Thermodynamics HW Solutions 545

# Thermodynamics HW Solutions 545 - 82 62 7321 10 407 4 332...

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Chapter 7 External Forced Convection 7-18E Air flows over a flat plate. The local friction and heat transfer coefficients at intervals of 1 ft are to be determined and plotted against the distance from the leading edge. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. Properties The properties of air at 1 atm and 60 ° F are (Table A-15E) 7321 . 0 Pr /s ft 10 1588 . 0 F Btu/h.ft. 01433 . 0 2 3 - = × = υ ° = k Air V = 7 ft/s T = 60 ° F Analysis For the first 1 ft interval, the Reynolds number is L = 10 ft 4 2 3 10 407 . 4 /s ft 10 1588 . 0 ft) ft/s)(1 7 ( Re × = × = υ = L L V which is less than the critical value of 51 0 5
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Unformatted text preview: 82 . 62 ) 7321 . ( ) 10 407 . 4 ( 332 . Pr Re 332 . 3 / 1 5 . 4 3 / 1 5 . = × = = = x x k hx Nu The local heat transfer and friction coefficients are F . Btu/h.ft 9002 . ) 82 . 62 ( ft 1 F Btu/h.ft. 01433 . 2 ° = ° = = Nu x k h x 00316 . ) 10 407 . 4 ( 664 . Re 664 . 5 . 4 5 . , = × = = x f C We repeat calculations for all 1-ft intervals. The results are x h x C f,x 2 4 6 8 10 0.5 1 1.5 2 2.5 3 0.002 0.004 0.006 0.008 0.01 0.012 x [ft] h x [Btu/h-ft 2-F] h x C f,x 1 0.9005 0.003162 2 0.6367 0.002236 3 0.5199 0.001826 4 0.4502 0.001581 5 0.4027 0.001414 6 0.3676 0.001291 7 0.3404 0.001195 8 0.3184 0.001118 9 0.3002 0.001054 10 0.2848 0.001 7-7...
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