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
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]
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]
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
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.
what is Geography? pliz help
Answer:
hope it's helpful please like and Follow me
Answer:
Geography is the science that studies and describes the surface of the Earth in its physical, current and natural aspect, or as a place inhabited by humanity.
Unit of rate of heat transfer
Answer:
The units on the rate of heat transfer are Joule/second, also known as a Watt.
Explanation:
Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.
hope it helps .
stay safe healthy and happy..The rate of heat transfer is measured in Joules per second, also known as Watts.
What is heat transfer?Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.
Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.
The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.
Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.
Thus, Joules per second or watts is the unit of rate of heat transfer.
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The object in ....................... shadow is not seen completely
Answer:
Dark shadow
Explanation:
Shadow is nothing but space when the light is blocked by an opaque object. It is just that part where light does not reach. When you stand in the sun, you are able to see your shadow behind you. ... This is because our body is opaque and does not allow the light to pass through it
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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
Which of the following are major components of a Class II BSC: A. Cabinet blower switch B. Foot support C. Drain spillage trough (catch basin) D. Rear grille E. Temperature control
Answer:
Cabinet blower switch ( A )
Explanation:
A major component of a class II BSC ( Biological safety cabinet ) is Cabinet blower switch because the Cabinet blower is an integral part of a class II BSC hence the switch is also a major component.
Class II BSC provides protection for the user, environment and sample to be manipulated in the laboratory ( mostly ; Pharmaceutical laboratories, Microbiology laboratories )
The National Weather Service has issued an alert for a severe storm that will bring 100 mm of rainfall in one hour. A farmer in the area is trying to decide whether to sand bag the creek that drains the 40 acres of row crops. The soil for the drainage area is a sandy clay loam and has a porosity of 0.398, effective porosity of 0.330, suction pressure of 52.3 cm, a hydraulic conductivity of 0.25 cm/hr and an effective saturation of 90%. Assuming that ponding occurs instantaneously, estimate the total depth of direct runoff in mm from the event using the Green-Ampt infiltration model.
a. 80
b. 89
c. 76
d. 72
The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet, calculate the number of turns on the input side, and the currentdrawn from the outlet.
Answer:
The current drawn from the outlet is 0.2 A
The number of turns on the input side is 350 turns
Explanation:
Given;
number of turns of the secondary coil, Ns = 35 turns
the output current, [tex]I_s[/tex] = 2 A
power supplied, [tex]P_s[/tex] = 24 W
the standard wall outlet in most homes = 120 V = input voltage
For an ideal transformer; output power = input power
the current drawn from the outlet is calculated;
[tex]I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A[/tex]
The number of turns on the input side is calculated as;
[tex]\frac{N_p}{N_s} = \frac{I_s}{I_p} \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns[/tex]
(8 pts.) Air in an Otto cycle engine is compressed to a temperature and pressure of 450 °C and 2.5 MPa. After the power stroke, the conditions are 600 °C and 0.45 MPa. Find the peak cycle temperature (°C), heat addition (kJ/kg), and efficiency
Answer:
a) [tex]Tb=1845.05K[/tex]
b) [tex]Q=1000.25KJ[/tex]
c) [tex]\mu=0.59[/tex]
Explanation:
From the question we are told that:
Temperature x [tex]Tx=450c=>723K[/tex]
Pressure x [tex]Px=2.5MPa[/tex]
Temperature y [tex]Ty=600c=>873K[/tex]
Pressure y [tex]Py=0.45MPa[/tex]
Let
Air atmospheric temperature be [tex]25c[/tex]
Therefore
Temperature [tex]Ta=25+273=298k[/tex]
Generally the equation for Otto cycle is mathematically given by
[tex]\frac{Tb}{Tx}=\frac{Ty}{Ta}[/tex]
[tex]Tb=\frac{873*723}{298}[/tex]
[tex]Tb=2118.05[/tex]
Therefore the peak cycle temperature (°C)
[tex]Tb=2118.05k[/tex]
[tex]Tb=2118.05-273[/tex]
[tex]Tb=1845.05K[/tex]
Generally the equation for Heat addition is mathematically given by
[tex]Q=Cv(Tb-Tx)[/tex]
[tex]Q=Cv(2118.05-723)[/tex]
[tex]Q=1000.25KJ[/tex]
Generally the equation for Thermal efficiency is mathematically given by
[tex]\mu=1-\frac{Ta}{Tx}[/tex]
[tex]\mu=1-\frac{298}{723}[/tex]
[tex]\mu=0.59[/tex]
Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by:
Answer: hello your question is incomplete attached below is the missing detail
answer :
Complex power = 2.5 ∠ 50° VA
apparent power = 2.5 VA
average power = 1.6 Watts
reactive power = 1.915 Var
power factor = 0.64 ( leading )
Explanation:
i) complex power
P = Vrms * Irms
= 17.67∠40° * 0.1414∠-10°
= 2.5∠50° VA
ii) Apparent power
s = Vrms * Irms
= 17.67 * 0.1414
= 2.5 VA
iii) Average power absorbed
Absorbed power ( p ) = Vrms * Irms * cos∅
= 17.67 * 0.1414 * cos ( 50 )
= 1.6 watt
iv) Reactive power
P = Vrms * Irms * sin∅
= 17.67 * 0.1414 * sin ( 50 )
= 1.915 VAR
v) power factor
P.F = cos ∅ = p /s
= 1.6 watt / 2.5 VA = 0.64.
Peter B. is admitted to a local hospital emergency room (ER) suffering from an anxiety attack. He tells the ER physician that he is anxious about a job promotion for which he is being considered. Peter’s secretary is worried about him and asks her father, Dr. K., who is on the medical staff at the hospital, to go to the ER and see how Peter is doing. Dr. K., who is often in the ER, knows all the staff and they willingly give him Peter’s chart when he asks for it. Dr. K. calls his daughter to tell her that Peter is being treated for anxiety with an anti-depressant drug and will probably be discharged. She relays this encouraging message to Peter’s boss. Peter does not receive the promotion.
a. Will it be an easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer.
b. What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?
c. In your opinion, should a diagnosis of anxiety be a concern for an employer? Why or why not?
a.No it will be not an easy matter for Peter to prove that the Er staff caused peter to lose his promotion
Here's why?
1)The boss may think he might not able to stretch himself in emergency situations
2)He can think he can't handle normal pressure
b.What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?
1)Peter should not let the medical personnel about his personal life
2)He might have not said extra information about his job promotion
3)Will it be easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer
No, it will not be easy, Peter has to prove in one or in another situation that he can face the difficulties and be able to handle anxiety attacks in the future.
Hence above answers are suitable for the given situation
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In a tension test of steel, the ultimate load was 13,100 lb and the elongation was 0.52 in. The original diameter of the specimen was 0.50 in. and the gage length was 2.00 in. Calculate (a) the ultimate tensile stress (b) the ductility of the material in terms of percent elongation
Answer:
a) the ultimate tensile stress is 66717.8 psi
b) the ductility of the material in terms of percent elongation is 26%
Explanation:
Given the data in the question;
ultimate load P = 13,100 lb
elongation δl = 0.52 in
diameter of specimen d = 0.50 in
gage length l = 2.00 inch
First we determine the cross-sectional area of the specimen
A = [tex]\frac{\pi }{4}[/tex] × d²
we substitute
A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²
A = 0.1963495 in²
a) the ultimate tensile stress σ[tex]_u[/tex]
tensile stress σ[tex]_u[/tex] = P / A
we substitute
tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495
tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi
Therefore, the ultimate tensile stress is 66717.8 psi
b) ductility of the material in terms of percent elongation;
percentage elongation of specimen = [change in length / original length]100
% = [ δl / l ]100
we substitute
% = [ 0.52 in / 2.00 in ]100
= [ 0.26 ]100
= 26
Therefore, the ductility of the material in terms of percent elongation is 26%
A small ship capable of making a speed of 6 knots through still water maintains a heading due east while being set to the south by an ocean current. The actual course of the boat is from A to B, a dis- tance of 10 nautical miles that requires exactly 2 hours. Determine the speed vC of the current and its direction measured clockwise from the north.
This question is incomplete, the missing diagram is uploaded along this answer below;
Answer:
the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively
Explanation:
Given the data in the question and as illustrated in the diagram below;
The absolute velocity of the ship Vs is 6 Knots due east
so we convert to meter per seconds
Vs = 6 Knots × [tex]\frac{0.51444 m/s}{1 Knots}[/tex] = 3.0866 m/s
Next we determine the relative velocity of the ship Vs/c
Vs/c = AB / t
given that distance between A to B = 10 nautical miles which requires 2 hours
so we substitute
Vs/c = 10 nautical miles / 2 hrs
Vs/c = [10 nautical miles × [tex]\frac{1852 m}{1 nautical-miles}[/tex] ] / [ 2 hrs × [tex]\frac{3600s}{1hr}[/tex] ]
Vs/c = 18520 / 7200
Vs/c = 2.572 m/s
Now, from the second diagram below, { showing the relative velocity polygon }
Now, using COSINE RULE, we calculate the velocity current.
Vc = √( V²s + V²s/c - 2VsSs/ccos10 )
we substitute
Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × cos10 ) )
Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × 0.9848 ) )
Vc = √( 9.527099 + 6.615184 - 15.6361 )
Vc = √0.506183
Vc = 0.71 m/s
Next, we use the SINE RULE to calculate the direction;
Vc/sin10 = Vs/c / sinθ
we substitute
0.71 / sin10 = 2.572 / sinθ
0.71 / 0.173648 = 2.572 / sinθ
4.0887 = 2.572 / sinθ
sinθ = 2.572 / 4.0887
sinθ = 0.62905
θ = sin⁻¹( 0.62905 )
θ = 38.98°
So, angle measured clock-wise will be;
θ = 270° - 38.98°
θ = 231.02°
Therefore, the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively
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]
True or false all workers who do class 1 asbestos work must be part of a medical surveillance program
Answer:
Yes
Explanation:
Answer:
true
Explanation:
hehehe
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
The term variation describes the degree to which an object or idea differs from others of the same type or from a standard.
a. True
b. False
Discuss typical advantages and disadvantages of an irrigation system?
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]
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
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:
Fill in the truth table for output A.
A = (x+y)(x'+z')(x'+z')
Answer:
1+1×1 multiplay then you get the answer
Analyze the boundary work done during the process having a rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively.
Complete Question
Analyze the boundary work done during the process having a rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively.
Determine the boundary work done during this process and heat Lose
Answer:
a) [tex]W=0[/tex]
b) [tex]dQ=-61.03KJ/kg[/tex]
Explanation:
From the question we are told that:
Pressure of air [tex]P_1=500kpa[/tex]
Temperature of Air [tex]T_2=150°C[/tex]
Pressure drop [tex]P_2=400kpa[/tex]
Temperature of drop [tex]T_2=65 \textdegree C[/tex]
Generally the Constant Volume Process is mathematically given by
[tex]V_1=V_2=V[/tex]
Therefore
a)
Generally the equation for boundary work w is mathematically given by
[tex]W=pdv[/tex]
[tex]W=P(V_2-V_1)[/tex]
[tex]W=P(V_V)[/tex]
[tex]W=0KJ[/tex]
b)
Generally the equation for Heat Change is mathematically given by
[tex]dQ=dU+dW[/tex]
[tex]dQ=dU[/tex]
[tex]dQ=C_v(T_2-T_1)[/tex]
Where
C_v=Specific Heat capacity of Air
[tex]C_v=0.718 kJ/kg K[/tex]
[tex]dQ=0.718(338-423)[/tex]
[tex]dQ=-61.03KJ/kg[/tex]
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]
Which option identifies the best way to reduce the environmental impact in the following scenario?
Sheryl has been assessing a factory where cheese is made and packaged. She has determined that the
product's packaging is the component that causes the highest environmental impact.
Sheryl should implement steps to create a cheese that does not need to be packaged.
0 Sheryl should implement steps to educate the factory personnel on the subject of conservation.
O Sheryl should implement steps to optimize the packaging in an eco-friendly manner.
O Sheryl should implement steps to reduce costs in the packaging department.
Sheryl should implement steps to optimize the packaging in an eco-friendly manner.
dentify the recommended practices when putting a tip on a micropipette. Select one or more: Gently push the micropipette into the tip and tap lightly to load the tip. Hold the micropipette at a 45 degree angle to the tip rack. Use the tip size designed for the micropipette size in use. Remove the tip from the rack and place it on micropipette by hand.
Answer:
Gently push the micropipette into the tip box and tag tightly to load the tip.
Explanation:
The recommended practice when putting a tip on a micropipette is ; Gently push the micropipette into the tip box and tag tightly to load the tip.
Given that it is not advisable to remove tip from rack so as not to contaminate it, if we want to put a tip on a micropipette we should gently push the micropipette into the tip box.
Problem 1. Network-Flow Programming (25pt) A given merchandise must be transported at a minimum total cost between two origins (supply) and two destinations (demand). Destination 1 and 2 demand 500 and 700 units of merchandise, respectively. At the origins, the available amounts of merchandise are 600 and 800 units. USPS charges $5 per unit from origin 1 to demand 1, and $7 per unit from origin 1 to demand 2. From origin 2 to demand 1 and 2, USPS charges the same unit cost, $10 per unit, however, after 200 units, the unit cost of transportation increases by 50% (only from origin 2 to demand 1 and 2).
a) Formulate this as a network-flow problem in terms of objective function and constraint(s) and solve using Excel Solver.
b) How many units of merchandise should be shipped on each route and what is total cost?
Solution :
Cost
Destination Destination Destination Maximum supply
Origin 1 5 7 600
Origin 2 10 10 800
15, for > 200 15, for > 200
Demand 500 700
Variables
Destination 1 2
Origin 1 [tex]$X_1$[/tex] [tex]$$X_2[/tex]
Origin 2 [tex]$X_3$[/tex] [tex]$$X_4[/tex]
Constraints : [tex]$X_1$[/tex], [tex]$$X_2[/tex], [tex]$X_3$[/tex], [tex]$$X_4[/tex] ≥ 0
Supply : [tex]$X_1$[/tex] + [tex]$$X_2[/tex] ≤ 600
[tex]$X_3$[/tex] + [tex]$$X_4[/tex] ≤ 800
Demand : [tex]$X_1$[/tex] + [tex]$$X_3[/tex] ≥ 500
[tex]$X_2$[/tex] + [tex]$$X_4[/tex] ≥ 700
Objective function :
Min z = [tex]$5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$[/tex]
[tex]$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$[/tex]
Costs :
Destination 1 Destination 2
Origin 1 5 7
Origin 2 10 10
15 15
Variables :
[tex]$X_1$[/tex] [tex]$$X_2[/tex]
300 300
200 400
[tex]$X_3$[/tex] [tex]$$X_4[/tex]
Objective function : Min z = 10600
Constraints:
Supply 600 ≤ 600
600 ≤ 800
Demand 500 ≥ 500
700 ≥ 500
Therefore, the total cost is 10,600.