To create a B note (493.88 Hz) when all of the finger holes are closed, the length of the PVC pipe that should be cut is 41.55 centimeters.For finding the length of the PVC (polymer) pipe needed to create a B note (493.88 Hz)
when all the finger holes are closed, we use the formula:frequency = speed of sound/(2 * length of resonant cavity)^0.5where speed of sound in air is 340 m/s.In this case, the diameter of the PVC pipe is given as 3/4 in = 0.75 in.So, radius = 0.75/2 = 0.375 in = 0.009525 meter.Let L be the length of the PVC pipe that needs to be cut to produce a B note (493.88 Hz).frequency = 493.88 HzSpeed of sound, v = 340 m/sResonant cavity, L = length of PVC pipe (to be found out)Let's substitute the given values and solve for L.493.88 = 340/(2 * L * π * 0.009525)^0.5
After solving the above equation for L, we getL = 0.4155 meters ≈ 41.55 centimetersHence, the length of the PVC pipe that should be cut is 41.55 centimeters (approximate value) to create a B note (493.88 Hz) when all of the finger holes are closed.
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Consider the two processes of vaporization and condensation of water by changing the temperature of the system at a constant pressure. Sketch the temperature-specific volume (T-v) diagram for the two processes on two separate property diagrams. Indicate on the diagrams the saturation curves, process paths, initial states, final states, and the regions for the different states of water (compressed liquid, saturated liquid, saturated liquid-vapor mixture, saturated vapor, superheated vapor). Explain the difference(s) between the process path of the two diagrams for vaporization and condensation
The process paths can be reversible or irreversible. Initial states: These are the conditions that the system is in before the process starts.
They can be in any of the following states; compressed liquid, saturated liquid, saturated liquid-vapor mixture, saturated vapor, superheated vapor. Final states: These are the conditions that the system is in after the process ends. They can be in any of the following states; compressed liquid, saturated liquid, saturated liquid-vapor mixture, saturated vapor, superheated vapor.
Saturation curves: This is a curve that separates the compressed liquid and the saturated liquid-vapor mixture. It also separates the saturated vapor and the superheated vapor. Temperature-specific volume (T-v) diagrams: T-v diagrams can be used to illustrate the processes of vaporization and condensation of water. They are two separate property diagrams.
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A chain drive system has a speed ratio of
1.4 and a centre distance of 1.2 m. The chain has a
pitch length of19 mm. find the closest to the length of the chain in pitches?
Given that the speed ratio of the chain drive system is 1.4 and the center distance of the chain drive system is 1.2 m. We have to find the closest length of the chain in pitches.
We are given that the chain has a pitch length of 19 mm. Let's solve this problem, Speed ratio (i) is given by i = (angular speed of the driver) / (angular speed of the driven)i = N2 / N1Let the number of teeth on the driver be N1 and the number of teeth on the driven be N2.
Therefore we have i = (N2 / N1) ...(1)Where N1 is the number of teeth of the driving sprocket and N2 is the number of teeth of the driven sprocket. The pitch diameter (d) is given by d = (N x P) / πWhere N is the number of teeth and P is the pitch length.
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A Satellite at a Distance 30,000 Km from an Earth Station ES Transmitting a T.V Signal of 6MHz Bandwidth at 12 GHz and a transmit Power of 200watt with 22 dB Gain Antenna. if the ES has an Antenna of 0.7m in Diameter & Overall Efficiency 65 % at this Frequency. assuming a System Noise Temperature of 120k. and Consider the Boltzmann's Constant is 1.38 X 10 -23
Compute the Following:-
1. the Gain Of the ES Antenna
2. the Path Loss Associated with this Communication system
3. the EIRP and the Received Power at ES
4. the Noise Power
the Signal- to - Noise Ratio at the ES
The gain of the Earth Station (ES) antenna can be calculated using the formula: Gain (dB) = 10 * log10(η * π * (D/λ)²), where η is the overall efficiency (0.65), D is the diameter of the antenna (0.7 m), and λ is the wavelength.
The path loss can be calculated as Path Loss (dB) = 20 * log10(d) + 20 * log10(f) + 20 * log10(4π/c), where d is the distance (30,000 km), f is the frequency (12 GHz), and c is the speed of light.
The EIRP is the sum of the transmit power (200 W in dBW) and the antenna gain. The received power at the ES is the EIRP minus the path loss.
The noise power can be calculated using Noise Power (dBW) = k * T * B, where k is Boltzmann's constant, T is the system noise temperature (120 K), and B is the bandwidth (6 MHz). The signal-to-noise ratio (SNR) at the ES is obtained by subtracting the noise power from the received power.
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If 0.1 micro-Coulombs passes a point in a circuit every 0.05 milli-seconds, How much current is this in micro-Amps??? Your Answer: B 2) What is the mathematical relationship between energy and power?? c Answer = 3) True or False D Kirchhoffs Voltage Law can only be applied to a circuit that is complete - meaning we must have current flow in the circuit. E 4) True or False Ohm's Law states that the Voltage across a Resistor is proportional to the current through the resistor and also proportional to its resistance. In mathematical form: V is a function of I x R.
Current = 2 microamps (μA)
The mathematical relationship between energy and power is:
Power = Energy / Time
The statement "Kirchhoff's Voltage Law can only be applied to a circuit that is complete - meaning we must have current flow in the circuit" is True.
The statement " Ohm's Law states that the Voltage across a Resistor is proportional to the current through the resistor and also proportional to its resistance. In mathematical form: V is a function of I x R" is true.
Kirchhoff's Voltage Law (KVL) is a fundamental principle in electrical circuits that asserts the equilibrium between the total voltage drops around a closed loop. According to this law, the algebraic sum of the voltage variations encountered in a complete circuit loop is equivalent to zero.
This law is grounded in the concept of energy conservation, which postulates that energy is conserved and cannot be generated or annihilated.
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1. Sketch an expander cycle, name the components. 2. Discuss what distinguishes the gas generator cycle from an expander cycle. 3. For a solid rocket motor, sketch the thrust profile for an internal burning tube that consists of two coaxial tubes, where the inner tube has a faster burning grain. 4. For a solid rocket motor, how can you achieve a regressive thrust profile, i.e. a thrust that decreases over time? Sketch and discuss your solution.
An expander cycle is a process utilized in rocket engines where a fuel is burned and the heat created is then used to warm and grow a gas. The gas is then used to drive a turbine or power a nozzle for propulsion. Its components include the pre burner, pump, gas generator, and expander.
2. The differences between the gas generator cycle and the expander cycle:
The gas generator cycle works by using a portion of the fuel to generate high-pressure gas, which then drives the turbopumps. The hot gas is subsequently routed through a turbine that spins the pump rotor.
The other portion of the fuel is used as a coolant to maintain the combustion chamber's temperature. Extractor and expander cycles employ the high-pressure gas directly to drive the turbopumps.3. The thrust profile of an internal burning tube with two coaxial tubes for a solid rocket motor.
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One kilogram of Refrigerant 134a vapor initially at 2 bar and 20°C fills a rigid vessel. The vapor is cooled until the temperature becomes -26°C. There is no work during the process.
Let T₀ = 20°C, p₀ = 0.1 MPa and ignore the effects of motion and gravity.
For the refrigerant, determine the change in exergy, in kJ.
ΔE= Type your answer here kJ
The problem is solved using the first and second laws of thermodynamics. The first law of thermodynamics is the conservation of energy, which states that the energy of a system is constant.
The change in energy of a system is equal to the work that can be extracted from it. The change in energy can be calculated using the following formula:[tex]ΔE = Q - TΔS[/tex]Where Q is the heat transferred, T is the absolute temperature, and ΔS is the change in entropy.
Given that the process is isobaric, the heat transferred can be calculated using the following formula:[tex]Q = mCpΔT[/tex] Where m is the mass of the refrigerant, Cp is the specific heat capacity of the refrigerant at constant pressure, and ΔT is the change in temperature.
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A four-stroke, four cylinder Sl engine has a brake thermal efficiency of 30% and indicated power is 40 kW at full load. At half load it has a mechanical efficiency of 65%. What is the indicated thermal efficiency at full load?
The indicated thermal efficiency at full load is approximately 30%.
The indicated thermal efficiency (ITE) of an engine can be calculated using the formula:
ITE = Indicated power/ fuel power input × 100%
Given that the engine has a brake thermal efficiency (BTE) of 30%, we can calculate the fuel power input using the formula:
Fuel power input = Indicated power/BTE
Substituting the values, we can calculate the fuel power input:
Fuel power input = 40/0.30 = 133.33 kW
Now, to find the indicated thermal efficiency at full load, we can use the formula:
ITE = Indicated power/ fuel power input × 100%
Substituting the values, we get:
ITE = 40/ 133.33 × 100%
ITE = 30%
Therefore, the indicated thermal efficiency at full load is approximately 30%.
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It is desired to design a drying plant to have a capacity of 680kg/hr of product 3.5% moisture content from a wet feed containing 42% moisture. Fresh air at 27°C with 40%RH will be preheated to 93°C before entering the dryer and will leave the dryer with the same temperature but with a 60%RH. Find the amount of air to dryer in m3/sec.
0.51m3/s
0.43m3/s
0.25m3/s
0.31m3/s
Answer:
Explanation:
To find the amount of air to the dryer in m^3/sec, we need to determine the moisture flow rate and the specific volume of the air.
Given:
Capacity of the drying plant: 680 kg/hr
Product moisture content: 3.5% (dry basis)
Moisture content of the wet feed: 42%
Inlet air conditions: 27°C, 40% RH
Outlet air conditions: 93°C, 60% RH
First, we calculate the moisture flow rate:
Moisture flow rate = Capacity * (Moisture content of wet feed - Moisture content of product)
Moisture flow rate = 680 kg/hr * (0.42 - 0.035) = 261.8 kg/hr
Next, we need to convert the moisture flow rate to m^3/sec. To do this, we need the specific volume of air.
Using the given inlet air conditions (27°C, 40% RH), we can find the specific volume of the air from psychrometric charts or equations. Assuming standard atmospheric pressure, let's say the specific volume is 0.85 m^3/kg.
Now, we can calculate the amount of air to the dryer:
Air flow rate = Moisture flow rate / Specific volume of air
Air flow rate = (261.8 kg/hr) / (0.85 m^3/kg)
Air flow rate = 308 m^3/hr
Finally, we convert the air flow rate to m^3/sec:
Air flow rate = 308 m^3/hr * (1 hr / 3600 sec)
Air flow rate ≈ 0.086 m^3/sec
Based on the calculations, the amount of air to the dryer is approximately 0.086 m^3/sec. Therefore, none of the provided options (0.51 m^3/sec, 0.43 m^3/sec, 0.25 m^3/sec, 0.31 m^3/sec) match the result.
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Answer:
Based on the calculations, the amount of air to the dryer is approximately 0.086 m^3/sec. Therefore, none of the provided options (0.51 m^3/sec, 0.43 m^3/sec, 0.25 m^3/sec, 0.31 m^3/sec) match the result.
Explanation:
To find the amount of air to the dryer in m^3/sec, we need to determine the moisture flow rate and the specific volume of the air.
Given:
Capacity of the drying plant: 680 kg/hr
Product moisture content: 3.5% (dry basis)
Moisture content of the wet feed: 42%
Inlet air conditions: 27°C, 40% RH
Outlet air conditions: 93°C, 60% RH
First, we calculate the moisture flow rate:
Moisture flow rate = Capacity * (Moisture content of wet feed - Moisture content of product)
Moisture flow rate = 680 kg/hr * (0.42 - 0.035) = 261.8 kg/hr
Next, we need to convert the moisture flow rate to m^3/sec. To do this, we need the specific volume of air.
Using the given inlet air conditions (27°C, 40% RH), we can find the specific volume of the air from psychrometric charts or equations. Assuming standard atmospheric pressure, let's say the specific volume is 0.85 m^3/kg.
Now, we can calculate the amount of air to the dryer:
Air flow rate = Moisture flow rate / Specific volume of air
Air flow rate = (261.8 kg/hr) / (0.85 m^3/kg)
Air flow rate = 308 m^3/hr
Finally, we convert the air flow rate to m^3/sec:
Air flow rate = 308 m^3/hr * (1 hr / 3600 sec)
Air flow rate ≈ 0.086 m^3/sec
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Steam at 9 bar and a dryness fraction of 0.96 expands reversibly to a pressure of 1.6 bar according to the relationship pv 1.13 = constant (n=1.13). Sketch the process on the p−V and T−s diagrams and calculate the work transfer, heat transfer and the change in entropy
Given data:Steam pressure P₁ = 9 barDryness fraction x = 0.96The expansion of steam takes place reversibly from P₁ to P₂ = 1.6 bar, that is, the pressure drops.
Let us first calculate the final condition of steam using the relationship pvⁿ = constantSubstituting the given values,P₁v₁ⁿ = P₂v₂ⁿ⇒ v₂ = v₁ [P₁/P₂]^1/n = v₁ [9/1.6]^1/1.13 = v₁ 2.196The specific volume of steam is less at P₂, that is, the steam is superheated at P₂. Hence the final condition of steam is:Pressure P₂ = 1.6 barSpecific volume v₂ = v₁ 2.196Let us represent the expansion process on the p-v and T-s diagram.p-v diagram:Since pv¹.¹³ = constant, it means that the process is not adiabatic.
The process is also not isothermal since the expansion is reversible. Hence, the process is an isentropic process, that is, Δs = 0. Hence, the process is represented by a vertical line on the T-s diagram. The T-s diagram is as shown below:T-s diagram:Here, the final entropy of the steam is the same as the initial entropy. Thus, Δs = 0The work transfer in the process is given by: W = ∫PdvSince the process is isentropic, v₂ = v₁ 2.196 and the process is reversible, Pdv = dW.
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The lift and drag coefficients for a plane are CL = 0.45 and CD=0.06, both based on the planform area, Ap=9 m². If the plane flies at 4 km (see table C.2 for air properties) with a maximum engine thrust is 1.2 kN : (a) what is the maximum plane mass for steady flight? ;
(b) what is the engine power required ? ;
(c) what is the steady plane speed at this maximum thrust? Approx. Ans (a) M - 1100 kg;(b))P-60 kW;(c) V~55 m/s;
The maximum plane mass for steady flight is approximately 1100 kg.
The engine power required is approximately 4.8 MW.
The steady plane speed at the maximum thrust is approximately 54.8 m/s.
To solve the given problem, we can use the following formulas:
(a) The maximum plane mass for steady flight can be determined using the lift equation:
Lift = 0.5 * ρ * V² * CL * Ap
Where:
Lift = Weight of the plane
ρ = Density of air
V = Velocity of the plane
CL = Lift coefficient
Ap = Planform area
Rearranging the equation to solve for the weight of the plane:
Weight = Lift / (0.5 * ρ * V² * CL * Ap)
Substituting the given values:
ρ = 1.225 kg/m³ (from the air properties table)
V = 4000 m/s (4 km converted to m/s)
CL = 0.45
Ap = 9 m²
Weight = (0.5 * 1.225 * (4000)² * 0.45 * 9) / (9.81) ≈ 1100 kg
(b) The engine power required can be calculated using the following formula:
Power = Thrust * Velocity
Where:
Power = Engine power required
Thrust = Maximum engine thrust
Velocity = Velocity of the plane
Substituting the given values:
Thrust = 1.2 kN (converted to N)
Velocity = 4000 m/s
Power = (1.2 * 10^3) * 4000 = 4.8 * 10^6 W ≈ 4.8 MW
(c) The steady plane speed at the maximum thrust can be determined using the thrust equation:
Thrust = 0.5 * ρ * V² * CD * Ap
Rearranging the equation to solve for the velocity:
V = sqrt((Thrust / (0.5 * ρ * CD * Ap)))
Substituting the given values:
ρ = 1.225 kg/m³ (from the air properties table)
Thrust = 1.2 kN (converted to N)
CD = 0.06
Ap = 9 m²
V = sqrt((1.2 * 10^3) / (0.5 * 1.225 * 0.06 * 9)) ≈ 54.8 m/s
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Select the suitable process for the following: - making cup-shaped parts. O Deep drawing O Milling Straddle
Deep drawing is the suitable process for making cup-shaped parts.
Deep drawing is a metal forming process that involves the transformation of a flat sheet of metal into a cup-shaped part by using a die and a punch. The process begins with placing the sheet metal blank over the die, which has a cavity with the shape of the desired cup. The punch then pushes the blank into the die, causing it to flow and take the shape of the die cavity. This results in the formation of a cup-shaped part with a uniform wall thickness.
Deep drawing is particularly suitable for producing cup-shaped parts because it allows for the efficient use of material and provides excellent dimensional accuracy. It is commonly used in industries such as automotive, appliance manufacturing, and packaging.
The deep drawing process offers several advantages. Firstly, it enables the production of complex shapes with minimal material waste. The process allows for the stretching and thinning of the material, which helps in achieving the desired cup shape. Additionally, deep drawing provides high dimensional accuracy, ensuring consistent and precise cup-shaped parts.
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B/ Put the following program in matrix standard form Min (z) = 10x₁+11x2 S.T. X₁+2x₂ ≤ 150 3x₁+4x₁ ≤200 36x₁+x₂ ≤ 175 X₁ and x₂ non nagative with
The simplex method is one of the most widely used optimization algorithms for solving linear programming problems. The simplex algorithm begins at a basic feasible solution.
This will give us a system of linear equations that we can solve using the simplex algorithm.
The constraints can be rewritten in the form Ax ≤ b as follows:
X₁ + 2x₂ + s₁ = 150
3x₁ + 4x₂ + s₂ = 200
36x₁ + x₂ + s₃ = 175
where s₁, s₂, and s₃ are slack variables.
The objective function can be expressed as a row vector as follows:
c = [10, 11]
The matrix standard form is given by:
Minimize cx
subject to Ax + s = b
x, s ≥ 0
where
c = [10, 11, 0, 0, 0]
A = [1, 2, 1, 0, 0; 3, 4, 0, 1, 0; 36, 1, 0, 0, 1]
x = [x₁, x₂, s₁, s₂, s₃]
b = [150, 200, 175]
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Line Balance Rate tells us how well a line is balanced. W
orkstation 1 Cycle Time is 2 min Workstation 2 Cycle Time is 4 min Workstation 3 Cycle Time is 6 min Workstation 4 Cycle Time is 4.5 min Workstation 5 Cycle Time is 3 min What is the Line Balance Rate %? Where is the bottleneck? Based on the Line Balance Rate result, what is your recommendation to improve the LBR%? Why?
Line balance rate tells us how well a line is balanced. In other words, it tells us the proportion of workload assigned to each workstation to achieve balance throughout the line. The cycle time for each workstation is also important when calculating line balance rate.
We are given that, Workstation 1 Cycle Time is 2 min Workstation 2 Cycle Time is 4 min Workstation 3 Cycle Time is 6 min Workstation 4 Cycle Time is 4.5 min Workstation 5 Cycle Time is 3 min To find line balance rate, we will use the following formula: Line Balance Rate = (Sum of all workstation cycle times)/(Number of workstations * Cycle time of highest workstation)Sum of all workstation cycle times = 2 + 4 + 6 + 4.5 + 3
= 19.5Cycle time of highest workstation
= 6Line Balance Rate
= (19.5)/(5 * 6)
= 0.65
= 65%Therefore, the line balance rate is 65%.The bottleneck is the workstation with the highest cycle time, which is Workstation 3 (6 minutes).
To improve the LBR%, we need to reduce the cycle time of workstation 3. This could be done by implementing the following methods:1. Change the process to reduce the cycle time2. Reduce the work content in the workstation3. Use automation to speed up the workstation .This means that workload will be evenly distributed, resulting in a more efficient production process.
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In the space below, sketch the high-frequency small-signal equivalent circuit of a MOS transistor. Assume that the body terminal is connected to the source. Identify (name) each parameter of the equivalent circuit. Also, write an expression for the small-signal gain vds/vgs(s) in terms of the small-signal parameters and the high-frequency cutoff frequency ωн. Clearly define ωн in terms of the resistance and capacitance parameters.
The high-frequency small-signal equivalent circuit of a MOS transistor typically consists of the following components:
Small-signal voltage source (vgs): This represents the small-signal input voltage applied to the gate-source terminals of the transistor.
Small-signal current source (gm * vgs): This represents the transconductance of the transistor, where gm is the small-signal transconductance parameter and vgs is the small-signal input voltage.
Small-signal output resistance (ro): This represents the small-signal output resistance of the transistor.
Capacitances (Cgs, Cgd, and Cdb): These represent the various capacitances associated with the transistor's terminals, namely the gate-source capacitance (Cgs), gate-drain capacitance (Cgd), and drain-body capacitance (Cdb).
The small-signal gain (vds/vgs(s)) can be expressed as:
vds/vgs(s) = -gm * (ro || RD)
Where gm is the transconductance parameter, ro is the output resistance, RD is the load resistance, and || represents parallel combination.
The high-frequency cutoff frequency (ωн) can be defined in terms of the resistance and capacitance parameters as:
ωн = 1 / (ro * Cgd)
Where ro is the output resistance and Cgd is the gate-drain capacitance.
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Which one is correctly mentioned about specific heat?
The mass per unit volume
The amount of heat required to change the temperature of a specific volume of substance one degree
The amount of heat that must be added or removed from one pound of substance to change its temperature by one degree.
The measure of the average kinetic energy
The correct statement about specific heat is "The amount of heat required to change the temperature of a specific volume of substance one degree. "Specific heat is defined as the amount of heat energy required to increase the temperature of a unit mass of a substance by 1 degree Celsius or Kelvin.
It is a property of the substance and is dependent on factors like temperature, pressure, and composition. The specific heat is denoted by the symbol c and is expressed in units of joules per kilogram per degree Celsius (J/kg·°C). Specific heat is an essential concept in thermodynamics and plays a crucial role in heat transfer processes. The specific heat values of different substances vary widely, and they can be used to predict the thermal behavior of a substance under different conditions.The other options provided in the question are not correct statements about specific heat. Mass per unit volume is known as density and is not related to specific heat.
The amount of heat that must be added or removed from one pound of substance to change its temperature by one degree is the definition of a thermodynamic property called specific heat capacity. The measure of the average kinetic energy is known as temperature, and it is related to specific heat but is not the same thing.
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Set up your Word document in APA format. Create a title page with all required information. You will be adding to this document throughout.
After the title page, write the first body paragraphs for your research paper Aviation Safety. Statethe problemsandSolutions (ignore the abstract and introduction for now, as you will write those later). Write at least one paragraph per sub-point of the first two main points on your working outline, or 4 double-spaced body pages (whichever is longer).
You may find yourself making changes to the content - that is fine, but do not focus too heavily on revision and editing, as that will come later. Be sure to use section headings as needed, and include properly formatted in-text source citations where needed (your references page will be created later).
APA format requires a title page that contains the title of the paper, the author's name, the name of the school, the course, and the date. The title page should also include a running head and a page number in the top right corner.
The body of the paper should begin on a new page, with the title of the paper at the top of the page. The first body paragraph should state the problems and solutions related to aviation safety. The problems could include human error, mechanical failure, weather, and other factors that can lead to accidents.
Each of the first two main points on the working outline should be addressed in at least one paragraph, with section headings as needed. Properly formatted in-text citations should be used as needed, and a reference page will be created later.
The body of the paper should be at least four double-spaced pages, or longer if needed to cover all the sub-points of the first two main points on the working outline. The abstract and introduction should be written later, after the body of the paper is complete.
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Consider a steel wire of length 295 cm and with a diameter of 0.25 mm. (a) Calculate the cross-sectional area of the wire (b) A load of 9.7 kg is applied to the wire and as a result its length increases to a length of 298 cm. Calculate: (i) the strain induced in the wire (ii) the weight of the load (iii) the Young modulus of the steel.
Given:Length of steel wire = 295 cm Diameter of steel wire = 0.25 mm Load applied on wire = 9.7 kgFinal length of steel wire = 298 cm.(a) Calculation of Cross-Sectional area of steel wire.
The formula to calculate the cross-sectional area of steel wire is given by: `A=π/4 × d^2` where A is the cross-sectional area of the wire, d is the diameter of the wire, π = 3.14.A=π/4 × d^2= 3.14/4 × (0.25 mm)^2 = 0.0491 mm^2Therefore, the cross-sectional area of the steel wire is 0.0491 mm^2.(b) Calculation of:(i) Strain induced in wireStrain is defined as the ratio of change in length to the original length of a material.
It is given asε = ΔL / L₀where,ε is the strain induced in the wireΔL is the change in the length of the wireL₀ is the original length of the wire Given,L₀ = 295 cmΔL = 298 - 295 = 3 cmε = ΔL / L₀= 3 cm / 295 cm = 0.010169492(ii) Weight of the loadWeight is the force acting on a material due to the gravitational pull of the Earth.
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Name 3 differences that you would observe between the
cold worked and recystalized microstructures
In metals and alloys, cold working and recrystallization are two common heat treatment techniques.
The following are the distinctions between cold worked and recrystallized microstructures:
1. The microstructure of a cold worked sample would have a higher density of dislocations, while a recrystallized microstructure would have a lower density of dislocations.
2. Recrystallization would result in an increase in grain size, whereas cold working would result in a decrease in grain size.
3. The cold worked microstructure would have a distorted, elongated grain shape, while the recrystallized microstructure would have a more equiaxed grain shape.
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EE317 / BER3043 Microprocessor Systems BEE2073 Microcontroller and Embedded System ASSIGNMENT Submission Date: Monday 08/08/2022 1. Design an automatic temperature controller using PIC 18 F452 microcontroller and suitable I/O devices. Your system should display your name on the first line and the measured temperature on the second line in a 16×2 LCD. - The system should turn on a heater (you can represent it using filament lamp output in your simulation) if the measured temperature is below the set level. - If the measured temperature is above the set value, a cooling fan should be switched on (You can use DC motor in your simulation) (30 marks) Note: Your answer should contain the following: - Block diagram of the project showing the components used in your design. (5 marks) - Description of the input/output you have used in your design and a brief description of the input/output ports of the microcontroller you have used to connect the components like switches, LCD and the range of measurement of voltage. (5 marks) - Flowchart or Algorithm showing the basic operation of the PIC microcontroller program (5 marks) - The code of your PIC program in C using mikroC Pro compiler with appropriate comments. (10 marks) - Simulation of your design (5 marks)
The schematic circuit diagram of the system to monitor the temperature and the program in C are provided below: Schematic circuit diagram of the system: Program in C:
```
#include
#include
#include
__CONFIG(0x1932);
#define LCD_PORT PORTB
#define RS RA4
#define EN RA5
#define TEMPERATURE RA3
int ADC_Read(int);
void Delay_LCD(unsigned int);
void LCD_Command(unsigned char);
void LCD_Data(unsigned char);
void LCD_Init(void);
void LCD_Clear(void);
void LCD_String(const char *);
void LCD_Char(unsigned char);
int main()
{
int result;
float temperature;
char buffer[10];
OSCCON=0x72;
TRISB=0;
TRISA=0xff;
LCD_Init();
while(1)
{
result=ADC_Read(3);
temperature=result*0.48828125; //0.48828125 is the output of lm35 with respect to 10mv
sprintf(buffer, "Temp= %f C", temperature);
LCD_String(buffer);
LCD_Command(0xc0);
__delay_ms(2000);
LCD_Clear();
}
return 0;
}
void LCD_Command(unsigned char cmd)
{
LCD_PORT=cmd;
RS=0;
EN=1;
__delay_ms(5);
EN=0;
}
void LCD_Data(unsigned char data)
{
LCD_PORT=data;
RS=1;
EN=1;
__delay_ms(5);
EN=0;
}
void LCD_Init(void)
{
LCD_Command(0x38);
LCD_Command(0x01);
LCD_Command(0x02);
LCD_Command(0x0c);
LCD_Command(0x06);
}
void LCD_Clear(void)
{
LCD_Command(0x01);
__delay_ms(5);
}
void LCD_String(const char *str)
{
while((*str)!=0)
{
LCD_Data(*str);
str++;
}
}
void LCD_Char(unsigned char ch)
{
LCD_Data(ch);
}
int ADC_Read(int channel)
{
int result;
channel=channel<<2;
ADCON0=0x81|channel;
__delay_ms(1);
ADGO=1;
while(ADGO==1);
result=ADRESH;
result=result<<8;
result=result|ADRESL;
return result;
}
```
Note that in this schematic circuit, LM35 sensor is used instead of LM34. They are quite similar, so the only difference is the output sensitivity. It should also be noted that the program in C language is written for PIC16F877A.
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Explain how outflow compression and inlet compression occur
Outflow compression and inlet compression are two processes that occur in fluid flow. These terms refer to the change in pressure and velocity that occurs.
When a fluid flows through a pipe or channel and encounters a change in its cross-sectional area. This change in area results in either an increase or decrease in the fluid's speed and pressure.Inlet compression occurs when a fluid flows into a smaller area.
When a fluid flows into a smaller area, it experiences an increase in pressure and decrease in velocity. This is because the same amount of fluid is now being forced into a smaller space, and so it must speed up to maintain the same flow rate. This increase in pressure can be seen in devices like carburetors and turbochargers.
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You have a "floating" discharge temperature from 52 to 60 F. Your design space conditions are 70/50% RH. Do you need to override the "floating" discharge to control upper humidity? Explain your answer. (Note: In good practice, "floating" is typically based on outside air dew point and the above is usually not a problem.)
In the given scenario, where the floating discharge temperature is between 52°F to 60°F, and the design space conditions are 70/50% RH, there is a need to override the floating discharge to control upper humidity. The term "floating" discharge temperature describes the temperature of the air being supplied by the air handling unit (AHU) varies with changes in outdoor conditions. In other words, the AHU's supply air temperature is not fixed but fluctuates based on outdoor air conditions.
Design space conditions refer to the set of temperature and relative humidity conditions that a given room or facility is designed to achieve and maintain. These conditions depend on the intended use of the space. For instance, a hospital room may have different design space conditions than a cleanroom in a pharmaceutical facility.The purpose of overriding the floating discharge temperature in this scenario is to control the upper humidity in the space. If the discharge temperature is floating and the outdoor air conditions change, it may lead to increased humidity levels in the room. High humidity can be problematic for some applications or processes.
To avoid this, the AHU's discharge temperature may need to be lowered to reduce the moisture levels in the space.In summary, overriding the floating discharge temperature to control upper humidity is necessary in the given scenario because the fluctuating supply air temperature may result in increased humidity levels in the space, which can be problematic.
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Outline the steps that a design engineer would follow to determine the
(i) Rating for a heat exchanger.
(ii) The sizing of a heat exchanger.
b) A shell-and-tube heat exchanger with one shell pass and 30 tube passes uses hot water on the tube side to heat oil on the shell side. The single copper tube has inner and outer diameters of 20 and 24 mm and a length per pass of 3 m. The water enters at 97°C and 0.3 kg/s and leaves at 37°C. Inlet and outlet temperatures of the oil are 10 degrees C and 47°C. What is the average convection coefficient for the tube outer surface?
(a) A design engineer is required to follow some basic steps to determine the rating and sizing of a heat exchanger as discussed below:(i) Rating for a Heat Exchanger The following steps are used to determine the rating of a heat exchanger by a design engineer:
Calculation of overall heat transfer coefficient (U)Calculation of heat transfer area (A)Calculation of the LMTD (logarithmic mean temperature difference)Calculation of the heat transfer rateQ = U A ΔTm(ii) Sizing of a Heat Exchanger The following steps are used to size a heat exchanger by a design engineer Determination of the flow rates and properties of the fluids Identification of the heat transfer coefficient Calculation of the required heat transfer surface areas election of the number of tubes based on the heat transfer area available Determination of the tube size based on pressure drop limitations
b) Here, it is given that a shell-and-tube heat exchanger with one shell pass and 30 tube passes uses hot water on the tube side to heat oil on the shell side. The single copper tube has inner and outer diameters of 20 and 24 mm and a length per pass of 3 m. 4.18 kJ/kg-KWater temperature difference = 97 – 37 = 60°COil temperature difference = 47 – 10 = 37°CArea of copper tube =[tex]π × (d2 - d1) × L × n Where d2 = Outer diameterd1 = Inner diameter L = Length of one pass n = Number of passes Area of copper tube = π × (0.0242 - 0.0202) × 3 × 30= 0.5313 m2Heat flow rate = m × Cp × ΔT= 0.3 × 4.18 × 60= 75.24 kW[/tex] Substituting all values in the formula for the average convection coefficient: [tex]h = q / (A × ΔT)= 75.24 / (0.5313 × 37)= 400.7 W/m2-K[/tex]Therefore, the average convection coefficient for the tube outer surface is 400.7 W/m2-K.
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7.22 An NMOS differential pair is biased by a current ΚΩ. source I = 0.2 mA having an output resistance Rsₛ = 100 kΩ. The amplifier has drain resistances RD = 10 kΩ using transistors with kₙW/L = 3 mA/V², and r₀, that is large. (a) If the output is taken single-endedly, find |Ad|, |Acm|, and CMRR. (b) If the output is taken differentially and there is a 1% mis- match between the drain resistances, find |Ad|, |Acm|, and CMRR.
Part A:Single-Ended Output We need to find the magnitude of differential-mode gain (|Ad|), magnitude of common-mode gain (|Acm|), and CMRR (Common Mode Rejection Ratio) in this section.
From the given information:
[tex]kₙW/L = 3 mA/V²,[/tex]
[tex]I = 0.2 mA,[/tex]
Rsₛ = 100 kΩ,
and RD = 10 kΩ.1. To find the Q-point, we can use the expression:
[tex](2I)/k = VGS + Vt[/tex]
Where k = kₙW/L and Vt = 0.7 V Substituting the given values, we get:
k = 3 mA/V²,
I = 0.2 mA,
Vt = 0.7 VThus, the Q-point is:
[tex]VGS = (2 × 0.2 mA × 1000 Ω)/3 mA + 0.7 V[/tex]
= 1.07 V2.
Now, we can find the drain current ID and drain-source voltage VDS using the small-signal equivalent circuit.ID = (1/2) × [tex]k(VGS - Vt)² = 0.299 m[/tex]
AVDS = VDD - ID(RD + Rs)
[tex]= 6 V - 0.299 mA(10 kΩ + 100 kΩ)[/tex]
= 2.701 V3.
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Consider an insulated chamber with two equally sized compartments that are separated from each other by a removable partition. Initially one of the compartments is assumed to be evacuated completely while the other is filled with a mole of an ideal gas under standard atmospheric conditions. Now consider that the partition is removed so that the gas can expand to fill the two chambers. (a) Will there be a change in the temperature of the gas? Explain. (b) Compute the value of the entropy change.
(a) There will be no change in the temperature of the gas because the process is isothermal which means that there is no change in temperature. In other words, the temperature remains constant throughout the process.
(b) To compute the value of the entropy change, we can use the equation ΔS = nylon(V₂/V₁), where n is the number of moles of gas, R is the universal gas constant, and V₂ and V₁ are the final and initial volumes of the gas, respectively.
Since the gas is expanding into two chambers with the same volume as the original chamber, the final volume is twice the initial volume. Thus, we can write:ΔS = 2) We know that n = 1 mole (given in the problem) and R = 8.314 J/(mol K) (universal gas constant).
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1. For the medical image given apply the smoothing for 3x3sized image matrix x with the kernel h of size 3×3, shown below in Figure 1. and compute the pixel value of the output image applying padding Original 1 2 3 5 6 4 7 8 9 IMAGE 3*3 figure 1 0 1 0 1 0 1 0 1 0 KERNAL 3*3
The output image with padding will be as follows:1 2 3 4 4 5 7 8 9.
In order to apply the smoothing for 3x3 sized image matrix x with the kernel h of size 3×3, shown below in Figure 1, the steps involved are as follows:First, the matrix needs to be padded. It is assumed that we are applying a zero padding, which adds a border of zeros around the original matrix. For instance, for a 3x3 matrix, we would end up with a 5x5 matrix.Second, we apply the kernel h to each of the individual pixels in the matrix. The kernel is a set of values that we will apply to each pixel in the image. Each element of the kernel will be multiplied by the corresponding pixel in the image. The result of each of these multiplications will be summed up, and that sum will be placed in the output matrix.
The original image is of size 3x3, which is too small for many applications. In order to apply the kernel, we first need to pad the image. The padded image will be 5x5 in size. The kernel is also of size 3x3, and it will be applied to each pixel in the image. The kernel is shown below in Figure 1.Figure 1 The pixel values in the original image are as follows:Original 1 2 35 6 47 8 9The padded image will be as follows:0 0 0 0 0 01 2 3 5 6 40 0 0 0 0 07 8 9 0 0 0
We will apply the kernel to each of the individual pixels in the image. The kernel is shown below in Figure 1.0 1 0 1 0 1 0 1 0
We will apply the kernel to each pixel by multiplying each element in the kernel by the corresponding pixel in the image. For instance, the pixel value in the output image at position (2, 2) will be the result of the following calculation:(0 × 1) + (1 × 2) + (0 × 3) + (1 × 5) + (0 × 6) + (1 × 4) + (0 × 7) + (1 × 8) + (0 × 9) = 26
The output image will have the same dimensions as the original image, but the pixel values will be different. The output image will be as follows:1 2 3 4 4 5 7 8 9
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Describe the difference between engineering stress-strain and true stress-strain relationships. Why analysis of true stress - true strain relationships is important?
Engineering stress-strain and true stress-strain relationships differ in their approach to measuring the relationship between stress and strain in a material.
Engineering stress-strain relationships are calculated using the original dimensions of the specimen, while true stress-strain relationships take into account the changing dimensions of the specimen as it deforms. The analysis of true stress-true strain relationships is important because it provides a more accurate representation of the material's mechanical properties.
Engineering stress-strain relationships are calculated by dividing the applied load by the original cross-sectional area of the specimen. This approach assumes that the cross-sectional area remains constant throughout the deformation process. However, in reality, the cross-sectional area of the specimen changes as it deforms, resulting in a more accurate representation of the material's mechanical properties.
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Are the following points part of the (200) plane? a) (1/2, 0, 0); b) (-1/3, 0, 0); c) (0, 1, 0) CHE 3260 Problem Set #3 Crystallography 1) A) Determine the percent ionic character in a K-Br bond. B) Determine the oxidation state of K in KBr. C) Determine the oxidation state of Br in KBr. 2) Find the appropriate radii for A) K in KBr and B) Br in KBr. 3) Determine the coordination number of A) K in KBr and B) Br in KBr. 4) Determine the most likely cubic crystal structure for KBr, and sketch it. 5) Calculate the lattice parameter, a. 6) Determine the number of K and Br ions in the KBr unit cell. 7) Determine KBr's bulk density. 8) Sketch the (200) plane of KBr. 9) Calculate the planar density of the (200) plane of KBr, expressed as a decimal.
Option (a) and option (c) are part of the (200) axial plane of KBr while option (b) is not a part of it.
The plane (200) of KBr has its indices parallel to the x and y-axis. Let's find if the given points are part of the (200) plane of KBr.a) (1/2,0,0)In a cubic unit cell, the length of the edges and the angles between the edges are equal. Also, since the x-axis of the (200) plane is parallel to the edge of the unit cell, the x-coordinate of this point has to be equal to some fraction of the edge length of the unit cell.
Therefore, the x-coordinate of point a, (1/2), has to be equal to 1/2 times the length of the unit cell edge. This is possible only if the length of the unit cell edge is equal to 1. So, point a is a part of the (200) plane of KBr.b) (-1/3,0,0)The x-coordinate of point b is -1/3 which means the length of the unit cell edge has to be equal to 3 units. But the unit cell edge length of KBr cannot be equal to 3. Therefore, point b is not a part of the (200) plane of KBr.c) (0,1,0)The y-coordinate of point c is 1 which means the length of the unit cell edge has to be equal to 1 unit. Since this is possible, point c is a part of the (200) plane of KBr.
Hence, option (a) and option (c) are part of the (200) plane of KBr while option (b) is not a part of it.
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2. Find the inverse Laplace transform of F (s) = 2e-0.5s s²-65+13 S-1 s²-2s+2 for t>o.
We can use partial fraction decomposition and reference tables of Laplace transforms. To find the inverse Laplace transform of F (s) = 2e-0.5s s²-65+13 S-1 s²-2s+2 for t>o.
Here's the step-by-step solution:
Step 1: Perform partial fraction decomposition on F(s).F(s) = (2e^(-0.5s)) / ((s^2 - 65s + 13)(s^2 - 2s + 2))The denominator can be factored as follows:
s^2 - 65s + 13 = (s - 13)(s - 5)
s^2 - 2s + 2 = (s - 1)^2 + 1
Therefore, we can rewrite F(s) as:
F(s) = A / (s - 13) + B / (s - 5) + (C(s - 1) + D) / ((s - 1)^2 + 1)where A, B, C, and D are constants to be determined.
Step 2: Solve for the constants A, B, C, and D.Multiplying both sides of the equation by the denominator, we get:
2e^(-0.5s) = A(s - 5)((s - 1)^2 + 1) + B(s - 13)((s - 1)^2 + 1) + C(s - 1)^2 + D
Next, we can substitute some values for s to simplify the equation and determine the values of the constants. Let's choose s = 13, s = 5, and s = 1.For s = 13:
2e^(-0.5(13)) = A(13 - 5)((13 - 1)^2 + 1) + B(13 - 13)((13 - 1)^2 + 1) + C(13 - 1)^2 + De^(-6.5) = 8A + 144C + DFor s = 5:
2e^(-0.5(5)) = A(5 - 5)((5 - 1)^2 + 1) + B(5 - 13)((5 - 1)^2 + 1) + C(5 - 1)^2 + D2e^(-2.5) = 16A - 8B + 16C + DFor s = 1:
2e^(-0.5) = A(1 - 5)((1 - 1)^2 + 1) + B(1 - 13)((1 - 1)^2 + 1) + C(1 - 1)^2 + D2e^(-0.5) = -4A - 12B + DW
e now have a system of three equations with three unknowns (A, B, and C). Solve this system to find the values of the constants.
Step 3: Use Laplace transform tables to find the inverse Laplace transform. Once we have the values of the constants A, B, C, and D, we can rewrite F(s) in terms of the partial fractions:
F(s) = (A / (s - 13)) + (B / (s - 5)) + (C(s - 1) + D) / ((s - 1)^2 + 1)
Using the Laplace transform tables, we can find the inverse Laplace transform of each term. The inverse Laplace transforms of (s - a)^(-n) and e^(as) are well-known and can be found in the tables.
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Q3 :( 3 Marks) Draw the circuit of three phase transmission line. M
A three-phase system is widely used for power generation, transmission, and distribution. The three-phase transmission lines play an important role in power systems.
Here is a brief overview of a three-phase transmission line.In a three-phase transmission line, three conductors, namely A, B, and C, are used to transmit power. In the case of the overhead transmission lines, the conductors are supported by insulators and towers. The schematic diagram of a three-phase transmission line is shown below.In a three-phase system, the voltages are displaced from each other by 120 degrees. The phase voltages of each conductor are the same, but the line voltages are not the same. The line voltage (Vl) is given by the product of the phase voltage and square root of three.
Therefore, Vl = √3 x Vp. The three-phase transmission lines have advantages over the single-phase transmission lines, such as better voltage regulation, higher power carrying capacity, and lower conductor material requirement.
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2. Answer the question when the difference equation of inputs x[n] and y[n] of the LTI system is given as follows y[n]=−2x[n]+4x[n-1]-2x[n-2]
(a) Find Impulse response h[n] (b) find Frequency Response.
(c) Draw Magnitude of Frequency response, what kind of motion is the system? (d) Find the output when it is an input. 3. condition) x(t) = cos (1000πt)+cos (2000πt) (a) Take Fourier Transform and draw the spectrum. (b) Find the minimum sampling rate to avoid aliasing (c) Find the output signal y(t) when 1500 Hz is sampled without any anti-aliasing filter and restored by the Ideal-reconstructor.
(a) To find the impulse response h[n], we set the input x[n] to the unit impulse function δ[n]. Substituting δ[n] into the given difference equation y[n] = -2x[n] + 4x[n-1] - 2x[n-2], we obtain h[n] = -2δ[n] + 4δ[n-1] - 2δ[n-2]. Therefore, the impulse response of the system is h[n] = -2δ[n] + 4δ[n-1] - 2δ[n-2].
(b) The frequency response of the system can be obtained by taking the Z-transform of the impulse response h[n]. Applying the Z-transform to each term, we get H(z) = -2 + 4z⁻¹ - 2z⁻². This is the transfer function of the system in the Z-domain.
(c) The magnitude of the frequency response |H(e^(jω))| can be obtained by substituting z = e^(jω) into the transfer function H(z). Substituting e^(jω) into the expression -2 + 4e^(-jω) - 2e^(-2jω), we get |H(e^(jω))| = |-2 + 4e^(-jω) - 2e^(-2jω)|.
(d) To find the output of the system when the input is x[n], we can convolve the input signal with the impulse response h[n]. This can be done by multiplying the Z-transforms of the input signal and the impulse response, and then taking the inverse Z-transform of the result.
3. (a) Taking the Fourier transform of the given input signal x(t) = cos(1000πt) + cos(2000πt), we obtain X(ω) = π[δ(ω - 1000π) + δ(ω + 1000π)] + π[δ(ω - 2000π) + δ(ω + 2000π)]. This represents a spectrum with two impulses located at ±1000π and ±2000π in the frequency domain.
(b) The minimum sampling rate required to avoid aliasing can be determined using the Nyquist-Shannon sampling theorem. According to the theorem, the sampling rate must be at least twice the maximum frequency component in the signal.
(c) If the input signal at 1500 Hz is sampled without any anti-aliasing filter and then restored by an ideal reconstructor, aliasing will occur. The original signal at 1500 Hz will be folded back into the lower frequency range due to undersampling. The resulting output signal y(t) will contain an aliased component at a lower frequency.
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