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umka2103 [35]
3 years ago
6

The outer surface of a grill hood is at 87 °c and the emissivity is 0.93. the heat transfer coefficient for convection between t

he hood and surroundings at 27 °c is 10 w/m2 /k. determine the net rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kw per m2 of surface area.
Physics
1 answer:
Tju [1.3M]3 years ago
4 0
The net flux, or rate of heat transfer per area, for convection and radiation are the following:

Convection:
Q/A = hΔT
Radiation:
Q/A = [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
where
h is the heat transfer coefficient: 10 W/m²·K or 0.01 kW/m²·K
ε is the emissivity: 0.93
σ is the Stephen-Boltzmann constant: 5.67×10⁻¹¹ kW/m²·K⁴
A₁ = A₂ because they have the same surface area
T is absolute temperature in Kelvin: K = °C + 273

Thus,

Q/A = hΔT + [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
Q/A = (0.01 kW/m²·K)(360 K - 300 K) + [0.93*5.67×10⁻¹¹ kW/m²·K⁴*(360⁴ - 300⁴)]/{1 - [1*(1-0.93)²]}
Q/A =  0.6 + 0.461
Q/A = 1.061 kW/m²
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Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
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Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

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putting the values in the expression

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= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

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