CSU MECH 344 - HW3 - Solutions (10 pages)

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HW3 - Solutions



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HW3 - Solutions

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Pages:
10
School:
Colorado State University- Fort Collins
Course:
Mech 344 - Heat and Mass Transfer
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PROBLEM 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 q T i T o 1 ho q L k 1 hi 0 004 m 40o C 10o C 1 2 65 W m K 1 4 W m K 50o C 0 0154 0 0029 0 0333 m 2 K W 1 30 W m 2 K 969 W m 2 Hence with q h i T i T o the inner surface temperature is Ts i T i q hi 40o C 969 W m 2 30 W m 2 K 7 7o C Similarly for the outer surface temperature with q h o Ts o T o find Ts o T o q 10o C 969 W m 2 2 4 9o C 65 W m K 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 ho Continued Surface temperatures Tsi or Tso C PROBLEM 3 3 Cont 40 30 20 10 0 10 20 30 30 25 20 15 10 5 0 Outside air temperature Tinfo C Tsi ho 100 W m 2 K Tso ho 100 W m 2 K Tsi ho 65 W m 2 K Tso ho 65 W m 2 K Tsi 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 R32 L k As R43 1 hi As Convection thermal resistance K W outer surface Conduction thermal resistance K W glass 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 area PROBLEM 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 q q q T o T i 1 hi Lp k p Li ki Lp k p 1 h o 25 4 C 2 1 5 W m 2 K 2 0 003m 60 W m K 0 050m 0 046 W m K 21 C 0 4 0 0001 1 087 m2 K W 14 1 W m 2 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 q Ts 1 T Ts 1 T R tot L kA 1 hA Therefore L 1 k h windy q calm q windy L 1 k h calm Applying a surface energy balance to the outer surface it also follows that q cond q conv Continued PROBLEM 3 10 Cont Hence k Ts 1 Ts 2 h Ts 2 T L k T Ts 1 hL Ts 2 k 1 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 q Ts 1 T Ts 1 T L 1 L 1 k h windy k h calm From these relations we can now find the results sought 0 003 m 1 q calm 0 2 W m K 65 W m 2 K 0 015 0 0154 a 0 003 m 1 q windy 0 015 0 04 0 2 W m K 25 W m 2 K q calm 0 553 q windy 15 C b Ts 2 calm 0 2 W m K 25 W m 2 K 0 003 m 22 1 C 0 2 W m K 1 25 W m 2 K 0 003 m c windy 0 2 W m K 36 C 65 W m 2 K 0 003m 1 15 C Ts 2 36 C 0 2 W m K 10 8 C 65 W m2 K 0 003m T …


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