There have been many attempts to manufacture and market plastic bicycles. All have been too flexible and soft. Which design-limi
1 answer:
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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
Answer:
a. = 40 psf
b. L ≈ 30.80 psf
c. The uniformly distributed total load for the beam = 812.8 ft./lb
d. The alternate concentrated load is more critical to bending , shear and deflection
Explanation:
The given parameters of the beam the beam are;
The span of the beam = 26 ft.
The width of the tributary, b = 16 ft.
The dead load, D = 20 psf.
a. The basic floor live load is given as follows;
The uniform floor live load, = 40 psf
The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²
Therefore, the uniform live load, = 40 psf
b. The reduced floor live load, L in psf. is given as follows;
For the school, = 2
Therefore, we have;
The reduced floor live load, L ≈ 30.80 psf
c. The uniformly distributed total load for the beam, = b ×
=
∴ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb
The uniformly distributed total load for the beam, = 812.8 ft./lb
d. For the uniformly distributed load, we have;
= 812.8 × 26/2 = 10566.4 lbs
= 812.8 × 26²/8 = 68,681.6 ft-lbs
= 5×812.8×26⁴/348/EI = 4,836,329.333/EI
For the alternate concentrated load, we have;
= 1000 lb
= 20 × 16 = 320 lb/ft.
= 1,000 + 320 × 26/2 = 5,160 lbs
= 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs
= 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI
Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load