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Stella [2.4K]
3 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.
Engineering
1 answer:
bonufazy [111]3 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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6 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
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Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

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σ_{fs} = F_{f} L / \pi  R^{3}

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Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

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  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

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Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

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  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

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  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

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  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

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F =  8849 N

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3 years ago
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A consulting firm submitted a bid for a large research project. The firm's management initially felt there was a 50/50 chance of
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