Respuesta :
Answer:
The steady-state surface temperature = 905.5 K
The time required to come within 10°C of its steady-state value is 137.4 sec.
Explanation:
Given data
D = 10 mm
k = 240[tex]\frac{W}{m K}[/tex]
[tex]\rho[/tex] = 2700 [tex]\frac{Kg}{m^{3} }[/tex]
[tex]C_{p}[/tex] = 900 [tex]\frac{J}{kg K}[/tex]
Ambient temperature [tex]T_{o}[/tex] = 27 °c = 300 K
Heat transfer per unit length Q = 2000 [tex]\frac{W}{m}[/tex]
(a). We know that at steady state the heat transfer per unit length is given by
Q = h ([tex]\pi[/tex] d l ) ([tex]T_{s}[/tex] - [tex]T_{o}[/tex]) ----- (1)
Since the convection heat transfer coefficient at 450 K is
h = 105.2 [tex]\frac{W}{m^{2} K }[/tex]
Put all the values in equation (1), we get
2000 = 105.2 (3.14 × 0.01 × 1) ([tex]T_{s}[/tex] - 300)
[tex]T_{s}[/tex] = 905.5 K
Therefore the steady-state surface temperature = 905.5 K
(b).The time required for the surface temperature to come within 10°C of its steady-state value is given by
[tex]\frac{T - T_{o} }{T_{s} - T_{o} } =[/tex] [tex]e^{-at}[/tex] ------- (2)
Here
[tex]a = \frac{h A}{\rho V C}[/tex]
Put all the values in above equation we get
[tex]a = \frac{6h}{\rho D C}[/tex]
[tex]a = \frac{(6)(105.2)}{(2700) (0.01) (900)}[/tex]
a = 0.026
From equation (2)
[tex]\frac{300-283}{905-300} = e^{-0.026t}[/tex]
0.028 = [tex]e^{-0.026t}[/tex]
-3.57 = -0.026 t
t = 137.4 sec
Therefore the time required to come within 10°C of its steady-state value is 137.4 sec.
