An aluminum cylinder bar ( 70 GPa E m = ) is instrumented with strain gauges and is subject to a tensile force of 5 kN. The diameter of the bar is 10 cm. The Poisson’s ratio of the bar is 0.33. A Wheatstone bridge is constructed to measure the axial strain. Gauge 1 measures the axial strain and gauge 2 measures the lateral strain.
Answers
Find the complete solution in the given attachments.
Note: The complete Question is attached in the first attachment as the provided question was incomplete
Related Questions
A liquid flows down an inclined plane surface in a steady, full developed laminar film of thickness h. Simplify the Navier-Stokes equations to model this flow field. Obtain expressions for the liquid velocity profile, the shear stress distribution, the volumetric flow rate, and the average velocity. Calculate the volumetric flow rate in a film of water h = 1 mm thick, flowing on a surface b = 1 m wide, inclined at 15 degrees to the horizontal.
Answers
Answer:
Explanation:
The solutions to these questions can be seen in the following screenshots from a solution manual;
11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine how the valve will operate. Both the model and prototype will use water as the fluid. The model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve). If the prototype flow rate is to be 700 ft3 /s, determine the model flow rate. Use Reynolds scaling for the velocity.
Answers
Answer:
7.94 ft^3/ s.
Explanation:
So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.
Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.
Therefore; kp/ks = 1/6.
Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.
Reynolds scaling==> Hp/ 700 = (1/6)^2.5.
= 7.94 ft^3/ s
A pump operating at steady state receives saturated liquid water at 50°C with a mass flow rate of 30 kg/s. The pressure of the water at the pump exit is 1.5 MPa. If the pump operates with negligible internal irreversibilities and negligible changes in kinetic and potential energy, determine power required in kW. (Moran, 01/2018, p. P-67) Moran, M. J., Shapiro, H. N., Boettner, D. D., Bailey, M. B. (01/2018). Fundamentals of Engineering Thermodynamics, Enhanced eText, 9th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781119391388 Always check citation for accuracy before use.
Answers
Given Information:
Temperature = T₁ = 50 °C
Mass flow rate = m = 30 kg/s
Exit Pressure = P₂ = 1.5 MPa = 1500 kPa
Required Information:
Power = P = ?
Answer:
Power = 45.16 kW
Explanation:
The power of the pump can be found using,
P = m*W
Where m is the mass flow rate and W is the work done by pump.
Work done is given by
W = vf*(P₂ - P₁)
Where vf is the specific volume and its value is found from the saturated water temperature table.
at T = 50 °C
vf = 0.001012 m³/kg
P₁ = 12.352 kPa
P = m*W
P = m*vf*(P₂ - P₁)
P = 30*0.001012*(1500 - 12.352)
P = 45.16 kW
Look at the home page of the Internet Society (www.internetsociety.org) and read about one of the designers of the original ARPANET—Larry Roberts, Leonard Kleinrock, Vinton Cerf, Robert Kahn, John Postel, or others. Learn about the early days of networking and the contributions that these individuals made to the ultimate development of the Internet. The home page of the Internet Society has links to many other places that provide a wealth of fascinating information about networks in general and the Internet and the web in particular.
Answers
Answer:
ARPANET is the direct precedent for the Internet, a network that became operational in October 1969 after several years of planning.
Its promoter was DARPA (Defense Advanced Research Projects Agency), a US government agency, dependent on the Department of Defense of that country, which still exists.
Originally, it connected research centers and academic centers to facilitate the exchange of information between them in order to promote research. Yes, being an undertaking of the Department of Defense, it is understood that weapons research also entered into this exchange of information.
It is also explained, without being without foundation, that the design of ARPANET was carried out thinking that it could withstand a nuclear attack by the USSR and, hence, probably the great resistance that the network of networks has shown in the face of major disasters and attacks.
It was the first network in which a packet communication protocol was put into use that did not require central computers, but rather was - as the current Internet is - totally decentralized.
Explanation:
Below I present as a summary some of the most relevant aspects exposed on the requested website about the origin and authors of ARPANET:
1. Licklider from MIT in August 1962 thinking about the concept of a "Galactic Network". He envisioned a set of globally interconnected computers through which everyone could quickly access data and programs from anywhere. In spirit, the concept was very much like today's Internet. He became the first head of the computer research program at DARPA, and from October 1962. While at DARPA he convinced his successors at DARPA, Ivan Sutherland, Bob Taylor and MIT researcher Lawrence G. Roberts, of the importance of this network concept.
2.Leonard Kleinrock of MIT published the first article on packet-switching theory in July 1961 and the first book on the subject in 1964. Kleinrock convinced Roberts of the theoretical feasibility of communications using packets rather than circuits, That was an important step on the road to computer networking. The other key step was to get the computers to talk together. To explore this, in 1965, working with Thomas Merrill, Roberts connected the TX-2 computer in Mass. To the Q-32 in California with a low-speed phone line creating the first wide-area (albeit small) computer network built . The result of this experiment was the understanding that timeshare computers could work well together, running programs and retrieving data as needed on the remote machine, but that the circuitry switching system of the phone was totally unsuitable for the job. Kleinrock's conviction of the need to change packages was confirmed.
3.In late 1966 Roberts went to DARPA to develop the concept of a computer network and quickly developed his plan for "ARPANET", and published it in 1967. At the conference where he presented the document, there was also a document on a concept of UK packet network by Donald Davies and Roger Scantlebury of NPL. Scantlebury told Roberts about NPL's work, as well as that of Paul Baran and others at RAND. The RAND group had written a document on packet switched networks for secure voice in the military in 1964. It happened that work at MIT (1961-1967), in RAND (1962-1965) and in NPL (1964-1967) all they proceeded in parallel without any of the investigators knowing about the other work. The word "packet" was adopted from the work in NPL and the proposed line speed to be used in the ARPANET design was updated from 2.4 kbps to 50 kbps.
Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has been determined that fracture results at a stress of 250 MPa when the maximum (or critical) internal crack length is 2.0 mm. For this same component and alloy, will fracture occur at a stress level of 325 MPa when the maximum internal crack length is 1.0 mm? Why or why not?;
Answers
Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).
Explanation:
in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.
let us begin,
from the question we have that;
stress = 325 MPa
fracture toughness (KIc) = 35 MPa√m
the max internal crack length = 1.0 m
using the formula;
Y = KIc/σ√(πα) ---------------(1)
solving for Y we have;
Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)
Y = 2.50
so to calculate the fracture roughness;
Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m
Kc = 32.2 MPa√m
From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.
cheers i hope this helps!!!!
Once an engineer has a sketch, the next step is to build the product.
True or false
Answers
Answer:
False
Explanation:
cost Evaluation
assembly of equipment and materials used
The given statement is false.
The following information should be considered:
In the case when the engineer should have sketch so the next step is to evaluate the cost, the assembly of an equipment and material should be used. No product should be build up.Learn more; brainly.com/question/16911495
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a pressure ratio of 8, and the temperature of the gases at the turbine inlet is 2200 F. Utilizing the air-standard assumptions, determine (a) the temperature and pressure of the gases at every point of the cycle, (b) the velocity of the gases at the nozzle exit
Answers
Answer:
Pressure = 115.6 psia
Explanation:
Given:
v=800ft/s
Air temperature = 10 psia
Air pressure = 20F
Compression pressure ratio = 8
temperature at turbine inlet = 2200F
Conversion:
1 Btu =775.5 ft lbf, [tex]g_{c}[/tex] = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²
Air standard assumptions:
[tex]c_{p}[/tex]= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R
k= 1.4
Energy balance:
[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\[/tex]
As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible
hence [tex]v_{a} ^{2} = 0[/tex]
[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }[/tex]
[tex]T_{1}[/tex] = 20+460 = 480°R
[tex]T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}[/tex]= 533.25°R
Pressure at the inlet of compressor at isentropic condition
[tex]P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}[/tex]
[tex]P_{a}[/tex] = [tex](10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}[/tex]= 14.45 psia
[tex]P_{2}= 8P_{a} = 8(14.45) = 115.6 psia[/tex]
Answer:
a) The temperature and pressure of the gases at every point of the cycle are
T = 38.23 K
P = 2.91 kpa
Respectively
b) The velocity V of the gasses at the nozzle exit = 3590 m/s
Explanation: Please find the attached files for the solutions
2. A fluid at 14.7 psi (lb-f per square inch) with kinematic viscosity (????????) 1.8 x10-4 ft2/sec and density(????????) 0.076 lb/ft3 enters a 10 inch diameter pipe with a uniform velocity and a Reynolds number 1000. Determine the decrease in pressure going from the entrance to 100 inch downstream the entrance. The entrance length, LLee is given by, LLee = 0.0288DD. RRRRDD. (Hint: calculate the pressure drop separately between 1 and 2 and between 2 and 3 because the region 1-2 shows developing flow and region 2-3 shows developed flow) region. The flow becomes fully developed after the entrance length, LLee. The thickness of the boundary layer 1 −� given as ????????(xx) = 5.0xxRRRR 2. Show that entrance length for this flow can be expressed LL = 0.01DD. RRRR . xxeeDD
Answers
Answer:
See explaination
Explanation:
We are going to define Pressure drop as the difference in total pressure between two points of a fluid carrying network. A pressure drop usually occurs when frictional forces, caused by the resistance to flow, act on a fluid as it flows through the tube.
See attachment for the step by step solution of the given problem.
Computer-controlled instrument panel dimming is being discussed. Technician A says the body computer dims the illumination lamps by varying resistance through a rheostat that is wired in series to the lights. Technician B says the body computer can use inputs from the panel dimming control and photo cell to determine the illumination level of the instrument panel lights on certain systems.Who is correct?A. A onlyB. B only.C. Both A and B.D. Neither A nor B.
Answers
Answer:
Answer : C ( Both A&B)
Explanation:
The level of illumination in the instrument panel lights will be determined by the panel dimming control and photocell. This panel dimming control consist of a potentiometer. The diameter positions existing in the diameter control acts like variable resistor. Based on its voltage drop BCM (Body Control Module) selects the intensity level by comparing the signal captured from photocell.
Another type is body control module receives the signal from head light rheostat which will be sent to instrument cluster. The instrument cluster controls the lamp intensity. Therefore, the statements said by both the technicians are correct.
Then, the correct option is C
what is the most important source of error in the Brinell test
Answers
Answer:
The measurement of the indentation.
Explanation:
Due to disparities in operators making the measurements, the results will vary even under perfect conditions. Less than perfect conditions can cause the variation to increase greatly.
_____________ is when a member is being pulled apart.
Group of answer choices
tension
shear
torsion
compression
Answers
Por que razão os engenheiros biomédicos devem ter familiaridade com a composição, as propriedades e o processamento de materiais?
Answers
Answer: Material selection
Explanation: As a biological engineer you need to be conversant with the composition, properties and processing of materials. This is important because when you design equipments and machines using your 3D model. During the fabrication process your knowledge of the composition of materials e.g 5% iron,
Their properties e.g ductility, non corrosive e.t.c and the way each and every materials being used in fabrication were processed plays a big role on how the 3D model would appear in real word. It’s durability and performance, so as a biological engineer this knowledge helps you make the right decision in materials selection.
a speed reducer has 20 deg full depth teeth and consisct of a 20 tooth steel spur pinion driving a 50 tooth cast iron gear. the horse power transmitted is 12 at a pinion speed of 1200 rev/min. for a dimateral pitch of 8 teeth/ in and a face width of 1.5 in find the contact stress. the yuong's modulus of the steel is 30E6 and young's modulus of cast iron is 15E6. use a poisson's ratio of 0.3 for both materials. use equation 14-14 to calc the contact stress in the gear only.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has emissivity of 0.90 and is maintained at 950 K. The side surface is black and is maintained at 450 K. Heat is supplied from the base surface at a rate of 340 kW. Determine the emissivity of the top surface and the net rates of heat transfer between the top and bottom surfaces, and between the bottom and side surfaces. Answers: 0.44; 54.4 kW; 285.6 kW
Answers
Answer:
Check the explanation
Explanation:
Assumptions.
1. The surfaces are diffuse, may and opaque
2. steady operating conditions exist
3. Heat transfer from and to the surfaces is only due to Radiation
Consider the base surface to be surface 2 the top surface to be surface and the side surfaces to surface 3 1. cubical furnace can be considered to be three-surface enclosure. the areas and black body emissive powers of surfaces can be calculated as seen in the attached images below.
A current I flows in the inner conductor of an infinitely long coaxial line and returns via the outer conductor. The radius of the inner conductor is a, and the inner and outer radii of the outer conductor is b and c, respectively. Find the magnetic flux density B for all regions and plot |B| versus r.
Answers
Answer:
See explaination
Explanation:
By definition, we can say that Magnetic flux density is defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction. An example of magnetic flux density is a measurement taken in teslas.
Please kindly check attachment for the step by step solution of the given problem.
The following two questions refer to the circuit below. Consider a non-ideal op amp where the output can saturate. The open loop gain A = 2 x 10^{4}10 4 , where v_{o}v o =−Av_{s}v s . The positive supply voltage for the op-amp is +V_S = 15+V S =15V. The negative supply voltage for the op-amp is -V_S = -10−V S =−10V. What is the most positive value v_{s}v s can take before the amplifier saturates? Express your answer in mV and omit units from your answer.
Answers
Answer:
The most positive is value of Vs is 0.5mV
When we utilize a visualization on paper/screen, that visualization is limited to exploring: Group of answer choices Relationship among an infinite number of variables Relationship between a dependent and an independent variable As many variables as we can coherently communicate in 2 dimensions Relationship between two variables, x
Answers
Answer:
As many variables as we can coherently communicate in 2 dimensions
Explanation:
Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.
When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).
Now, suppose that you have a balanced stereo signal in which the left and right channels have the same voltage amplitude, 500 mVpp. This time, however, you want to be able to mix these two channels into a single inverted output while independently varying the gain of the two channels. Design and build an op amp circuit with potentiometers so that you can independently vary the gain of the left and right channels. Choose resistors so that the overall output of your circuit ranges between 0.1Vpp (when both channels are set to minimum gain) and 20Vpp (when both channels are set to maximum gain).
Answers
Answer:
R₁ = 32kΩ
Explanation:
See attached image
Continuous and aligned fiber-reinforced composite with cross-sectional area of 300 mm2 (0.47 in.2) is subjected to a longitudinal load of 46900 N (10500 lbf). Assume Vf = 0.3, Vm = 0.7, Ef = 131 GPa and Em = 2.4 GPa. (a) Calculate the fiber-matrix load ratio. (b) Calculate the actual load carried by fiber phase. (c) Calculate the actual load carried by matrix phase. (d) Compute the magnitude of the stress on the fiber phase. (e) Compute the magnitude of the stress on the matrix phase. (f) What strain is expected by the composite?
Answers
Answer:
a) 23.39
b) 44977.08 N
c) 1922.92N
d) 454.31 MPa
e) 8.32 MPa
f) [tex] 3.47*10^-^3 [/tex]
Explanation:
a) fiber-matrix load ratio:
Let's use the formula :
[tex] \frac{F_f}{F_m} = \frac{V_f E_f}{V_m E_m}[/tex]
[tex] = \frac{0.3 * 131 GPa}{0.7 * 2.4 GPa} = 23.39 [/tex]
b & c)
Total load is given as:
Fc = Ff + Fm
46900 = Fm(23.39) + Fm
46900 = 24.39 Fm
Actual load carried by matrix=
[tex]F_m = \frac{46900}{24.39}[/tex]
= 1922.92N=> answer for option c
Actual load carried by fiber, Ff:
Ff = 46900 - 1922.92
Ff = 44977.08 N => answer option b
d)
Let's find area of fiber, A_f.
[tex] A_f = V_f * A_c[/tex]
Ac = Cross sectional area =300mm²
= 0.3 * 300 = 99 mm²
Area of matrix=
[tex] A_m = V_m * A_c[/tex]
= 0.7 * 300 = 231 mm²
Magnitude of the stress on the fiber phase:
[tex] \sigma _f= \frac{F_f}{A_f} [/tex]
[tex] \sigma _f= \frac{44977.08}{99} = 454.31 MPa [/tex]
e) Magnitude of the stress on the matrix phase.
[tex] \sigma _m = \frac{F_m}{A_m} [/tex]
[tex] \sigma _m = \frac{1922.92}{231} = 8.32 MPa[/tex]
f) Strain in fiber = [tex] \frac{\sigma _f}{E_f} [/tex]
[tex]= \frac{454.31*10^6}{131*10^9} = 3.47*10^-^3[/tex]
Strain in matrix = [tex] \frac{\sigma _m}{E_m} [/tex]
[tex]= \frac{8.32*10^6}{2.4*10^9} = 3.47*10^-^3[/tex]
Composite strain = [tex] (E_f *V_f) + (E_m * V_m) [/tex]
[tex] (3.47*10^-^3 * 0.3) + (3.47*10^-^3 * 0.7) = 3.47*10^-^3 [/tex]
A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is estimated to be 3,300 kJ/h and the heat rejection in the condenser is 4,800 kJ/h. If the refrigeration cycle was as efficient as the Carnot cycle, how much power (in kW) would be required to remove 3,300 kJ/h heat from the cooled space?
Answers
Answer: P = 0.416 kW
Explanation:
taken a step by step process to solving this problem.
we have that from the question;
the amount of heat rejected Qn = 4800 kJ/h
the cooling effect is Ql = 3300 kJ/h
Applying the first law of thermodynamics for this system gives us
Шnet = Qn -Ql
Шnet = 4800 - 3300 = 1500 kJ/h
Next we would calculate the coefficient of performance of the refrigerator;
COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2
COPr = 2.2
The Power as required gives;
P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416
P = 0.416 kW
cheers i hope this helps!!!!1
There is a metal plate with thickness, length, width, thermal conductivity, density, and specific heat of 10mm, 2 m, 1m, 200 W/m·°C, 2500 kg/m3, and 900 J/kg.K, respectively. It has an initial uniform temperature of 22°C and is being heated in an oven for 2 minutes. Air in the oven is maintained at 900 C with convection heat transfer coefficient of 190 W/m2.K. Determine the temperature of the plate when it is removed from the oven.\
Answers
Answer:
788°C
Explanation:
Metal palte details
Thickness = t=10 mm =0.01 m.
Thermal conductivity= k =200 W/mC
Heat capacity C=900 J/kg-k.
Initial temp Ti =22 C.
Time for which it has been kept in oven =t=2 min =120 sec.
Air properties.
Temp of air in oven =Ta =900 C
Heat transfer coefficient =h =190 W/m-k.
Temp of block after 2 min= T =?
In problem, it is not given block dimension. Let us assume that Block is 1 m in legth and 1 m in width.
Volume of block =V= 1*1*0.01 =0.01 m3 .(We know thckness =0.01 m).
Surface area of block = A=2(1*1+1*0.01+0.01*1)=2.04 m2.
Biot number for this problem is 0.004656 which is less than 0.1. So Lumped capacitance method is applicable to this problem, according to formulae,
T - Tair /Ti - Tair =e(-hAt/density* C* V) .
T - 900/(22-900) = e(-190*2.04*120 /2500*900*0.01) .
T - 900/(22-900) =e-2.0672 .
T -900 = - 110.09
Temp of block after 2 min =788 °C.
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 1.1. If, after 129 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?
Answers
Answer:
3.305 * 10 ^ ⁻4
Explanation:
Solution
Given:
We calculate the value of k for which is the dependent variable in Avrami equation
y = 1 - exp (-kt^n)
exp (-kt^n) =1-y
-kt^n = ln (1-y)
so,
k =ln(1-y)/t^n
Now,
we substitute 1.1 for n, 0.50 for y, and 129 s for t
k = ln (1-0.50)/129^1.1
k= 3.305 * 10 ^ ⁻4
You want to amplify a bio-potential signal that varies between 2.5 V and 2.6 V. Design an amplifier circuit for this signal such that the output spans 0 V to +10 V. The signal cannot be inverted. You can use any number of op amps and any number of resistors (with any values). But you can use only one +10 V DC voltage source (for powering the op amps as well as for any other needs). Clearly draw the complete circuit and show all component values.
Answers
Answer:
See attachment
Explanation:
Gain= Vo/Vin
If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7
Using above value of gain, let's design non-inverting op-amp configuration
Gain= 1+Rf/Rin
3.7= 1= Rf/Rin
2.7= Rf/Rin
If Rin=100Ω then Rf= 270Ω
A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 10 in2 and that approximately 2.6 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. (convert the time from hours to years). The density of steel is 7.9 g/cm3. K
Answers
The time of submersion of the steel plate in years is; 10 years
Corrosion Penetration Rate Calculation
We are given;
Corrosion penetration rate; CPR = 200 mpyWeight of the corroded portion of plate; W = 2.6 kg = 2.6 × 10⁶ mgExposed surface area; A = 10 in²Density of the metal plate; ρ = 7.9 g/cm³The time of submersion of the steel plate is given by the formula;
t = KW/(ρA*CPR)
Now K is a constant and is equal to 534 provided CPR is in mpy and
A is in square inches.
Thus;
t = (534 * 2.6 × 10⁶)/(7.9 * 10 * 200)
t = 8.8 × 10⁴ hours
Now, 24 hours makes one day and there are 365 days in a year. Thus;
number of hours in a year = 24 * 365
Thus;
t in years = (8.8 × 10⁴)/(24 * 365)
t ≈ 10 years
Read more about corrosion at; https://brainly.com/question/5168322
A 5% upgrade on a six-lane freeway (three lanes in each direction) is 1.25 mi long. On thissegment of freeway, there is 3% SUTs and 7% TTs, and the peak hour factor is 0.9. The lanes are 12ft wide, there is no lateral obstructions within 6ft from the roadway, and the total ramp density is 1.0 ramps per mile. What is the maximum directional peak-hour volume that can be accommodated without exceeding LOS C operating conditions
Answers
Answer:
3.586.543veh/hr
Explanation:
On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to answer the questions in the subsequent parts of this GO exercise.(a) The angle -4.45° locates a principal stress plane. Give the stress value including sign if any.(b) The angle 85.6° locates the other in-plane principal stress plane. Give the stress value including sign if any.(c) What stress or stresses occur on a plane at 40.6°? Give the stress value or values including sign if any.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step and very detailed solution of the given problem
A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which is at 7 °C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.
Answers
Answer:
No.
Explanation:
The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
[tex]COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}[/tex]
[tex]COP_{HP} = 3.339[/tex]
The power required to make the heat pump working is:
[tex]\dot W = \frac{300\,kW}{3.339}[/tex]
[tex]\dot W = 89.847\,kW[/tex]
The heat absorbed from the exterior air is:
[tex]\dot Q_{L} = 300\,kW - 89.847\,kW[/tex]
[tex]\dot Q_{L} = 210.153\,kW[/tex]
According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:
[tex]\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0[/tex]
[tex]\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}[/tex]
[tex]\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}[/tex]
[tex]\dot S_{gen} = 0.259\,\frac{kW}{K}[/tex]
Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.
Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rate of 1.8 kg/s. Water is heated by steam condensing at 120°C (hfg = 2203 kJ/kg) in the shell. If the overall heat transfer coefficient of the heat exchanger is 700 W/m2 ·°C, determine the length of the tube required in order to heat the water to 80°C using (a) the LMTD method, and (b) the ????????–NTU method. Answers: 129.5 m; 129.6 m
Answers
Answer:
Length = 129.55m, 129.55m
Explanation:
Given:
cp of water = 4180 J/kg·°C
Diameter, D = 2.5 cm
Temperature of water in = 17°C
Temperature of water out = 80°C
mass rate of water =1.8 kg/s.
Steam condensing at 120°C
Temperature at saturation = 120°C
hfg of steam at 120°C = 2203 kJ/kg
overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C
U = 700 W/m2 ·°C
Since Temperature of steam is at saturation,
temperature of steam going in = temperature of steam out = 120°C
Energy balance:
Heat gained by water = Heat loss by steam
Let specific capacity of steam = 2010kJ/Kg .°C
Find attached the full solution to the question.
A spherical seed of 1 cm diameter is buried at a depth of 1 cm inside soil (thermal conductivity of 1 Wm-1K-1) in a sufficiently large planter. There is a 1 cm thick layer of mulch (thermal conductivity of 2 Wm-1K-1) on top of the soil. The planter has top surface dimensions of 10 cm by 10 cm and is exposed to 200 Wm-2 of heat. You find the top surface temperature of the mulch (Ts,1) to be uniform at 50˚C. What is the surface temperature of the seed (Ts,2) in ˚C
Answers
Answer:
Find the attachment for the answer
Practice Problem: Large-Particle CompositesThe mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elasticity of the metal and oxide are, respectively, 60 GPa and 380 GPa, what is the (a) upper-bound, and (b) lower-bound modulus of elasticity values (in GPa) for a composite that has a composition of 33 vol% of oxide particles.
Answers
Answer: (a). Ec(μ) = 165.6 GPa
(b). Ec(∝) = 83.09 GPa
Explanation:
this is quite straightforward, so we will go step by step.
from the data we have that,
Moduli of elasticity of the metal -(Em) is 60 Gpa
Moduli of elasticity of oxide is (Ep) is 380 Gpa
volume Vp = 33% = 0.33
(a). To solve the upper bound-modulus of the elasticity is calculate thus;
Ec (μ) = EmVm + EpVp ----------------(1)
where E rep the modulus of elasticity
v rep the volume fraction
c rep the composite
Vm = 100% - Vp
Vm = 100% - 33% = 67%
Vm = 0.67
substituting the valus of Em, Vm, Ep, Vp from equation (1) we have;
Ec(μ) = (60×0.67) + (380×0.33)
Ec(μ) = 40.2 + 125.4 = 165.6 GPa
Ec(μ) = 165.6 GPa
(b). The lower bound modulus of elasticity can be calculated thus;
Ec(∝) = EmVp / EpVm + EmVp -------------- (2)
substituting values Em,Vm,Ep,Vp.
Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)
Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa
Ec(∝) = 83.09 GPa
cheers i hope this helps!!!!