Answer:
d)1241W
Explanation:
We know that drag force given as
[tex]F_D=\dfrac{1}{2} \rho C_D AV^2[/tex]
Given that [tex]F_D=0.1N[/tex]
[tex]A=1 m^2[/tex]
V=3 m/s
[tex]\rho =1.204 kg/m^3[/tex]
Now by putting the values
[tex]0.1=\dfrac{1}{2} \times 1.204 C_D\times 1\times 3^2[/tex]
[tex]C_D=0.018[/tex]
Now from Reynold's analogy
[tex]St\times Pr^{\frac{2}{3}}=\dfrac{C_D}{2}[/tex]
Now by putting the values
[tex]St\times 0.7309^{\frac{2}{3}}=\dfrac{0.018}{2}[/tex] (Pr=0.7309 given)
So St=0.0111
We know that [tex]St=\dfrac{h}{\rho VC_p}[/tex]
⇒[tex]0.0111=\dfrac{h}{1.204\times 3\times 1007}[/tex]
So [tex]h=41.37 \dfrac{W}{m^2-k}[/tex]
We know that heat transfer given as
Q=hAΔT
Here given that ΔT=30°C
So now by putting the values
[tex]Q=41.37 \times 1\times 30[/tex]
So heat rate from plate=1241.1W
So our answer is d.