What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances
1 answer:
Explanation:
Air fuel ratio:
Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that
As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.
When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.
When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.
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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