Respuesta :
Answer:
summer case Q = 52.25 W/m²
winter case Q = 119.37 W/m²
Explanation:
Given data
Air temperature T = 20 degree = 293 K
Environment temperature T1 = 27 degree = 300 K
Environment temperature T2 = 14 degree = 287 K
Surface temperature Ts = 32 degree = 305 K
Emissivity ε = 0.90
Coefficient of heat transfer h = 2 W/m²-K
Boltzman constant σ = 5.67 × 10-8 W/m²-[tex]k^4[/tex]
Solution:
We will apply here first Newton law for the and get here
q = h × A ( Ts – T) ……………………..1
[tex]\frac{q}{A}[/tex] = 2 × (305 – 293)
[tex]\frac{q}{A}[/tex] = 24 W/m²
Now we will apply here stefan Boltzam law for summer case
q = ε × A × σ × [tex]( (Ts)^4- (T1)^4 )[/tex] ……………..2
put her value and we get
[tex]\frac{q}{A} = 0.9 \times 5.67 \times 10^{-8} \times ( (305)^4 - (300)^4 )[/tex]
Solve it and we get
[tex]\frac{q}{A}[/tex] = 28.25 W/m²
Q = q(radiation) + q(convection) ……………….3
Q = 28.25 + 24
Q = 52.25 W/m²
And
For winter it will be
q = ε × A × σ × [tex]( (Ts)^4 - (T2)^4 )[/tex] ……………..4
put her value and we get
[tex]\frac{q}{A} = 0.9 \times 5.67 \times 10^{-8} \times ( (305)^4 - (287)^4 )[/tex]
Solve it and we get
[tex]\frac{q}{A}[/tex] = 95.37 W/m²
Now put value in equation 3
Q = 95.37 + 24
Q = 119.37 W/m²
As we can say that here difference between the summer and winter time radiations flux and chilled conditions are attributes the effect of colder wall on radiations
