Answer:
a) he rate of heat loss from the steam line is 18.413588 kW
b) the annual cost of heat loss from line is $12904.25
Explanation:
a)
first we find the area
A = πdL
d is the diameter (0.1m) and L is the length (25m)
so
A = π × 0.1 × 25
A = 7.85 m²
Now rate of heat loss through convection
qconv = hA(Ts -Ta)
h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)
so we substitute
qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)
qconv = 9817.477 J/s
Now heat lost through radiation
qrad = ∈Aα ( Ts⁴ - Ta⁴)
∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),
first we shall covert our temperatures from Celsius to kelvin scale
Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)
so we substitute
qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )
qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰
qrad = 8596.112 J/s
Now to get the total rate of heat loss through convection and radiation, we say
q = qconv + qrad
q = 9817.477 + 8596.112
q = 18413.588 J/s ≈ 18.413588 kW
Therefore the rate of heat loss from the steam line is 18.413588 kW
b)
annual cost of heat lost rate
A = C × q/n × ( 3600 × 24 × 365 )
C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)
so we substitute
A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )
A = $12904.25
Therefore the annual cost of heat loss from line is $12904.25
