Thermodynamics HW Solutions 67

Thermodynamics HW Solutions 67 - Chapter 1 Basics of Heat...

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

View Full Document Right Arrow Icon
Chapter 1 Basics of Heat Transfer 1-127 A standing man is subjected to high winds and thus high convection coefficients. The rate of heat loss from this man by convection in still air at 20°C, in windy air, and the wind-chill factor are to be determined. Assumptions 1 A standing man can be modeled as a 30-cm diameter, 170-cm long vertical cylinder with both the top and bottom surfaces insulated. 2 The exposed surface temperature of the person and the convection heat transfer coefficient is constant and uniform. 3 Heat loss by radiation is negligible. Analysis The heat transfer surface area of the person is Windy weather A s = π DL = (0.3 m)(1.70 m) = 1.60 m² The rate of heat loss from this man by convection in still air is Q still air = hA s Δ T = (15 W/m²· ° C)(1.60 m²)(34 - 20) ° C = 336 W In windy air it would be Q windy air = hA s T = (50 W/m²· ° C)(1.60 m²)(34 - 20) ° C = 1120 W To lose heat at this rate in still air, the air temperature must be 1120 W = ( hA s T ) still air = (15 W/m²· ° C)(1.60 m²)(34 - T effective ) ° C which gives T effective = -12.7 ° C
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

Ask a homework question - tutors are online