Answer:
D
Explanation:
the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.
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 minor road intersects a major 4-lane divided road with a design speed of 50 mph and a median width of 12 ft. The intersection is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 5%.
Answer:
minimum sight distance = 699 ft
Explanation:
given data
road lane = 4 divided road
median width = 12 ft
grade road = 5%
solution
we take here time gap factor for minor road vehicle when enter to major road from table
time gap = 8.1 sec
and for median width of 12 ft
time gap = 8.2 + 0.7 ( 1 + [tex]\frac{12}{12}[/tex] )
time gap = 9.5 second
so minimum sight distance will be
minimum sight distance = 1.47 × design speed × time gap
minimum sight distance = 1.47 × 50 × 9.5
minimum sight distance = 699 ft
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!!!!
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.
A system executes a power cycle while receiving 1050 kJ by heat transfer at a temperature of 525 K and discharging 700 kJ by heat transfer at 350 K. There are no other heat transfers. a. Using Eq. 5.13, determine whether the cycle is internally reversible, irreversible, or impossible. b. Determine the thermal efficiency using Eq. 5.4 and the given heat transfer data. Compare this value with the Carnot efficiency calculated using Eq. 5.9 and comment.
The answer explains how to determine if a power cycle is reversible or irreversible based on heat transfer data and how to calculate thermal efficiency and compare it with Carnot efficiency.
Explanation:Question: A system executes a power cycle while receiving 1050 kJ by heat transfer at a temperature of 525 K and discharging 700 kJ by heat transfer at 350 K. Determine whether the cycle is internally reversible, irreversible, or impossible. Determine the thermal efficiency using the given heat transfer data and compare it with the Carnot efficiency.
Answer:a. Using the given data, we can calculate the change in entropy for the system to determine if the cycle is reversible or irreversible. If the total change in entropy is zero, the cycle is reversible. If not, it is irreversible.
b. To calculate thermal efficiency, use the formula η = W/Qh, where W is the work done and Qh is the heat input. Compare this efficiency value with the Carnot efficiency, which gives the maximum possible efficiency for the given temperature range.
A six-lane urban freeway (three lanes in each direction) is on rolling terrain with 11-ft. lanes, obstructions 2 ft. from the right edge of the traveled pavement, and nine ramps within three miles upstream and three miles downstream of the midpoint of the analysis segment. A directional weekday peak-hour volume of 2300 vehicles is observed, with 700 vehicles arriving in the most congested 15-min period. If the traffic stream has 15% large trucks and buses and no recreational vehicles, determine the level of service
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
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
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;
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 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
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).
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.
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 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.
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.
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
The semiconductor is a homogeneous, p-type material in thermal equilibrium for t ≤ 0. The excess minority carrier lifetime is 10^-6 ns.At t = 0, an external source is turned on which produces excess carriers uniformly at rate of g'=1020 cm^-3s^-1. At t = 2*10^-6 s , the external source is turned off.
(a) Derive the expression for the excess-electron concentration as a function of time for 0 ≤ t ≤[infinity].
(b) Determine the value of excess electron concentration at (i) t = 0, (ii) t = 2*10^-6 s , (iii) t = 3*10^-6 s and (iv) t = [infinity].
(c) Plot the excess electron concentration as a function of time.
Answer:
Find the attachments for complete solution
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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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]
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
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.
What historical event allowed both aerospace fields to make enormous strides forward?
Answer:Aerospace History Timeline
Explanation:
The primary transformer voltage is 2.8 kV and the secondary is 230 V. The transformer is connected to a variable load (0 to 300 kW) with a lagging power factor of 0.83 and a load voltage equal to the rated transformer secondary. Determine: (a) the total input impedance of the transformer when the secondary is shorted; and (b) the input current, voltage, power and power factor at full load (150 kW). (c) Plot the voltage regulation versus load, and determine the load
Answer:
a) πa⁵ Po² / 6E
Vacuum has infinite volume that's why I didn't consider that for total energy calculations
Explanation:
Check attached images for explanation and solutions to b and c
A coil with an average diameter of 5 inch will have an area of ""blank"" square meters
Answer:
19.64 square inches
Explanation:
Area will be (¶d^2)/4
= (3.142 x 5^2)/4
= 19.64 square inches
A four-cylinder, four-stroke spark-ignition engine operates at 2800 RPM. The processes within each cylinder can be modeled as an air-standard Otto cycle with a temperature of 280 K and a pressure of 70 kPa at the beginning of the compression process. The compression ratio is 10, and the total engine volume is 2555.6 cm3 . Suppose heat generated from the combustion process (i.e. heat addition per unit mass) is 1080 kJ/kg and air is an ideal gas having variable specific heats. Determine: (a) Total mass of air in the engine (in kg) (6 Points) (b) Determine the power developed by the engine (in hp). (24 Points)
Answer:
a) Total mass of air in the engine (in kg) = 0.00223 kg
b) power developed by the engine (in hp) = 180.61 hp
Explanation:
Initial temperature, [tex]T_{1} = 280 K\\[/tex]
Initial pressure, [tex]P_{1} = 70 kPa[/tex]
Compression ratio, r = 10
The initial total engine volume, [tex]V_{1} = 2555.6 cm^{3} = 2555.6 * 10^{-6} m^{3}[/tex]
a) Total mass of air in the engine
Using the gas equation, [tex]P_{1} V_{1} = mRT_{1}[/tex]
Where R = 0.287 kJ/kg-K
70 * 2555.6 * 10⁻⁶ = m * 0.287 * 280
[tex]m = \frac{70 * 2555.5 * 10^{-6} }{0.287 * 280}[/tex]
m = 0.00223 kg
b) Power developed by the engine
Heat generated due to combustion, [tex]Q_{in} = 1080 kJ/kg[/tex]
[tex]\frac{T_{2} }{T_{1} }= (\frac{V_{1} }{V_{2} })^{\gamma -1}[/tex]
Compression ratio, [tex]r = \frac{V_{2} }{V_{1} } = 10[/tex]
[tex]\frac{T_{2} }{T_{1} }=10^{0.4}\\\frac{T_{2} }{280 }=10^{0.4}\\T_{2} = 280 * 10^{0.4}\\T_{2} = 703.328 K[/tex]
[tex]Q_{in} = c_{v} (T_{3} -T_{2} )[/tex]
Where Specific capacity of air, [tex]c_{v} = 0.718 kJ/kg-K[/tex]
[tex]1080 = 0.718 (T_{3} -703.328 )\\1504.18 + 703.328 = T_{3}\\ T_{3} = 2207.51 K[/tex]
[tex]\frac{T_{4} }{T_{3} } = (\frac{V_{2} }{V_{1} } )^{\gamma -1} \\\frac{T_{4} }{T_{3} } = (1/10 )^{0.4} \\\frac{T_{4} }{2207.51 } = (1/10 )^{0.4} \\T_{4} = 2207.51 * (1/10 )^{0.4} \\T_{4} = 878.82 K[/tex]
[tex]Q_{out} = c_{v} (T_{4} - T_{1} )\\Q_{out} = 0.718 (878.82 - 280 )\\Q_{out} = 430 kJ/kg[/tex]
[tex]w_{net} = Q_{in} - Q_{out}\\w_{net} = 1080 - 430\\w_{net} = 650 kJ/kg[/tex]
There are 4 cylinders, k = 4
N = 2800/2
N = 1400
Power developed by the engine,
[tex]P =\frac{ mw_{net} Nk}{60} \\P =\frac{ 0.00223*650* 1400*4}{60} \\P = 134.68 kW[/tex]
1 kW = 1.34102 hp
P = 134.68 * 1.34102
P = 180.61 hp
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
In tin plating of a continuous steel pipe of outer diameter (OD) 8 cm, it is required to deposit the coating only on the outer surface. Plating is carried out from an electrolyte (valence charge = 2) contained in a 10 m long tank. The speed of the pipe through the tank is 200 m/min., and the current efficiency is 100%. Find the thickness of the coating deposited if a current density of 3,500 A.m-2 is used (density of tin = 7.31 g/cc, atomic weight of tin = 118.7).
Answer:
0.0198 mm/sec
Explanation:
Given:
d = 8 cm
L = 10 m = 1000 cm
v = 200 m/min
E = 100%
density = 3500 A.m^-2
Using Faraday's law of electrolysis, we have:
Q = n(e) * F
Where,
Q = electric charge
F = Faraday's constant=96500Cmol
n(e) = electron charge = 2
Therefore
Q= 96500 * 2 = 193000
Let's find the area of pipe, A = πr²(L)
But radius, r = [tex] \frac{d}{2} = \frac{4}{2} = 4cm [/tex]
Therefore,
A = 3.142 * 4² * 1000
A = 50272
To find the thickness of the coating deposited, we have :
[tex] = \frac{193000 * 7.31 * 118.7 * 200}{96500 * 50272 * 3500} = 1.98 * 10^-^3[/tex]
= 0.0198 mm/sec
_____________ is when a member is being pulled apart.
Group of answer choices
tension
shear
torsion
compression
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