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2 years ago

10

Under conditions for which the same roojm temperature is mainteined bt a heating or cooling system, it is not uncommon for a per

son to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) y by considering a room whose air temperature is maintained at 20 Dagree C throughout the year, while the walls of the room are nominally at 27 Dagree C and 14 Dagree C in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of 32 Dagree C throughout the year and to have an emissivity of 0.90. The coefficient associated with heat transfer by natural convection between the person and the room air is approximately 2 W/m2 Middot K.

1 answer:

bonufazy [111]2 years ago

7 0

Answer:

Net heat transfers: during summer = 52.253 W/m^2 during winter = 119.375 w/m^2

Explanation:

Given data :

Room temperature throughout the year = 20⁰c = 293 k

Room temperature during summer = 27⁰c = 300 k

Room temperature during winter = 14⁰c = 287 k

surface temperature of a person throughout the year = 32⁰c = 305 k

coefficient of heat transfer by natural convection (h)= 2 w /m^2 k

emissivity = 0.9

An explanation to the condition of feeling chilled during the winter and comfortable during summer can be explained with the calculation below

The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as

q = hΔt = 2 * ( 305 - 293) = 2 * 12 = 24 w /m^2

also calculate heat transfer through radiation using this formula

$q_{rad} =$ εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]

ε = 0.90,(emissivity) σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)

during summer :

$q_{rad}$ = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2

therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2

During winter :

$q_{rad}$ = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3

therefore the net heat transfer during the winter

= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2

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$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

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