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ASE 3333 - Test No. 3 Solutions: Thermodynamics and Heat Transfer Problems - Prof. Pasqual, Exams of Aerospace Engineering

The solutions to problem 1 to problem 5 from test no. 3 of ase 3333 thermodynamics and heat transfer course. The problems involve calculating thermal resistances, heat losses, and finding temperatures using given data.

Typology: Exams

2014/2015

Uploaded on 12/04/2015

cooperboy55
cooperboy55 🇺🇸

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Download ASE 3333 - Test No. 3 Solutions: Thermodynamics and Heat Transfer Problems - Prof. Pasqual and more Exams Aerospace Engineering in PDF only on Docsity! ASE 3333 - Test No. 3 - Solutions (last updated 11/24/2013) kJ 1000J Problem 1 Known quantities: T_inlet 45 °C T_1 30 °C T_2 15 °C D_1 4 cm D_2 4.5cm mdot 0.5 kg s  k_PVC 0.092 W m K  c_P 4.179 kJ kg K  L_pipe 90m Let's find the thermal resistance of the pipe: R_pipe ln D_2 D_1     2 π k_PVC L_pipe 2.264 10 3  K W  Then the heat lost by the pipe becomes: Qdot_lost T_1 T_2 R_pipe 6.626 kW This heat lost must be equal to the enthalpy loss along the pipe (see 1st Law for Control volumes): Given T_exit 40 °C <== initial guess! Qdot_lost mdot c_P T_inlet T_exit( )= Solution Find T_exit( ) Solution 41.829 °C Solution 314.979 K Problem 2 Known values: k_ins 0.03 W m K  Height 3 m Length 5m Δx 2cm Qdot_lost_before 2kW T_in 22 °C T_out 5 °C Area Height Length 15 m 2  The thermal resistance before the insulation is added can be found: R_before T_in T_out( ) Qdot_lost_before 8.5 10 3  K W  This resistance is the sum of inner and outer convection, plus the conduction resistance of the wall. None of these values change when insulation is added! The new thermal resistance is then found by simply adding to the previous value the thermal resistance of the insulation. R_ins Δx Area k_ins 0.044 K W  R_after R_before R_ins 0.053 K W  Finally the new heat loss can be estimated: Qdot_lost_after T_in T_out( ) R_after 321.091 W Problem 5 Surface 1: flat disk Diam_1 95cm Area_1 π 4 Diam_1 2  0.709 m 2  Surface 2: cube minus flat disk Side_2 1m Area_2 6 Side_2( ) 2  Area_1 5.291 m 2  We have 4 view factors, and we need F_21. We can start easily: F_11 0 <== Flat surface F_12 1 F_11 1 <== Conservation of energy F_21 F_12 Area_1 Area_2 0.134 <== Reciprocity
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