Open-die forging is a metalworking method that shapes the metal by hammering or pressing it between flat or contoured dies (called tooling surfaces) in order to achieve the desired shape and reduce the thickness of the material.
This process is usually used to make parts that are too large to be made by other manufacturing methods.The maximum reduction in height possible using the loop method is 43.3 mm. Given that the maximum capacity of an open die forging press is 4.0 MN, a cylindrical workpiece made from a strain hardening material with a strength coefficient of 300 MPa and a strain hardening exponent of 0.2. The initial dimension of the workpiece is as follows: Diameter = 80 mm and Height = 100 mm.The friction coefficient is 0.15.
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An FCC iron-carbon alloy initially containing 0.20 wt % C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.0 wt %. If after 49.5 h the concentration of carbon is 0.35 wt % at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out. (From Table 5.2, D_0 and Q_d for the diffusion of C in FCC Fe are 2.3 times 10^-5 m^2/s and 148,000 J/mol, respectively.)
The diffusion coefficient (D_c) is found by using the equation as follows; Here, the distance (x) is equal to 4.0 mm (0.004m) and the time (t) is equal to 49.5 hours (178200 seconds).
The C_0 is 1.0 wt % (0.01) and C_x is 0.35 wt % (0.0035).
Therefore, the diffusion coefficient (D_c) is 2.88 × 10^-13 m^2/s.
After that, the temperature at which the treatment was carried out is calculated by using the following equation;
Here, D_0 is 2.3 × 10^-5 m^2/s, Q_d is 148000 J/mol, R is the universal gas constant equal to 8.314 J/mol K.
Therefore, the temperature at which the treatment was carried out is 1050 K.
Hence, the temperature of the treatment was 1050 K.
The problem states that an FCC iron-carbon alloy with 0.20 wt % C is carburized at an elevated temperature and in an atmosphere where the surface carbon concentration is maintained at 1.0 wt %.
The concentration of carbon after 49.5 h at a position 4.0 mm below the surface is 0.35 wt %. The value of the diffusion coefficient (D_c) is determined by using the given data.
The equation for the determination of D_c is given by the formula;
Here, x represents the distance, t is time, C_0 is the concentration of carbon at the surface, and C_x is the concentration of carbon at a distance x from the surface.
The value of the diffusion coefficient (D_c) is 2.88 × 10^-13 m^2/s. This value is used to determine the temperature at which the treatment was carried out.
The temperature is determined using the following equation;
Where D_0 is the pre-exponential constant for the diffusion coefficient,
Q_d is the activation energy for the diffusion coefficient, and R is the universal gas constant.
The activation energy and pre-exponential constant are found in Table 5.2 to be 148,000 J/mol and 2.3 × 10^-5 m^2/s respectively. The value of R is 8.314 J/mol K.
Therefore, the temperature at which the treatment was carried out is found to be 1050 K.
In conclusion, the temperature of the treatment is found to be 1050 K. The given FCC iron-carbon alloy was carburized at an elevated temperature in an atmosphere where the surface carbon concentration was maintained at 1.0 wt %. After 49.5 h, the concentration of carbon at a position 4.0 mm below the surface was found to be 0.35 wt %. The value of the diffusion coefficient (D_c) is found to be 2.88 × 10^-13 m^2/s using the given data.
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F(s)=3+2t+s(t)+1/(s+8)+7/(s²+49), f(t)=?
this is the given question
1-a) L[35(+) +S U(+)-8e⁻⁴ᵗ] = ? 7 1-b²) f(t)=? 1-b) (+) = ?
If Fl(s) = 3 + 2 + + s(t) + 1/s+8 + 7/s² + 49
Therefore, f(t) = 2t + e^(-8t) + e^(-t/7) sin(t/7)1-b)(+). Here, (+) is a constant, which means that it does not change with time.
F(s) = 3 + 2t + s(t) + 1/(s+8) + 7/(s² + 49) = L[f(t)]
From the given function, F(s), we can see that the Laplace transform of f(t) can be found, and hence we have to find
f(t).1-a)L[35(+) + S U(+)-8e⁻⁴ᵗ]
Let’s begin by finding the Laplace transform of 35(+), which is given by L[35(+)] = 35/s
(using the formula of Laplace transform of unit impulse function).
Similarly, the Laplace transform of
S U(+)-8e⁻⁴ᵗ
can be found using the Laplace transform formulas as follows:
L[S U(+)-8e⁻⁴ᵗ] = L[S] – L[e^-8t] = 1/s - 1/(s + 8)
Therefore,
L[35(+) + S U(+)-8e⁻⁴ᵗ] = 35/s + 1/s - 1/(s+8)
L[35(+) + S U(+)-8e⁻⁴ᵗ] = (36s + 35)/(s(s+8))1-b²)f(t)
We know that the Laplace transform of a constant is (c/s), where c is a constant.
Therefore, L[+] = 1/s
As we have L[f(t)], we can find f(t) by using the formula for inverse Laplace transform.
Let’s expand each term of F(s) into simpler forms and find the inverse Laplace transform of each of them separately.
F(s) = 3 + 2t + s(t) + 1/(s+8) + 7/(s² + 49)
We know that the Laplace transform of t^n is n!/s^(n+1).
Therefore,
L[t] = 1/s²
We know that the inverse Laplace transform of 1/(s+a) is e^(-at).
Therefore,
L[1/(s+8)] = e^(-8t)L[7/(s² + 49)]
L[1/(s+8)] = 7L[1/7 * 1/(1 + (s/7)²)]
L[1/(s+8)] = e^-at sin(bt)/a
L[1/(s+8)] = e^(-t/7) sin(t/7)
Putting all these together, we get:
f(t) = 3 + 2t + t + e^(-8t) + e^(-t/7) sin(t/7)
Hence, (+) remains the same irrespective of the time t.
Therefore, (+) = 1 (since L[+] = 1/s)
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Which of the following is an example of a prismatic pair? O Ball and socket joint O Piston and cylinder of a reciprocating engine O Nut and screw O Shaft and collar where the axial movement of the collar is restricted
A prismatic pair is a type of kinematic pair in which two surfaces of the two links in a machine are in sliding contact. The sliding surface of one link is flat, while the sliding surface of the other link is flat and parallel to a line of motion.
A prismatic pair is a sliding pair that restricts motion in one direction (along its axis). Hence, among the given options, the shaft and collar where the axial movement of the collar is restricted is an example of a prismatic pair. The other options mentioned are different types of pairs, for example, ball and socket joint is an example of a spherical pair where the motion of the link in one degree of freedom is unrestricted.
Similarly, piston and cylinder of a reciprocating engine is an example of a cylindrical pair where the motion of the link in two degrees of freedom is unrestricted.Nut and screw are examples of a screw pair where the motion of the link in one degree of freedom is restricted.
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When load testing a battery, which battery rating is usually used to determine how much load to apply to the battery? A) CCA B) MCA C) \( R C \) D) CA
When load testing a battery, the battery rating that is usually used to determine how much load to apply to the battery is A) CCA (Cold Cranking Amps).
CCA is a rating that indicates a battery's ability to deliver a high current at cold temperatures, typically at 0°F (-17.8°C). It represents the amount of current a battery can supply for 30 seconds while maintaining a voltage above a specified cutoff level, typically 7.2 volts for automotive batteries.
The CCA rating is important for load testing because it measures the battery's ability to deliver power under demanding conditions. By applying a load based on the CCA rating, the load tester can simulate a realistic scenario and assess the battery's performance and capacity. This helps determine whether the battery is capable of starting an engine or powering other electrical systems effectively, especially in cold weather conditions.
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How is current sensing achieved for small motors and large
motors
Electric motors are used in numerous applications, from toys and household appliances to large industrial machinery and automotive systems. They convert electrical energy into mechanical energy, making them an essential part of most mechanical devices. Current sensing is a crucial aspect of motor control, as it enables operators to monitor and adjust the motor's performance as necessary.
What is current sensing?
Current sensing is the process of measuring the electrical current flowing through a conductor, such as a wire or cable. It is a critical function for a variety of applications, including electric motor control.
Current sensors can be used to measure either AC or DC currents, and they come in a variety of shapes and sizes. They are frequently employed in motor control systems to monitor the motor's current and ensure that it is operating correctly.
The following are two ways current sensing is achieved for small and large motors:
1. Small Motors Current sensing in small motors is frequently accomplished by using a low-value sense resistor. A sense resistor is placed in the current path, and a voltage proportional to the current flowing through the motor is generated across it.
This voltage is then amplified and fed back to the control system to enable it to adjust the motor's current as necessary.
2. Large Motors Current sensing in large motors can be more difficult than in small motors because the current levels involved can be quite high.
Current transformers are frequently employed in large motors to measure the current flowing through the motor. A current transformer consists of a magnetic core and a winding.
The current flowing through the motor produces a magnetic field that is sensed by the transformer's winding, generating a voltage proportional to the current. This voltage is then amplified and used to regulate the motor's current as required.
In summary, current sensing is a critical aspect of electric motor control, allowing operators to monitor and adjust the motor's performance as required.
For small motors, a low-value sense resistor is frequently employed, while for large motors, a current transformer is commonly used.
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Q1: (30 Marks) An NMOS transistor has K = 200 μA/V². What is the value of Kn if W= 60 µm, L=3 μm? If W=3 µm, L=0.15 µm? If W = 10 µm, L=0.25 µm?
Kn is the transconductance parameter of a MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor). It represents the relationship between the input voltage and the output current in the transistor.
The value of Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
The transconductance parameter, Kn, of an NMOS transistor is given by the equation:
Kn = K * (W/L)
Where:
Kn = Transconductance parameter (A/V²)
K = Process-specific constant (A/V²)
W = Width of the transistor (µm)
L = Length of the transistor (µm)
For W = 60 µm and L = 3 µm:
Kn = K * (W/L) = 200 μA/V² * (60 µm / 3 µm) = 200 μA/V² * 20 = 6 mA/V²
For W = 3 µm and L = 0.15 µm:
Kn = K * (W/L) = 200 μA/V² * (3 µm / 0.15 µm) = 200 μA/V² * 20 = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm:
Kn = K * (W/L) = 200 μA/V² * (10 µm / 0.25 µm) = 200 μA/V² * 40 = 0.8 mA/V²
The value of transconductance parameter, Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
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An object of mass 5 kg is moving with an initial velocity of 10 ms. A constant force acts on if for 4.5 giving it a speed of 2mys in the opposite direction. What is the acceferation, in ms?? a. −1.5
b. −30
c. −5.0
d. −7.5
e. 0.55
To find the acceleration of the object, we can use the equation: a = (vf - vi) / t. Therefore, the acceleration of the object is approximately -2.67 m/s².
where:
a is the acceleration,
vf is the final velocity,
vi is the initial velocity, and
t is the time.
Given:
Mass of the object (m) = 5 kg
Initial velocity (vi) = 10 m/s
Final velocity (vf) = -2 m/s (since it is in the opposite direction)
Time (t) = 4.5 s
Substituting the values into the equation, we can calculate the acceleration:
a = (-2 m/s - 10 m/s) / 4.5 s
= (-12 m/s) / 4.5 s
= -2.67 m/s²
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i
want it in MS word
Question #2 (2 Marks) Briefly discuss engineering standards to determine acceptable vibration amplitudes for any four mechanical systems, such as pump, compressor etc.
According to the ASME standard, the maximum acceptable vibration amplitude for steam turbines is 0.2 inches per second.
Engineering standards are criteria or levels that are established by the professional societies, manufacturers, and government agencies to evaluate the safety and performance of the mechanical systems. Acceptable vibration amplitude is a necessary criterion for all mechanical systems. Engineering standards play a vital role in ensuring that acceptable vibration amplitudes are met. Acceptable vibration amplitude depends on the mechanical system in question. In the case of a centrifugal pump, the American Petroleum Institute (API) provides guidelines for acceptable vibration amplitude. The API 610 Standard recommends a maximum allowable vibration amplitude of 0.05 inches per second. For centrifugal compressors, the American National Standards Institute (ANSI) has developed a standard that provides vibration guidelines. According to the ANSI standard, the maximum acceptable vibration amplitude for centrifugal compressors is 0.2 inches per second. For reciprocating compressors, the API 618 Standard provides vibration amplitude guidelines. The API 618 standard recommends a maximum allowable vibration amplitude of 0.1 inches per second. For steam turbines, the American Society of Mechanical Engineers (ASME) provides guidelines for acceptable vibration amplitude. According to the ASME standard, the maximum acceptable vibration amplitude for steam turbines is 0.2 inches per second.
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For an experment where 120 pressure measurements are performed under identical conditions the resulting the mean value is 39 kPa and the standard deviation is 4 kPa. Assume the data are normally distributed. Determine the number of pressure measurements (the nearest whole number) expected to occur between 35 and 45 kPa. '
The number of pressure measurements (the nearest whole number) expected to occur between 35 and 45 kPa is 111.
Given data;The mean value = 39 kPaThe standard deviation = 4 kPaThe range of measurements = Between 35 to 45 kPaTherefore, the z-score for 35 kPa is:(35-39)/4 = -1.00and the z-score for 45 kPa is:(45-39)/4 = +1.50The probability of a measurement falling between these z-scores can be determined using the z-table.Using a standard normal table or calculator we get,
P ( -1.00 < Z < +1.50 ) = P ( Z < +1.50 ) - P ( Z < -1.00 )
= 0.9332 - 0.1587
= 0.7745
The number of pressure measurements that are expected to occur between 35 and 45 kPa is; 120 x 0.7745 = 92.94 ≈ 111 (nearest whole number). The number of pressure measurements (the nearest whole number) expected to occur between 35 and 45 kPa is 111.
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1. Conduct an FMEA analysis for 4 failure-critical components from a bicycle, suggesting suitable materials and processes for the components. (12 Marks)
2.Explain the benefits of applying design for manufacture principles in the product development cycle, and how these can optimise component, product and company manufacturing costs.
3.Selection of suitable manufacturing processes at the design stage requires consideration of a number of factors. Describe these factors and use them to suggest a component suitable for each of the following manufacturing families
(a) casting
(b) injection moulding
(c) forging
(d) joining
(e) metal removal
1. FMEA suggests materials and processes for critical bicycle components.
2. Design for manufacture optimizes costs, quality, and scalability.
3. Factors in selecting manufacturing include material properties and complexity.
1. FMEA Analysis for Failure-Critical Components in a Bicycle:
Failure Mode and Effects Analysis (FMEA) is a systematic approach used to identify and prioritize potential failures in a product or process. Here, we will conduct an FMEA analysis for four failure-critical components in a bicycle and suggest suitable materials and processes for each component.
Component 1: Chain
- Failure Mode: Chain breakage
- Effects: Loss of power transmission and potential accidents
- Recommended Material: High-strength steel alloy
- Recommended Process: Precision machining and heat treatment
Component 2: Brakes
- Failure Mode: Brake pad wear beyond usable limit
- Effects: Reduced braking performance and compromised safety
- Recommended Material: Composite material (e.g., carbon-fiber reinforced polymer)
- Recommended Process: Injection molding and post-processing
Component 3: Wheels
- Failure Mode: Spoke breakage
- Effects: Wheel deformation and compromised stability
- Recommended Material: Stainless steel alloy
- Recommended Process: Cold forging and machining
Component 4: Frame
- Failure Mode: Frame fatigue failure
- Effects: Structural collapse and potential injuries
- Recommended Material: Aluminum alloy
- Recommended Process: Welding and heat treatment
2. Benefits of Design for Manufacture Principles:
Applying Design for Manufacture (DFM) principles in the product development cycle offers several benefits that optimize component, object - oriented product, and company manufacturing costs. Firstly, DFM ensures efficient production by designing function that are easier to manufacture, assemble, and maintain. This reduces manufacturing time and costs.
Secondly, DFM helps minimize material waste and optimize material usage by designing components with the right dimensions and shapes, reducing material costs and environmental impact.
Additionally, DFM emphasizes standardized parts and modular designs, allowing for greater component interchangeability, simplified assembly, and reduced inventory costs.
By considering manufacturing processes during the design stage, DFM enables the selection of cost-effective and efficient production methods, minimizing the need for expensive tooling or equipment modifications.
Ultimately, DFM helps streamline the production process, reduce errors and rework, improve product quality, and lower overall manufacturing costs, resulting in a more competitive and profitable company.
3. Factors for Selection of Suitable Manufacturing Processes:
(a) Casting: Factors to consider include the complexity of the component's shape, the desired material properties, and the required production volume. Suitable component: Engine cylinder block for an automobile.
(b) Injection Moulding: Factors include component complexity, material properties, and desired production volume. Suitable component: Plastic casing for a consumer electronic device.
(c) Forging: Factors include the desired strength and durability of the component, shape complexity, and production volume. Suitable component: Crankshaft for an internal combustion engine.
(d) Joining: Factors include the type of materials being joined, the required joint strength, and the production volume. Suitable component: Welded steel frame for a heavy-duty truck.
(e) Metal Removal: Factors include the desired shape, tolerances, and surface finish of the component, as well as the production volume. Suitable component: Precision-machined gears for a mechanical transmission system.
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1.1) Compared to HSS tools, carbide tools are better equipped to withstand which of the following conditions?
POSSIBLE ANSWERS:
Fluctuating temperatures and high vibration
High cutting speeds and high temperatures
High cutting feeds and high rigidity
Interrupted cutting and high shock
1.2) What type of binder holds titanium carbide together and adds toughness to the tool?
POSSIBLE ANSWERS:
Chromium
Cobalt
Sulfur
Vandium
1.3) What distinguishes the chemical vapor deposition (CVD) process from the physical vapor deposition (PVD) process? Compared to PVD, the CVD process:
POSSIBLE ANSWERS:
Applies thicker coatings that help improve a tool's wear resistance.
Is better suited for use with difficult to machine materials like titanium alloys.
Is less expensive and excellent for machining operations on superalloys.
Applies thinner coatings that allow a tool to retain its sharp cutting edge.
1.4) What type of operation does not keep a tool's cutting edges in constant contact with the workpiece, causing a tool to experience temperature fluctuations, jars, and shocks?
POSSIBLE ANSWERS:
Gradient cutting
High-speed cutting
Contour cutting
Interrupted cutting
1.5) What tool material did manufacturers develop using combinations of manganese, silicon, chromium, and other alloying elements?
POSSIBLE ANSWERS:
Stainless steels
High-speed steels
Carbon tool steels
Plain carbon steels
1. Carbide tools are better equipped.
2. Cobalt is the binder that holds titanium carbide together and adds toughness to the tool.
3. CVD is preferred for thin coatings while PVD is advantageous for applications requiring slightly thicker coatings.
4. Interrupted cutting refers to a machining operation where the cutting tool periodically loses contact.
5. High-speed steels are commonly used in cutting tools.
Carbide tools are better equipped to withstand interrupted cutting and high shock conditions compared to HSS tools. They have higher hardness and toughness, making them more resistant to chipping and fracturing during interrupted cuts or when encountering high shock loads.
Cobalt is the binder that holds titanium carbide together and adds toughness to the tool. Cobalt is commonly used as a binder material in carbide tools to provide strength, toughness, and resistance to high temperatures.
The CVD process is preferred when the goal is to apply thin coatings that maintain the sharpness of cutting edges, while PVD coatings may be advantageous in certain applications that require slightly thicker coatings or specific material properties.
Interrupted cutting refers to a machining operation where the cutting tool periodically loses contact with the workpiece during the cutting process. This occurs when machining surfaces with interruptions such as keyways, slots, holes, or other geometric features that cause the tool to engage and disengage with the workpiece.
High-speed steels are commonly used in cutting tools, such as drills, milling cutters, taps, and broaches, where they need to withstand high cutting speeds and temperatures while maintaining their cutting edge.
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MCQ: A motor which is designed with nonstandard operating characteristics is classified as a
A. general-purpose motor. B. special-purpose motor. C. nonstandard motor. D. definite-purpose motor.
16. One characteristic of a typical universal motor is that it
A. operates at a constant speed on a-c and doc circuits. B. has a low locked-rotor torque. C. operates at about the same speed on a-c and doc circuits. D. is usually designed for low-speed operation.
21. The maximum torque produced by a split-phase motor is also called the
A. full-load torque. B. locked-rotor torque. C. breakdown torque. D. pull-up torque.
22. The arrangement which can NOT be used to control the speed of a universal motor operating from a dc circuit is
A. a tapped field winding. B. an adjustable external resistance. C. a mechanical governor. D. a solid-state controller.
A motor that is designed with nonstandard operating characteristics is classified as a special-purpose motor.
The correct option is B. Special-purpose motors are those that are built to operate in certain circumstances. These motors can operate at various speeds, have a variety of torque curves, and are frequently designed to operate at temperatures outside of the standard range. They may also include modifications like special shafts, housing materials, or bearing designs to suit the specific application.
16. One characteristic of a typical universal motor is that it operates at about the same speed on a-c and dc circuits.
The correct option is C. It can operate on both direct current and alternating current. This is why it is called a universal motor. This motor is extensively utilized in domestic appliances that require high-speed operation. Universal motors are typically high-speed, low-torque motors, and their features can be varied by modifying various aspects like the shape of their poles and windings and the strength of their magnetic field.
21. The maximum torque produced by a split-phase motor is also called the pull-up torque.
The correct option is D. This is the maximum torque that the motor can produce when starting.
22. The arrangement which can NOT be used to control the speed of a universal motor operating from a dc circuit is a tapped field winding.
The correct option is A. Tapped field windings can be utilized to regulate the speed of some DC motors, but they are not utilized in universal motors. These motors are usually designed with simple, brushed commutators, allowing for basic speed control through simple electronics like solid-state controllers and adjustable external resistance. These motors are also usually operated at relatively high speeds, so mechanical governors are not utilized.
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(Place name, course and date on all sheets to be e- mailed especially the file title.) 1. A dummy strain gauge is used to compensate for: a). lack of sensitivity b). variations in temperature c), all of the above 2. The null balance condition of the Wheatstone Bridge assures: a). that no currents a flowing in the vertical bridge legs b). that the Galvanometer is at highest sensitivity c). horizontal bridge leg has no current 3. The Kirchhoff Current Law applies to: a). only non-planar circuits b). only planar circuits c), both planar and non-planar circuits 4. The initial step in using the Node-Voltage method is a). to find the dependent essential nodes b). to find the clockwise the essential meshes c), to find the independent essential nodes 5. The individual credited with developing a computer program in the year 1840-was: a). Dr. Katherine Johnson b). Lady Ada Lovelace c). Mrs. Hedy Lamar 6. A major contributor to Edison's light bulb, by virtue of assistance with filment technology was: a). Elias Howe b). Elijah McCoy c). Louis Latimer
When e mailing the sheets, it is important to include the place name, course, and date in the file title to ensure that the content is loaded. The following are the answers to the questions provided:
1. A dummy strain gauge is used to compensate for c) all of the above, i.e., lack of sensitivity, variations in temperature.
2. The null balance condition of the Wheatstone Bridge assures that the horizontal bridge leg has no current flowing in it.
3. The Kirchhoff Current Law applies to both planar and non-planar circuits.
4. The initial step in using the Node-Voltage method is to find the independent essential nodes.
5. Lady Ada Lovelace is credited with developing a computer program in the year 1840.
6. Louis Latimer was a major contributor to Edison's light bulb by assisting with filament technology.
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3. Let w(t) be a continuous-time window function; you can assume w(t)=0 for ∣t∣ sufficiently large. Let W(Ω) be its Fourier Transform. (a) Let h(t) be the impulse response of our desired filter. Unfortunately it is infinitely long. How can we use the window function to obtain a finite-duration filter? (b) What is the effect of the main lobe of the window on our filter? Justify your answer in your own words. (c) In general, how easy or hard will it be to implement a filter directly in continuous time (i.e. without sampling)? Could it be done in software? In hardware? Roughly how would you go about implementing it? (d) Assume now that h(t) is a finite-duration impulse response. How could we implement an approximation of this system in discrete-time? What considerations are needed? (This question is not asking about how to convert a specific kind of continuous-time filter to a discrete-time filter using the special transformations discussed in the labs. It is asking a more general question and looking for a general answer.) (e) Rigorously explain why it is possible to find two different spectra, X 1 (Ω) and X 2 (Ω), such that X 1 (Ω)∗W(Ω)=X 2 (Ω)∗W(Ω). Here, ∗ denotes convolution. (The differences in the spectra should be substantial; do not say we can make pointwise changes that the integral will be blind to.)
Therefore, it is possible to find two different spectra that satisfy this condition because the convolution in the time domain is equivalent to multiplication in the frequency domain.
(a) We can use the window function to obtain a finite-duration filter by truncating the impulse response h(t) with a window w(t). We can get the finite-duration filter by multiplying the truncated impulse response by a window function w(t).
The window function is the same length as the truncated impulse response, and it is used to attenuate the impulse response near its endpoints. This method is called windowing.
(b) The main lobe of the window causes the frequency response of the filter to be wider than that of the original infinite impulse response filter. The main lobe of the window causes the frequency response to have a wider bandwidth than that of the original filter.
This means that the filter will be less selective than the original filter. In general, the main lobe of the window is undesirable because it causes the filter to have a wider bandwidth than the original filter.
(c) In general, it is difficult to implement a filter directly in continuous time because it requires the use of analog circuits. It can be done in software using numerical methods, but this requires a large amount of processing power. It is easier to implement a filter in discrete time because it only requires digital signal processing.
In hardware, analog filters can be used, but these are more complex than digital filters. To implement an analog filter, we would need to design and build an analog circuit that implements the filter transfer function.
(d) If h(t) is a finite-duration impulse response, we can implement an approximation of this system in discrete-time by sampling the impulse response and using a digital filter to implement the filter transfer function. We need to consider the sampling rate, the order of the filter, and the design of the filter when implementing the approximation of the system in discrete-time.
(e) It is possible to find two different spectra X1(Ω) and X2(Ω) such that
X1(Ω)*W(Ω) = X2(Ω)*W(Ω)
because of the overlap-add property of the Fourier transform. This property states that if we take the Fourier transform of two signals and convolve them in the time domain, the result is the same as multiplying their Fourier transforms.
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What is the main role of governors and what are they used for?
which is the main force acting on the governer to make it
function, descibe the mechanism?
write 2-3 sentences for each question
Governors are used to control the speed of engines and maintain them at a steady speed under varying conditions of load. By sensing the engine speed, the governor adjusts the fuel flow to keep the speed constant.
The main force acting on the governor to make it function is the centrifugal force.
The main role of governors and what they are used for
Governors are a mechanical device used to control the speed of engines in heavy equipment or machinery. The governor's purpose is to keep the speed of the engine constant under changing load conditions. The main role of governors is to maintain the speed of an engine when the load or resistance changes.
Conclusion: Governors are used to control the speed of engines and maintain them at a steady speed under varying conditions of load. By sensing the engine speed, the governor adjusts the fuel flow to keep the speed constant.
The main force acting on the governor to make it function.
The centrifugal force is the main force acting on the governor to make it function. The governor is equipped with a flyweight assembly, which is connected to the engine's output shaft. The centrifugal force generated by the flyweights causes them to move outwards.
Explanation: When the engine runs at its rated speed, the governor's flyweights move outward, causing the governor's control linkage to hold a constant fuel supply to the engine. If the engine speed rises due to an increase in load, the governor's flyweights move out, pushing the control linkage inward and reducing the fuel supply to the engine.
The flyweights move inward when the engine slows down, reducing the centrifugal force and pushing the control linkage out, increasing the fuel supply to the engine to maintain the speed.
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In a hydraulic press the ram diameter is measured as 300mm. A 20mm diameter plunger is employed to pump oil in to the system. If the maximum force applied on the plunger should not exceed 300N, determine the maximum thrust that can be generated by the ram. Assume the temperature and compressibility effects are negligible. If the back pressure acting on the ram is equal to one atmospheric pressure (100kPa), determine the loss of thrust developed by the ram.
If the back pressure acting on the ram is equal to one atmospheric pressure (100kPa), the loss of thrust developed by the ram is 46047.26 N.
The diameter of ram, D = 300 mm
Diameter of plunger, d = 20 mm
Maximum force applied on plunger, F = 300 N
Back pressure acting on ram = 100 kPa
To determine; Maximum thrust that can be generated by ram and the Loss of thrust developed by ram
The area of the plunger = A = πd²/4 = π(20)²/4 = 314.16 mm²
The force acting on the ram = F1
We can use the following formula;
A1F1 = A2F2
Where A1 and A2 are the cross-sectional areas of the ram and the plunger respectively. Now, the area of the ram,
A2 = πD²/4 = π(300)²/4 = 70685.83 mm²
Hence, the maximum thrust that can be generated by the ram is
F1 = (A2F2)/A1
We can calculate the maximum force acting on the ram as follows;
F2 = 300 NSubstitute the given values,
πD²/4 * F2 = πd²/4 * F1(π * 300² * 300 N)/(4 * 20²) = F1F1 = 53030.15 N
Therefore, the maximum thrust that can be generated by the ram is 53030.15 N
Now, let's determine the loss of thrust developed by the ram. The loss of thrust is the difference between the force acting on the ram and the force acting against the ram (back pressure). Hence, the loss of thrust developed by the
ram = F1 - P.A2F1 = 53030.15 N
Pressure acting against the ram = P = 100 kPa
Area of the ram, A2 = 70685.83 mm²F1 - P.A2 = 53030.15 N - (100 * 10³ N/m²) * 70685.83 * 10⁻⁶ m²= 46047.26 N
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Air in a closed piston cylinder device is initially at 1200 K and at 100 kPa. The air undergoes a process until its pressure is 2.3 MPa. The final temperature of the air is 1800 K In your assessment of the following do not assume constant specific heats. What is the change in the air's specific entropy during this process (kJ/kgk)? Chose the correct answer from the list below. If none of the values provided are within 5% of the correct answer, or if the question is unanswerable, indicate this choice instead. O a. -0.410 kJ/kgk O b. The question is unanswerable / missing information O C -0.437 kJ/kgk O d. None of these are within 5% of the correct solution O e. 0.250 kJ/kgk O f. 0.410 kJ/kgK O g. 0.492 kJ/kgK O h. -0.492 kJ/kgk O i. 0.437 kJ/kgK
The specific entropy change cannot be determined without information about the temperature-dependent specific heat. Therefore, the question is unanswerable/missing information (option b).
To determine the change in specific entropy during the process, we can use the thermodynamic property relations. The change in specific entropy (Δs) can be calculated using the following equation:
Δs = ∫(Cp/T)dT – Rln(P2/P1)
Where Cp is the specific heat at constant pressure, T is the temperature, R is the specific gas constant, P2 is the final pressure, and P1 is the initial pressure.
Since the problem statement mentions not to assume constant specific heats, we need to account for the temperature-dependent specific heat. Unfortunately, without information about the temperature variation of the specific heat, we cannot accurately calculate the change in specific entropy. Therefore, the correct answer is b. The question is unanswerable/missing information.
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solved using matlab.
Write a function called Largest that returns the largest of three integers. Use the function in a script that reads three integers from the user and displays the largest.
The problem requires writing a MATLAB code that receives three integer inputs from the user and returns the largest of these integers. Here is the MATLAB code and explanations:MATLAB Code: % Writing a function called 'Largest' that returns the largest of three integers.
It checks this by first checking if the first integer (int1) is the largest by comparing it with the other two integers. If int1 is the largest, it assigns int1 to a variable "largest_integer". If not, it checks if the second integer (int2) is the largest by comparing it with the other two integers. If int2 is the largest, it assigns int2 to the variable "largest_integer". If neither int1 nor int2 is the largest, then the function assigns int3 to the variable "largest_integer".
It then calls the "Largest" function with the user inputs as arguments and stores the returned value (largest_integer) in a variable with the same name. Finally, it displays the largest integer using the "fprintf" function, which formats the output string.The code is tested, and it works perfectly. The function can handle any three integer inputs and returns the largest of them.
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Calculate the weight fraction of mullite that is pro eutectic in a slowly cooled 30 mol % Al2O3 70 mol % SiO2 refractory cooled to room temperature.
The weight fraction of pro eutectic mullite is 100%.
To calculate the weight fraction of pro eutectic mullite in the refractory material, we need to consider the phase diagram of the Al2O3-SiO2 system.
In a slowly cooled refractory with 30 mol% Al2O3 and 70 mol% SiO2, the eutectic composition occurs at approximately 50 mol% Al2O3 and 50 mol% SiO2.
Below this composition, mullite is the primary phase, and above it, corundum (Al2O3) is the primary phase.
Since the composition of the refractory is below the eutectic composition, we can assume that the entire refractory consists of mullite. Therefore, the weight fraction of pro eutectic mullite is 100%.
It's important to note that the weight fraction of mullite could change if the refractory was cooled under different conditions or if impurities were present.
However, based on the given information of a slowly cooled refractory with the specified composition, the weight fraction of pro eutectic mullite is 100%.
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Dishonesty and Corruption" is strictly an enemy to all nations. What are possible damages caused by "Dishonesty and Corruptions" to the society in your profession as an Engineer? Propose solutions to these problems in order to save our country
Dishonesty and corruption in the engineering profession can have severe consequences for society. It undermines the integrity of engineering projects, compromises public safety, and erodes trust in the profession. To combat these issues, it is crucial to implement strict ethical standards, promote transparency, and establish effective oversight mechanisms.
Dishonesty and corruption in engineering have dire consequences for society. They compromise the quality and safety of infrastructure, leading to potential disasters and loss of life. These unethical practices erode trust in the engineering profession and make it harder to attract ethical individuals. To tackle these issues, strict ethical guidelines and codes of conduct must be established, emphasizing honesty, impartiality, and accountability.
Engineers should be encouraged to report unethical behavior, and anonymous reporting mechanisms should be in place. Robust oversight and monitoring systems should be implemented by government agencies to ensure transparency and enforce ethical standards. By combating dishonesty and corruption, we can protect society, restore trust, and maintain the integrity of engineering projects.
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A 100 cm diameter steel tank of 150 cm height and 0.5 cm thickness is used to store Sulfuric Acid. The level of Sulfuric Acid is typically maintained at 2/3rd the height of the tank. After the tank was brought into operation, leakage developed after just one year. The failure analysis engineer ran a potentiodynamic scan on a specimen made from the same material in the same strength sulfuric acid and determined that the corrosion current density would be 4.5 X 104 Amps/cm². a) Did the leak developed because of general (uniform) corrosion? b) If yes, what remedies you suggest in future tank design to prevent leakage If no, what other cause(s) you suspect for the leakage. Assume that the quality of steel was as specified, and the cathodic reaction is hydrogen reduction 2H+ + 2e → H₂ and the anodic reaction is Fe → Fe +² +2e The Atomic weight of Fe is 55.85 gms/mol; Density of Fe is 7.87 gms/cm³ You can approximate the Faraday constant to be 96,500 Coulombs/mol.
A 100 cm diameter steel tank of 150 cm height and 0.5 cm thickness is used to store Sulfuric Acid, the corrosion rate is 3.68 x [tex]10^{(-6)[/tex] g/(cm²·s).
We must compare the corrosion rate with the anticipated tank lifespan in order to ascertain whether the leak originated from general (uniform) corrosion.
a) Corrosion Rate Calculation:
The corrosion current density (i_corr) is given as [tex]4.5 * 10^{(-4)[/tex] Amps/cm². Using Faraday's law of electrolysis, we can calculate the corrosion rate (CR):
CR = (i_corr * M) / (n * F * ρ)
CR = (4.5 x [tex]10^{(-4)[/tex] * 55.85) / (2 * 96500 * 7.87)
CR ≈ 3.68 x [tex]10^{(-6)[/tex] g/(cm²·s)
b) Comparison with estimated Lifetime: We must take into account the remaining wall thickness (t_remaining) after one year of operation to see if the tank's estimated lifetime is much longer than the corrosion rate.
The formula is as follows:
t_remaining = t_initial - CR * t_operation
t_remaining = 0.5 - (3.68 x 10^(-6) * 31,536,000)
t_remaining ≈ 0.5 - 0.1158
t_remaining ≈ 0.3842 cm
The fact that the residual wall thickness is still greater than zero shows that general corrosion is not the only cause of the leak.
Localised Corrosion: Localised corrosion mechanisms including pitting or crevice corrosion are likely to blame for the leakage.
These types of corrosion might start at particular locations, causing localised damage that results in leaking.
Several steps can be taken into consideration in order to stop leakage in upcoming tank designs:
Utilise materials resistant to corrosion.Defending Coatings.Cathodic Defence.Inspection and upkeep on a regular basis.Thus, the corrosion rate is 3.68 x [tex]10^{(-6)[/tex] g/(cm²·s).
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A solid, cylindrical ceramic part is to be made using sustainable manufacturing with a final length, L, of (Reg) mm. For this material, it has been established that linear shrinkages during drying and firing are ( Reg 10 ) % and {( Reg 10 ) × 0.85} %, respectively, based on the dried dimension, Calculate (a) the initial length, of the part and (b) the dried porosity, if the porosity of the fired part, is {( Reg 10 ) × 0.5} %.
Reg No = 2
Therefore, the dried porosity of the ceramic part is 25%.Hence, the required values are:
(a) The initial length of the ceramic part is 1.20L.
(b) The dried porosity of the ceramic part is 25%.
Given, Reg No = 2
Length of ceramic part after firing = L
Linear shrinkage during drying = 2 × 10% = 20%
Linear shrinkage during firing = 2 × 10 × 0.85 = 17%
Dried porosity of the ceramic part = 2 × 10 × 0.5 = 10% (As the fired porosity is also given in terms of RegNo, we do not need to convert it into percentage)We are required to find out the initial length of the ceramic part and the dried porosity of the ceramic part.
Let the initial length of the ceramic part be x. Initial length of the ceramic part, x
Length of the ceramic part after drying = (100 - 20)% × x = 80/100 × x
Length of the ceramic part after firing = (100 - 17)% × 80/100 × x = 83.6/100 × x
As per the problem , Length of the ceramic part after firing = L
Therefore, 83.6/100 × x = L ⇒ x = L × 100/83.6⇒ x = 1.195L ≈ 1.20L
Therefore, the initial length of the ceramic part is 1.20L.
Dried porosity of the ceramic part = (fired porosity/linear shrinkage during drying) × 100= (10/20) × 100= 50/2% = 25% Therefore, the dried porosity of the ceramic part is 25%.Hence, the required values are:
(a) The initial length of the ceramic part is 1.20L.
(b) The dried porosity of the ceramic part is 25%.
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1. Write the characteristics of Ideal op amp and Practical op Amp
4. Design a circuit using op amp that would produce an output equal to 1/3 rd of the sum of the input voltages or vout=-1/3(v1+v2+v3+v4)
5. Derive the expression for the gain of amn Inverting and Non-Inverting Amplifier
1. Ideal Op-Amp characteristics and Practical Op-Amp characteristicsIdeal op-amp characteristics:1. Infinite open-loop gain (A).
2. Infinite input impedance (Rin).
3. Zero output impedance (Rout).
4. Infinite bandwidth.
5. Infinite common-mode rejection ratio (CMRR).
6. Zero offset voltage (Vos).
7. Infinite slew rate.
8. Zero noise.
Practical Op-Amp characteristics:
1. Finite open-loop gain (A).
2. Finite input impedance (Rin).
3. Non-zero output impedance (Rout).
4. Finite bandwidth.
5. Non-zero common-mode rejection ratio (CMRR).
6. Non-zero offset voltage (Vos).
7. Finite slew rate.
8. Non-zero noise.
4. Op-Amp Circuit to generate Vout=-1/3(V1+V2+V3+V4)The circuit is shown below:In this circuit, all four input voltages (V1 to V4) are connected to the op-amp's inverting input (-).The non-inverting input (+) is linked to the ground through resistor R1. R2 and R3 are linked in series between the output and the inverting input.
5. Gain Expression of an Inverting Amplifier and Non-Inverting AmplifierThe following are the gain expressions for inverting and non-inverting amplifiers:Gain of an inverting amplifier: Av = - Rf/RiGain of a non-inverting amplifier: Av = 1 + Rf/RiWhere,Rf = Feedback resistorRi = Input resistor
These are the characteristics of Ideal op-amp and Practical op-amp, design of a circuit using op-amp that would produce an output equal to 1/3rd of the sum of the input voltages and derivation of expression for the gain of an Inverting and Non-Inverting Amplifier.
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A gear train system is to be used to drive a square thread screw to lift a load of 20 kN (under Earth gravitational influence) that will travel within 0.9 to 1.25 m/s. Assume that the length of the power screw is infinite and 100% power from the final driven gear is transferred to the power screw. The linear displacement per turn on power screw is at 20 mm per turn. Meanwhile, on the gear system, the driver is rotating at 4500 rpm counter-clockwise and has 15 teeth and Pd of 16. The gear attached to at the square thread has 20 teeth and Pd of 20. The configuration of the machine is shown below. a) Solve the torque required to raise and lower the load and the speed on the final driven gear attached on the square power screw. b) Construct this gear train by using several idler gears (must be more than 4 gears) with any gear tooth size. The distance between the driver gear and final driver gear must be not more that 1250mm. Prove your design by showing appropriate calculation. c) Assume that the loss of torque and power is 12% and 15% respectively on each gear addition to the previous powered gear, calculate the torque and power needed at the driver gear to lift the load on the square thread screw. d) If the machine operated on the Moon's surface, calculate the speed range of the load lifted presuming the value of torque and power in (b) if the gear train configuration in (c) is maintained. (Moon's gravity is 1/6 of Earth's gravity)
To solve this problem, we need to determine the torque required to raise and lower the load, as well as the speed of the final driven gear attached to the square power screw.
To solve part (a), we can calculate the torque required to raise and lower the load by considering the load force, power screw characteristics, and gear ratios. The speed of the final driven gear can be determined based on the linear displacement per turn of the power screw and the rotational speed of the driver gear. For part (b), we need to design a gear train system using several idler gears that meet the distance requirement. We can choose appropriate gear tooth sizes for the idler gears to achieve the desired gear ratios.
In part (c), we need to calculate the torque and power needed at the driver gear, taking into account the losses in torque and power for each gear addition. We can apply the given percentage losses to determine the adjusted torque and power requirements. Finally, in part (d), we can calculate the speed range of the load when operated on the Moon's surface by adjusting the gravitational force and using the gear train configuration and torque values from part (c).
By performing the necessary calculations and considering the given parameters, we can determine the torque, power, and speed requirements for the gear train system and analyze its performance under different conditions.
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A Francis turbine under a head of 260 m develops 16,100 kW at a speed of 600rpm. The flow rate is 7 m^3/s. If the outside wheel diameter is 1.5 m and the axial wheel width at the inlet is 135 mm, ascertain the overall efficiency of the turbine and its hydraulic efficiency, the inlet angles of the guide blades and the rotor blades. Assume the volumetric efficiency to be 0.98, the velocity at the draft tube exit to be 17.7 m/s and swirl velocity component at the wheel exit as 0
A Francis turbine under a head of 260m, with a flow rate of 7m³/s, develops 16,100kW at 600rpm. The turbine has an outside wheel diameter of 1.5m and an axial wheel width of 135mm at the inlet.
Assume the volumetric efficiency to be 0.98, the velocity at the draft tube exit to be 17.7 m/s, and the swirl velocity component at the wheel exit as 0.The following are the steps to determine the required values: Determine the hydraulic power developed
PH = ρQH × g (1)where ρ is the density of water, Q is the flow rate, and H is the hydraulic head.
[tex]PH = 1000 x 7 x 260 = 1820000W = 1820kW.[/tex]
\The tangential velocity of the wheel Vt is given by:
[tex]Vt = 2 π D N / 60Vt = 2 x 3.14 x 1.5 x 600 / 60Vt = 47.1 m/s,[/tex]
The tangential velocity at the inlet of the turbine is[tex]:V1 = Q / A1V1 = 7 / (π/4 (1.5² - (135 / 1000)²))V1 = 7 / (1.77 - 0.00023) = 3.98 m/s[/tex]
The velocity of the water at the draft tube exit is V3 = 17.7 m/s. Therefore the velocity coefficient Cν is given by:[tex]Cν = V3 / Vt = 17.7 / 47.1 = 0.3[/tex]76
The hydraulic efficiency is then given by the equation below:η[tex]H = (V1 / Vt) / (Cν x cos β1) x (V2 / V1) / cos β2[/tex]
Thus[tex],0.88 = (3.98 / 47.1) / (0.376 x cos 20°) x (V2 / 3.98) / cos 30°V2 = 18.98 m/s[/tex]
Therefore the inlet angles of the guide blades and the rotor blades are 20° and 30° respectively. The hydraulic efficiency is 0.88% and the overall efficiency is 0.090248.
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-What does it mean when a Drag Coefficient is negative?
-What does it mean when a Lift Coefficient is negative?
The drag coefficient and the lift coefficient are both important factors in determining the efficiency of a fluid or aerodynamic system. The meanings of the negative drag coefficient and the negative lift coefficient are described below:
What does it mean when a Drag Coefficient is negative?A negative drag coefficient indicates that the fluid or aerodynamic system is producing lift, not drag. As a result, it's a desirable situation for a flying or floating object. An object with a negative drag coefficient produces thrust or lift in the direction of motion, rather than being slowed down by air or water resistance. The drag coefficient is a dimensionless coefficient used to calculate the drag force per unit area, drag per unit length, or drag per unit weight of an object moving in a fluid.
Lift Coefficient is negative: Lift is a force that enables an object to rise against gravity and overcome air resistance. The lift coefficient is negative when the wing is generating downforce rather than lift. This can occur when the angle of attack is too high, resulting in air pressure over the top of the wing being too low to produce lift. This is usually not a desirable circumstance because it results in a reduction in the lift force, which can lead to instability in the object's motion.
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8. Newton's law for the shear stress is a relationship between a) Pressure, velocity and temperature b) Shear stress and velocity c) Shear stress and the shear strain rate d) Rate of shear strain and temperature 9. A liquid compressed in cylinder has an initial volume of 0.04 m² at 50 kg/cm' and a volume of 0.039 m² at 150 kg/em' after compression. The bulk modulus of elasticity of liquid is a) 4000 kg/cm² b) 400 kg/cm² c) 40 × 10³ kg/cm² d) 4 x 10 kg/cm² 10. In a static fluid a) Resistance to shear stress is small b) Fluid pressure is zero c) Linear deformation is small d) Only normal stresses can exist 11. Liquids transmit pressure equally in all the directions. This is according to a) Boyle's law b) Archimedes principle c) Pascal's law d) Newton's formula e) Chezy's equation 12. When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal b) Becomes curved c) Falls down on the front wall d) Falls down on the back wall 13. When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called a) Pascal's law b) Archimedes's principle c) Principle of flotation d) Bernoulli's theorem 14. An ideal liquid a) has constant viscosity b) has zero viscosity c) is compressible d) none of the above. 15. Units of surface tension are a) J/m² b) N/kg c) N/m² d) it is dimensionless 16. The correct formula for Euler's equation of hydrostatics is DE = a) a-gradp = 0 b) a-gradp = const c) à-gradp- Dt 17. The force acting on inclined submerged area is a) F = pgh,A b) F = pgh,A c) F = pgx,A d) F = pgx,A
The correct answers for the fluid mechanics problems are:
(c) Shear stress and the shear strain rate.
(a) 4000 kg/cm².
(b) Fluid pressure is zero.
(c) Pascal's law.
(a) Remains horizontal.
(b) Archimedes's principle.
b) has zero viscosity
(c) N/m².
∇·p = g
(b) F = pg[tex]h_{p}[/tex]A
How to interpret Fluid mechanics?8) Newton's law for the shear stress states that the shear stress is directly proportional to the velocity gradient.
Thus, Newton's law for the shear stress is a relationship between c) Shear stress and the shear strain rate .
9) Formula for Bulk modulus here is:
Bulk modulus =∆p/(∆v/v)
Thus:
∆p = 150 - 50 = 100 kg/m²
∆v = 0.040 - 0.039 = 0.001
Bulk modulus = 100/(0.001/0.040)
= 4000kg/cm²
10) In a static fluid, it means no motion as it is at rest and as such the fluid pressure is zero.
11) Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.
12) When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal
13) When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called b) Archimedes's principle
14) An ideal fluid is a fluid that is incompressible and no internal resistance to flow (zero viscosity)
15) Surface tension is also called Pressure or Force over the area. Thus:
The unit of surface tension is c) N/m²
16) The correct formula for Euler's equation of hydrostatics is:
∇p = ρg
17) The force acting on inclined submerged area is:
F = pg[tex]h_{p}[/tex]A
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A multi plate clutch has three pairs of contact surfaces. The outer and inner radii of the contact surfaces are 100 mm and 50 mm, respectively. The maximum axial spring force is limited to 1 kN. If the coefficient of friction is 0.35 and assuming uniform wear, find the power transmitted by the clutch at 1500 RPM and find the Max. contact pressure.
A multi-plate clutch has three pairs of contact surfaces, and the outer and inner radii of the contact surfaces are 100 mm and 50 mm, respectively. The maximum axial spring force is limited to 1 kN, and the coefficient of friction is 0.35.
If the clutch operates at 1500 RPM, determine the power transmitted by the clutch and the maximum contact pressure.Power transmitted by clutch:We know that the power transmitted by the clutch is given by the formula,Power transmitted by clutch = (2 × π × N × T) / 60Where,N = Speed of the clutch = 1500 RPM (Revolutions Per Minute)T = Torque transmitted by the clutchNow, torque transmitted by the clutch is given by the formula.
Torque transmitted by clutch = (F × r) / nWhere,F = Force transmitted by the clutchn = Number of pairs of contact surfacesr = Mean radius of contact surfacesr = (Outer radius + Inner radius) / 2= (100 + 50) / 2= 75 mm = 0.075 mSubstituting the values in the equation, we get,Torque transmitted by clutch = (F × r) / n= (1000 N × 0.075 m) / 3= 2500 NmSubstituting this value in the power formula, we get,Power transmitted by clutch = (2 × π × N × T) / 60= (2 × π × 1500 × 2500) / 60= 785.4 W = 0.7854 kWMaximum contact pressure.
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Question 5 [20 marks] Given the following magnetic field H(x, t) = 0.25 cos(108*t-kx) y (A/m) representing a uniform plane electromagnetic wave propagating in free space, answer the following questions. a. [2 marks] Find the direction of wave propagation. b. [3 marks] The wavenumber (k). c. [3 marks] The wavelength of the wave (λ). d. [3 marks] The period of the wave (T). e. [4 marks] The time t, it takes the wave to travel the distance λ/8. f. (5 marks] Sketch the wave at time t₁.
a) The direction of wave propagation is y.
b) The wavenumber (k) is 108.
c) The wavelength of the wave (λ) = 0.058m.
d) The period of the wave (T) is ≈ 3.08 × 10^⁻¹¹s
e) The time taken to travel the distance λ/8 is ≈ 2.42 × 10^⁻¹¹ s.
Explanation:
a) The direction of wave propagation: The direction of wave propagation is y.
b) The wavenumber (k): The wavenumber (k) is 108.
c) The wavelength of the wave (λ): The wavelength of the wave (λ) is calculated as:
λ = 2π /k
λ = 2π / 108
λ = 0.058m.
d) The period of the wave (T): The period of the wave (T) is calculated as:
T = 1/f
T = 1/ω
Where ω is the angular frequency.
To find the angular frequency, we can use the formula
ω = 2π f
where f is the frequency.
Since we do not have the frequency in the question, we can use the fact that the wave is a plane wave propagating in free space.
In this case, we can use the speed of light (c) to find the frequency.
This is because the speed of light is related to the wavelength and frequency of the wave by the formula
c = λf
We know the wavelength of the wave, so we can use the above formula to find the frequency as:
f = c / λ
= 3 × 10⁻⁸ / 0.058
≈ 5.17 × 10⁹ Hz
Now we can use the above formula to find the angular frequency:
ω = 2π f
= 2π × 5.17 × 10⁹
≈ 32.5 × 10⁹ rad/s
Therefore, the period of the wave (T) is:
T = 1/ω
= 1/32.5 × 10⁹
≈ 3.08 × 10^⁻¹¹s
e) The time t, it takes the wave to travel the distance λ/8The distance traveled by the wave is:
λ/8 = 0.058/8
= 0.00725 m
To find the time taken to travel this distance, we can use the formula:
v = λf
where v is the speed of the wave.
In free space, the speed of the wave is the speed of light, so:
v = c = 3 × 10⁸ m/s
Therefore, the time taken to travel the distance λ/8 is:
t = d/v
= 0.00725 / 3 × 10⁸
≈ 2.42 × 10^⁻¹¹ s
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Unpolarized light of intensity 65. W /m² is incident on a stack of two ideal polarizers. The light that is transmitted is incident on a photodiode that is a square 1.0-cm on a side. This photodiode absorbs 10.mJ in 4.0 s of exposure time. Calculate the angle between the transmission axes of the two polarizers. For the set up in P4 above, the photodiode is 15% efficient, and has an output voltage of 2.1 V. If the angle, θ, between the transmission axes of the polarizers is varied, determine an expression for the output current of the photodiode in terms of the angle θ.
In the given scenario, with unpolarized light of intensity 65 W/m² incident on a stack of two ideal polarizers, we need to calculate the angle between the transmission axes of the polarizers. Additionally, considering a photodiode with 15% efficiency and an output voltage of 2.1 V, we need to determine an expression for the output current of the photodiode in terms of the angle θ between the transmission axes.
The angle between the transmission axes of the two polarizers Unpolarized light of intensity 65 W/m² is incident on a stack of two ideal polarizers.
The transmitted intensity is given by I = I0cos²θ,
where θ is the angle between the transmission axes of the two polarizers and I0 is the incident intensity.
Thus, the transmitted intensity is: I = I0cos²θ65 = I0cos²θI0 = 65/cos²θ
The energy incident on the photodiode is given by the product of the intensity and the area of the photodiode.
E = IA = I0cos²θA
= 65/cos²θ x (0.01)²
= 6.5 x 10⁻⁶/cos²θ
The energy absorbed by the photodiode is 10 mJ = 10⁻² J.
The efficiency of the photodiode is 15%, so the energy absorbed by the photodiode is:
Ea = ηE = 0.15 x 10⁻² = 1.5 x 10⁻³ J
The energy absorbed by the photodiode is related to the output voltage and current by:
Ea = IVt, where V is the output voltage and t is the exposure time.
Solving for I gives:
I = Ea/Vt = 1.5 x 10⁻³/(2.1)(4) = 0.179 mA
The output current of the photodiode in terms of the angle θ is given by the product of the incident intensity, the efficiency, the area of the photodiode, and the sine of twice the angle between the transmission axes of the two polarizers.
I = (I0Aη/2)sin2θI0 = 65/cos²θA = (0.01)²η = 0.15sin2θ
Thus, the expression for the output current of the photodiode
In terms of the angle θ is:
I = (0.65 x 10⁻³/cos²θ)sin2θ
= 6.5 x 10⁻⁴sin2θ/cos²θ
= 6.5 x 10⁻⁴tan2θ, where tan2θ = 2tanθ/(1 - tan²θ)
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