Respuesta :
Answer:
a) [tex]h_c = 0.1599 W/m^2-K[/tex]
b) [tex]H_{loss} = 5.02 W[/tex]
c) [tex]T_s = 302 K[/tex]
d) [tex]\dot{Q} = 25.125 W[/tex]
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature, [tex]T_a = 25^0 C = 298 K[/tex]
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
[tex]\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec[/tex]
Rate of heat transfer,
[tex]\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W[/tex]
a) To calculate the convection coefficient relationship for heat transfer by convection:
[tex]\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K[/tex]
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe, [tex]k_c = 0.8[/tex]
[tex]\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K[/tex]
b) Heat loss from the pipe to the environment:
[tex]H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W[/tex]
d) The required fan control power is 25.125 W as calculated earlier above
