The number of poles required for a 3-phase stepping motor to achieve 0.05 mm spatial control depends on the increments per revolution, which was not provided. To move the axis at 90 mm/sec, the pulses per second from the controller will be calculated based on the screw pitch and the number of increments per revolution. Additional information on the stepper motor is needed for precise calculations.
Explanation:To answer the student's questions regarding a 3-phase stepping motor for spatial control in a robotics application:
Poles of the motor: The number of poles in a stepper motor determines the resolution of movement. To achieve a spatial control of at least 0.05 mm with a screw pitch of 1 mm, the motor needs to have enough poles to allow for a fractional turn equal to that precision. Given that stepper motors can have a range of 5,000 to 10,000 increments in a 90-degree rotation, the exact number of poles required can be calculated based on the needed increments per mm.Pulses needed for motion: To move the axis at a speed of 90 mm/sec, the number of pulses required from the controller per second will depend on the revolutions per minute (rpm) the screw thread needs to turn, which is determined by the number of poles or increments per revolution of the motor. Generally, to calculate this, we would use the formula (speed in mm/sec) / (screw pitch in mm) * (number of increments per revolution), giving us the pulses per second.Without the specific number of increments per revolution of the stepping motor, it's impossible to provide an exact answer. However, typically stepper motors with more poles can provide finer control, and thus would be preferred in this application to meet the 0.05 mm spatial control requirement.
What historical event allowed both aerospace fields to make enormous strides forward?
Answer:Aerospace History Timeline
Explanation:
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 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.
In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation.
Answer:
(a) The shear plane angle is: ∅=26. 86°
(b) the shear strain for the operation is: γ = 2.155234
Explanation:
In orthogonal cutting, the wedge-shaped tool used in the cutting edge is perpendicular to the direction of motion.
(a) The shear plane angle is:
shear plane angle is the angle between shear plane and the cutting velocity in orthogonal cutting.
r= t[tex]1[/tex] ÷ t[tex]2[/tex]
Where, t[tex]1[/tex]= Initial thickness before the cut
t[tex]2[/tex]= deformed thickness
r = cutting ratio, which signifies the ratio of chip thickness of metal before cutting to the thickness after cutting.
r = 0.30 mm ÷ 0.65 mm
= 0.461538
∅ = tan∧[tex]^{-1}[/tex] ( 0.461538 cos 15 ÷ ( 1 - 0.461538 sin 15) )
= tan∧[tex]^{-1}[/tex] (0.506289)
∅=26. 86°
(b) the shear strain for the operation is:
Shear strain is the stress tangent of an angle, that acts parallel to a material cross section.
γ = cot 26.86 + tan (26.86 - 15)
=1.974523 + 0.210004
γ = 2.155234
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
An etchant was used to electively etch the grain boundaries. Mention whether these grain boundaries would appear brighter or darker after etching when observed through optical microscope and explain the reason for such appearance.
Answer:
Explanation:
a. Polishing or fining of a metal sample is done to get a mirror finish of the metal in order to get a clearer or good view of the materials composing of the metal under the microscope view. Polishing will not give the whole structural view. Etching is done to give the microstructure of metal by applying selective chemical attack. Polishing are also removed through the process of etching, thin grinding layer which are brought by polishing.
2. Grain boundaries are been harmed by chemicals more than the entire grain are been affected. In an optical microscope view, scattering of light is base on how the surface is etched. As the grain boundaries becomes more etched, more light is been reflected therefore given a dark appearance of the grain boundaries in the microscope view.
11. The gas field known as the Pinedale Anticline is expected to stop the energy boom. a. True b. False 12. The oil wells are not expected to disrupt the sensitive economic system in the Valley’s Basin a. True b. False 13. Each human generates ________ of garbage each day. a. 45 pounds b. 14.5 pounds. c. 4.5 pounds d. None of the above. 14. The hole in the stratospheric ozone layer over the Antarctic expected to not recover until sometime between the years 2060 to 2075. a. True b. False 15. BHP had a tailings containment plan study prepared to deal with the OK Tedi Copper Mine river pollution. Of the following approaches, which one was the least total likely cost. a. Mine and dredge. b. Mine only. c. Mine and dredge and store d. Early Closure
Answer:
11. b. False
12. b. False
13. c. 4.5 pounds
14. a. True
15. b. Mine only
Explanation:
The aforementioned are associated with pollution and resource depletion. Pollution refers to a process of contamination of the environment through undesirable and unintended production and usage of resources.
Resource depletion describes the usage of limited or scarce resources.
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
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 .
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.
A six-lane freeway (three lanes in each direction) currently operates at maximum LOS C conditions. The lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and there are two ramps within three miles upstream of the segment midpoint and one ramp within three miles downstream of the segment midpoint. The highway is on rolling terrain with 10% heavy vehicles, and the peak hour factor is 0.90. Determine the hourly volume for these conditions.
Answer:
4.071 veh/h.
Explanation:
Step one: calculate the estimated free flow speed by using the formula below;
= 75.4 - F(L) - F(C) - 3.22T^ 0.84.
The value of F(L) = 1.9 m/h and F(C) = 0.8 m/h.
Hence,
free flow speed = 75.4 - 1.9 - 0.8 - 3.22(3/6)^0.84.
free flow speed= 75.4 - 1.9 - 0.8 - 1.799
free flow speed= 70.901 mi/h = 70 mi/h.
Step two: determine the adjustment factor by using the formula below;
The adjustment factor = 1/ [1 + Pm (Em - 1) + Pn (En - 1)].
The adjustment factor = 1/[ 1+ 10/100 ( 2.5 - 1) + 0(2.0 - 1] = 0.869.
Step three;
Calculate the hourly volume;
The value corresponding to the LOS C conditions and free flow speed at 70 mi/h is 1735.
Therefore,
hourly volume, V = 1735 × 0.9 × 0.869 × 3.
Hourly volume, V = 4.071 veh/h.
The calculated hourly volume under maximum Level of Service C conditions on a six-lane freeway with a peak hour factor of 0.90 and 10% heavy vehicles is approximately 11,220 vehicles. This considers factors like the number of lanes, peak hour factor, and percentage of heavy vehicles.
Explanation:The calculation of the hourly traffic volume on a highway is a complex process that involves the consideration of many factors. For this scenario, you have a six-lane freeway (three lanes each way) that's operating at maximum Level of Service (LOS) C conditions, which implies moderate traffic density where the operation speeds are about 50% of free flow speed with restricted lane changing.
Using the Highway Capacity Manual (HCM) 2010 specifications for multilane highways, the maximum service flow rate for LOS C on rolling terrain is approximately 1,700 passenger cars per hour per lane (pc/hpl). The total volume would then be found by multiplying this number by the number of lanes and then adjusting for the peak hour factor and percentage of heavy vehicles.
Therefore, Hourly Volume = (1700 pc/hpl x 6 lanes) x (0.90 peak hour factor) x (1 + 0.1 heavy vehicles) which totals to approximately 11,220 vehicles. Thus, under these conditions, the hourly volume on this segment of the freeway would be around 11,220 vehicles.
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Old cars used to have a purely mechanical device called a carburetor to draw fuel into the engine for combustion. (Modern cars accomplish this with electronic fuel injection systems.) To prepare for this discussion, do some online research about how carburetors function.
1.How does the Bernoulli Effect allow the carburetor to function?
2.What advantages and disadvantages do you see with only using a carburetor?
3.Why do you think modern cars use electronic fuel injection systems instead of carburetors?
Answer:
Explanation:
1. Air flows into the top of the carburetor from the car's air intake, passing through a filter that cleans it of debris. ... When the throttle is open, more air and fuel flows to the cylinders so the engine produces more power and the car goes faster. The mixture of air and fuel flows down into the cylinders
2.Advantages :
Carburetor parts are not expensive as that of fuel injectors, especially EFI, which would give you large savings.
With the use of carburetor you get more air and fuel mixture.
In terms of road test, carburetors have more power and precision.
Carburetors are not restricted by the amount of gas pumped from the fuel tank which means that cylinders may pull more fuel through the carburetor that would lead to denser mixture in the chamber and greater power as well.
Disadvantages :
At a very low speed, the mixture supplied by a carburetor is so weak that it will not ignite properly and for its enrichment, at such conditions some arrangement in the carburetor is required.
The working of carburetor is affected by changes of atmospheric pressure.
It gives the proper mixture at only one engine speed and load, therefore, suitable only for engines running at constant speed increase or decrease.
More fuels are consumed since carburetors are heavier than fuel injectors.
More air emissions than fuel injectors.
Maintenance costs of carburetor is higher than with fuel injection system.
3. The electronics used in the system will calculate this information and constantly adjust. This type of controlled fuel injection results in a higher power output, greater fuel efficiency and much lower emissions. One of the main issues is that these systems are sophisticated and will cost much more than a carburetor
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 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
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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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.
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
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
The following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area: Diameter of specimen, 50 mm; Gauge length (the length of a specimen used for testing purposes), 100 mm; Minimum diameter after fracture, 10 mm. Load (kN) 2.2 4.5 6.8 9.2 11.8 14.4 17 19.6 22.2 24.7 Extension (μm) 5.6 11.9 18.2 24.5 31.5 38.5 45.5 52.5 59.5 66.5 Load (kN) 27.13 28.6 29.1 29.6 30.2 31 31.7 32 32.9 34.8 Extension (μm) 73.5 81.2 89.6 112 224 448 672 840 1120 1680 Load (kN) 35.8 37 38.7 39.5 40 39.6 35.7 28 Extension (μm) 1960 2520 3640 5600 7840 11200 13440 14560 (a) Generate engineering stress vs engineering strain plot (use a computer tool for accurate plotting. Hand sketch is Not allowed). (b) Determine Young’s modulus of elasticity. (c) Determine the ultimate tensile stress. (d) Determine the percentage reduction of area. (e) Determine the percentage elongation. (f) Determine true stress at fracture.
Answer:
See Explaination
Explanation:
1)here for given stress strain curve graph is given as follows
where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)
for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m
2)so graph is as follows
3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as
E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2
4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as
UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2
5)percentage reduction in area is given by
percentage reduction in area=[original area-final area/original area]*100
Percent reduction=[5062-10^2]*100/50^2=96%
6)percentage elongation is given by
percent elongation=[final length-original length/original length]*100
final length at fractureis=14.56+100=114.56 mm
so we get percent elongation as=[114.56-100/100]*100=14.56%
7)true fracture stress is given by load at fracture devided by true area at fracture
Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2
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 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;
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]
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
_____________ is when a member is being pulled apart.
Group of answer choices
tension
shear
torsion
compression
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
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.
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.
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
From experimental results, the hydraulic conductivity of a sand at a void ratio of 0.75 is 0.030 cm/sec. Estimate the hydraulic conductivity of this sand at a void ratio of 0.43. Use Casagrande and Kozeny-Carman relationships.
Answer:
0.00692 cm/Sec
Explanation:
Please kindly check attachment for the step by step solution of the given problem.