mechanically cleaned wastewater bar screen is constructed using 6.5-mm-wide bars spaced 5.0 cm apart center to center; the bars
1 answer:
Answer: The head loss is 1.5 cm
Explanation: Equation needed to solve is:
where C is the friction coefficient. To determine the area of flow, we nee to realize that the area available for flow over an area of 1*1 m, Aₓ= 0.87 m², observed from the correlation that for every width of 5.0 cm, we lose 6.5 mm, roughly about 12.9 or 13%.
For a fresh clean screen, the velocity will tend to increase by a factor of (1/0.87).
The velocities through the screen equals to (0.4/0.87) = 0.46 m/s and (0.8/0.87) = 0.92 m/s.
Apply the above mentioned equation:
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Answer:
Length = 129.55m, 129.55m
Explanation:
Given:
cp of water = 4180 J/kg·°C
Diameter, D = 2.5 cm
Temperature of water in = 17°C
Temperature of water out = 80°C
mass rate of water =1.8 kg/s.
Steam condensing at 120°C
Temperature at saturation = 120°C
hfg of steam at 120°C = 2203 kJ/kg
overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C
U = 700 W/m2 ·°C
Since Temperature of steam is at saturation,
temperature of steam going in = temperature of steam out = 120°C
Energy balance:
Heat gained by water = Heat loss by steam
Let specific capacity of steam = 2010kJ/Kg .°C
Find attached the full solution to the question.
Answer:
The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh
However, the average power P, required for a year, t = ? Kw
Expressing their relationship, we will have
U = P x t
Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)
t = 8760 hours
P =
P = 7134.7Kw
Hence, the average power required during the year is 7,135Kw
Now to calculate the energy used by the power plant in a year (in quads)?
Recall, Efficiency, η = Power Output/Power Input (100)
so, we have
η = P₀/P₁, given
0.45 =
P₁ = 15,855Kw
the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules
So to convert this to quads, Note;
1 quads of energy = 10¹⁵ Joules
The total energy used is 0.000000139 quads
Now to find the cubic feet of natural gas required to generate this power?
Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used
Since, the power plant generated = 62500000000Kwh
The cubic feet of natural gas used =
Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa