Answer:
540 W
Explanation:
The dry weather average flow rate for a river is 8.7 m3/s. During dry weather flow, the average COD concentration in the river is 32 mg/L. An industrial source continuously discharges 18,000 m3/d of wastewater contains an average 342 mg/L COD concentration into the river. What is the COD mass loading in the river upstream of the industrial source discharge
Answer:
6156 kg /day
Explanation:
Determine the COD mass loading in the river upstream of the industrial source discharge
Given data:
Flow rate of river = 8.7 m^3/s
Average COD concentration in river = 32 mg/L
Industrial source continuous discharge ( Qw )= 18,000 m^3/d
Yw = 342 mg/l
since :
1 m^3 = 1000 liters
Qw = 18 * 10^6 liters = ( 18 million per day )
Hence the COD mass loading
= Yw * Qw
= 342 * 18 liters
= 6156 kg /day
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Reporters and Correspondents? Check all that apply.
Answer:
Acef
Explanation:
Edginuity 2021
Answer:
2,3,4,5
Explanation:
guy above me is wrong
Question 1: What is the power observed in the energy analyzer when the rated voltage(U1) is applied to the primary of the transformer, and there is no load at the secondary?
Question 2: Find the transformation ratio of the transformer using the values U1,U2 recorded in the experiment.
Question 3: Sketch the no-load operation graph of the transformer using the values U1, I2 and the values read in the energy analyzer.
Question 4: How can we find the number of turns of transformer?
Question 5: Explain the operation principle of the transformer.
Question 6: State your final observations about the experiment.
Answer:
preguntas a parte o no???????
A demand factor of _____ percent applies to a multifamily dwelling with ten units if the optional calculation method is used.
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is 3 m thick. Assume that the data for the sample at 7.0 m are representative of the entire clay profile. Also assume that the clay is heavily over consolidated and that the danse sands at the surface of the profile are so stiff that they do not contribute to the settlement. Find the settlement of the surface due to compression of the clay layer
Answer:
hello your question lacks some information attached below is the complete question with the required information
answer : 81.63 mm
Explanation:
settlement of the surface due to compression of the clay ( new consolidated )
= 81.63 mm
attached below is a detailed solution to the given problem
A workpiece in the form of a bar 100 mm in diameter is to be turned down to 70 mm diameter for 50 mm of its length. A roughing cut using maximum power and a depth of cut of 12 mm is to be followed by a finishing cut using a feed of 0.1 mm and a cutting speed of 1.5 m/s. It takes 20 s to load and unload the workpiece and 30 s to set the cutting conditions, set the tool at the beginning of the cut and engage the feed. The specific cutting energy for the material is 2.3 GJ/m3 and the lathe has a 3-kW motor and a 70 percent efficiency. Estimate:
Answer:
Hello your question has some missing part below is the missing part
Estimate The matching time for the Finish cut
answer: 79.588 seconds
Explanation:
Calculate the matching time for the Finish cut
Diameter of workpiece before cutting = 100 mm
hence
Diameter of workpiece before Finishing cut ( same as after rough cut )
D2 = D1 - 2 * d = 100 - (2 * 12) = 76mm
step 1 : determine spindle speed
V = [tex]\frac{\pi D_{2}N_{s} }{60}[/tex] -------- ( 1 ). where : D2 = 76 mm , Ns = ? , V = 1.5 ( input values into equation 1 )
therefore Ns = 376.94 rpm
Finally : Determine the matching time
T = [tex]\frac{L + A}{Ft * Ns}[/tex] ---- ( 2 ) . where : Ft = 0.1 mm , Ns = 376.94 rpm, L = 50 mm , A = 0 ( input values into equation 2 )
note : A = allowance length ( not given ) , ft = feed rate
T = 50 / ( 0.1 * 376.94 )
= 1.3265 minutes ≈ 79.588 seconds
Some of our modern kitchen cookware is made of ceramic materials. (a) List at least three important characteristics required of a material to be used for this application. (b) Make a comparison of three ceramic materials as to their relative properties and, in addition, to cost. (c) On the basis of this comparison, select the material most suitable for the cookware.
Answer:
A)
It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity
B)
stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stonesporcelain: mostly used for mugs and it is non-toxic Pyrex : posses good thermal conductivity used in ovenC) All the materials are suitable because they serve different purposes when making modern kitchen cookware
Explanation:
A) characteristics required of a ceramic material to be used as a kitchen cookware
It should be Non- toxicIt should possess high Thermal conductivityIt should have the Required Thermal diffusivityB) comparison of three ceramic materials as to their relative properties
stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stonesporcelain: mostly used for mugs and it is non-toxic Pyrex : posses good thermal conductivity used in ovensC) material most suitable for the cookware.
All the materials are suitable because they serve different purposes when making modern kitchen cookware
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the 4160V side of the transformer. The load has leading power factor of 0.9. It issupplied by 1 p.u. voltage on the 13.8kV side. The transformer per unit impedance is j0.12 referred to thesecondary side.
a. Find the load impedance.
b. Find the input current on the primary side in real units.
c. Find the input power factor
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
[tex]I_{load}[/tex] = 0.75 < 25.84°
attached below is the remaining part of the solution
B) Find the input current on the primary side in real units
load current in primary = 31.38 < 25.84 A
C) find the input power factor
power factor = 0.9323 leading
attached below is the detailed solution
What is the importance of ethics in emerging technologies?
Explanation:
the Ethics of emerging Technology can only make use of speculative data about future products,uses and impacts.
how do we succeed in mechanical engineering?
Yeah order for a firm voltage dividers to operate properly The load resistance value should be at least Times greater than resistance value of the voltage divider bleeder resistor
Answer:
A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors.
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The temperature inside the refrigerator is 38°F. Room temperature is 72°F. [answer in BTUs]
Answer:
The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU
Explanation:
The given parameters are;
The duration of the heat transfer, t = 24 hours = 86,400 seconds
The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²
The material of the insulator in the door = Cellulose fiber
The thickness of the insulator in the door, d = 2.0" = 0.0508 m
The temperature inside the fridge = 38° F = 276.4833 K
The temperature of the room = 78°F = 298.7056 K
The thermal conductivity of cellulose fiber = 0.040 W/(m·K)
By Fourier's law, the heat flow through a by conduction material is given by the following formula;
[tex]\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}[/tex]
[tex]Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t[/tex]
Therefore, we have;
[tex]Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522[/tex]
The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU
Answer the question on the image and a brianiest will be given to the person that provided the right answer to it.
Answer:
(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m
(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m
(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)
Explanation:
The wagon motion parameters are;
The mass of the wagon, m = 7,200 kg
The initial velocity with which the wagon is projected along the horizontal rail, v = U
The length of the horizontal portion of the rail = 100 m
The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)
The exerted frictional resistance to motion of the rail, [tex]F_f[/tex] = 140 N
∴ θ = sin⁻¹(0.01)
The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J
(a) If U = 3 m/s, we have;
Kinetic energy = 1/2·m·v²
The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;
K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J
The energy, E, required to move a distance, 'd', up the slope is given as follows;
E = [tex]F_f[/tex] × d + m·g·h
Where;
[tex]F_f[/tex] = The friction force = 140 N
m = The mass of the wagon = 7,200 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height reached = d × sin(θ) = d × 0.01
Therefore;
E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d
The energy, [tex]E_{NET \ horizontal}[/tex], remaining from the horizontal portion of the rail is given as follows;
[tex]E_{NET \ horizontal}[/tex] = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail
∴ [tex]E_{NET \ horizontal}[/tex] = 32,400 J - 14,000 J = 18,400 J
[tex]E_{NET \ horizontal}[/tex] = 18,400 J
Therefore, for the wagon with energy, [tex]E_{NET \ horizontal}[/tex] to move up the train, we get;
[tex]E_{NET \ horizontal}[/tex] = E
∴ 18,400 J = 846.32N × d
d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m
d₁ ≈ 21.74 m
The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m
(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m
The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;
P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J
The work done, 'W', on the frictional force on the return of the wagon is given as follows;
W = [tex]F_f[/tex] × d₂
Where d₂ = the distance moved by the wagon
By conservation of energy, we have;
P.E. = W
∴ 15,355.3968 = 140 × d₂
d₂ = 15,355.4/140 = 109.681405714
Therefore;
The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m
The distance the wagon comes to rest from the starting point, d₃, is given as follows;
d₃ = Horizontal distance + d₁ - d₂
d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m
The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m
(c) For the wagon to come finally to rest at it starting point, we have;
The initial kinetic energy = The total work done
1/2·m·v² = 2 × [tex]F_f[/tex] × d
∴ 1/2 × 7,200 × U² = 2 × 140 × d₄
d₄ = 100 + (1/2·m·U² - 140×100)
(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01
∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)
d₄ = 100 + d₁
∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)
∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))
3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)
= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32
= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²
3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32
U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)
U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649
U = √(10.3319363649) = 3.21433295801
The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s
Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)
The difference in value can come from difference in calculating methods