Answer:
One
Explanation:
The basic rule for equipment rated 1000V, nominal, or less states that at least one entrance of sufficient area shall be provided to give access to and egress from working space about electrical equipment.
The minimum width and height for the entrance regardless of the width and rated amperes of the equipment. At least one entrance to enclosures for electrical installations should not be less than 24 inches wide and 6 ft 6 inches high. This is the minimum size opening into and out of the working space; the actual opening can be larger.
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.
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.
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
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).
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?
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?
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
Once an engineer has a sketch, the next step is to build the product.
True or false
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 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.
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.
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).
Answer:
R₁ = 32kΩ
Explanation:
See attached image
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.\
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.
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.
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!!!!
What is a microwave transmitter?a) A technology that uses active or passive tags in the form of chips or smart labels that can store unique identifiers and relay this information to electronic readers.b) Commonly used to transmit network signals over great distances.c) A means of linking computers using infrared or radio signals.d) An omnidirectional wireless technology that provides limited-range voice and data transmission over the unlicensed 2.4-GHz frequency band, allowing connections with a wide variety of fixed and portable devices that normally would have to be cabled together.
Answer:
b) Commonly used to transmit network signals over great distances.
Explanation:
The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.
The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,
Answer:
d) An omnidirectional wireless technology that provides limited-range voice and data transmission over the unlicensed 2.4-GHz frequency band, allowing connections with a wide variety of fixed and portable devices that normally would have to be cabled together.
Explanation:
Microwave transmission is a technology widely used in the 1950s and 1960s for transmitting signals, such as long-distance telephone calls and television programs between two terrestrial points on a narrow beam of microwaves. In microwave radio relay, microwaves are transmitted on a line of sight path between relay stations using directional antennas, forming a fixed radio connection between the two points. The requirement of a line of sight limits the separation between stations to the visual horizon, about 30 to 50 miles (48 to 80 km). Before the widespread use of communications satellites, chains of microwave relay stations were used to transmit telecommunication signals over transcontinental distances.
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.
Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
Final temperature: [tex]\( 336.575 \, \text{K} \)[/tex] . Vessel volume: [tex]\( 100 \, \text{L} \).[/tex]
To solve this problem, we can use the principle of conservation of mass and energy.
1. **Conservation of Mass:** Since the membrane ruptures and the entire volume gets filled, the mass of steam remains constant.
2. **Conservation of Energy:** We can use the first law of thermodynamics (energy conservation) to solve for the final temperature.
Given:
- Initial state: 1 kg of steam at 400°C and 200 bar.
- Final pressure: 100 bar.
Let's solve:
1. Conservation of Mass:
Since mass is conserved, the final mass of steam will remain 1 kg.
2. Conservation of Energy:
Using the first law of thermodynamics:
[tex]\[ Q = m \cdot c_v \cdot \Delta T \][/tex]
where:
- ( Q ) is the heat transferred,
- ( m )is the mass of the substance (1 kg in this case),
[tex]- \( c_v \)[/tex] is the specific heat capacity at constant volume, and
[tex]- \( \Delta T \)[/tex] is the change in temperature.
For steam, we can consider it to be an ideal gas at these high temperatures and pressures. The specific heat capacity at constant volume [tex](\( c_v \))[/tex] can be taken as a constant value.
Now, we need to find the final temperature [tex](\( T_f \))[/tex] . We can use the ideal gas law to relate the initial and final states:
[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \][/tex]
Where:
[tex]- \( P_1 \) and \( T_1 \)[/tex] are the initial pressure and temperature,
[tex]- \( P_2 \) and \( T_2 \)[/tex] are the final pressure and temperature,
[tex]- \( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes, respectively.
Solving:
Given:
[tex]- \( P_1 = 200 \, \text{bar} = 20,000 \, \text{kPa} \)\\- \( T_1 = 400 \, \text{°C} = 673.15 \, \text{K} \)\\- \( P_2 = 100 \, \text{bar} = 10,000 \, \text{kPa\\} \)\\- \( m = 1 \, \text{kg} \)[/tex]
Let's assume specific heat capacity at constant volume [tex](\( c_v \))[/tex] for steam to be approximately [tex]\( 2.0 \, \text{kJ/(kg·K)} \)[/tex] (This is a rough estimation).
From the ideal gas law:
[tex]\[ T_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot V_2}} \][/tex]
Given that initially, the other side of the vessel is evacuated, so [tex]\( V_2 \)[/tex] is negligible compared to [tex]\( V_1 \).[/tex]
[tex]\[ T_2 \approx \frac{{P_2 \cdot T_1}}{{P_1}} \]\[ T_2 \approx \frac{{10,000 \times 673.15}}{{20,000}} \]\[ T_2 \approx 336.575 \, \text{K} \][/tex]
Now, we can calculate the final volume using the ideal gas law:
[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \]\[ V_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot T_2}} \]\[ V_2 = \frac{{10,000 \times 1 \times 673.15}}{{20,000 \times 336.575}} \]\[ V_2 = \frac{{673,150}}{{6731.5}} \]\[ V_2 = 100 \, \text{L} \][/tex]
So, the final temperature of the steam is approximately
[tex]\( 336.575 \, \text{K} \)[/tex] and the volume of the vessel is [tex]\( 100 \, \text{L} \).[/tex]
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.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step and very detailed solution of the given problem
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.
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
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?;
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!!!!
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
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.
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
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.
Por que razão os engenheiros biomédicos devem ter familiaridade com a composição, as propriedades e o processamento de materiais?
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 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.
Answer:
Explanation:
The solutions to these questions can be seen in the following screenshots from a solution manual;
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.
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
A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm (0.320 in.), and the fractured gauge length is 74.17 mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.
Find the complete solution in the given attachments
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.
Answer:
The most positive is value of Vs is 0.5mV
what is the most important source of error in the Brinell test
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.
An ideal Otto cycle has a compression ratio of 10.5, takes in air at 90 kPa and 40°C, and is repeated 2500 times per minute. Using constant specific heats at room temperature, determine the thermal efficiency of this cycle and the rate of heat input if the cycle is to produce 90 kW of powe
Find the attachments for complete solution
The Otto cycle has a thermal efficiency of approximately 56%. Given the power output, the required rate of heat input to the system is approximately 161 KW.
Explanation:The first step to determining the thermal efficiency of an Otto cycle is to calculate it using the formula
ηth = 1 - (1/rγ-1)
, where ηth is the thermal efficiency, γ is the heat capacity ratio (1.4 for air), and r is the compression ratio.
In this case, substituting the given values r = 10.5 and γ = 1.4, we obtain ηth ≈ 0.56 or 56%.
The second step is to use the basic principle of power, which states power = energy/time. From the given repetition rate, the cycle time is 1/2500 minute = 0.0004 minute.
The total rate of heat input is therefore power/ ηth = 90kW / 0.56 ≈ 161 kW.
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You wish to use an 8-bit A/D converter with Vref = 5 V to measure a signal (amplitude = 2 Vpp, frequency = 1 kHz) that includes substantial noise (amplitude ≤ 0.5 Vpp, frequency = 5 kHz). What is the minimum sampling frequency required to ensure that aliasing will not inhibit your ability to accurately interpret the data? What would you do to prevent the high frequency noise from affecting your readings if you do not want to sample at such a hig
Answer:
The answer is explained below for the sampling frequency
Explanation:
Solution
Recall that:
The frequency Fm1 = 1 kHz
The frequency Fm2 = 5 kHz
Fs = n (2Fm2)
so,
[Fs = 8 * ( 2 * 5) = 80 8kHz], This the sampling frequency for 8-bit A/D converter
Now,
By increasing quantization level we can also increase the substantial noise ratio
Design an arithmetic circuit with two select variablesS1 and S2 and two n-bit data inputs A and B. The circuit generates the following eight arithmetic operations in conjunction with carry Cin: S1 S0 0 0 0 1 1 0 1 1 Cin = 0 F = A + B’ F = A’ + B F=A (transfer) F = A’ (1’s compliment) Cin = 1 F = A + B’ + 1 (Subtract A - B) F = A’ + B + 1 (Subtract A - B) F=A+1(Increment) F = A’ + 1 (2’s compliment)
An arithmetic circuit with select variables S1 and S2 generates eight different operations involving n-bit data inputs A and B, with the output being conditional on the values of S1, S2, and carry in Cin. These operations include various forms of addition, complementing, and incrementing. The circuit design integrates basic digital circuits like adders and incrementers.
Explanation:Arithmetic Circuit Design for Various Operations
The design of an arithmetic circuit for executing multiple operations based on select variables S1 and S2, as well as input carry Cin, encompasses the creation of a multi-functional digital circuit. This circuit needs to handle eight specific functions with two n-bit inputs, A and B, with the results being contingent on the values of S1, S2, and Cin. The operations are as follows:
When Cin is 0 and S1S2 is 00, F equals A plus the complement of B.When Cin is 0 and S1S2 is 01, F equals the complement of A plus B.When Cin is 0 and S1S2 is 10, F is simply A.When Cin is 0 and S1S2 is 11, F equals the complement of A.When Cin is 1 and S1S2 is 00, F equals A plus the complement of B plus one, effectively calculating A - B.When Cin is 1 and S1S2 is 01, F equals the complement of A plus B plus one, also representing A - B.When Cin is 1 and S1S2 is 10, F equals A plus one, which increments A.When Cin is 1 and S1S2 is 11, F equals the complement of A plus one, which is A's two's complement.Each operation can be implemented by combining basic digital circuits like adders, incrementers, and complement generators within the arithmetic circuit design.
_____________ is when a member is being pulled apart.
Group of answer choices
tension
shear
torsion
compression
Estimate the design-stage uncertainty in determining the voltage drop across an electric heating element. The device has a nominal resistance of 30 V and power rating of 500 W. Available is an ohmmeter (accuracy: within 0.5%; resolution: 1 V) and ammeter (accuracy: within 0.1%; resolution: 100 mA). Recall E ¼ IR.
Answer:
Find the attachment for solution
Note: The resistance is measured in ohms (Ω). In the question Ω will replace V, where ever the resistance unit is required .
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.
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.
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
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
6.27 Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 4 bar, 40°C and a volume of 0.2 m^3, is stirred until its temperature is 353°C. Assuming the ideal gas model with k = 1.4 for the air, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, in kJ/K. Ignore kinetic and potential energy.
The question pertains to the calculation of final properties of air after an adiabatic process, using principles of thermodynamics applicable to an ideal gas in a closed, adiabatic system. Final pressure, work done, and entropy produced must be calculated, with specific formulas applying to each aspect.
Explanation:The question is asking to calculate the final properties of air inside a rigid, insulated tank after being stirred by a paddle wheel. This is a thermodynamics problem specifically regarding a closed system where work is done without heat transfer (adiabatic process).
Part (a) - Final PressureSince the tank is rigid, the volume remains constant, and since it is adiabatic and insulated, we know no heat is exchanged with the surroundings. For an ideal gas undergoing a process in a closed system, the relationship between pressure and temperature is given by Gay-Lussac's law, which states that P1/T1 = P2/T2 where P is pressure and T is temperature in Kelvin. The pressure will increase as the temperature increases.
Part (b) - Work DoneFor a rigid container, there is no change in volume; hence no boundary work is done. However, work is done by the paddle wheel, and this is equivalent to the change in the internal energy of the gas since it is an adiabatic process. Using the first law of thermodynamics, the work can be found. Since the internal energy of an ideal gas depends only on temperature, the work done by the gas would be the difference in internal energy, which is a function of the difference in temperatures and the specific heat at constant volume.
Part (c) - Entropy ProducedSince the process is adiabatic and there is work done on the gas, entropy is produced within the system. However, without interaction with the surroundings, the entropy exchanged with the environment is zero. The change in entropy of the system will thus be a function of the initial and final states of the gas, considering that entropy is a state function.