Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon
tal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.
1 answer:
The complete Question is:
Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.
What is the Rayleigh number for free convection on the outer sides of the duct?
What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?
What is the Rayleigh number for free convection on the top of the duct?
What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?
What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?
What is the total heat gain to the duct per unit length, in W/m?
Answers:
- 7709251 or 7.709 ×10⁶
- 4.87
- 965073
- 5.931 W/m² K
- 2.868 W/m² K
- 69.498 W/m
Explanation:
Find the given attachments for complete explanation
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Answer:
# kelvin_to_celsius function is defined
# it has value_kelvin as argument
def kelvin_to_celsius(value_kelvin):
# value_celsius is initialized to 0.0
value_celsius = 0.0
# value_celsius is calculated by
# subtracting 273.15 from value_kelvin
value_celsius = value_kelvin - 273.15
# value_celsius is returned
return value_celsius
# celsius_to_kelvin function is defined
# it has value_celsius as argument
def celsius_to_kelvin(value_celsius):
# value_kelvin is initialized to 0.0
value_kelvin = 0.0
# value_kelvin is calculated by
# adding 273.15 to value_celsius
value_kelvin = value_celsius + 273.15
# value_kelvin is returned
return value_kelvin
value_c = 0.0
value_k = 0.0
value_c = 10.0
# value_c = 10.0 is used to test the function celsius_to_kelvin
# the result is displayed
print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')
value_k = 283.15
# value_k = 283.15 is used to test the function kelvin_to_celsius
# the result is displayed
print(value_k, 'is', kelvin_to_celsius(value_k), 'C')
Explanation:
Image of celsius_to_kelvin function used as guideline is attached
Image of program output is attached.
