Question

For heat transfer purposes, a standing man can be modeled as 35cm diameter and 175.5cm long vertical cylinder with only the bottom surface insulated. The remaining surface is at average ambient and surface temperature of 13.52°c and 35.5°c respectively. If the time taken in the open air by the man is 0.052 hours. Find the rate of heat lost by convection from this man. And the amount of heat energy lost in the open air.

Note: the convection heat transfer coefficient=15kw/m^2-k

Answer 1

Ans) We know,

Q = h A dT

where, Q = heat transferred per unit time

h = heat transfer coefficient

A = heat transfer area of surface

dT = change in temperature

Surface area = 2rH + r²

= 2 x 3.14 x 0.175 x 0.175 x 1.755 + ( 3.14 x 0.175²)

= 0.434 m²

dT = (273 + 35.5) - (273 + 13.52)

= 21.98 K

Q = 15000 x 0.434 x 21.98

= 143089.8 W

or 143.09 kW

Heat energy lost in atmosphere = Q t

= 143089.8 x 0.052 x 60 x 60

= 26.78 x 10⁶ J

or 26.78 MJ