Answer:
I dont kno
Explanation:
Im so sorry
4 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is 83.33 veh-s/cycle and the saturation flow rate is 1000 veh/h. If the capacity of the approach equals the number of arrivals per cycle, determine the approach flow rate and cycle length.
Answer:
Following are the responses to the given question:
Explanation:
Effective red duration is applied each cycle r=30 second D/D/1 queuing
In total, its approach delay is 83.33 sec vehicle per cycle
Flow rate(s) of saturated = 1,000 vehicles each hour
Total vehicle delay per cycle[tex]= \frac{v \times 30^2}{2(1-\frac{v}{0.2778})}[/tex]
[tex]\to \frac{v\times 30^2}{2(1-\frac{v}{0.2778})}= 83.33\\\\\to 900v=166.66-599.928v\\\\\to v=0.111 \frac{veh}{sec}\\\\[/tex]
The flow rate for such total approach is 0.111 per second.
The overall flow velocity of the approach is 400 cars per hour
The approach capacity refers to the number of arrivals per cycle.
Environmentally friendly time ratio to cycle length:
[tex],\frac{g}{C} \ is = \frac{400}{1000}=0.4\\\\r= c-g\\\\30\ sec =C - 0.4 C\\\\C=50 \ sec[/tex]
A main cable in a large bridge is designed for a tensile force of 2,600,000 lb. The cable consists of 1470 parallel wires, each 0.16 in. in diameter. The wires are cold-drawn steel with an average ultimate strength of 230,000 psi. What factor of safety was used in the design of the cable
Answer:
the factor of safety was used in the design of the cable is 2.6146
Explanation:
Given the data in the question;
Load on the main capable [tex]P_{initial[/tex] = 2600000 lb
number of parallel wires n = 1470
Diameter d = 0.16 in
average ultimate strength [tex]S_{ultimate[/tex] = 230000 psi
First we calculate the Load acting on each cable;
[tex]P_{initial[/tex] = P × n
P = [tex]P_{initial[/tex] / n
we substitute
P = 2600000 lb / 1470
P = 1768.70748 lb
Next we determine the working stress acting in a member;
[tex]S_{working[/tex] = P/A
{ Area A = [tex]\frac{\pi }{4}[/tex]d² }
[tex]S_{working[/tex] = P / [tex]\frac{\pi }{4}[/tex]d²
we substitute
[tex]S_{working[/tex] = 1768.70748 / [tex]\frac{\pi }{4}[/tex](0.16)²
[tex]S_{working[/tex] = 1768.70748 / 0.02010619298
[tex]S_{working[/tex] = 87968.29 psi
Now we calculate the factor of safety F.S
F.S = [tex]S_{ultimate[/tex] / [tex]S_{working[/tex]
we substitute
F.S = 230000 psi / 87968.29 psi
F.S = 2.6145785 ≈ 2.6146
Therefore, the factor of safety was used in the design of the cable is 2.6146
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.
This question is incomplete, the complete question is;
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.
Use the cold air standard assumptions.
Answer:
a) The compression ratio is 18.48
b) The maximum temperature of the cycle is 1893.4 K
c) The cutoff ratio, v₃/v₂ is 1.946
Explanation:
Given the data in the question;
Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K
Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K
Net work per cycle [tex]W_{net[/tex] = 590.1 kJ/kg
Heat transfer input per cycle Qs = 925 kJ/kg
a) compression ratio;
As illustrated in the diagram below, 1 - 2 is adiabatic compression;
so,
Tγ[tex]^{Y-1[/tex] = constant { For Air, γ = 1.4 }
hence;
⇒ V₁ / V₂ = [tex]([/tex] T₂ / T₁ [tex])^{\frac{1}{Y-1}[/tex]
so we substitute
⇒ V₁ / V₂ = [tex]([/tex] 973 K / 303 K [tex])^{\frac{1}{1.4-1}[/tex]
= [tex]([/tex] 3.21122 [tex])^{\frac{1}{0.4}[/tex]
= 18.4788 ≈ 18.48
Therefore, The compression ratio is 18.48
b) maximum temperature of the cycle
We know that for Air, Cp = 1.005 kJ/kgK
Now,
Heat transfer input per cycle Qs = Cp( T₃ - T₂ )
we substitute
925 = 1.005( T₃ - 700 )
( T₃ - 700 ) = 925 / 1.005
( T₃ - 700 ) = 920.398
T₃ = 920.398 + 700
T₃ = 1620.398 °C
T₃ = ( 1620.398 + 273 ) K
T₃ = 1893.396 K ≈ 1893.4 K
Therefore, The maximum temperature of the cycle is 1893.4 K
c) the cutoff ratio, v₃/v₂;
Since pressure is constant, V ∝ T
So,
cutoff ratio S = v₃ / v₂ = T₃ / T₂
we substitute
cutoff ratio S = 1893.396 K / 973 K
cutoff ratio S = 1.9459 ≈ 1.946
Therefore, the cutoff ratio, v₃/v₂ is 1.946
1. Using the formula above, complete this task.
Concrete is currently sold in increments of 12 yd'. Concrete current cost is $45 per V2 yd. How much would it cost a Shameka's Construction Company to cement a pad size 18ft x 20 ft. floor, 4 inches thick?
Answer:
gfbhjdskmlgtg
Explanation:
grtwhjyeywhtagsfdd
Discuss typical advantages and disadvantages of an irrigation system?
What is the Code requirement for receptacles connected to circuits of different voltages and installed in one building
Answer:
National Electrical Code
Explanation:
The National Electrical Code (NEC) is the foundational code that establishes the minimum requirements for electrical installation work. The NEC protects individuals and properties from being discriminated against dangers associated with the use of electricity. For electrical installation work, local and state electrical standards and codes should be consulted in addition to the NEC.
Section 406.4 of the NEC refers to On the same premises, receptacles linked to circuits with distinct voltages frequencies, or kinds of current (AC or DE) must be designed in such a way that the attachment plugs utilized on these circuits cannot be interchanged.
In how many ways 2450 can be written as a product of 2 factors?
Select one:
O a. 9
O b. 6
C. 18
O d. 10
Answer:
search up factors of 2450 and divide by two
The number of ways 2450 can be written as a product of 2 factors is 9 ways
How to find factors of numbers?Factors of 2,450 = 1, 2, 5, 7, 10, 14, 25, 35, 49, 50, 70, 98, 175, 245, 350, 490, 1225, 2450
2450 has 18 factors.The number of ways 2450 can be written as a product of 2 factors = number of factors / 2
= 18/2
= 9 wats
This means, taking the product of two factors such as;
1 × 2450
2 × 1225
5 × 490
7 × 350
10 × 245
14 × 175
25 × 98
35 × 70
50 × 98
Learn more about factors:
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Complete the following two statements regarding quantum-mechanical concepts associated with the Bohr model of the atom and refinements that resulted from the wave- mechanical atomic model.
Two important quantum-mechanical concepts associated with the Bohr model of the atom are that eledtrons are particles moving in _________ electron energy is quantized into
(b) Two important refinements resulting from the wave-mechanical atomic model are that the electron position is described in terms of a ______ electron energy is quantized into both shells and subshells--each electron is characterized by ________, and ______ , and quantum numbers.
tech a says that a slightly lean mixture offers good fuel economy and low exhaust emissions. Tech b says that a mixture that is too rich fouls spark plugs and causes incomplete burning. Who is correct
If the same type of thermoplastic polymer is being tensile tested and the strain rate is increased, it will: g
Answer:
It would break I think need to try it out
Explanation:
an amplifier has an input resistance of 100k a short-circuit transconductance of 10 mA/V and an output resistance of 100k. Find the open-circuit voltage gain
Answer:
The open circuit voltage gain is [tex]A_{vo}=-10^{3}[/tex]
Explanation:
Given data is input resistance of an amplifier is [tex]R_{in}=100k[/tex]Ω and output resistance of an amplifier is [tex]R_{o} =100k[/tex]Ω.
Trans conductance of an amplifier is [tex]g_{m}=10mA/V[/tex]
Thus Open circuit voltage gain is
[tex]A_{vo} =-g_{m}R_{o}[/tex]
[tex]A_{vo}=-10[/tex]×[tex]10^{-3}[/tex]×100×[tex]10^{3}[/tex]
Since 1m=[tex]10^{-3}[/tex] and 1k=[tex]10^{3}[/tex]
Thus,
[tex]A_{vo}=-1000[/tex]
[tex]A_{vo}=-10^{3}[/tex]
Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air. State the assumptions made in solving this problem
Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS[tex]_{air[/tex] = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS[tex]_{air[/tex] = -40 kW / 298.15 K
ΔS[tex]_{air[/tex] = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
6.D1 A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 13,300 N (3000 lbf). Determine the minimum required wire diameter assuming a factor of safety of 2 and a yield strength of 860 MPa (125,000 psi).
Answer:
6.33 mm
Explanation:
Stress is an internally resistive force produced by the molecules of the object to resist the deformation when an amount of load acts on the object. The SI unit of stress is measured in Pascal.
Given that force (F) = 13300 N, factor of safety (λ) = 2, yield strength (σy) = 860 MPa
The working stress [tex](\sigma_w)=\frac{\sigma_y}{\lambda}=\frac{860\ MPa}{2}=430\ MPa[/tex]
Let d be the minimum diameter, hence it is calculated using:
[tex]\sigma_w=\frac{F}{A}\\\\430 *10^6=\frac{13300}{A}\\\\A=30.93*10^{-6}\ m^2\\\\A=area=\frac{\pi d^2}{4} \\\\30.93*10^{-6}\ m^2=\frac{\pi d^2}{4} \\\\d=\sqrt{\frac{4*30.93*10^{-6}}{\pi } } \\\\d=0.0063\ m=6.3\ mm[/tex]
1. A hydro facility operates with an elevation difference of 50 m and a flow rate of 500 m3/s. If the rotational speed is 90 RPM, find the most suitable type of turbine and estimate the power output of the arrangement
Answer:
a) Pelton Turbine
b) [tex]P=2.42*10^{5}KW[/tex]
Explanation:
From the question we are told that:
Height [tex]h=50[/tex]
Flow Rate [tex]R= 500 m^3/s[/tex]
Rotational speed [tex]\omega=\90 RPM[/tex]
Let
Density of water
[tex]\rho=1000[/tex]
Generally the equation for momentum is mathematically given by
[tex]P=\rho gRh[/tex]
[tex]P=1000*9.81*500*50[/tex]
[tex]P=2.42*10^{5}KW[/tex]
What happens to resistance in the strain gage and voltage drop from a connected Wheatstone bridge if you were to pull the strain gauge along the long axis
7
Which wire can carry a higher current?
ered
tof
Select one:
A. AWG 6
B. AWG 18
C. AWG 12
D. AWG 24
Answer:
the wire that can carry much current is AWG 24
AWG 24 is the wire can carry a higher current. The diameter of the wire increases with decreasing gauge. The electrical resistance to the signals decreases with increasing wire diameter. Hence, option D is correct.
What is meant by high current?Any current above 10 mA has the potential to deliver an unpleasant to severe shock, while 200 mA and above is considered lethal. Despite the fact that currents above 200 mA might cause serious burns and unconsciousness, they usually do not cause death if the sufferer receives timely medical attention.
Higher voltage and lower current are frequently more effective combinations. The amount of current that a wire is rated to handle is frequently indicated by its capacity rating. People can use the same cable to drive a load twice as large by doubling the voltage and maintaining the same current.
Thus, option D is correct.
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#SPJ2
explain "columnar transposition" for the transposition ciphers using the following plaintext as an example:
abcd
efgh
ijkl
mnop
Answer:
wa 4tefaw gfawe f
Explanation:
so you do e Fse gaewr gware gfawe agfawe gwar g
What happens to resistance in the strain gauge and voltage drop from a connected Wheatstone bridge if you were to pull the strain gauge along the long axis
Answer:
Resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis.
Explanation:
Pulling the strain gauge in the long axis causes the wires to elongate/thin, the effect of this is that it will lead to an increase in resistance and voltage drop (V = I*R).
As a result of the resultant effect, resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis, such as the long axis.
The roof of a house has three layers: (1) 2 inch thick pine, (2) 4 inches of fiberglass insulation and (3) 0.1 inch thick Asphalt shingles. Calculate the rate of flow of heat (BTU/hour) through the roof. The temperature inside the house is 700F and the temperature outside is 300F. The R/inch for pine is 1.28, the R/inch for fiberglass is 3.0 and the R/inch for Shingles is 4.0. Take the area of the roof to be 500 square feet. The Unit of R is mixed: ft2 . 0F.h/BTU
Answer:
the rate of flow of heat through the roof is 45616.858 BTU/hr
Explanation:
Given the data in the question;
pin thickness [tex]t_p[/tex] = 2 in
fiber glass thickness [tex]t_f[/tex] = 4 in
Asphalt shingles thickness [tex]t_a[/tex] = 0.1 in
R/inch for pine = 1.28
R/inch for fiberglass = 3.0
R/inch for Shingles = 4.0
Temperature inside the house [tex]T_{inside[/tex] = 700 F
Temperature outside the house [tex]T_{outside[/tex] = 300 F
area of the roof A = 500 ft²
we calculate the total Resistance;
R = ( 2 × 1.28 ) + ( 4 × 3.0 ) + ( 0.1 × 4.0 )
R = 2.56 + 12 + 0.4
R = 14.96
Now, we determine the rate of heat flow;
dQ/dt = ΔT(A) / R
⇒ ( [tex]T_{inside[/tex] - [tex]T_{outside[/tex] )A / R
we substitute
⇒ (( 700 - 300 ) × 500 ) / 14.96
⇒ ( 400 × 500 ) / 14.96
⇒ 200000 / 14.96
⇒ 13368.98 watt
we know that 1 watt = 3.412142 BTU/hr
⇒ ( 13368.98 × 3.412142 ) BTU/hr
⇒ 45616.858 BTU/hr
Therefore, the rate of flow of heat through the roof is 45616.858 BTU/hr
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a constant temperature of 100C, find:
Complete Question
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:
The total heat transfer rate from the plate to the air
Answer:
[tex]q=1.7845[/tex]
Explanation:
From the question we are told that:
Air Temperature [tex]T_1=40c[/tex]
Length [tex]l=2m[/tex]
Velocity [tex]v=7m/s[/tex]
Width [tex]w=0.5[/tex]
Constant temperature [tex]T_t= 100C[/tex]
Generally the equation for Total heat Transfer is mathematically given by
[tex]q=hA(T_s-T_\infty)[/tex]
Where
h=Convective heat transfer coefficient
[tex]h=29.9075w/m^2k[/tex]
Therefore
[tex]q=h(L*B)(T_s-T_\infty)[/tex]
[tex]q=29.9075*(2*0.5)(100+273-(40+273))[/tex]
[tex]q=1794.45w[/tex]
[tex]q=1.7845[/tex]
A 2400-lb rear-wheel drive tractor carrying 900 lb of gravel starts from rest and accelerates forward at 3ft/s2. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.
Answer:
Explanation:
The missing diagram attached to the question is shown in the attached file below:
The very first thing we need to do in other to solve this question is to determine the mass of both the tractor and the mass of the gravel
For tractor, the mass is:
[tex]m_1 = \dfrac{2400 \ lb }{32.2 \ ft/s^2}[/tex]
[tex]m_1 = 74.53 \ lb.s^2/ft[/tex]
For gravel, the mass is:
[tex]m_2 = \dfrac{900 \ lb}{32.2 \ft/s^2}[/tex]
[tex]m_2 = 27.95 \ lb.s^2/ft[/tex]
From the diagram, let's consider the force along the horizontal components and vertical components;
So,
[tex]\sum F_x = ma_x \\ \\ 2F = (m_1+m_2) a \\ \\ F = \dfrac{1}{2}(74.53 4 + 27.950)lb.s^2/ft(2 \ ft/s^2) \\ \\ F = 102.484 \ lb[/tex]
[tex]\sum F_y = 0 \\ \\ 2N_A+2N_B - 2400 -900 = 0 \\ \\ N_A +N_B = 1650 \ lb[/tex]
Consider the algebraic sum of moments in the plane of A, with counter-clockwise moments being positive.
[tex]\sum M_A = I_o \alpha + \sum ma (d) \\ \\ = -2400 (20) + 2N_B (60) -900(110) = 0 - (74.534)(2)(20) - (27.950)(2)(40)[/tex]
[tex]=-48000 + 2N_B (60) -99000 = -2981.36-2236 \\ \\ = + 2N_B (60) = -2981.36-2236+48000+99000 \\ \\ = + 2N_B (60) = 141782.64 \\ \\ N_B = \dfrac{141782.64}{120} \\ \\ N_B = 1181.522 \ lb[/tex]
Replacing the value of 1181.522 lb for [tex]N_B[/tex] in equation (1)
[tex]N_A[/tex] + 1181.522 lb = 1650 lb
[tex]N_A[/tex] = (1650 - 1181.522)lb
[tex]N_A[/tex] = 468.478 lb
The net reaction on each of the rear wheels now is:)
[tex]F_R = \sqrt{N_A^2 +F^2}[/tex]
[tex]F_R = \sqrt{(468.478)^2 + (102.484)^2}[/tex]
[tex]\mathbf{F_R =479.6 \ lb}[/tex]
Now, we can determine the angle at the end of the rear wheels at which the resultant reaction force is being made in line with the horizontal
[tex]\theta = tan ^{-1}( \dfrac{468.478 }{102.484})[/tex]
[tex]\theta = 77.7^0[/tex]
Finally, the net reaction on each of the front wheels is:
[tex]F_B = N_B[/tex]
[tex]F_B =[/tex] 1182 lb
Find the equation of the output voltage as a function of time assuming the switch closes at t = 0 and the capacitor is fully discharged for t < 0.
Answer: Hello your question is incomplete attached below is the complete question
answer : V(out) (t) = 1 - e^-100t
Explanation:
The equation of the output voltage as a function of time assuming at t = 0 switch closes and capacitor will be discharged when t < 0
V(out) (t) = 1 - e^-100t
attached below is the step by step explanation
Consider a single crystal of silver oriented such that tensile stress is applied along a (0 0 1) direction. If slip occurs on a (1 1 1) plane and in a (1 0 1) direction and is initiated at an applied tensile stress of 1.1 MPa (160 psi), compute the critical resolved shear stress.
Answer:
i don't know
Explanation:
sorry
How do Machine Learning and Artificial Intelligence (AI) technologies help businesses use their enterprise data effectively?
Answer:
Explanation:
Application of artificial intelligence in business
You can use AI technologies to:
Improve customer services - eg use virtual assistant programs to provide real-time support to users (for example, with billing and other tasks).
Automate workloads - eg collect and analyse data from smart sensors, or use machine learning (ML) algorithms to categorise work, automatically route service requests, etc.
Optimise logistics - eg use AI-powered image recognition tools to monitor and optimise your infrastructure, plan transport routes, etc.
Increase manufacturing output and efficiency - eg automate production line by integrating industrial robots into your workflow and teaching them to perform labour-intensive or mundane tasks.
Prevent outages - eg use anomaly detection techniques to identify patterns that are likely to disrupt your business, such as an IT outage. Specific AI software may also help you to detect and deter security intrusions.
Predict performance - eg use AI applications to determine when you might reach performance goals, such as response time to help desk calls.
Predict behaviour - eg use ML algorithms to analyse patterns of online behaviour to, for example, serve tailored product offers, detect credit card fraud or target appropriate adverts.
Manage and analyse your data - eg AI can help you interpret and mine your data more efficiently than ever before and provide meaningful insight into your assets, your brand, staff or customers.
Improve your marketing and advertising - for example, effectively track user behaviour and automate many routine marketing tasks.
Machine learning is an Artificial intelligence-powered system that is based on a similar concept and able to learn from the intelligence provided by humans.
The AI systems are used to perform complex tasks in a way that is similar to humans but with precision. The AI-enabled machinery learning can boost sales and enhance the marketing campaign of any business or organization. It can develop a faster road map.Learn more about Machine Learning and Artificial Intelligence (AI) technologies.
brainly.in/question/32566751.
What would the Select lines need to be to send data for the fifth bit in an 8-bit system (S0 being the MSB and S2 being the LSB)?
A. S0 = 1, S1 = 0, S2 = 0
B. S0 = 0, S1 = 0, S2 = 0
C. S0 = 0, S1 = 1, S2 = 0
D. S0 = 0, S1 = 1, S2 = 1
Answer:
A. S0 = 1, S1 = 0, S2 = 0
lines need to send data for the fifth bit in an 8 bit system
Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The channel is having a bed slope of 1 in 2000. Take Chezy's coefficient, C=55
Answer:
V = 1.5062 m/s
Explanation:
look to the photos
A cylinder is internally pressurized to a pressure of 100 MPa. This causes tangential and axial stresses in the outer surface of 400 and 200 MPa, respectively. Make a Mohr circle representation of the stresses in the outer surface. What maximum normal and shear stresses are experienced by the outer surface?
Answer:
[tex]\mu_{max}=200Mpa[/tex]
Explanation:
From the question we are told that:
Internally pressurized [tex]P_i=100MPa[/tex]
Tangential Stress [tex]P_t=400mpa[/tex]
Axial stress [tex]P_a=200mpa[/tex]
Generally the equation for maximum normal and shear stresses are experienced by the outer surface is mathematically given by
[tex]\mu_{max}=|\frac{P_t-P_a}{2}|,|\frac{P_t}{2}|,|\frac{P_t}{2}|[/tex]
Therefore
[tex]\mu_{max}=|\frac{400-200}{2}|,|\frac{400}{2}|,|\frac{200}{2}|[/tex]
[tex]\mu_{max}=200Mpa[/tex]
A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of , an effective filtration rate of 7.70 m/h, and a production efficiency of 96 percent. A complete filter cycle duration is 52 h and the filter is rinsed for 20 minutes at the start of each cycle
Required:
a. What flow rate (m3/s) does the filter handle during production?
b. What volume of water is needed for backwashing plus rinsing the filter in each filter cycle?
Answer: hello there is a Missing information below is the missing information
surface dimensions of 10m * 8m
answer :
a) 640 m^3/hr
b) 1334.66 m^3
Explanation:
a) Determine the flow rate ( m^3/s ) of the filter handle
Va = Q /A[tex]_{f}[/tex]
where ; Va = filtration rate ( 8.00 m/h ) , Q = flow rate of filter handle , A[tex]_{f}[/tex] = surface area ( 10 m * 8 m )
Q = 8 * ( 10m * 8m ) = 640 m^3 / hr
b) Determine the volume of water needed for backwashing plus rinsing the filter in each filter cycle
Лf = 0.96 ( production efficiency )
Vb + Vr = 0.04 Vf
∴ Vf = ( Vb + Vr ) / 0.04 ------ ( 1 )
next step ; determine the volume of filtered water making use of effective filtration rate
Ref = ( Vf - Vb - Vr ) / A[tex]_{f}[/tex]T[tex]_{c}[/tex]
therefore : Vb + Vr = Vf - ( 80 * 52 * 7.7 ) ---- ( 2 )
Input equation 1 into 2
Vb + Vr = ( ( Vb + Vr ) / 0.04 ) - 32032 ---- ( 3 )
Resolve equation 3
hence the Volume of water needed for Backwashing and rinsing the filter in each filter cycle
= 1334.66 m^3
A frost free, 17 cu. ft. refrigerator-freezer uses energy at a rate of 500. watts when you hear the compressor running. If the fridge runs for 200. hours per month, how many kilowatt-hours of energy does the refrigerator use each month
Answer:
100 kWh
Explanation:
Since the freezer has a rating of 500 watts and runs for 200 hours in a month, the energy consumption can be gotten by getting the product of the rating of the freezer in kilowatts and the amount of time the fridge is on per month.
The rating of the freezer = 500 watts = 0.5 kW, time = 200 h
Energy consumption = rating * time = 0.5 kW * 200 h
Energy consumption = 100 kWh
Therefore the refrigerator uses 100 kWh per month
Assuming you determine the required section modulus of a wide flange beam is 200 in3, determine the lightest beam possible that will satisfy this condition.
Answer:
W18 * 106
Explanation:
Given that the section modulus of the wide flange beam is 200 in^3 the lightest beam possible that can satisfy the section modulus must have a section modulus ≥ 200 in^3. also the value of the section modulus must be approximately closest to 200in^3
From wide flange Beam table ( showing the section modulus )
The beam that can satisfy the condition is W18 × 106 because its section modulus ( s ) = 204 in^3