Answer:
a) The flow capacity of the fan is 3150 L/min
d) the minimum diameter is 0.11 m
Explanation:
given data:
A = area of residence = 200 m²
h = height of building = 2.7 m
Percentage of air that must be replaced by fresh air is 35%
v = velocity of air in the duct = 5.5 m/s
a) The volume of the entire building is:
The flow capacity of the fan is equal to:
b) The volume flow rate of fresh air is equal to:
Answer:a) recovery factor = 53.3% ,bi) volume of oil per acre-ft = 11090.64bbl/acre-ft
bii) bulk volume of the reservoir in acre-ft = 1000 acre-ft
c) 62214.74 cubic feet
Explanation: a) recovery factor is the percentage amount of oil that can recovered from a reservoir, it is the oil produced divided by oil initially in place
Recovery factor= 1 - (Soi/1-Swi)
= 1 - (0.35/1-0.25)
= 0.533 × 100%
= 53.3%
bi) volume of oil which may be recovered per acre-ft = 7758.porosity.(1- Swi-Soi)
= 7758 x 0.2 x (1-0.25-0.35)
= 620.64 bbl/acre-ft
bii) bulk volume of the reservoir in acre-ft= Area x thickness
= 100 acres x 10 ft
= 1000 acre-ft
c) total volume of oil in cubic feet
since we have gotten volume as 1000 acre-ft we simply multiply it by the volume of oil gotten in answer bi)
= 1000 acre-ft x 620.64 bbl/ acre-ft
= 620640bbl
So we convert from barrel(bbl) to cubic feet, and 1 barrel is equal to 5.609 cubic feet, so to convert the answer from barrel to cubic feet we multiply the answer 620640 bbl by 5.609
= 620640 bbl x 5.609
= 3481580.711 cubic feet
The pressure at the point on the roof where the speed is 60mph is 14.66 psi
<u>Explanation:</u>
The question follows Bernoulli's equation
P1 + 1/2 ρ V1² = P2 + 1/2 ρ V2²
V1 = 40 X 5280 / 3600 = 58.7 ft/s
V2 = 60 X 5280 / 3600 = 88 ft/s
P2 = P1 + 1/2 ( 0.00238 ) [(58.7)² - (88)²]
P2 - P1 = -5.12lb/ft²
The negative pressure created tends to lift the roof.
We know,
P1 = 14.7 psi = 2116.8 lb/ft²
P2 = 2116.8 - 5.12 lb/ft²
P2 = 2111.68 lb/ft²
P2 = 14.66 psi
Therefore, the pressure at the point on the roof where the speed is 60mph is 14.66 psi
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Answer: required tensile stress is 0.889 MPa
Explanation:
Given that;
tensile load is oriented along the [1 1 1] direction
shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane
first we determine
λ which is Angle between [1 1 1] and [1 0 1]
so
cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]
= 2 / √3√2 = 2/√6
Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]
so,
cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]
cos∅ = [ 2-1] / [√3√3 ]
cos∅ = 1/3
Now, we know that;
σ = T_stress/cosλcosθ
so we substitute
σ = 0.242 / ( 2/√6 × 1/3 )
σ = 0.242 / 0.2721
σ = 0.889 MPa
Therefore, required tensile stress is 0.889 MPa
