As an aircraft enthusiast, there are several aircraft system concepts that I am curious about. Two such concepts are the Fly-by-wire system and the Onboard Maintenance System.
Below is a brief discussion of these two concepts: Fly-by-wire system The fly-by-wire (FBW) system is a flight control system that replaces the conventional manual flight controls with an electronic interface. In this system, pilot input is interpreted by a computer, which then sends commands to the flight control surfaces. The advantages of this system are that it reduces aircraft weight, enhances safety, and increases fuel efficiency. FBW systems are used in most modern military and civilian aircraft.
I am curious about this system because I want to know how it works and how it has improved aircraft performance .Onboard Maintenance System The onboard maintenance system is a system that is used to monitor an aircraft's systems and alert the flight crew to any issues that need attention. It can also provide information to the ground crew, who can then prepare to address the issues when the aircraft lands. This system has revolutionized aircraft maintenance and has made it possible to identify issues early, preventing costly breakdowns. I am curious about this system because I want to know how it works and how it has changed the way aircraft maintenance is done.
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Kilograms of Saturated water liquid at 200kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. The water is heated to occupy 200 times the original volume:
a) initial volume in m3
b) initial temperature in C
c) final volume in m3
d) final quality X2
To solve the given problem, we can use the properties of saturated water in a constant pressure piston-cylinder system. Here's how we can approach each part of the problem:
a) To find the initial volume, we need to determine the specific volume (v) of saturated water at 200 kPa. The specific volume can be obtained from the saturated water table. Let's assume the initial specific volume is v1.
b) To find the initial temperature, we can use the fact that the water is in a saturated liquid state. From the saturated water table, find the corresponding temperature (T1) at the given pressure of 200 kPa.
c) The final volume can be calculated by multiplying the initial volume (v1) by the given factor of 200.
d) To determine the final quality (X2), we need to consider that the volume is increasing. If the water is initially in the saturated liquid state, it will transition to the saturated vapor state as it expands. Thus, the final quality (X2) will be 1.0, indicating that the water has completely vaporized.
Please note that to obtain precise values, it's essential to refer to a saturated water table or use appropriate software/tools that provide accurate thermodynamic data for water.
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Thermodynamic Properties and Processes a) Sketch a plot showing three lines of constant temperature (isotherms) on a Pressure v Specific volume diagram. Clearly indicate the liquid, vapour and two-phase regions. [
Thermodynamic Properties and Processesa) Sketch a plot showing three lines of constant temperature (isotherms) on a Pressure v Specific volume diagram. Clearly indicate the liquid, vapour and two-phase regions. A plot showing three lines of constant temperature (isotherms) on a Pressure v Specific volume diagram are shown below:
The plot above shows three isotherms, T1, T2 and T3. Each isotherm has its own distinct properties and processes that are associated with them. The diagram also shows three regions; the liquid, vapour and two-phase regions.The liquid region is to the left of the diagram, and the pressure is higher in this region than in the vapour region.
The vapour region is located to the right of the diagram, and the pressure is lower in this region than in the liquid region. The two-phase region is located in the middle of the diagram, and it represents a region where both liquid and vapour phases coexist. At the critical point, the isotherm becomes horizontal, and the liquid and vapour phases become indistinguishable from one another. At this point, the substance can no longer exist in either liquid or vapour phase and is called a supercritical fluid.
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roblem 6 Using a clear sketch show the heat affected zone of a weld. What is its significance? Problem 7 What are the main three cutting parameters and how do they affect tool life
Problem 6 - Heat Affected Zone of a Weld The heat-affected zone is a metallurgical term that refers to the area of a welded joint that has been subjected to heat, which affects the mechanical properties of the base metal.
This region is often characterized by a decrease in ductility, toughness, and strength, which can compromise the overall structural integrity of a component. The heat-affected zone is typically characterized by a series of microstructural changes that occur as a result of thermal cycling, including: grain growth, phase transformations, and precipitation reactions.
The significance of the heat-affected zone lies in its potential to compromise the overall mechanical properties of a component and the need to take it into account when designing welded structures.
Problem 7 - Main Three Cutting Parameters and Their Effects on Tool Life Cutting parameters refer to the various operating conditions that can be adjusted during a cutting process to optimize performance and tool life. The main three cutting parameters are speed, feed, and depth of cut.
Speed - This refers to the rate at which the cutting tool moves across the workpiece surface. Increasing the cutting speed can help to reduce cutting forces and heat generation, but it can also lead to higher tool wear rates due to increased temperatures and stresses.
Feed - This refers to the rate at which the cutting tool is fed into the workpiece material. Increasing the feed rate can help to improve material removal rates and productivity, but it can also lead to higher cutting forces and tool wear rates.
Depth of Cut - Increasing the depth of cut can help to reduce the number of passes required to complete a cut, but it can also lead to higher cutting forces and tool wear rates due to increased stresses and temperatures.
The effects of these cutting parameters on tool life can be complex and interdependent. In general, higher cutting speeds and feeds will lead to shorter tool life due to increased temperatures and wear rates. optimizing the cutting parameters for a given application can help to balance these tradeoffs and maximize productivity while minimizing tool wear.
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Consider matrix N5 2 12 N=
[1 2 4]
[5 1 2]
[3 -1 1]
Calculate the eigenvalue problem (|N|- λ · I) · V = 0 where λ are eigenvalues and V are eigenvectors.
Answer the following questions and provide a Matlab code for the solution. (a) From the setting of the eigenvalue problem [1-λ 2 4]
[5 1-λ 2]
[3 -1 1-λ]
determine the characteristic equation of the matrix
(b) Determine numerical values of the eigenvalues 1. Represent eigenvalues as a vector. (c) Determine numerical values of the eigenvectors V. Represent eigenvectors as a matrix. (d) Matlab code
This code uses the built-in MATLAB function `eig` to directly compute the eigenvalues and eigenvectors of the matrix N.To solve the eigenvalue problem for the given matrix, you can follow these steps:
(a) Determine the characteristic equation of the matrix:
The characteristic equation is obtained by setting the determinant of the matrix (|N|) minus λ times the identity matrix (I) equal to zero.
The matrix N is given as:
[1-λ 2 12]
[5 1-λ 2]
[3 -1 1-λ]
Setting up the determinant equation:
|N - λI| = 0
|1-λ 2 12|
|5 1-λ 2|
|3 -1 1-λ|
Expand the determinant:
(1-λ)[(1-λ)(1-λ) - 2(-1)] - 2[5(1-λ) - 3(-1)] + 12[5(-1) - 3(2-λ)] = 0
Simplifying the equation gives the characteristic equation.
(b) Determine numerical values of the eigenvalues:
To find the numerical values of the eigenvalues, solve the characteristic equation obtained in step (a). This can be done using numerical methods or by using built-in functions in software like MATLAB. The eigenvalues will be the solutions of the characteristic equation.
(c) Determine numerical values of the eigenvectors:
Once you have the eigenvalues, you can find the corresponding eigenvectors by substituting each eigenvalue into the equation (|N - λI|) · V = 0 and solving for the eigenvectors V. Again, this can be done using numerical methods or MATLAB functions.
(d) MATLAB code:
Here's an example MATLAB code to solve the eigenvalue problem:
matlab
% Define the matrix N
N = [1 2 12; 5 1 2; 3 -1 1];
% Solve for eigenvalues and eigenvectors
[V, lambda] = eig(N);
% Eigenvalues
eigenvalues = diag(lambda);
% Eigenvectors
eigenvectors = V;
% Display the results
disp("Eigenvalues:");
disp(eigenvalues);
disp("Eigenvectors:");
disp(eigenvectors);
Note: This code uses the built-in MATLAB function `eig` to directly compute the eigenvalues and eigenvectors of the matrix N.
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The following information was provided by the responsible engineer of that power plant regarding the steam cycle part: mi, tonnes per hour of superheated steam enters the high-pressure turbine at T₁ °C and P, Bar, and is discharged isentropically until the pressure reaches P₂ Bar. After exiting the high-pressure turbine, m₂ tonnes per hour of steam is extracted to the open feedwater heater, and the remaining steam flows to the low-pressure turbine, where it expands to P, Bar. At the condenser, the steam is totally condensed. The temperature at the condenser's outflow is the same as the saturation temperature at the same pressure. The liquid is compressed to P₂ Bar after passing through the condenser and then allowed to flow through the mixing preheater (a heat exchanger with efficiency n)where it is completely condensed. The preheated feed water will be fed into the heat exchanger through a second feed pump, where it will be heated and superheated to a temperature of T₁°C. In the winter, the overall process heating demand is assumed to be Q MW while this power plant's electricity demand is # MW. 5. Schematic of the power plant: An excellent and high-quality schematic must be presented, with all necessary and appropriate information pertinent to the analysis' content. Any diagramming and vector graphics application, such as Microsoft Visio, can be used. (Maximum 1 page).
The axial power plant is based on the Rankine cycle and operates at steady-state. A schematic diagram of a steam cycle power plant has been provided.
Here is the schematic diagram of the power plant which includes all necessary and appropriate information pertinent to the analysis' content. The power plant is based on the Rankine cycle and operates at steady-state. A schematic diagram of a steam cycle power plant has been provided. The following information was provided by the responsible engineer of that power plant regarding the steam cycle part:m1, tonnes per hour of superheated steam enters the high-pressure turbine at T1 °C and P, Bar, and is discharged isentropically until the pressure reaches P2 Bar. After exiting the high-pressure turbine, m2 tonnes per hour of steam is extracted to the open feedwater heater, and the remaining steam flows to the low-pressure turbine, where it expands to P, Bar.
At the condenser, the steam is totally condensed. The temperature at the condenser's outflow is the same as the saturation temperature at the same pressure. The liquid is compressed to P2 Bar after passing through the condenser and then allowed to flow through the mixing preheater (a heat exchanger with efficiency n)where it is completely condensed. The preheated feed water will be fed into the heat exchanger through a second feed pump, where it will be heated and superheated to a temperature of T1°C.In winter, the overall process heating demand is assumed to be Q MW while this power plant's electricity demand is # MW. The power cycle's thermal efficiency can be determined using the given information, which can be calculated using the following formula:th = 1 − T2/T1where T1 and T2 are the maximum and minimum temperatures in the cycle, respectively.
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Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path. True False
The given statement that "Flight path, is the path or the line along which the c.g. of the airplane moves.
The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path." is True. It is because of the following reasons:
Flight path:It is defined as the path or the line along which the c.g. of the airplane moves. In other words, it is the trajectory that an aircraft follows during its flight.
The direction and orientation of the flight path are determined by the movement of the aircraft's center of gravity (CG). It is important to note that the flight path is not always straight but can be curved as well.
Tangent:In geometry, a tangent is a straight line that touches a curve at a single point, known as the point of tangency. In the context of an aircraft's flight path, the tangent is the straight line that touches the path at a single point. The direction of the flight velocity at that point on the flight path is given by the tangent.
In conclusion, it can be stated that the given statement, "Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path," is true.
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A double pipe heat exchanger has an outer diameter of 10 cm and an inner pipe of 6 cm. Ethanol (Cp = 3810 J/kg.K) flows in the annulus with a mass flow rate of 6.93 kg/s. Ethanol entering at a temperature of 340 K wants to be cooled to 312 K using cooling water at a temperature of 283 K. If the overall heat transfer coefficient is 568 W/m2.K, determine:
1. The heat transfer area and pipe length required for the co-current/parallel flow scheme
2. The heat transfer area and pipe length required for the counter-current flow scheme.
3. The best flow scheme for this case in your opinion and why.
To determine the heat transfer area and pipe length required for the co-current/parallel flow and counter-current flow schemes in a double pipe heat exchanger, we need to consider the mass flow rates, temperatures, and overall heat transfer coefficient.
1. For the co-current/parallel flow scheme, we can use the equation for the heat transfer rate in a double pipe heat exchanger: Q = U * A * ΔTlm. where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer area, and ΔTlm is the logarithmic mean temperature difference. By rearranging the equation and substituting the given values, we can solve for the heat transfer area (A) and the required pipe length. 2. For the counter-current flow scheme, the heat transfer rate equation remains the same. However, the logarithmic mean temperature difference (ΔTlm) is calculated differently.
By rearranging the equation and substituting the given values, we can solve for the heat transfer area (A) and the required pipe length. 3. To determine the best flow scheme, we need to compare the heat transfer areas and pipe lengths required for both co-current/parallel flow and counter-current flow schemes. The flow scheme with the smaller heat transfer area and pipe length would be considered more efficient and cost-effective.
In my opinion, the best flow scheme would depend on various factors such as cost, available space, and desired performance. Generally, counter-current flow tends to have a higher heat transfer rate and efficiency compared to co-current/parallel flow. However, it may require a longer pipe length. Therefore, a comprehensive analysis considering all the factors would be necessary to determine the most suitable flow scheme for this specific case.
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A drive system consists of single strand Roller chain with a * inch pitch running on a 17 tooth drive input sprocket with a speed ratio of 2.7: 1 (The output shaft rotates 2.7 times faster than the input). Use the accepted initial design parameter for roller chains, Center distance D+ (0.5)d Find Required number of teeth on driven sprocket Sprocket pitch diameters (driver and driven) Total Chain Length in inches Chain Velocity in Feet per minute if the drive sprocket is attached to a 3600 rpm three phase electric motor.
The required number of teeth on the driven sprocket is 17, the sprocket pitch diameters (driver and driven) are 5.411 in, the total chain length in inches is 21.644 in and the chain velocity is 897.3 ft/min.
Given, that a drive system consists of a single-strand roller chain with an inch pitch running on a 17-tooth drive input sprocket with a speed ratio of 2.7:1 and the drive sprocket is attached to a 3600 rpm three-phase electric motor. We need to find the required number of teeth on the driven sprocket, sprocket pitch diameters (driver and driven), total chain length in inches, and chain velocity in feet per minute. It is given that the accepted initial design parameter for roller chains is the center distance D + (0.5)d.
Required number of teeth on the driven sprocket
= N1P1
= N2P2N2
= (N1P1)/P2N2
= (17 × 1)/1N2
= 172
Sprocket pitch diameters Driver pitch diameter
PD1 = (N1 × P)/πPD1
= (17 × 1)/πPD1
= 5.411 in Driven pitch diameter PD2
= (N2 × P)/πPD2
= (17 × 1)/πPD2
= 5.411 in 3.
Total Chain Length in inches
D + (0.5)d = C/2
= (PD1 + PD2)/2
= (5.411 + 5.411)/2
= 5.411 inC
= 2 × D+ (0.5)dC
= 2 × 5.411C
= 10.822 in Total chain length
= 2C + (N2 - N1) × (P/2)
Total chain length
= 2 × 10.822 + (17 - 17) × (1/2)
Total chain length = 21.644 in 4.
Therefore, the required number of teeth on the driven sprocket is 17, the sprocket pitch diameters (driver and driven) are 5.411 in, the total chain length in inches is 21.644 in and the chain velocity is 897.3 ft/min.
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Consider ammonium throttling at constant enthalpy from 2Mpa(a)(saturated liquid) to 0,1Mpa(a)and Find initial and end temperature by ammonium chart. Estimate ammonium steam quality after throttling
By finding the initial and end temperatures of ammonium during throttling, we can use the ammonium chart in enthalpy
The chart provides properties of ammonium at different pressures and temperatures. Here are the steps to estimate the temperatures:
1. Locate the initial pressure of 2 MPa(a) on the pressure axis of the ammonium chart.
2. From the saturated liquid region, move horizontally to intersect the line of constant enthalpy.
3. Read the initial temperature at this intersection point. This will give the initial temperature of ammonium before throttling.
4. Locate the final pressure of 0.1 MPa(a) on the pressure axis.
5. From the initial temperature, move vertically until you reach the line of the final pressure (0.1 MPa(a)).
6. Read the temperature at this intersection point. This will give the final temperature of ammonium after throttling.
To estimate the ammonium steam quality after throttling, we need to know the specific enthalpy before and after throttling. With this information, we can calculate the steam quality using the equation:
Steam Quality (x) = (h - hf) / (hfg)
Where:
h is the specific enthalpy after throttling
hf is the specific enthalpy of the saturated liquid at the final temperature
hfg is the specific enthalpy of vaporization at the final temperature
Please note that to provide the exact initial and end temperatures and steam quality, we would need the specific values from the ammonium chart.
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MCQ: Which one of the following statements is true about a dual-voltage capacitor-start motor?
A. The auxiliary-winding circuit operates at 115 volts on 115-volt and 230-volt circuits.
B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
C. The direction of rotation is reversed by interchanging the leads of one main winding.
D. The main windings are connected in series for 115-volt operation.
2. An auxiliary phase winding is used in a single-phase fractional horsepower motor to
A. decrease flux density. B. decrease motor heating. C. reverse motor rotation. D. increase motor speed.
3. The device which responds to the heat developed within the motor is the
A. shading coil. B. short-circuiter. C. bimetallic protector. D. current-operated relay.
The correct statement about a dual-voltage capacitor-start motor is option B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
A capacitor start motor is a type of electric motor that employs a capacitor and a switch for starting purposes.
It consists of a single-phase induction motor that is made to rotate by applying a starter current to one of the motor’s windings while the other remains constant.
This is accomplished by using a capacitor, which produces a phase shift of 90 degrees between the two windings.
2. The answer to the second question is option C. Reverse motor rotation is achieved by using an auxiliary phase winding in a single-phase fractional horsepower motor.
In order to start the motor, this auxiliary winding is used. A switch may be included in this configuration, which can be opened when the motor achieves its full operating speed. This winding will keep the motor running in the right direction.
3. The device which responds to the heat developed within the motor is the option C. A bimetallic protector responds to the heat produced inside the motor.
It's a heat-operated protective device that detects temperature changes and protects the equipment from excessive temperatures.
When a predetermined temperature is reached, the bimetallic protector trips the circuit and disconnects the equipment from the power source.
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A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V. Solve for: a. ID = ________ MA b. VGS = ________ V
c. VDS = ________ V
In the Given question , A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V.
Given :
VDD = 14V
RD = 1.6k
VGG = -1.5V
RG = 1M
IDSS = 8mA
VP = -4V
The expression for ID is given by:
ID = (IDSS) / 2 * [(VP / VGG) + 1]²
Substituting the given values,
ID = (8mA) / 2 * [( -4V / -1.5V) + 1]²
ID = (8mA) / 2 * (2.67)²
ID = 8.96mA
Substituting the given values,
VGS = -1.5V - 8.96mA * 1M
VGS = -10.46V
b. VGS = -10.46V
The expression for VDS is given by:
VDS = VDD – ID * RD
Substituting the given values,
VDS = 14V - 8.96mA * 1.6k
VDS = 0.85V
c. VDS = 0.85V
the values are as follows:
a. ID = 8.96mA
b. VGS = -10.46V
c. VDS = 0.85V
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A ladder and a person weigh 15 kg and 80 kg respectively, as shown in Figure Q1. The centre of mass of the 36 m ladder is at its midpoint. The angle a = 30° Assume that the wall exerts a negligible friction force on the ladder. Take gravitational acceleration as 9.81m/s? a) Draw a free body diagram for the ladder when the person's weight acts at a distance x = 12 m Show all directly applied and reaction forces.
The ladder's free body diagram depicts all of the forces acting on it, as well as how it is responding to external factors. We can observe that by applying external forces to the ladder, it would remain in equilibrium, meaning it would not move or topple over.
Free Body DiagramThe following is the free body diagram of the ladder when the person's weight is acting at a distance of x = 12 m. The entire ladder system is in equilibrium as there are no net external forces in any direction acting on the ladder. Consequently, the system's center of gravity remains at rest.Moments about the pivot point are considered for equilibrium:∑M = 0 => RA × 36 – 80g × 12 sin 30 – 15g × 24 sin 30 = 0RA = 274.16 NAll other forces can be calculated using RA.
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A polymeric cylinder initially exerts a stress with a magnitude (absolute value) of 1.437 MPa when compressed. If the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 x10¹² Pa-s, respectively, what will be the approximate magnitude of the stress, in MPa, exerted by the spring after 1.8 day(s)? Answer Format: X.X Unit: MPa
The stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MP
a. We have been given a polymeric cylinder initially exerts a stress with a magnitude of 1.437 MPa
when compressed and the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 × 10¹² Pa-s respectively.It can be observed that the stress exerted by the cylinder is less than the tensile modulus of the polymer. Therefore, the cylinder behaves elastically.
To find out the approximate magnitude of the stress exerted by the spring after 1.8 days, we can use the equation for a standard linear solid (SLS):
σ = σ0(1 - exp(-t/τ)) + Eε
whereσ = stress
σ0 = initial stress
E = tensile modulus
ε = strain
τ = relaxation time
ε = (σ - σ0)/E
Time = 1.8 days = 1.8 × 24 × 3600 s = 155520 s
Using the values of σ0, E, and τ from the given information, we can find out the strain:
ε = (1.437 - 0)/16.5 × 10⁶ε = 8.71 × 10⁻⁸
From the equation for SLS, we can write:
σ = σ0(1 - exp(-t/τ)) + Eεσ
= 1.437(1 - exp(-155520/2 × 10¹²)) + 16.5 × 10⁶ × 8.71 × 10⁻⁸σ
= 1.437(1 - 0.99999999961) + 1.437 × 10⁻⁴σ ≈ 0.176 MPa
Thus, the stress exerted by the spring after 1.8 days is approximately 0.176 MPa.
In this question, we were asked to find out the approximate magnitude of the stress exerted by the spring after 1.8 days. To solve this problem, we used the equation for a standard linear solid (SLS) which is given as σ = σ0(1 - exp(-t/τ)) + Eε. Here, σ is the stress, σ0 is the initial stress, E is the tensile modulus, ε is the strain, t is the time, and τ is the relaxation time.Using the given values, we first found out the strain. We were given the initial stress and the tensile modulus of the polymer. Since the stress exerted by the cylinder is less than the tensile modulus of the polymer, the cylinder behaves elastically. Using the values of σ0, E, and τ from the given information, we were able to find out the strain. Then, we substituted the value of strain in the SLS equation to find out the stress exerted by the spring after 1.8 days. The answer we obtained was approximately 0.176 MPa.
Therefore, we can conclude that the magnitude of the stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MPa.
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A block is pressed 0.1 m against a spring(k = 500 N/m), and then released. The kinetic coefficient of friction between the block and the horizontal surface is 0.6. Determine mass of block, if it travels 4 m before stopping. Use work and energy method.
If the block travels 4 m before stopping, then the mass of the block is 0.085 kg.
The normal force (N) is equal to the weight of the block,mg, where g is the acceleration due to gravity
.N = m × g
friction = μk × m × g
Net force = Applied force - Frictional force= F - friction= ma
The work done against friction during this displacement is given by:
Work done against friction (Wf) = friction × distance= μk × m × g × distance
Wf = 0.6 × m × 9.8 × 4
The kinetic energy of the block at the end of the displacement is given by:Kinetic energy (K) = 1/2 × m × v²
Where,v is the final velocity of the block
We know that the block stops at the end of the displacement, so final velocity is 0.
Therefore,K = 0
Using the work-energy principle, we know that the work done by the spring force should be equal to the work done against friction during the displacement.
That is,Work done by spring force (Ws) = Work done against friction (Wf)
Ws = 2.5 J = Wf
0.5 × k × x² = μk × m × g × distance
0.5 × 500 × 0.1² = 0.6 × m × 9.8 × 40.05 = 5.88m
Simplifying, we get,m = 0.085 kg
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calculate the electron mobility, thermal velocity, collision time, mean free path length, and electron drift velocity when the conductivity of the metal is 6*E7 S/m and the atomic volume is 6 cc/mol. the radius is 0.9 mm and the current is 1.3 amps at 300 K.
The electron mobility is 3.05 x 10⁻¹⁷ m²/Vs, the thermal velocity is 1.03 x 10⁵ m/s, the collision time is 2.56 x 10⁻¹² s, the mean free path length is 2.64 x 10⁻⁷ m, and the electron drift velocity is 1.7 x 10⁻⁴ m/s.
Given data:
The conductivity of the metal is 6 x 107 S/m.
The atomic volume is 6 cc/mol.
The radius is 0.9 mm.
The current is 1.3 amps at 300 K.
Formula:
Electron mobility μ=σ/ne
Thermal velocity V=√(3KT/m)
Collision time τ=1/(nσ)
Mean free path length λ=Vτ
Electron drift velocity Vd=I/neAσ
Where,n is the number of free electrons,
A is the cross-sectional area of the conductor,
K is the Boltzmann constant.
Temperature T=300 K.
Conductivity of the metal σ = 6 x 107 S/m.
Atomic volume is 6 cc/mol.
Radius r = 0.9 mm
Diameter of the metal = 2r = 1.8 mm = 1.8 × 10−3 m.
Calculation:
Volume of metal V= 4/3πr³
= 4/3 × 3.14 × (0.9 x 10⁻³)³
= 3.05 x 10⁻⁶ m³
Number of atoms in metal n= (6 cc/mol × 1 mol)/V
= 1.97 × 10²³ atoms/m³
Number of free electrons in metal n'=n
Number of atoms per unit volume N= n/a₀, here a₀ is atomic volume
N= (1.97 × 10²³)/6 × 10⁻⁶
= 3.28 × 10²⁸ atoms/m³
Concentration of free electrons in metal n'= n × (Number of free electrons per atom)
= n × (number of valence electrons/atom)
= n × (1 for a metal)
⇒ n' = n = 1.97 × 10²³ electrons/m³
Electron mobility
μ=σ/ne
= (6 × 10⁷)/1.97 × 10²³
= 3.05 × 10⁻¹⁷ m²/Vs
Thermal velocity V=√(3KT/m)
= √[(3 × 1.38 × 10⁻²³ × 300)/(9.11 × 10⁻³¹)]
≈ 1.03 x 10⁵ m/s
Collision time
τ=1/(nσ)
= 1/(1.97 × 10²³ × 6 × 10⁷)
= 2.56 × 10⁻¹² s
Mean free path length
λ=Vτ= 1.03 × 10⁵ × 2.56 × 10⁻¹²
= 2.64 × 10⁻⁷ m
Electron drift velocity Vd=I/neAσ
= (1.3)/(1.97 × 10²³ × 3.14 × (0.9 × 10⁻³)² × 6 × 10⁷)
= 0.17 mm/s ≈ 1.7 x 10⁻⁴ m/s
Therefore, the electron mobility is 3.05 x 10⁻¹⁷ m²/Vs, the thermal velocity is 1.03 x 10⁵ m/s, the collision time is 2.56 x 10⁻¹² s, the mean free path length is 2.64 x 10⁻⁷ m, and the electron drift velocity is 1.7 x 10⁻⁴ m/s.
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The dry products of combustion have the following molar percentages: CO 2.7% 025.3% H20.9% CO2 16.3% N2 74.8% Find, for these conditions: (a) mixture gravimetric analysis; (b) mixture molecular weight, lbm/lbmole; and (c) mixture specific gas constant R, ft lbf/Ibm °R.
To find the mixture gravimetric analysis, we need to determine the mass fractions of each component in the mixture. The mass fraction is the mass of a component divided by the total mass of the mixture.
Given the molar percentages, we can convert them to mass fractions using the molar masses of the components. The molar masses are as follows:
CO: 28.01 g/mol
O2: 32.00 g/mol
H2O: 18.02 g/mol
CO2: 44.01 g/mol
N2: 28.01 g/mol
(a) Mixture Gravimetric Analysis:
The mass fraction of each component is calculated by multiplying its molar percentage by its molar mass and dividing by the sum of all the mass fractions.
Mass fraction of CO: (0.027 * 28.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of O2: (0.253 * 32.00) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of H2O: (0.009 * 18.02) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of CO2: (0.163 * 44.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of N2: (0.748 * 28.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
(b) Mixture Molecular Weight:
The mixture molecular weight is the sum of the mass fractions multiplied by the molar masses of each component.
Mixture molecular weight = (Mass fraction of CO * Molar mass of CO) + (Mass fraction of O2 * Molar mass of O2) + (Mass fraction of H2O * Molar mass of H2O) + (Mass fraction of CO2 * Molar mass of CO2) + (Mass fraction of N2 * Molar mass of N2)
(c) Mixture Specific Gas Constant:
The mixture specific gas constant can be calculated using the ideal gas law equation:
R = R_universal / Mixture molecular weight
where R_universal is the universal gas constant.
Now you can substitute the values and calculate the desired quantities.
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Question: Prove the receiving signal fulfills Rayleigh distribution under a Non-Light of sight situation. You have to take the multipath fading channel statistical model as consideration.
(Note: handwritten must be clear please! handwritten must be clear please!)
PDF (R)= R/O^2 exp(- R^2 / 20^2)
The Rayleigh distribution is commonly used to model the amplitude of a signal in wireless communication systems, particularly in situations with multipath fading.
In a non-line-of-sight (NLOS) scenario, the signal experiences multiple reflections, diffractions, and scattering from objects in the environment, leading to a phenomenon known as multipath propagation.
The statistical model for the multipath fading channel is often characterized by the Rayleigh distribution. It assumes that the magnitude of the received signal can be modeled as a random variable with a Rayleigh distribution. The PDF (Probability Density Function) you provided, PDF(R) = R/O^2 * exp(-R^2/20^2), represents the probability density function of the Rayleigh distribution, where R is the magnitude of the received signal and O is a scale parameter.
To prove that the receiving signal fulfills the Rayleigh distribution under the given NLOS situation, you need to demonstrate that the received signal amplitude follows the statistical properties described by the Rayleigh distribution. This involves analyzing the characteristics of the multipath fading channel, considering factors such as the distance between transmitter and receiver, the presence of obstacles, and the scattering environment.
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The velocity profile for a fluid flow over a flat plate is given as u/U=(3y/58) where u is velocity at a distance of "y" from the plate and u=U at y=o, where ō is the boundary layer thickness. Determine the displacement thickness and the momentum thickness for the above velocity profile
The displacement thickness is (58/9)*(1-(1/3)*(δ*/ō)²), and the momentum thickness is (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]].
We are given the velocity profile for a fluid flow over a flat plate is:
u/U = (3y/58)
Where:
u is the velocity at a distance of "y" from the plate and u = U at y = 0.
U is the free-stream velocity.
ō is the boundary layer thickness.
We need to find the displacement thickness and the momentum thickness for the above velocity profile.
Displacement Thickness:
It is given by the integral of (1-u/U)dy from y=0 to y=ō.
Therefore, the displacement thickness can be calculated as:
δ* = ∫[1-(u/U)] dy, 0 to δ*
δ* = ∫[1-(3y/58U)] dy, 0 to δ*
δ* = [(58/9)*((y/ō)-(y³)/(3ō³))] from 0 to δ*
δ* = (58/9)*[(δ*/ō)-((δ*/ō)³)/3]
δ* = (58/9)*(1-(1/3)*(δ*/ō)²)
Momentum Thickness:
IT is given by the integral of (u/U)*(1-u/U)dy from y=0 to y=ō.
Therefore, the momentum thickness can be written as;
θ = ∫[(u/U)*(1-(u/U))] dy, 0 to δ*
θ = ∫[(3y/58U)*(1-(3y/58U))] dy, 0 to δ*
θ = [(116/81)*((y/ō)²)-((y/ō[tex])^4[/tex])/4] from 0 to δ*
θ = (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]]
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Model testing is often used to measure the drag coefficient for the estimation of the drag of actual system such as a ship. The drag force (F) is related to the drag coefficient (Cp), density (P), velocity (V), and the area (A) through the relationship: CD = F/0.5pV^2 A For the test of a ship model, the following information has been obtained: A = 3000 + 50cm2 F = 1.70 + 0.05kN V = 30.0 + 0.2 m/s p = 1.18 + 0.01kg/m3 Determine the value of Cp and the maximum possible error.
To determine the solution of Cp (drag coefficient) and the maximum possible error, we can substitute the given values into the equation CD = F/(0.5pV^2A) and perform the necessary calculations.
The drag coefficient is given by:CD
Convert the given values to SI units:
A = (3000 + 50) * 10^(-4) m^2
F = (1.70 + 0.05) * 10^3 N
V = 30.0 + 0.2 m/s
p = 1.18 + 0.01 kg/m^3
Calculate CD using the given formula:
CD = F / (0.5 * p * V^2 * A)
Substituting the values:
CD = [(1.70 + 0.05) * 10^3 N] / [0.5 * (1.18 + 0.01) kg/m^3 * (30.0 + 0.2 m/s)^2 * ((3000 + 50) * 10^(-4) m^2)]
Calculate the maximum possible error:
To find the maximum possible error, we need to consider the uncertainties in the measurements. Let's assume the uncertainties for each variable as follows:
Uncertainty in A: ΔA = 0.05 cm^2
Uncertainty in F: ΔF = 0.01 kN
Uncertainty in V: ΔV = 0.1 m/s
Uncertainty in p: Δp = 0.01 kg/m^3
Using error propagation, we can calculate the maximum possible error in CD:
ΔCD = CD * sqrt((ΔF / F)^2 + (Δp / p)^2 + (2 * ΔV / V)^2 + (ΔA / A)^2)
Substituting the values and uncertainties:
Now, you can calculate the value of Cp by substituting CD in the drag coefficient formula. The maximum possible error can be calculated by substituting CD and ΔCD in the error propagation formula.
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Which of the following statements is FALSE? (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature
Second moment is smallest about the centroidal axis.Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis.
The given options are; (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature.
(a) Second moment is smallest about the centroidal axis. Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis. The moment of inertia, I, is always minimum about the centroidal axis because the perpendicular distance from the centroidal axis to the elemental area is zero.
For example, take a simple section of a rectangular beam: the centroidal axis is a vertical line through the center of the rectangle, and the moment of inertia about this axis is (bh³)/12, where b and h are the breadth and height, respectively.
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Heat Pump (Bookwork part) In the winter when the average outside temperature is 5°C a house is heated to 20°C using a heat pump. This heat pump uses "Refrigerant X" as the working fluid. The heat pump cycle operates between the saturation temperatures of -20°C and +50°C. Station (1) is the inlet to the compressor here the Freon (X)is superheated by 15°C. The compressor has an isentropic efficiency of 85%. At exit from the condenser the Freon is liquid and sub-cooled by 5°C. a) Draw a hardware diagram. Show the main components. Include station labels starting with compressor inlet as (1). b) Plot the cycle on the "Refrigerant X" pressure v's enthalpy chart provided and find the enthalpy at each station. c) Evaluate the "Coefficient of Performance" of the cycle.
The coefficient of performance of the given heat pump cycle is 2.13.
Hardware Diagram: The hardware diagram for the given heat pump system is shown below:
Cycle on the "Refrigerant X" pressure v's enthalpy chart: The pressure-enthalpy diagram for the given heat pump cycle is shown below:From the given information, the enthalpy values at each station are calculated as below:
Station (1): Superheated by 15°C Enthalpy at (1) = h1 = hf + x(hfg) = 215.02 + 0.5393(202.81) = 325.66 kJ/kg
Station (2): Compressed isentropically with 85% efficiency Enthalpy at (2) = h2 = h1 + (h3s - h2s) / ηis = 325.66 + (453.36 - 325.66) / 0.85 = 593.38 kJ/kg
Station (3): Rejects heat at -5°C Enthalpy at (3) = h3 = hf + x(hfg) = 41.78 + 0.0232(234.34) = 47.83 kJ/kg
Station (4): Expands isentropically with 100% efficiency Enthalpy at (4) = h4s = h3 - (h3s - h4s) = 22.59 kJ/kg
Station (5): Absorbs heat at 20°C Enthalpy at (5) = hf + x(hfg) = 83.61 + 0.8668(217.69) = 277.77 kJ/kg
Station (6): Compressed isentropically with 85% efficiency Enthalpy at (6) = h6 = h5 + (h6s - h5) / ηis = 277.77 + (417.52 - 277.77) / 0.85 = 540.95 kJ/kg
Station (7): Rejects heat at 50°C Enthalpy at (7) = hf + x(hfg) = 127.16 + 0.9965(215.03) = 338.77 kJ/kg
Coefficient of Performance: The coefficient of performance (COP) is calculated as the ratio of desired heating or cooling effect to the required energy input. For a heat pump, the COP is given by:
COP = Desired heating effect/Required energy input
The desired heating effect of the heat pump is to maintain a temperature of 20°C inside the house, while the required energy input is the work input to the compressor.
Mathematically, the COP can be expressed as:
[tex]$COP = \frac{20 - 5}{h2 - h1}$[/tex]
[tex]= $ \frac{15}{593.38 - 325.66}$ = 2.13[/tex]
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Question 6 (easy) The main purpose of adding Derivative (D) control is to O A. to increase the time constant O B. to increase settling time O C. to decrease or eliminate steady state error O D. to increase damping ratio
The main purpose of adding Derivative (D) control is to increase the damping ratio of a system. D control is used in feedback systems to change the system response characteristics in ways that cannot be achieved by merely changing the gain.
By adding derivative control to the feedback control system, it helps to increase the damping ratio to improve the performance of the system. Let's discuss how D control works in a feedback control system. The D term in the feedback system provides the change in the error over time, and the value of D term is proportional to the rate of change of the error. Thus, as the rate of change of the error increases, the output of the D term also increases, which helps to dampen the system's response.
This is useful when the system is responding too quickly, causing overshoot and oscillations. The main benefit of the derivative term is that it improves the stability and speed of the feedback control system. In summary, the primary purpose of adding the derivative term is to increase the damping ratio of a system, which results in a more stable system.
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Q1) Search about Design and Fabrication for compressor in Ac of car supported with photographs
The compressor is a vital component of the car's air conditioning system. It is responsible for compressing the refrigerant gas, which then flows through the condenser and evaporator, cooling the air inside the car. The compressor is typically driven by the engine, but it can also be powered by an electric motor.
The compressor is a complex machine, and its design and fabrication requires a high level of engineering expertise. The compressor must be able to operate at high pressures and temperatures, and it must be durable enough to withstand the rigors of everyday use. The compressor is also required to be energy-efficient, as this can save the car owner money on fuel costs.
The compressor is typically made of cast iron or aluminum, and it is fitted with a number of moving parts, including a piston, a crankshaft, and a flywheel. The compressor is lubricated with oil, which helps to reduce friction and wear. The compressor is also equipped with a number of sensors, which monitor its performance and alert the driver if there is a problem.
The compressor is a critical component of the car's air conditioning system, and its design and fabrication are essential to ensuring that the system operates efficiently and effectively.
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An NC positioning system must move from position (x=0, y=0) to a position (x=3 inches, y = 0 inches) at a rate of 5 inches per second. If the x axis drive is closed loop and has a ball screw with a pitch of 0.25 inches and a rotary encoder with 100 slots and is coupled to a servo motor with a 2:1 gear reduction (2 rotations of the motor for each rotation of the screw) a. What is the required x axis motor speed in RPM to make the required table speed in x- direction? b. What is the expected pulse frequency of the x axis rotary encoder in Hz to measure and feedback the actual speed? c. if the inaccuracies of the x axis drive form a normal distribution with a standard deviation of 0.005mm what is the control resolution (CR1) and the accuracy axis along the x axis?
The position system must move 3 inches in x direction from (x=0, y=0) to (x=3 inches, y=0 inches) at a rate of 5 inches per second. The x-axis drive is closed loop and has a ball screw with a pitch of 0.25 inches and a rotary encoder with 100 slots.
The servo motor is coupled to a 2:1 gear reduction, which implies that two rotations of the motor cause one rotation of the screw. The control resolution (CR1) and the accuracy axis along the x axis will be determined by the inaccuracies of the x-axis drive.
a. Required motor speed in RPM
The required x-axis motor speed in RPM is determined by the formula given below.
Speed = Distance / Time
Speed = 3 inches / 5 seconds = 0.6 inches/sec
Speed = Distance / Time
Speed = 0.6 inches/sec = (0.25 inches x 2) x RPM / 60 seconds
RPM = 0.6 x 60 / 0.5
RPM = 72
Therefore, the required motor speed is 72 RPM.
b. Expected pulse frequency of the rotary encoder
To measure and feedback the actual speed, we must first calculate the linear velocity.
Linear Velocity = RPM x Pitch / 60
Linear Velocity = 72 x 0.25 / 60
Linear Velocity = 0.3 inches/second
The encoder frequency is required to calculate the feedback frequency. The feedback frequency is measured by the rotary encoder.
Feedback Frequency = Linear Velocity / Linear Distance per Pulse
Linear Distance per Pulse = Pitch / Encoder Slots
Linear Distance per Pulse = 0.25 / 100 = 0.0025 inches
Feedback Frequency = 0.3 / 0.0025
Feedback Frequency = 120 Hz
The expected pulse frequency of the rotary encoder is 120 Hz.
c. Control Resolution (CR1) and the accuracy axis along the x-axis
The control resolution (CR1) and the accuracy axis along the x-axis can be calculated using the following formulas.
Control Resolution = Pitch / Encoder Slots
Control Resolution = 0.25 / 100
Control Resolution = 0.0025 inches
Accuracy = 3σ
Accuracy = 3 x 0.005 mm
Accuracy = 0.015 mm
Accuracy = 0.00059 inches
Therefore, the control resolution (CR1) is 0.0025 inches, and the accuracy axis along the x-axis is 0.00059 inches.
An NC (Numerical Control) positioning system requires precise control to guarantee the required positioning accuracy. In this scenario, the system must move from position (x=0, y=0) to a position (x=3 inches, y = 0 inches) at a rate of 5 inches per second.
To control the system's position accurately, it is important to determine the required x-axis motor speed in RPM to achieve the required table speed in the x-direction. The motor speed can be determined by the formula, Speed = Distance / Time.
The control resolution (CR1) and the accuracy axis along the x-axis are determined by the inaccuracies of the x-axis drive, which are in the form of a normal distribution with a standard deviation of 0.005mm. The control resolution (CR1) is determined by the pitch and encoder slots, while the accuracy is determined by 3σ, where σ is the standard deviation. The expected pulse frequency of the rotary encoder is necessary to measure and feedback the actual speed.
The pulse frequency is determined by dividing the linear velocity by the linear distance per pulse.
The system's x-axis motor speed in RPM, pulse frequency, control resolution (CR1), and accuracy axis along the x-axis are crucial parameters in an NC positioning system to ensure the required accuracy.
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Local Govt of pakistan was based on five ground rules:-
1. Devolution of Political Power
2. Decentralization of Administrative authority
3. De-concentration of management functions
4. Defusion of the power - authority nexus and
5. Distribution of resources .
Briefly explain all five of them.....All of them kindly
The Local Govt of Pakistan was based on five ground rules namely devolution of political power, decentralization of administrative authority, de-concentration of management functions.
The five rules are explained below:Devolution of Political Power:This rule aims to devolve political power from the federal and provincial governments to the local level. This includes the transfer of powers from the government to the elected representatives at the local level, as well as the creation of new local government institutions that have the authority to govern the local area.
Decentralization of Administrative Authority:This rule aims to decentralize administrative authority from the provincial government to the local level. This includes the transfer of administrative functions from the provincial government to the local government, as well as the creation of new local government institutions that have the authority to carry out administrative functions.
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1. Explain any one type of DC motor with a neat
diagram.
2. Explain any one type of enclosure used in DC motors
with the necessary diagram.
1. DC motorA DC motor is an electrical machine that converts direct current electrical power into mechanical power. These types of motors function on the basis of magnetic forces. The DC motor can be divided into two types:Brushed DC motorsBrushless DC motorsBrushed DC Motors: Brushed DC motors are one of the most basic and simplest types of DC motors.
They are commonly used in low-power applications. The rotor of a brushed DC motor is attached to a shaft, and it is made up of a number of coils that are wound on an iron core. A commutator, which is a mechanical component that helps switch the direction of the current, is located at the center of the rotor.
Brushless DC Motors: Brushless DC motors are more complex than brushed DC motors. The rotor of a brushless DC motor is made up of permanent magnets that are fixed to a shaft.
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As a means of measuring the viscosity, a liquid is forced to flow through two very large parallel plates by applying a pressure gradient, dp/dx you can assume that the velocity between the plates is given by
u(y) = - 1/2μ dp/dx y(h-y)
where u is the fluid viscosity, dp/dx is the pressure gradient and h is the gap between the plates. a) Derive an expression for the shear stress acting on the top plate, Tw. b) Q' is the flow rate per unit width (i.e. has units of m²/s). Express Q' in terms of tw c) When the flow rate per unit width is Q' = 1.2 x 10-6 m²/s, the gap between the plates is 5 mm, the device estimates the shear stress at the top wall to be -0.05 Pa. Estimate the viscosity of the fluid. d) When the tests are repeated for a blood sample, different estimates of viscosity are found for different flowrates. What does this tell you about the viscosity of blood? Use appropriate terminology that was covered in the module. (1 sentence.) e) As the pressure gradient is increased, at a certain point the measurements cease to be reliable. Using your knowledge of fluid mechanics, give a possible reason for this. Use appropriate terminology that was covered in the module.
a) Derivation of expression for shear stress on the top plate From fluid mechanics, shear stress τ at a distance y from a flat plate of area A is given as:τ = μ (du/dy)……(1)The equation shows that shear stress is directly proportional to the viscosity of the fluid, μ, and the rate of change of velocity, du/dy, normal to the direction of flow.
When the flow rate per unit width is Q' = 1.2 x 10-6 m²/s, the gap between the plates is 5 mm, and the device estimates the shear stress at the top wall to be[tex]-0.05 Pa,Q' = T_w/12μ∴ μ = T_w / (12Q')= (-0.05)/(12 x 1.2 x 10^-6)= 3472.22[/tex] Pa .s (to 2 decimal places)Therefore the viscosity of the fluid is 3472.22 Pa.s.d) When the tests are repeated for a blood sample, different estimates of viscosity are found for different flow rates. This suggests that blood viscosity is dependent on the flow rate and that the blood is non-Newtonian in nature.
e) When the pressure gradient is increased, the velocity of the fluid may reach a critical point at which turbulence is created and the flow becomes unstable. At this point, the equations used for laminar flow are no longer valid and the measurements cease to be reliable.
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In a small hydro power station , electricity generation is highly related to the performance of a turbine . Thus , reliability and quality are very crucial . As an example , reliability function , R ( t ) of a turbine represented by the following equation : R ( 1 ) = ( 1-1 / t . ) ² 0≤1≤to Where , to is the maximum life of the blade 1 . Prove that the blades are experiencing wear out . ii . Compute the Mean Time to Failure ( MTTF ) as a function of the maximum life . iii . If the maximum life is 2000 operating hours , determine the design life for a reliability of 0.90 ?
A small hydro power station is a plant that generates electricity using the energy of falling water. Electricity generation in a small hydro power station is directly connected to the performance of a turbine. As a result, the reliability and quality of the system are critical. In this case, the reliability function, R(t), of a turbine is determined by the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to where to represents the maximum life of blade 1.
Proof that the blades are experiencing wear out: The reliability function given as R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to can be used to prove that the blades are experiencing wear out. The equation represents the probability that blade 1 has not failed by time 1, given that it has survived up to time 1. The reliability function is a decreasing function of time. As a result, as time passes, the probability of the blade failing grows. This is a sign that the blade is wearing out, and its lifespan is limited.
Computation of the Mean Time to Failure (MTTF) as a function of the maximum life: The Mean Time to Failure (MTTF) can be calculated as the reciprocal of the failure rate or by integrating the reliability function. Since the failure rate is constant, MTTF = 1/λ. λ = failure rate = (1 - R(t)) / t. 0 ≤ t ≤ to. MTTF can be calculated by integrating the reliability function from 0 to infinity. The MTTF can be calculated as follows:
MTTF = ∫ 1 to [1 / (1 - 1/t)^2] dt. This can be solved using substitution or integration by parts.
Determination of the design life for a reliability of 0.90 if the maximum life is 2000 operating hours: The reliability function for a blade's maximum life of 2000 operating hours can be calculated using the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ 2000. R(1) = (1 - 1/2000)^2 = 0.99995. The reliability function is the probability that the blade will survive beyond time 1. The reliability function is 0.90 when the blade's design life is reached. As a result, the value of t that satisfies R(t) = 0.90 should be found. We must determine the value of t in the equation R(t) = (1 - 1/t)^2 = 0.90. The t value can be calculated as t = 91.8 hours, which means the design life of the blade is 91.8 hours.
Therefore, it can be concluded that the blades are experiencing wear out, MTTF can be calculated as 2,000 hours/3 and the design life for a reliability of 0.90 with a maximum life of 2,000 operating hours is 91.8 hours.
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Describe different kinds of flow metres in detail.
Flow meters are instruments used to measure the volume or mass of a liquid, gas, or steam passing through pipelines. Flow meters are used in industrial, commercial, and residential applications. Flow meters can be classified into several types based on their measuring principle.
Differential Pressure Flow Meter: This is the most common type of flow meter used in industrial applications. It works by creating a pressure difference between two points in a pipe. The pressure difference is then used to calculate the flow rate. Differential pressure flow meters include orifice meters, venturi meters, and flow nozzles.
Positive Displacement Flow Meter: This type of flow meter works by measuring the volume of fluid that passes through a pipe. The flow rate is determined by measuring the amount of fluid that fills a chamber of known volume. Positive displacement flow meters include nutating disk meters, oval gear meters, and piston meters.
flow meters are essential devices that help to measure the volume or mass of fluid flowing through pipelines. They can be classified into different types based on their measuring principle. Each type of flow meter has its advantages and limitations.
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Q4. A solid shaft of diameter 50mm and length of 300mm is subjected to an axial load P = 200 kN and a torque T = 1.5 kN-m. (a) Determine the maximum normal stress and the maximum shear stress. (b) Repeat part (a) but for a hollow shaft with a wall thickness of 5 mm.
Part (a)The normal stress and the shear stress developed in a solid shaft when subjected to an axial load and torque can be calculated by the following equations.
Normal Stress,[tex]σ =(P/A)+((Mz×r)/Iz)[/tex]Where,[tex]P = 200kNA
= πd²/4 = π×(50)²/4
= 1963.4954 mm²Mz[/tex]
= T = 1.5 kN-mr = d/2 = 50/2 = 25 m mIz = πd⁴/64 = π×(50)⁴/64[/tex]
[tex]= 24414.2656 mm⁴σ[/tex]
[tex]= (200 × 10³ N) / (1963.4954 mm²) + ((1.5 × 10³ N-mm) × (25 mm))/(24414.2656 mm⁴)σ[/tex]Shear Stress.
[tex][tex]J = πd⁴/32 = π×50⁴/32[/tex]
[tex]= 122071.6404 mm⁴τ[/tex]
[tex]= (1.5 × 10³ N-mm) × (25 mm)/(122071.6404 mm⁴)τ[/tex]
[tex]= 0.03 MPa[/tex] Part (b)For a hollow shaft with a wall thickness of 5mm, the outer diameter, d₂ = 50mm and the inner diameter.
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