negligible, that due to convection is not inconsequential and is comparable to the thermal resistance of the insulation.PROBLEM 3.10 KNOWN: A layer of fatty tissue with fixed inside temperature can experience different outside convection conditions. FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surface temperature for different convection conditions, and (c) Temperature of still air which achieves same cooling as moving air (wind chill effect). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state conditions, (3) Homogeneous medium with constant properties, (4) No internal heat generation (metabolic effects are negligible), (5) Negligible radiation effects. PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/m⋅K. ANALYSIS: The thermal circuit for this situation is Hence, the heat rate is s,1 s,1totTT TTq.R L/kA 1/ hA∞∞−−= =+ Therefore, windycalmwindycalmL1khq.L1qkh+′′=′′+ Applying a surface energy balance to the outer surface, it also follows that cond convq q.′′ ′′= Continued …PROBLEM 3.10 (Cont.) Hence, ( ) ( )s,1 s,2 s,2s,1s,2kT T hT TLkTThLT.k1+hL∞∞−= −+= To determine the wind chill effect, we must determine the heat loss for the windy day and use it to evaluate the hypothetical ambient air temperature, ′∞T , which would provide the same heat loss on a calm day, Hence, s,1 s,1windy calmTT TTqL1 L1kh kh∞∞′−−′′= = ++   From these relations, we can now find the results sought: (a) 2calmwindy20.003 m 1q0.2 W/m K0.015 0.015465 W/m K0.003 m 1q 0.015 0.040.2 W/m K25 W/m K+′′⋅+⋅= =′′++⋅⋅ calmwindyq0.553q′′=′′ < (b) ()( )()( )2s,2calm20.2 W/m K15 C 36 C25 W/m K 0.003 mT 22.1 C0.2 W/m K125 W/m K 0.003 m⋅−+⋅= =⋅+⋅ < ()( )()( )2s,2windy20.2 W/m K15 C 36