Thermodynamics HW Solutions 607

Thermodynamics HW Solutions 607 - Chapter 7 External Forced...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 7 External Forced Convection For this staggered arrangement tube bank, the friction coefficient corresponding to Re D = 5294 and S L / D = 1.5/0.8 = 1.875 is, from Fig. 7-27b, f = 0.44. Also, χ = 1 for the square arrangements. Then the pressure drop across the tube bank becomes Pa 638.2 = = ρ χ = Δ 2 2 3 2 max m/s kg 1 N 1 2 m/s) 571 . 8 )( kg/m 316 . 1 ( ) 1 )( 44 . 0 ( 30 2 V f N P L Discussion The arithmetic mean fluid temperature is ( T i + T e )/2 = (0 -15.6)/2 = -7.8 ° C, which is fairly close to the assumed value of -5 ° C. Therefore, there is no need to repeat calculations. 7-72 Air is heated by hot tubes in a tube bank. The average heat transfer coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is constant. Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the assumed mean temperature of 70 ° C and 1 atm (Table A-15):
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online