Answer:
the output of the IF amplifier consist of 675 kHz
Explanation:
Given the data in the question;
AM signal carrier frequency [tex]_{RF[/tex] = 460 kHz
Local oscillator frequency[tex]_{lo[/tex] = 1135 kHz
Now, The output of the IF amplifier consists of difference of local oscillator frequency & AM carrier signal frequency;
FREQUECY[tex]_{IF[/tex] = FREQUECY[tex]_{lo[/tex] - FREQUECY[tex]_{RF[/tex]
so we substitute in our given values
FREQUECY[tex]_{IF[/tex] = 1135 kHz - 460 kHz
FREQUECY[tex]_{IF[/tex] = 675 kHz
Therefore, the output of the IF amplifier consist of 675 kHz
Which of the following is NOT part of a car's drive train?
A axle
B rotor
C differential
D transmission
Answer:
B. rotor
Explanation:
The correct answer Is rotor because the others are part of a cars drivetrain
The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.
Which is not part of the drive train?A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.
Therefore, the correct answer is option B rotors.
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Engineers design products or processes to meet desired needs. Your desired need or goal (hopefully) is to graduate with your Bachelor of Science degree in engineering. But what is the process you need to apply to be successful in achieving this goal?
explain all the characteristics of computer
Determine the voltage which must be applied to a 1 k 2 resistor in order that a current of
10 mA may flow.
Answer:
The correct solution is "20 volt".
Explanation:
Given that:
Current,
I = 10 mA
or,
= [tex]10\times 10^{-3} \ A[/tex]
Resistance,
R = 2 K ohm
or,
= [tex]1\times 10^3 \ ohm[/tex]
Now,
The voltage will be:
⇒ [tex]V=IR[/tex]
By putting the values, we get
[tex]=10\times 10^{-3}\times 2\times 10^{3}[/tex]
[tex]= 20 \ volt[/tex]
You are the public relations director of a nonprofit hospital in a competitive market in a midsized city located in a metro area of 350,000 people. It is sweeps week for broadcast media. One of the stations is running a series on HIV/AIDS in the community. Recent segments have included those listed below.
The need for a confidential clinic
Homelessness related to HIV/AIDS
How persons living with HIV/AIDS suffer from being outcasts
How understanding has increased in some circles but prejudice remains in many.
An appealing case of a hemophiliac who acquired HIV/AIDS through transfusion at the university hospital has been reported in the series. The university and its teaching hospital are located elsewhere in the state.
At a staff meeting this morning, you learn that petitions are circulating in the community to request your hospital to convert its former nurses’ residence into a clinic and residential shelter for HIV/AIDS patients.
The hospital no longer maintains a school of nursing. The three-story brick building has been used for miscellaneous administrative purposes since nurses’ training was phased out. The residence is connected to the hospital by an elevated corridor, similar to a skywalk. It does not have facilities for food service or laundry. Nurses always ate at the hospital and the hospital handled their laundry. The residence and hospital are served by common systems for hot water, steam heat, ventilating and sewer. The building predates central air conditioning.
The human resources director reports that the business agent for kitchen and laundry workers, who are represented by a union, has already made informal contacts about this proposal, suggesting that grievances will result, at the very least, and a strike could ensue.
Your director of volunteers expresses concern about reaction of volunteers who now handle many peripheral duties.
Your physical plant supervisor, who lives in the neighborhood, says neighbors are already anxious over the possibility of HIV/AIDS patients in their vicinity.
Your hospital’s five-year Strategic Plan, which was recently updated, has no mention of developing an HIV/AIDS specialty.
Individual Assignment
For this assignment, you are to write a letter (using your own name) to the hospital CEO, John Dolman and the Board of Governors outlining your plan to resolve this issue.
Remember, that your letter will be made public to the other audiences and publics. In your recommendations, you should provide both long and short-term strategies that would solve this problem. To solve this problem you should note if additional resources or expenditures would be need to achieve the goals that you define.
When writing the letter you should provide a brief description of the problems, you have found. What are the facts about major issues? Identify facts about key players in the case, the business problem(s) and then rank order the critical issues. Consider relevant information and underlying assumptions. Finally provide your recommendations.
As you write your recommendations, think about the following;
How do the cultural values at the hospital relate to communication, technology, information flow and openness?
As the hospital goes forward, should it stick to the espoused culture or should it change? How would you recommend that they change?
How would you suggest that they resolve this disparity? In other words what should they do?
Provide specific suggestions that will help the organization as they go forward.
In the closing, highlight benefits of your recommendations. As a PR director, you need to be honest but tactful in your recommendations.
Remember
Carefully read the "case study" and type a letter that details your specific analysis and recommendations about how problem presented in case might be approached and solved.
This document should be formatted for reading ease, not a page of text. Make sure you use address it correctly, provide a date, headings, and bullet points as needed to make it easy to read.
Answer:
Following are the application to the given question:
Explanation:
To,
The Chief Operating Officer,
Maidan Hospitals, USA
02.10.2020
Subject: HIV/AIDS disease spreading control
Respected Mr Dolman,
The media as well as the individuals living in the area have indeed been notified again that the spread of Hiv / Aids must be regulated and managed immediately. A proper environment must be created artificially to heal this sickness and help stop it from spreading.
Its fight for the sickness ought to be a forward end to bring it through to the high levels instead of being fearful of disease. The disease also is seen in the hospital as it has grown via infusion so that the nursing staff must keep a check mostly on the issue frequently.
This position taken through distributing petitions in the community proposes that a small number of clinics must be established, as this could help to stop the disease from becoming spread. They have also seen that hospital nurses eat, sleep, washing dishes, etc. in such a single place.
To stop this same nursing staff that lives in the health facilities shortly, I suggest that perhaps it should be broken into combinations of 4 individuals and relocated to newly built hospitals, in which they can stay individually and retain hygiene, preserve the atmosphere, help educate, etc. Those who should be accountable. Each clinic and physicians visit every day to examine their clients.
Many patients might well be healed without great faith using these specific HIV/AIDS measures because all norms and laws are observed and mainly social distance is maintained of each patient. Its clinics, as well as the environment, must be sanitized but kept from outside so that the community members recover the confidence and trust of the health departments beforehand. More importantly, a regulatory body should be established that can monitor the operations properly, handle cash, etc. to run smoothly.
Here are a few of my suggestions and promise that it could be successful if it is kept & rigorously implemented.
Thank you and regards,
Joy Roy.
(PR. Director)
The floor of a light storage warehouse is made of 6-in.-thick cinder concrete. The floor is a slab having a length of 16 ft and width of 14 ft.
A. Determine the resultant force caused by the dead load.
B. Determine the resultant force caused by the live load.
Explanation:
6/12 = 0.5ft
length = 16
width = 14
The volume of cinder concrete = 0.5 * 16 *14 = 112
the resultant force that is caused by the dead load
density of cinder concrete * volume
density is assumed to be 108
dead load = 108 * 112 = 12096 lb
resultant force caused by the live load
liveload = 125lb/ft2
= 125 * 14 * 16
= 28000 lb
What possible scenarios may happen if you do the task without using PPE?
Resistance depends on which three properties of a wire?
Color and texture are not directly related to a wire’s resistance.
1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture
Answer:
2
Explanation:
From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.
For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is
This question is incomplete, the missing diagram is uploaded along this answer below;
Answer:
from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D
Explanation:
Given the data in the question;
We write the expression for the local Nusselt number for Laminar flow over the flat plate;
Nu = [tex]C[/tex][tex](Re_x)^{0.5[/tex] [tex](Pr)^{1/3[/tex]
Nu = [tex]C(\frac{Vx}{v})^{0.5}[/tex] [tex](Pr)^{1/3[/tex]
[tex]\frac{h_xx}{k}[/tex] = [tex]C(\frac{V}{v})^{0.5}[/tex] [tex](Pr)^{1/3[/tex] [tex](x)^{0.5[/tex]
[tex]h_x[/tex] = [tex]\frac{1}{x^{1/2}}[/tex]
Next we write down the expression for the local heat flux from the plate with uniform surface temperature;
q = [tex]h_xA([/tex] T[tex]_s[/tex] - T∞ )
q ∝ [tex]h_x[/tex]
∴
q ∝ [tex]\frac{1}{x^{1/2}}[/tex]
The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.
Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D
State two factors that shows that light travels in a straight line
Explanation:
Light travels in straight lines
Once light has been produced, it will keep travelling in a straight line until it hits something else. Shadows are evidence of light travelling in straight lines. An object blocks light so that it can't reach the surface where we see the shadow.
Answer:
The two factors are Air and Object
What is the per capita GDP of China? Be sure to indicate the calendar year that this information represents.
The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.
What is GDP?GDP termed Gross Domestic Product, has been evaluated with the value producing the economy of the region with the values added with the used products formed to be the less of the economy produced. It has been termed as the measure of the income of a region and not the wealth.
The per capita GDP has been the total income earned by a person in a region during a specified period of time. The calculation has been made by dividing the total gross income of the region by the total population.
China has been the world's most populous country in the East Asian region. It has been found that the per capita GDP of China is low because of its large population. In the calendar year 2021, the per capita GDP of China was 12,359 U.S. dollars.
Learn more about the GDP, here:
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What is An ampere is
Answer:
the SI base unit of electrical current.
Answer:
An ampere is the SI base unit of electrical current
A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?
Answer:
The block will float with its axis vertical.
Explanation:
For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.
So, weight of cylindrical block, W = specific weight × volume
specific weight = 7500 N/m³
volume = πd²h/4 where d = diameter of block and h = height of block
volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³
W = 7500 N/m³ × 0.7854 m³ = 5890.5 N
Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³
Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced = 9800 N/m³ × 0.7854 m³ = 7696.92 N
Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.
A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?
Answer:
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
Explanation:
From the question we are told that:
Diameter [tex]d=2m[/tex]
Length [tex]l=2.5m[/tex]
Weight [tex]W=22kN[/tex]
Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]
Generally the equation for weight of cylinder is mathematically given by
Weight of cylinder = buoyancy Force
[tex]W=(pwg)Vd[/tex]
Where
[tex]V_d=\pi/4(d)^2y[/tex]
Therefore
[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]
Therefore
Center of Bouyance B
[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]
Center of Gravity
[tex]G=\frac{I.B}{2}=2.6m[/tex]
Generally the equation for\BM is mathematically given by
[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]
Therefore
[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]
Therefore
[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]
Therefore
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft above that of the first. the total length of pipes required is 1000 ft. All pipes are 12 in. in diameter and are made of galvanized iron with relative roughness equal to 0.0005 (you may assume fully-rough flow). the pump performance curves suggest that the H-Q relationship is of the form: H_pump=665-0.051Q^2 (Q in ft) the expected flow rate the brake horsepower required to drive the pump (assume an efficiency of 78%). the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).
Answer:
a) the expected flow rate is 31.4 ft³/s
b) the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation is -8.4 ft
Explanation:
Given the data in the question;
free surface elevation = 200 ft
total length of pipe required = 1000 ft
diameter = 12 inch
Iron with relative roughness ( k/D ) = 0.0005
H[tex]_{pump[/tex] = 665-0.051Q² [Qinft ]
a) the expected flow rate
given that;
k/D = 0.0005
k/2R = 0.0005
R/k = 1000
now, we determine the friction factor;
1/√f = 2log₁₀( R/k ) + 1.74
we substitute
1/√f = 2log₁₀( 1000 ) + 1.74
1/√f = 6 + 1.74
1/√f = 7.74
√f = 1/7.74
√f = 0.1291989
f = (0.1291989)²
f = 0.01669
Now, Using Bernoulli theorem between two reservoirs;
(p/ρq)₁ + (v²/2g)₁ + z₁ + H[tex]_p[/tex] = (p/ρq)₂ + (v²/2g)₂ + z₂ + h[tex]_L[/tex]
so
0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²
665-0.051Q² = 200 + flQ²/2gdA²
665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )
665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]
665 - 200 - 0.051Q² = 0.420138Q²
665 - 200 = 0.420138Q² + 0.051Q²
465 = 0.471138Q²
Q² = 465 / 0.471138
Q² = 986.97196
Q = √986.97196
Q = 31.4 ft³/s
Therefore, the expected flow rate is 31.4 ft³/s
b) the brake horsepower required to drive the pump (assume an efficiency of 78%).
we know that;
P = ρgH[tex]_p[/tex]Q / η
where; H[tex]_p[/tex] = 665 - 0.051(986.97196) = 614.7
we substitute;
P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )
P = 1204804.6236 / 429
P = 2808.4 bhp
Therefore, the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).
NPSH = ([tex]P_{atom[/tex] / ρg) - h[tex]_s[/tex] - ( P[tex]_v[/tex] / ρg )
we substitute
25 = ( 2116 / 62.42 ) - h[tex]_s[/tex] - ( 30 / 62.42 )
h[tex]_s[/tex] = 8.4 ft
Therefore, the location of pump inlet to avoid cavitation is -8.4 ft
The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension
Answer:
At the threads due to shear.
Explanation:
Given :
The allowable tensile stress = 207 MPa
The allowable shear stress = 103 MPa
If a tensile force is applied, the maximum shear stress occurs at the threads of the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.
The portion of the bolt having threads has the minimum cross sectional area.
So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.
. (5 points) Based on the Reynolds number for each of the following objects, identify each flow as either inertial or viscous force dominant and in which cases the flows are laminar or turbulent. Flow Re A. A bee larva in honey 0.2 B. A ball tossed on Mars 500 C. An elderly driver on i95 1,2000,000
Answer:
Part A
The flow of the bee lava is inertia dominant and laminar
Part B
The flow of a baseball in Mars is inertia dominated and the flow is laminar
Part C
The flow of an elderly on i95 is viscous force dominated and turbulent
Explanation:
The Reynold's number when inertia is dominant is low, and the flow is laminar
When viscous force is dominant, the Reynold's number is high, and we have turbulent flow
The Reynold's number of laminar flow is Re < 2,000
The Reynold's of unstable or intermediate flow is 2,000 < Re < 4,000
The Reynold's number of turbulent flow is Re > 4,000
Therefore, we have;
Part A
The Reynold's number of a bee lava, Re = 0.2, therefore, the flow is inertia and laminar
Part B
The Reynold's number of the ball in Mars is Re = 500, therefore, given that the Reynold's number is less than 2,000, the inertia is dominant, and the flow is laminar flow
Part C
The Reynold's number of the driver on i95 = 1,200,000 which is larger than 4,000, therefore, the flow is viscous force dominated and the flow is turbulent.
A cylindrical rod of copper originally 16.0 mm in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked yield strength in excess of 250 MPa and a ductility of at least 12% EL are desired. Furthermore, the final diameter must be 11.3 mm. Explain how this may be accomplished:
(1) show why it cannotbe realized by single drawing and (2) suggest proper procedures to fulfill the requirements.You will need to use following figuresto solve the problem.
Solution :
Given data:
Diameter of the copper cylindrical rod = 16 mm
Yield strength = 250 MPa
Calculating the percent cold work
[tex]$\text{Percentage Cold Work} = \frac{\pi\left(\frac{d_0}{2}\right)^2-\pi\left(\frac{d_d}{2}\right)^2}{\pi\left(\frac{d_0}{2}}\right)^2} \times 100$[/tex]
[tex]$ = \frac{\pi\left(\frac{16}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{16}{2}}\right)^2} \times 100$[/tex]
= 50% CW
Therefore, at [tex]50\% \ CW[/tex], the yield strength of copper will be of the order of 330 MPa.
The ductility will be 4% elongation (EL).
Rather than performing drawing in single operation, we draw some of the fraction of total deformation, and then the anneal them to recrystallize and also finally w do cold work on the material for the second time to achieve its final diameter, ductility and yield strength.
[tex]21\% \ CW[/tex]is required for a yield strength of [tex]250 \ MPa[/tex]. Similarly, a maximum of [tex]23\% \ CW[/tex] is required for[tex]12\% \ EL[/tex].
The average of the two values is [tex]22\% \ CW[/tex]. To achieve both the [tex]\text{specified yield strength and ductility,}[/tex] the copper should be deformed to[tex]22\% \ CW[/tex]. The[tex]\text{ final diameter}[/tex] after the first drawing and the initial diameter for the second drawing is [tex]$d_0'$[/tex] , then
[tex]$22\% \ CW = \frac{\pi\left(\frac{d_0'}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{d_0'}{2}}\right)^2} \times 100$[/tex]
[tex]$d_0'=\frac{11.3}{\sqrt{1-\frac{22\% \ CW}{100}}}$[/tex]
[tex]$d_0'=12.8 \ mm$[/tex]
Vince is trying to figure out the volume of two mystery matters. The volume of one of the substances needs to be measured by submerging it in water and the other needs to be measured using a graduated cylinder. Based on the properties of two mystery matters, what are they? (2 points)
Group of answer choices
A rock and orange juice
Helium and a golf ball
Lemonade and milk
Orange juice and helium
Answer:
Rock and orange juice
Explanation:
The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.
If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding
Answer:
5,465.4165939453
Explanation:
formula
A=P(1+r/n)^n(t)
p=1000
r=0.09
n=4
t=5
A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.
Answer:
The answer is "22.501,-22.899"
Explanation:
Just as in the previous problems find the angle the velocity makes with the x-axis and radius of curvature.
[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]
For the radius of curvature, we can use the expression from the last two problems, but first express the position and derivatives as y(x).
[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]
The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)
[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]
The statement that is correct about the relation between the velocity boundary layer and heat transfer for flow over a flat plate that is uniform in temperature is
Answer: the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.
Explanation:
The correct statement about the relation between the velocity boundary layer and heat transfer for flow over a flat plate that is uniform in temperature is that the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.
It should be noted that the heat goes in a streamline direction in a laminar flow, thereby the molecules less collide with each other. On the other hand, the direction is zig zag in a turbulent heat flux and this will bring about more collision of the molecules which leads to a rise in the heat flux.
can help me with this circuit question?
the last word is (cutoff)
Answer:
The answer is "[tex]25\times 10^{-9}[/tex]".
Explanation:
Using formula:
[tex]f_c=\frac{1}{2\pi RC}\\\\w_c= 4 \frac{krad}{sec}\\\\w_c=2\pi fc\\\\R=w\\\\c=\frac{1}{w_c\ R}\\\\[/tex]
[tex]=\frac{1}{4 \times 10^3 \times 10\times 10^3}\\\\=\frac{1}{40 \times 10^6 }\\\\=0.025 \times 10^{-6 }\\\\=25\times 10^{-9}[/tex]
Water enters and leaves a pump in pipelines of the same diameter and approximately the same elevation. If the pressure on the inlet side of the pump is 30 kPa and a pressure of 500 kPa is desired for the water leaving the pump, what is the head that must be added by the pump
Answer:
The head added by the pump is approximately 51.8 meters
Explanation:
The given parameters of the water are;
The initial diameter of the pump ≈ The final diameter of the pump
The inlet side pressure, p₁ = 30 kPa
The intended outlet side pressure, p₂ = 500 kPa
The power of the pump, P = ρ·g·Q·H
Where, the pressure added by the pump, Δp = p₂ - p₁ = ρ·g·H
ρ = The density of the water ≈ 997 kg/m³
g = The acceleration due to gravity ≈ 9.81 m/s²
H = The head added by the pump
Therefore, we have;
500 kPa - 30 kPa = 997 kg/m³ × 9.81 m/s² × H
H = (500 kPa - 30 kPa)/(997 kg/m³ × 9.1 m/s²) ≈ 51.8 m
The head added by the pump, H ≈ 51.8 meters.
Things to be done before isolation
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2. Derive the unit hydrograph by the inverse procedure. Assume a constant baseflow of 550 cfs.
Hour Day 1 Day 2 Day 3 Day 4
Midnight 550 5,000 19,000 550
6 am 600 4,000 1400
Noon 9000 3000 1000
6 pm 6600 2500 750
Answer:
33.56 ft^3/sec.in
Explanation:
Duration = 6 hours
drainage area = 185 mi^2
constant baseflow = 550 cfs
Derive the unit hydrograph using the inverse procedure
first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below
Vdrh = sum of drh * duration
= 29700 * 6 hours ( 216000 secs )
= 641,520,000 ft^3.
next step : Calculate the volume of runoff in equivalent depth
Vdrh / Area = 641,520,000 / 185 mi^2
= 1.49 in
Finally derive the unit hydrograph
Unit of hydrograph = drh / volume of runoff in equivalent depth
= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in
Hi, can anyone draw me an isometric image of this shape?
For a pipe system with a pump (pumping uphill), the change in elevation is 400 feet and the total head loss is 408.5 feet. Assuming gage pressure at the entrance and exit and no difference in velocity between the entrance and exit, determine the total energy transferred to the water. Estimate the required power input if the pump efficie
Answer:
Explanation:
From the given information;
There is no change or any difference in velocity in between the inlet and the outlet.
Therefore by using Bernoulli's equation, we have:
[tex]\dfrac{V_1^2}{2g}+ \dfrac{P_1}{\gamma}+ z_1 + Epump= \dfrac{V_2^2}{2g}+ \dfrac{P_2}{\gamma}+ z_2+ H_L[/tex]
By dividing like terms on both sides, the equation is reduced to:
[tex]z_1 + E_{pump} = z_2+H_L \\ \\ E_{pump} =(z_2-z_1)+H_L[/tex]
where;
[tex]\Delta z = 400[/tex]
[tex]\Delta z = z_2-z_1[/tex]
[tex]\text{total head loss}= 408.5[/tex]
[tex]E_{pump} =(400)+408.5[/tex]
[tex]E_{pump} = 808.5 \ ft[/tex]
The required power input can be determined by using the formula:
[tex]P= \dfrac{\gamma_wQH_{pump}}{\eta}[/tex]
Assuming the missing pump efficiency = 70% and the flow rate Q= 1.34
Then:
[tex]P= \dfrac{62.40\times 1.34 \times 808.5}{0.7}[/tex]
[tex]P = \dfrac{96576.48 \ ft.lb/s}{550\dfrac{ ft*lb/s}{hp}}[/tex]
P = 175.594 hp
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium.If the left and bottom edges of the plate are fixed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate.
Answer:
[tex]X_t=2.17391304*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]e_t=0.0026[/tex]
[tex]e_r=0.0035[/tex]
Explanation:
From the question we are told that:
Dimension [tex]12*12[/tex]
Thickness [tex]l_t=5mm=5*10^-3[/tex]
Normal tensile force on top side [tex]F_t= 15kN[/tex]
Normal tensile force on right side [tex]F_r= 20kN[/tex]
Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]
Generally the equation for Normal Strain X is mathematically given by
[tex]X=\frac{Force}{Area*E}[/tex]
Therefore
For Top
[tex]X_t=\frac{Force_t}{Area*E}[/tex]
Where
[tex]Area=L*B*T[/tex]
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]
[tex]X_t=2.17391304*10^{-4}[/tex]
For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]
Where
Area=L*B*T
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
Generally the equation for elongation is mathematically given by
[tex]e=strain *12[/tex]
For top
[tex]e_t=2.17391304*10^{-4}*12[/tex]
[tex]e_t=0.0026[/tex]
For Right
[tex]e_r=2.89855072*10^{-4} *12[/tex]
[tex]e_r=0.0035[/tex]
An aggregate blend consists of 65% of aggregate A and 35% of aggregate B. The bulk specific gravities of aggregate A and B are 2.45 and 3.25, respectively. What is the bulk specific gravity of the blend?
a) 2.45
b) 2.68
c) 2.73
d) 2.92
Answer:
2.68
Explanation:
Percentage by Mass of each Aggregate :
Pa = 65% ; Pb = 35%;
Bulk Specific gravity of each aggregate :
Ga = 2.45 ; Gb = 3.25
Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)
Gsb = (65 + 35) / (65/2.45 + 35/3.25)
Gsb = (65 + 35) / 37.299843
Gsb = 100 / 37.299843
Gsb = 2.68