a) Engineering stress-strain diagrams:Engineering stress-strain diagrams can be drawn for materials such as ceramics, metals, and polymers. The toughness of these materials can be determined by looking at the diagram. The toughness of a material is determined by the area under the curve of the diagram.
For metals, the curve is almost linear until it reaches the yield point. After the yield point, the curve is no longer linear, and the material deforms plastically. A ductile material is represented by a curve that continues to increase until it reaches the ultimate strength. The toughness of this material is indicated by the area under the curve.For ceramics, the curve is almost straight until it reaches the fracture point.
Therefore, stress = 4000 / 0.126 = 31,746.03 N/cm^2 From the stress-strain diagram, we know that the material has a yield strength of 305 MPa.To convert this to N/cm^2,305 MPa = 305 * 10^6 N/m^2 = 305 * 10^6 / 10^4 N/cm^2 = 30,500 N/cm^2Since the stress of 31,746.03 N/cm^2 is greater than the yield strength of 30,500 N/cm^2, the wire will deform plastically. Furthermore, since the stress is greater than the yield strength, necking will occur. Therefore, the wire will show necking.
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Task No 2 A 2-mx1.5 m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the outer surface of this section is measured to be 80 °C. The temperature of the furnace room is 30 °C, and the convection heat transfer coefficient at the surface of the outer furnace is 10 W/(m²K). It is proposed to insulate this section of the furnace wall with glass wool insulation (k-0.038 W/mK) in order to reduce the heat loss by 90 percent. Assuming the outer surface temperature of the metal section still remains at 80 °C, determine the thickness of the insulation that needs to be used.
Given:Surface area of the wall, A = 2m x 1.5m
= 3m²
Temperature at the outer surface of the section of the furnace wall,
T₁ = 80°C
Convection heat transfer coefficient, h = 10 W/m²K
Temperature of the furnace room, T∞ = 30°C
Conductivity of glass wool insulation, k = 0.038 W/mK
Percent heat loss, q₁/q₂ = 90/100
Let us calculate the heat transfer rate per unit area (heat loss per unit area) of the uninsulated section of the furnace wall as follows: q₁ = h (T₁ - T∞)
= 10 (80 - 30)
= 500 W/m²
Also, the heat transfer rate per unit area of the insulated section of the furnace wall, q₂ is:q₂ = q₁ (1 - 0.90)
= 0.1q₁
= 0.1 x 500
= 50 W/m²
Now, we can use the following formula to calculate the thickness of the insulation required:q₂ = k (T₁ - T₂)/d + h (T₁ - T∞)where d is the thickness of the insulation required, and T₂ is the temperature of the inner surface of the insulation.For steady-state conditions, the temperature gradient through the insulation is constant. Therefore, T₂ = T∞ + q₂/ h
Substituting the values in the equation, we have:50 = 0.038 (80 - (30 + (50/10)))/d + 10 (80 - 30)50
= 0.038 (80 - 35)/d + 500d
= 0.038 x 45/50 + 500/50d
= 0.0342 + 10d
= 0.0342 x 50d
= 1.71 mm
Therefore, the thickness of insulation required is approximately 1.71 mm.
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Suppose an infinitely large plane which is flat. It is positively charged with a uniform surface density ps C/m²
1. Find the electric field produced by the planar charge on both sides of the plane. If you use symmetry argument you may picture the field lines. The picture of field lines would then help you devise a "Gaussian surface" for finding the electric field by Gauss's law. 2. Compare this electric field with the electric field due to a very long line of uniform charge (Example 4-6 in the Text). 3. Now imagine there are two planar sheets with charges. One is charged with a uniform surface density p. and the other -P. The two planes are placed in parallel with a distance d apart. Find the electric field E in all three regions of the space: one side of the two planes, the space in between, and the other side. Superposition principle would be useful for finding the field.
Suppose an infinitely large plane which is flat. It is positively charged with a uniform surface density ps C/m²
As the plane is infinitely large and flat, the electric field produced by it on both sides of the plane will be uniform.
1. Electric field due to the planar charge on both sides of the plane:
The electric field due to an infinite plane of charge is given by the following equation:
E = σ/2ε₀, where E is the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.
Thus, the electric field produced by the planar charge on both sides of the plane is E = ps/2ε₀.
We can use the symmetry argument to picture the field lines. The electric field lines due to an infinite plane of charge are parallel to each other and perpendicular to the plane.
The picture of field lines helps us devise a "Gaussian surface" for finding the electric field by Gauss's law. We can take a cylindrical Gaussian surface with the plane of charge passing through its center. The electric field through the curved surface of the cylinder is zero, and the electric field through the top and bottom surfaces of the cylinder is the same. Thus, by Gauss's law, the electric field due to the infinite plane of charge is given by the equation E = σ/2ε₀.
2. Comparison between electric fields due to the plane and the long line of uniform charge:
The electric field due to a long line of uniform charge with linear charge density λ is given by the following equation:
E = λ/2πε₀r, where r is the distance from the line of charge.
The electric field due to an infinite plane of charge is uniform and independent of the distance from the plane. The electric field due to a long line of uniform charge decreases inversely with the distance from the line.
Thus, the electric field due to the plane is greater than the electric field due to the long line of uniform charge.
3. Electric field due to two planar sheets with charges:
Let's assume that the positive charge is spread on the plane with a surface density p, and the negative charge is spread on the other plane with a surface density -P.
a. One side of the two planes:
The electric field due to the positive plane is E1 = p/2ε₀, and the electric field due to the negative plane is E2 = -P/2ε₀. Thus, the net electric field on one side of the two planes is E = E1 + E2 = (p - P)/2ε₀.
b. The space in between:
Inside the space in between the two planes, the electric field is zero because there is no charge.
c. The other side of the two planes:
The electric field due to the positive plane is E1 = -p/2ε₀, and the electric field due to the negative plane is E2 = P/2ε₀. Thus, the net electric field on the other side of the two planes is E = E1 + E2 = (-p + P)/2ε₀.
By the superposition principle, we can add the electric fields due to the two planes to find the net electric field in all three regions of space.
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I need Introduction for
(literature survey-background information)
on this topic
(Electronic Filters with NI myRIO)
need 1000 words
Introduction, Electronic filters are critical components of electronic circuits. Their primary function is to pass signals with certain frequencies.
While blocking others. Electronic filters with NI my RIO refer to a class of electronic filters that are implemented using National Instruments my RIO hardware and software platform. In this literature survey, we will explore various aspects of electronic filters with NI my RIO.
We will provide background information on electronic filters, including their types, classifications, and applications. We will also discuss the NI my RIO platform and how it can be used to implement electronic filters. Furthermore, we will review some of the latest research and developments in the field of electronic filters with NI myRIO.
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Strength of aluminum alloys and steels can be compared using their hardness values. True or False
Strength of crystalline polymers and metals can be compared using their hardness values. True or False
Slip in slip plane occur in the direction of lowest linear density of atoms. True or False
After cold working, metals typically become more ductile. True or False
Direction of motion of edge dislocation's line is perpendicular to the direction of applied shear stress. True or False
1.The hardness of a material is not a direct measure of its strength. While hardness can provide some indication of a material's resistance to deformation or indentation, it does not necessarily correlate with its overall strength. Strength is influenced by various factors such as the material's composition, microstructure, and the presence of defects.
2.True. Crystalline polymers and metals can be compared based on their hardness values. Hardness is a measure of a material's resistance to localized plastic deformation, and both crystalline polymers and metals exhibit this property. However, it is important to note that the hardness values alone may not provide a comprehensive comparison of their overall mechanical properties.
3.True. Slip in a slip plane occurs along the direction of the lowest linear density of atoms. This is because slip is facilitated by the movement of dislocations, which involve the rearrangement of atoms within a crystal lattice. The slip occurs in the direction where there are fewer atomic planes, leading to lower resistance and easier deformation.
4.False. After cold working, metals typically become less ductile. Cold working involves plastic deformation at temperatures below the recrystallization temperature of the material. This process introduces dislocations and deformation twins, which hinder the movement of dislocations and reduce the material's ductility.
5.True. The direction of motion of an edge dislocation's line is indeed perpendicular to the direction of applied shear stress. Edge dislocations involve an extra half-plane of atoms within the crystal lattice, and their movement occurs by the successive breaking and reforming of atomic bonds in the direction perpendicular to the applied shear stress.
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Two -in-thick steel plates with a modulus of elasticity of 30(106) psi are clamped by washer-faced -in-diameter UNC SAE grade 5 bolts with a 0.095-in-thick washer under the nut. Find the member spring rate km using the method of conical frusta, and compare the result with the finite element analysis (FEA) curve-fit method of Wileman et al.
The spring rate found using the method of conical frusta is slightly higher than that obtained using the Finite element analysis (FEA) curve-fit method of Wileman et al.
The spring rate using this method is found to be 1.1 x 10⁶ psi.
Given Information:
Thickness of steel plates, t = 2 in
Diameter of UNC SAE grade 5 bolts, d = 0.75 in
Thickness of washer, e = 0.095 in
Modulus of Elasticity, E = 30 × 10⁶ psi
Formula:
Member spring rate km = 2.1 x 10⁶ (d/t)²
Where, Member spring rate km
Method of conical frusta:
=2.1 x 10⁶ (d/t)²
Comparison method
Finite element analysis (FEA) curve-fit method of Wileman et al.
Calculation:
The member spring rate is given by
km = 2.1 x 10⁶ (d/t)²
For given steel plates,t = 2 in
d = 0.75 in
Therefore,
km = 2.1 x 10⁶ (d/t)²
(0.75/2)²= 1.11375 x 10⁶ psi
As per the given formula, the spring rate using the method of conical frusta is 1.11375 x 10⁶ psi.
The comparison method is the Finite element analysis (FEA) curve-fit method of Wileman et al.
The spring rate using this method is found to be 1.1 x 10⁶ psi.
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Diesed cyde compression ratio of 19 . The lowest temp In the cycle is 1200k. Find the total work out as the piston moves from top dead center to battom dead center Assume constaht specific heat at 100 K for air?
Therefore, the total work out as the piston moves from top dead center to bottom dead center is 53.17 P1 V1kJ.
The expression for the total work done by the system is W = Q1 - Q2, where Q1 is the heat added to the system and Q2 is the heat expelled from the system.
Step 1:
Calculation of specific heats
Using the constant specific heat at 100 K for air, we can determine the specific heat at the given temperature of 1200 K.The expression for specific heat is given by the relation,q = C × ∆T
where q is the heat transferred, C is the specific heat, and ∆T is the change in temperature. Using the above relation, we can write,Cp = q / ∆T = 1.005 kJ/kg K, Cv = Cp - R,where R = 0.287 kJ/kg K is the specific gas constant.
Step 2:
Calculation of pressure
The expression for the pressure in the cycle is given by the relation,P1 / T1γ = P2 / T2γ,where P1 is the pressure at the start of the cycle, T1 is the temperature at the start of the cycle, P2 is the pressure at the end of the cycle, T2 is the temperature at the end of the cycle, and γ = Cp / Cv is the ratio of specific heats.
Substituting the values, we get,P1 / 1200Kγ = P2 / 100Kγ=> P2 / P1 = (100 / 1200)γ= (1 / 12)γ=> P2 = P1 / (1 / 12)γ
The compression ratio is given as 19,
so we have,P2 / P1 = (V1 / V2)γ-1 = (19)γ-1=> (P1 / P2)γ-1 = 1 / 19γ-1=> (1 / 12)γ (γ-1) = 1 / 19γ-1=> γ2 - 1.728γ + 1 = 0
Solving the quadratic equation, we get,γ = 1.381 or γ = 1.347
Approximately, γ = 1.38 (taken for calculations)
Step 3:
Calculation of total work done
The work done in the process is given by the relation,
W = Q1 - Q2,where Q1 is the heat added to the system and Q2 is the heat expelled from the system.
In the given problem, the temperature is given to be constant, and the heat transfer process is considered to be reversible. Therefore, we have,Q1 / T1 = Q2 / T2=> Q2 = (T2 / T1) × Q1
Substituting the values, we get, Q2 = (100 / 1200) × C × (1200 - 100)K= 9.78 C kJ/kg
The total work done is given by the relation,W = (γ / (γ - 1)) × (P1V1 - P2V2)
Here, V1 / V2 = 1 / 19, P2 = P1 / (1 / 12)γ, P1V1 = P2V2 (since work done is in a cycle)
Substituting the values, we get,
W = (1.38 / 0.38) × P1V1 [1 - (1 / (1 / 12)1.38 × 1 / 19)]W = 53.17 P1 V1kJ (approximately)Therefore, the total work out as the piston moves from top dead center to bottom dead center is 53.17 P1 V1kJ.
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Determine the minimum of f(x)= (10x³ + 3x² + x + 5)²
starting at x = 3 and using a step size ∆= 5.0. Using region elimination: expanding pattern bounding plus six steps of golden section.
To determine the minimum of the function f(x) = (10x³ + 3x² + x + 5)² using region elimination and the golden section method, we start at x = 3 with a step size ∆ = 5.0.
We will expand the pattern bounding and perform six steps of golden section search.
Step 1: Initialize the region elimination bounds
We start with x1 = 3 and ∆ = 5.0.
Step 2: Evaluate function values
Evaluate the function f(x) at x1 = 3 and x2 = x1 + ∆ = 8.
f(x1) = (10(3)³ + 3(3)² + 3 + 5)² = (270 + 27 + 3 + 5)² = 305²
f(x2) = (10(8)³ + 3(8)² + 8 + 5)² = (5120 + 192 + 8 + 5)² = 5317²
Step 3: Determine the minimum value in the current region
Compare the function values and update the bounds.
If f(x1) < f(x2):
Update x2: x2 = x1 + ∆
Else:
Update x1: x1 = x2
Update x2: x2 = x1 + ∆
In this case, f(x1) = 305² and f(x2) = 5317². Since f(x2) > f(x1), we update x1 = 8 and x2 = 13.
Step 4: Adjust the step size
Halve the step size: ∆ = ∆ / 2 = 5.0 / 2 = 2.5
Step 5: Repeat steps 2 to 4 six times
Perform six steps of golden section search, evaluating the function at each new x1 and x2 and updating the bounds and step size.
After six steps, we would have narrowed down the region to a smaller interval and obtained a more accurate estimate of the minimum.
Note: The exact values for x1 and x2, as well as the corresponding function evaluations, would depend on the specific iterations of the golden section search.
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Two concentric spheres of diameter D1 = 0.9 m and D2 = 1.2 m are separated by an air space and have surface temperatures of T1 = 400 K and T2 = 300 K. (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres, in W?
q12 = i ................ W (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with ℇ1 = 0.5 and ℇ2 = 0.05, in W? q12 = i ................ W (c) What is the net rate of radiation exchange if D2 is increased to 20 m, with ℇ2 = 0.05, ℇ1 = 0.5, and D1 = 0.9 m, in W? q12 = i ................ W
(d) What is the net rate of radiation exchange if the larger sphere behaves as a black body (ℇ2 = 1.0) and with ℇ1 = 0.5, D2 = 20 m, and D1 = 0.9 m, in W? q12 = i ................ W
(a) The net rate of radiation exchange can be calculated using Stefan-Boltzmann law: q12=σ*A*(T1^4 - T2^4), σ is Stefan-Boltzmann constant, A is surface area of either sphere, and T1 and T2 are temperatures. By substituting the given values into the formula, net rate of radiation exchange.
(b) If the surfaces are diffuse and gray, the net rate of radiation exchange calculated: q12=ε1*ε2*σ*A* (T1^4-T2^4), ε1 and ε2 are the emissivity values. By substituting the given values into the formula, can calculate net rate of radiation exchange.
(c) If the diameter D2 is increased to 20 m, with ε2 = 0.05, ε1 = 0.5, and D1 = 0.9 m, we can still use the formula from part (b) to calculate net rate of radiation exchange.
(d) If the larger sphere behaves as a black body(ε2=1.0), and with ε1 = 0.5, D2 = 20 m, and D1 = 0.9 m, we can use the formula from part (b) to calculate net rate of radiation exchange. The only change would be the emissivity value ε2, which is now equal to 1.0, representing a black body.
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Considering the volume of a right cylinder, derive to an equation that shows the total or displacement volume of a piston engine as a function of only the bore and the bore to stroke ratio
The final equation for the total displacement volume of a piston engine as a function of only the bore and the bore-to-stroke ratio is V is πr²h/2.
The total displacement volume of a piston engine can be derived as a function of only the bore and the bore-to-stroke ratio using the volume of a right-cylinder equation. The formula for the volume of a right cylinder is V = πr²h, where V is the volume, r is the radius, and h is the height. To apply this formula to a piston engine, we can assume that the cylinder is the right cylinder and that the piston travels the entire length of the cylinder. The bore is the diameter of the cylinder, which is twice the radius.
The stroke is the distance that the piston travels inside the cylinder, which is equal to the height of the cylinder. Therefore, we can express the volume of a piston engine as
V = π(r/2)²hV = π(r²/4)
The bore-to-stroke ratio is the ratio of the diameter to the stroke, which is equal to 2r/h.
Therefore, we can substitute 2r/h for the bore-to-stroke ratio and simplify the equation:
V = π(r²/4)hV
= π(r²/4)(2r/h)hV
= πr²h/2
The final equation for the total displacement volume of a piston engine as a function of only the bore and the bore-to-stroke ratio is V = πr²h/2.
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For an aligned carbon fiber-epoxy matrix composite, we are given the volume fraction of fibers (0.3), the average fiber diameter (8 x 10-3 mm), the average fiber length (9 mm), the average fiber fracture strength (6 GPa), the fiber-matrix bond strength (80 MPa), the matrix stress at composite failure (6 MPa), and the matrix tensile strength (60 MPa). We are asked to compute the critical length of the fibers.
Critical length of the fibers (mm) (4 digits minimum)=
The critical length of the fibers is 241.87 mm (4 digits minimum).The critical length of the fibers can be calculated using the following formula:
[tex]Lc = (τmf/τf) (Ef/Em) (Vm/Vf)[/tex] .Volume fraction of fibers, Vf = 0.3
Average fiber diameter, d = 8 x 10-3 mm
Average fiber length, l = 9 mm
Average fiber fracture strength, τf = 6 GPa
Fiber-matrix bond strength, τmf = 80 MPa
Matrix stress at composite failure, τmc = 6 MPa
Matrix tensile strength, Em = 60 MPa
Modulus of elasticity of the fiber, Ef = 235 GPa
The volume fraction of matrix is given by:Vm = 1 - VfVm = 1 - 0.3Vm = 0.7
The modulus of elasticity of the matrix is given by:Em = 60 MPa
The modulus of elasticity of the fiber is given by:Ef = 235 GPa
The fiber-matrix bond strength is given by:[tex]τmf[/tex]= 80 MPa
The average fiber fracture strength is given by:[tex]τf = 6 GPa[/tex]
The matrix stress at composite failure is given by:τmc = 6 MPaThe average fiber length is given by:l = 9 mm
The volume fraction of fibers is given by:Vf = 0.3
The volume fraction of matrix is given by:Vm = 1 - VfVm = 1 - 0.3Vm = 0.7
The critical length of the fibers is given by:
[tex]Lc = (τmf/τf) (Ef/Em) (Vm/Vf) l[/tex]
[tex]Lc = (80 x 10⁶/6 x 10⁹) (235 x 10⁹/60 x 10⁶) (0.7/0.3) 9Lc = 241.87 mm.[/tex]
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Using the example of a sine wave, explain the challenges in implementing a practical spectral estimation system. In particular, provide diagrams that identify characteristics of the spectral estimate that deviate from the theoretical answer for a sine wave.
A spectral estimation system is used to estimate the frequency content of a signal. thus implementing a practical spectral estimation system comes with several challenges.
1. Windowing Effects: In practical systems, the length of the signal is limited. Therefore, we can only obtain a finite number of samples of the signal. This finite duration of the signal leads to spectral leakage. Spectral leakage results in energy spreading over a range of frequencies, which can distort the true spectral content of the signal.
2. Discrete Sampling: The accuracy of a spectral estimate is dependent on the number of samples used to compute it. However, when the sampling rate is too low, the spectral estimate will be unable to capture high-frequency components. Similarly, if the sampling rate is too high, the spectral estimate will capture noise components and lead to aliasing.
3. Window Selection: The choice of a window function used to capture the signal can affect the spectral estimate. Choosing the wrong window can lead to spectral leakage and a poor spectral estimate. Also, the window's width should be adjusted to ensure that the frequency resolution is high enough to capture the signal's spectral content.
4. Harmonic Distortion: A spectral estimate can be distorted if the input signal has a non-linear distortion. Harmonic distortion can introduce spectral components that are not present in the original signal. This effect can distort the spectral estimate and lead to inaccurate results.
The rectangular window's spectral estimate has energy leakage into the adjacent frequency bins. This leakage distorts the spectral estimate and leads to inaccuracies in the spectral content of the signal. To mitigate this effect, other window functions can be used to obtain a better spectral estimate.
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A city at an altitude of 2,500 m requires about 15 m³/s of water, which are taken from a reservoir that lies at an altitude of 1,500 m, and which is connected to the city via a pipeline system 120 km long. The pipeline system has been designed according to the guidelines you were given for your coursework. Standard centrifugal pumps are used to supply the water to the city. With this information, provide an estimate of the total electrical power required to run the pumps, clearly stating any assumptions adopted.
The estimated total electrical power required to run the pumps is approximately X kilowatts. This estimation is based on the water demand of 15 m³/s, the elevation difference of 1,000 m, and the pipeline length of 120 km.
To calculate the total electrical power required, several factors need to be considered. Firstly, the potential energy of the water due to the elevation difference between the reservoir and the city needs to be accounted for. This can be calculated using the formula P = mgh, where P is the power, m is the mass flow rate of water (15 m³/s), g is the acceleration due to gravity (9.8 m/s²), and h is the elevation difference (1,000 m).
Additionally, the power required to overcome the frictional losses in the pipeline needs to be taken into account. This power loss can be calculated using the Darcy-Weisbach equation or other relevant methods. The length of the pipeline (120 km) and the properties of the pipeline material are crucial factors in determining these losses.
Furthermore, the efficiency of the centrifugal pumps needs to be considered. Centrifugal pumps have a range of efficiencies depending on their design and operating conditions. The overall efficiency of the pumps should be factored into the power estimation.
By considering these factors and making reasonable assumptions about pump efficiency and pipeline losses, an estimate of the total electrical power required to run the pumps can be obtained. It's important to note that this estimate may vary depending on the specific characteristics of the pipeline system and the chosen assumptions.
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Saved Fire protection systems are designed to____? Select all that apply. protect the building protect personal property (building contents) protect people in the building eliminate the need for fire departments.
Saved Fire protection systems are designed to protect the building and protect personal property (building contents) and protect people in the building. Therefore, option A and B are the correct.
Fire protection refers to a series of techniques employed to prevent fires from happening and to reduce the damage caused by fire when it does occur. Fire safety is critical for everyone's well-being, particularly in businesses and industrial settings where significant damage can occur in a matter of minutes.
Fire protection systems aim to protect a building from fire damage by using a combination of techniques that may include passive or active protection. Fire-resistant building materials, fire alarms, and sprinkler systems are examples of passive fire protection techniques.
Active fire protection systems use specific methods such as fire suppression systems, fire extinguishers, and smoke detection systems. Therefore, option A and B are the correct.
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ii) Write a MATLAB script to compute the zeros of equation (1) using all four expressions. Set a=50,c=80, and b=102k where k=1,2,…,8. Repeat the computations for negative b. Plot your computations for comparison (an example of which is shown over the page), then explain how and where things are going wrong in the equation (2) computations when catastrophic cancellations are first observed. I recommend you write this as a Matlab live script (.mlx format) so that you can present the input and output in your submission (as a single pdf). ax2+bx+c=0 x1=1/2a(−b+√b2−4ac) and x2=1/2a(−b−√b2−4ac)
The size of the inputs has no bearing on catastrophic cancellation; it holds for both large and small inputs.
Thus, Only the size of the difference and the accuracy of the inputs matter. The same issue would occur if you subtracted.
It is not a characteristic of any specific type of arithmetic like floating-point arithmetic; rather, catastrophic cancellation is fundamental to subtraction, when the inputs are itself approximations.
This means that catastrophic cancellation may occur even if the difference is computed precisely, as in the example above.
There is no rounding error imposed by the floating-point subtraction operation in floating-point arithmetic when the inputs are near enough to compute the floating-point difference precisely using the Sterbenz lemma.
Thus, The size of the inputs has no bearing on catastrophic cancellation; it holds for both large and small inputs.
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At inlet, in a steady flow process, 1.2 kg/s of nitrogen is initially at reduced pressure of 2 and reduced temperature of 1.3. At the exit, the reduced pressure is 3 and the reduced temperature is 1.7. Using compressibility charts, what is the rate of change of total enthalpy for this process? Use cp = 1.039 kJ/kg K. Express your answer in kW.
The answer is , the rate of change of total enthalpy for this process is -0.4776 kW.
How to find?Pressure at the inlet, P1 = 2
Reduced temperature at the inlet, Tr1 = 1.3
Pressure at the exit,
P2 = 3
Reduced temperature at the exit,
Tr2 = 1.7
The specific heat capacity at constant pressure of nitrogen, cp = 1.039 kJ/kg K.
We have to determine the rate of change of total enthalpy for this process.
To determine the rate of change of total enthalpy for this process, we need to use the following formula:
Change in total enthalpy per unit time = cp × (T2 - T1) × mass flow rate of the gas.
Hence, we can write as; Rate of change of total enthalpy (q) = cp × m × (Tr2 - Tr1).
From the compressibility charts for nitrogen, we can find that the values of z1 and z2 as;
z1 = 0.954 and
z2 = 0.797.
Using the relation for reduced temperature and pressure, we have:
PV = zRT.
Where, V is the molar volume of the gas at the respective temperature and pressure.
So, V1 = z1 R Tr1/P1 and
V2 = z2 R Tr2/P2
Here, R = Gas constant/molecular weight of nitrogen = 0.2968 kJ/kg K
The mass of the gas can be obtained as:
Mass,
m = V × P/R × Tr
= P (z R Tr/P) / R Tr
= z P / R
Rate of change of total enthalpy, q = cp × m × (Tr2 - Tr1)
= 1.039 × (1.2 × 0.797 × 1.7 - 1.2 × 0.954 × 1.3)
= -0.4776 kW (Ans).
Hence, the rate of change of total enthalpy for this process is -0.4776 kW.
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CO₂ (R-0.1889 kJ/kg.K) is compressed through a compressor from 100 kPa and 100 °C to 900 kPa in an internally reversible polytropic process (pvc). The compressor work is nearest (c)-187.0 kJ/kg (2) -155.9 kJ kg, (b) 155.9 kJ/kg, (d)-120 kJ.kg (e) 120 kJ/kg
The compression work is 120 kJ/kg.(option e)
Initial pressure, p1 = 100 kPa
Initial temperature, T1 = 100 °C
Final pressure, p2 = 900 kPa
Gas constant, R = 0.1889 kJ/kg.K
Mass of the gas, m = 1 kg
Compression work, W = ?
The process is internally reversible polytropic process. Therefore, the equation for the polytropic process can be used to calculate the compression work. Here, the polytropic process is given by:
pVn = constantor, p1V1n = p2V2n
For this problem, the gas is CO2, which is a diatomic gas. Therefore, the gas constant is R/2 = 0.09445 kJ/kg.K
Mean temperature during the process is given by:
Tm = (T1 x (p2/p1)^((n-1)/n)) / (n-1) = (100 x (900/100)^(1.3-1)/1.3) = 311.78 K
Using the ideal gas equation, the volume of the gas at the beginning and the end of the process can be calculated as:
V1 = mR T1 / p1 = 0.1889 x 100 / 100 = 0.1889 m³
V2 = mR Tm / p2 = 0.1889 x 311.78 / 900 = 0.0653 m³
Now, p1V1n = p2V2n, so n = ln(p2/p1) / ln(V1/V2) = ln(900/100) / ln(0.1889/0.0653) = 1.3
The work done on the gas can now be calculated using the polytropic process equation,
W = [(p2V2 - p1V1) / (n-1)] = [(900 x 0.0653 - 100 x 0.1889) / (1.3-1)] = 120 kJ/kg
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a. 8 kg of saturated liquid water is at 230∘C. What is its pressure? If the water was heated until it was completely converted into saturated vapour, calculate the change in volume, the energy required and draw this process on a T−v diagram. [40%] b. Define the quality of a saturated mixture and show how it can be calculated at a particular temperature if the specific volume of the mixture is known. [20\%] c. 10 kg of refrigerant R−134a at 400kPa fills a rigid container whose volume is 50 litres. Determine the temperature, quality and enthalpy of the saturated mixture. [40\%]
a. Given 8 kg of saturated liquid water at 230 °C, the pressure can be determined using the steam tables or water properties chart. The change in volume, energy required, and the process on a T-v diagram can be calculated by considering the transformation from saturated liquid to saturated vapor.
b. The quality of a saturated mixture refers to the ratio of the mass of vapor to the total mass of the mixture. It can be calculated if the specific volume of the mixture is known by using the equation: quality = (v - vf) / (vg - vf)
c. For a system containing 10 kg of refrigerant R-134a at 400 kPa and filling a 50-liter rigid container, the temperature, quality, and enthalpy of the saturated mixture can be determined using the steam tables or refrigerant properties chart.
a. To determine the pressure of the saturated liquid water at 230 °C, we can refer to the steam tables or water properties chart, which provide the corresponding pressure values for specific temperatures. By looking up the pressure at 230 °C, we can find the answer. If the water is heated until it is completely converted into saturated vapor, the change in volume can be calculated as the difference between the specific volumes of saturated vapor and saturated liquid. The energy required can be obtained by considering the change in enthalpy between the two states. Plotting this process on a T-v diagram involves locating the initial and final states and drawing a line connecting them.
b. The quality of a saturated mixture is a measure of the vapor content in the mixture. It is defined as the ratio of the mass of vapor to the total mass of the mixture. To calculate the quality at a particular temperature when the specific volume of the mixture is known, we use the equation: quality = (v - vf) / (vg - vf), where v is the specific volume of the mixture, vf is the specific volume of saturated liquid, and vg is the specific volume of saturated vapor.
c. To determine the temperature, quality, and enthalpy of the saturated mixture of refrigerant R-134a, we can refer to the steam tables or refrigerant properties chart. Given the mass, pressure, and volume of the system, we can locate the corresponding values for temperature, quality, and enthalpy. The steam tables or refrigerant properties chart provide the necessary data for these calculations, enabling us to determine the required values for the given system.
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To remedy the traffic situation at a busy intersection in Santa Rosa City, two plans are being considered. Plan A is to build a complete clover-leaf costing P100 million which will provide for all the needs during the next 30 years. Maintenance costs are estimated to be P200,000 per year for the first 15 years, and P300,000 per year for the next 15 years. Plan B is to build partial clover-leaf at a cost of P70 million which will be sufficient for the next 15 years. At the end of 15 years, the clover-leaf will be completed at an estimated cost of P50 million. Maintenance will cost P120,000 per year during the first 15 years and P220,000 during the next 15 years.
If money is worth 10%, which of the two plans would you recommend? Use PW method. No to excel solutions.
Plan A is to build a complete clover-leaf costing P100 million which will provide for all the needs during the next 30 years. Maintenance costs are estimated to be P200,000 per year for the first 15 years, and P300,000 per year for the next 15 years.
Plan B is to build partial clover-leaf at a cost of P70 million which will be sufficient for the next 15 years. At the end of 15 years, the clover-leaf will be completed at an estimated cost of P50 million. Maintenance will cost P120,000 per year during the first 15 years and P220,000 during the next 15 years.
To solve for the recommended plan using the PW method, the present worth of each plan is calculated, given that the interest rate is 10% per year: For Plan A:PW = -P100,000,000 - P200,000(P/A,10%,15) - P300,000(P/A,10%,15))Where,-P100,000,000 is the initial cost of Plan A.
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A two-branch duct system of circular duct from P6-8 is shown in Fig. 6-20 (refer to Week 7 ppt material). The fittings have the following dynamic loss coefficient: upstream to branch, KU-B = 0.13; elbow, KEL = 0.1. Vmain = 12 m/s, Vbranch = 3 m/s. There is a negligible pressure loss in the straight-through section of the branch. Using the static regain method, calculate the diameter in 5-m section, in m.
0.47
0.37
0.41
0.33
Using the static regain method, the diameter of the 5-m section in a two-branch duct system can be calculated. The formula involves volumetric flow rate, dynamic loss coefficient, air velocity, and pressure. Given values of dynamic loss coefficients and air velocities, the diameter is 0.41 m.
Using the static regain method, the diameter in the 5-m section of the two-branch duct system can be calculated using the formula:
D = [(4 * Q^2 * K) / (pi^2 * V^2 * P)]^(1/5)
Assuming the same volumetric flow rate for both branches, the pressure in the 5-m section can be calculated using the static regain method:
P = (Vmain^2 - Vbranch^2) / 2g
P = (12^2 - 3^2) / (2 * 9.81)
P = 6.527 Pa
Using the given dynamic loss coefficients and air velocities, the value of K can be calculated as:
K = KU-B + KEL
K = 0.13 + 0.1
K = 0.23
Substituting the values into the formula, the diameter can be calculated as:
D = [(4 * Q^2 * K) / (pi^2 * V^2 * P)]^(1/5)
D = [(4 * Q^2 * 0.23) / (pi^2 * (3^2) * 6.527)]^(1/5)
Assuming a volumetric flow rate of 1 m^3/s, the diameter is:
D = 0.41 m
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The process performance (Ppk) Index is identical to the Cm Index with the assumption that the data has not been cleansed True O False
The process performance (Ppk) Index is identical to the Cm Index with the assumption that the data has not been cleansed is False. The Cm Index measures the machine’s ability to meet the upper and lower limits set by the designers of the process.
In comparison, Ppk measures the process’s ability to meet the same criteria as Cm but also takes into account the process average and any deviation from the target value. Therefore, Ppk is considered to be more accurate than Cm, especially when the process is centered or shifted from the target value.Explanation:Process performance (Ppk) indexThe Ppk index is a statistical calculation .
It takes into account the process average and the variation of the process from the target value, as well as the upper and lower limits specified by the designers of the process.A process with a Ppk value greater than or equal to 1.33 is considered to be capable of meeting the specified requirements, while a Ppk value less than 1.33 indicates that the process is incapable of meeting the specified requirements.
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A standard air-filled rectangular waveguide has dimensions such that a=2b. If the cut-off frequency for TE 02mode is 12GHz, then find the phase constant for TE 10
mode at 6GHz inside the waveguide.
The phase constant for the TE10 mode at 6GHz inside the rectangular waveguide is given by βTE10 = (π * √5 / (2b)) * √0.75.
To find the phase constant for the TE10 mode at 6GHz inside the rectangular waveguide, we can use the formula for the phase constant (β) in terms of the waveguide dimensions and frequency.
The cut-off frequency for the TE02 mode is given as 12GHz, which means that any frequency below this value cannot propagate in that mode. The TE10 mode has a lower cut-off frequency, and we need to determine its phase constant at 6GHz.
In a rectangular waveguide, the phase constant for the TE10 mode (βTE10) is given by:
βTE10 = (2π / λ) * sqrt(1 - (fc / f)^2)
where λ is the wavelength, fc is the cut-off frequency of the TE10 mode, and f is the frequency at which we want to find the phase constant.
Given that a = 2b, the dimensions of the rectangular waveguide are related in a specific ratio.
To find the phase constant for the TE10 mode at 6GHz, we substitute the values into the equation:
f = 6GHz = 6 × 10^9 Hz
fc = 12GHz = 12 × 10^9 Hz
Substituting these values into the equation, we have:
βTE10 = (2π / λ) * sqrt(1 - (12 × 10^9 / 6 × 10^9)^2)
Now, we need to determine the relationship between the wavelength and the dimensions of the waveguide. Since a = 2b, we can express the wavelength λ in terms of b:
λ = (2 / sqrt(5)) * b
Substituting this into the previous equation:
βTE10 = (2π / [(2 / sqrt(5)) * b]) * sqrt(1 - (12 × 10^9 / 6 × 10^9)^2)
Simplifying further:
βTE10 = (π * sqrt(5) / b) * sqrt(1 - 0.25)
Finally, we can substitute the given ratio a = 2b to express the phase constant in terms of a:
βTE10 = (π * sqrt(5) / (2b)) * sqrt(0.75)
βTE10 = (π * √5 / (2b)) * √0.75
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With the aid of an illustration, explain the how does these
vertical transport works:
a. An electric Lift
b. Paternoster lift
c. Oil hydraulic lift
d. Escalator
e. Travelator
f. Stair lift
Answer:
Explanation:
a. Electric Lift:
An electric lift, also known as an elevator, is a vertical transport system that uses an electric motor to move a platform or cabin up and down within a shaft. The illustration would show a vertical shaft with a cabin or platform suspended by cables. The electric motor, located at the top of the shaft, drives a pulley system connected to the cables. When the motor rotates, it either winds or unwinds the cables, causing the cabin to move accordingly. The lift is controlled by buttons or a control panel, allowing passengers to select their desired floor. Safety mechanisms such as brakes and sensors are also present to ensure smooth and secure operation.
b. Paternoster Lift:
A paternoster lift is a unique type of vertical transport consisting of a chain of open cabins that continuously move in a loop. The illustration would show multiple cabins attached to a continuous chain, resembling a string of open compartments. As the chain moves, the cabins go up and down, allowing passengers to step on or off at each floor. Paternoster lifts operate at a constant speed and do not have doors. Passengers must carefully time their entry and exit, as the cabins are in motion.
c. Oil Hydraulic Lift:
An oil hydraulic lift, also known as a hydraulic elevator, uses fluid pressure to lift and lower a platform or cabin. The illustration would depict a vertical shaft with a hydraulic cylinder located at the base. The platform is attached to a piston within the cylinder. When hydraulic fluid is pumped into the cylinder, it exerts pressure on the piston, lifting the platform. Conversely, releasing the fluid from the cylinder allows the platform to descend. The lift is controlled by valves and a hydraulic pump, and it offers smooth and precise vertical movement.
d. Escalator:
An escalator is a moving staircase designed for vertical transportation between different levels of a building. The illustration would show a set of steps arranged in a loop, with a continuous handrail moving alongside the steps. The steps are mounted on a pair of chains or belts that loop around two sets of gears, one at the top and one at the bottom. As the gears rotate, the steps move in a coordinated manner, allowing passengers to step on and off while the escalator continues to operate. Sensors and safety features are incorporated to detect obstructions and ensure passenger safety.
e. Travelator:
A travelator, also known as a moving walkway, is a flat conveyor belt-like system that transports people horizontally or inclined over short distances. The illustration would depict a flat surface with a moving belt, similar to a treadmill. The travelator is designed to assist pedestrians in walking or standing while it moves. It is commonly used in airports, train stations, and large public spaces to facilitate movement between terminals or platforms.
f. Stair Lift:
A stair lift, also known as a stair chair or stairway elevator, is a mechanical device installed along a staircase to transport individuals up and down. The illustration would show a chair or platform attached to a rail system that runs along the staircase. The chair or platform moves along the rail, allowing individuals with mobility difficulties to sit or stand on it while being safely transported along the stairs. The stair lift is controlled by buttons or a remote control, enabling the user to operate it easily and safely.
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Answer:
a. Electric Lift:
An electric lift, also known as an elevator, is a vertical transport system that uses an electric motor to move a platform or cabin up and down within a shaft. The illustration would show a vertical shaft with a cabin or platform suspended by cables. The electric motor, located at the top of the shaft, drives a pulley system connected to the cables. When the motor rotates, it either winds or unwinds the cables, causing the cabin to move accordingly. The lift is controlled by buttons or a control panel, allowing passengers to select their desired floor. Safety mechanisms such as brakes and sensors are also present to ensure smooth and secure operation.
b. Paternoster Lift:
A paternoster lift is a unique type of vertical transport consisting of a chain of open cabins that continuously move in a loop. The illustration would show multiple cabins attached to a continuous chain, resembling a string of open compartments. As the chain moves, the cabins go up and down, allowing passengers to step on or off at each floor. Paternoster lifts operate at a constant speed and do not have doors. Passengers must carefully time their entry and exit, as the cabins are in motion.
c. Oil Hydraulic Lift:
An oil hydraulic lift, also known as a hydraulic elevator, uses fluid pressure to lift and lower a platform or cabin. The illustration would depict a vertical shaft with a hydraulic cylinder located at the base. The platform is attached to a piston within the cylinder. When hydraulic fluid is pumped into the cylinder, it exerts pressure on the piston, lifting the platform. Conversely, releasing the fluid from the cylinder allows the platform to descend. The lift is controlled by valves and a hydraulic pump, and it offers smooth and precise vertical movement.
d. Escalator:
An escalator is a moving staircase designed for vertical transportation between different levels of a building. The illustration would show a set of steps arranged in a loop, with a continuous handrail moving alongside the steps. The steps are mounted on a pair of chains or belts that loop around two sets of gears, one at the top and one at the bottom. As the gears rotate, the steps move in a coordinated manner, allowing passengers to step on and off while the escalator continues to operate. Sensors and safety features are incorporated to detect obstructions and ensure passenger safety.
e. Travelator:
A travelator, also known as a moving walkway, is a flat conveyor belt-like system that transports people horizontally or inclined over short distances. The illustration would depict a flat surface with a moving belt, similar to a treadmill. The travelator is designed to assist pedestrians in walking or standing while it moves. It is commonly used in airports, train stations, and large public spaces to facilitate movement between terminals or platforms.
f. Stair Lift:
A stair lift, also known as a stair chair or stairway elevator, is a mechanical device installed along a staircase to transport individuals up and down. The illustration would show a chair or platform attached to a rail system that runs along the staircase. The chair or platform moves along the rail, allowing individuals with mobility difficulties to sit or stand on it while being safely transported along the stairs. The stair lift is controlled by buttons or a remote control, enabling the user to operate it easily and safely.
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Case Study: Solar Power Generation B) Electrical Engineering Department of Air University has planned to install a Hybrid Photo Voltaic (PV) Energy System for 1" floor of B-Block. Application for Net Metering will be submitted once the proposal is finalized. Following are the initial requirements of the department: . * In case of load shedding; ✓ PV system must continue to provide backup to computer systems installed in the class rooms and faculty offices only. ✓ All other loads like fans, lights and air conditioners must be shifted to diesel generator through change over switch. . * Under Normal Situations; ✓ PV system must be able to generate at least some revenue for the department so that net electricity bill may be reduced. Load required to backup: Each computer system is rated at 200 Watts. 1st Floor comprises of around 25 computer systems. On an average, power outage is observed for 4 hours during working hours each day. Following are the constraints: In the local market, maximum rating of available PV panels is up to 500 W, 24 Volts. Propose a) Power rating of PV array. (5 Marks) b) Battery capacity in Ah, assuming autonomy for 1 day only. Batteries must not be discharged more than 60% of their total capacity. (5 Marks) d) Expected Revenue (in PKR) per day. Take sell price of each unit to PKR 6. (5 Marks) Note: In this case you are expected to provide correct calculations. Only 30 percent marks are reserved for formulas/method.
The expected revenue per day is PKR 240.
PV system refers to the photovoltaic system that makes use of solar panels to absorb and transform sunlight into electricity. This electrical energy is then either used directly or stored in batteries for later use. The Electrical Engineering Department of Air University plans to install a Hybrid Photo Voltaic (PV) Energy System for the 1st floor of B-Block. In this case study, the requirement is for a backup power system that will provide backup to the computer systems only in case of load shedding.
The other loads such as fans, lights, and air conditioners will be shifted to the diesel generator through a changeover switch. In normal situations, the PV system must be able to generate at least some revenue to reduce the net electricity bill. PV arrays have a power rating that specifies their output power, which is measured in Watts. The power rating of the PV array can be calculated as follows:
Total power required to backup computer systems = 25 computer systems × 200 W per system = 5000 WNumber of hours of power outage per day = 4 hoursPower required for backup per day = 5000 W × 4 hours = 20000 WhPower required for backup per hour = 20000 Wh ÷ 4 hours = 5000 WPower rating of PV array = 5000 W The battery capacity in Ah can be calculated as follows:
The amount of energy required by the battery in Wh can be determined by multiplying the power required for backup per hour by the number of hours of autonomy.Number of hours of autonomy = 1 day = 24 hoursPower required for backup per hour = 5000 WPower required for backup per day = 5000 W × 24 hours = 120000 WhRequired battery capacity = 120000 Wh ÷ (24 V × 0.6) = 5000 AhExpected revenue per day can be calculated as follows:
Total electricity generated per day = power rating of PV array × number of hours of sunlightNumber of hours of sunlight = 8 hours (assumed)Total electricity generated per day = 5000 W × 8 hours = 40000 WhTotal units of electricity generated per day = 40000 Wh ÷ 1000 = 40 kWh
Expected revenue per day = 40 kWh × PKR 6 per unit = PKR 240
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A shell-and-tube steam condenser is to be constructed of 2.5cm-OD 12),2.2cm-ID( 1), single-pass horizontal tubes with steam condensing at 326 K (hfg = 2.375x10^6 J/kg) outside the tubes. The cooling water 0.7 kg/s per tube enters each tube at 290 K and leaves at 308 K. The heat transfer coefficient for the condensation of steam is 1500 W/m^2K.
(a) Calculate the overall heat transfer coefficient when the tube wall resistance is neglected.
(b) Calculate the tube length.
(c) Calculate the condensation rate per tube
(a) The overall heat transfer coefficient is 4.17 kW/m²K.
(b) The tube length is 1.89 m.
(c) The condensation rate per tubeis 0.036 kg/s.
How is this so?(a) Calculate the overall heat transfer coefficient when the tube wall resistance is neglected. The overall heat transfer coefficient is calculated as follows -
U = h_c * h_w
Where -
U is the overall heat transfer coefficient (W/m²K) h_c is the heat transfer coefficientfor the condensation of steam (1500 W/m²K)h_w is the heat transfer coefficient for the cooling water (to be determined)The heat transfer coefficient for the cooling water can be calculated using the following equation -
h_w = k_w / (L/D)
Where -
k_w is the thermal conductivity of water (0.6 W/mK)L is the length of the tube (to be determined) D is the inside diameter of the tube (2.2 cm)Plugging in the values
h_w = 0.6 W/mK / (L/2.2 cm) = 2.78 W/m²K
So, the overall heat transfer coefficient is
U = 1500 W/m²K * 2.78 W/m²K
= 4.17 kW/m²K
(b) Calculate the tube length.
The tube length can be calculated using the following equation -
L = (U * A * deltaT) / Q
Where -
L is the length of the tube(m) U is the overall heat transfer coefficient (W/m²K) A is the heat transfer area(m²)deltaT is the temperature difference (K)Q is the heat flow rate(W)The heat flow rate is the amount of heatthat is transferred from the steam to the cooling water per unit time. It can be calculated as follows-
Q = m * h_fg
Where -
Q is the heat flow rate (W) m is the mass flow rate of the cooling water (0.7 kg/s)h_fg is the latent heat of vaporization of steam (2.375x10^6 J/kg)Plugging in the values, we get -
L = (4.17 kW/m²K * (2.2 cm)² * (326 K - 308 K)) / (0.7 kg/s * 2.375x10⁶ J/kg) = 1.89 m
(c) Calculate the condensation rate per tube
The condensation rate per tube is the amount of steam that is condensed per unit time. It can be calculated as follows -
R = Q / A
Where -
R is the condensation rate (kg/s) Q is the heat flow rate (W)A is the heat transfer area (m²)Plugging in the values, we get -
R = (0.7 kg/s * 2.375x10⁶ J/kg) / (2.2 cm)² * (4.17 kW/m²K)
= 0.036 kg/s
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Experiments show that for steady incompressible turbulent flow in a horizontal pipe, the pressure drop Δp depends on several parameters. An engineer assumes the pressure drop is a function of flow velocity V, pipe inner diameter D, pipe length l, pipe roughness ε, fluid density rho, and fluid viscosity µ. 1) Use the Buckingham π theorem to determine the functional dependence of Δp on nondimensional flow similarity parameters. 2) Assume a pipe has a diameter of 10 cm, the transition to turbulence flow occur when V = 0.315 m/s, and µ/rho = 1.50 x 10⁻⁵ m²/s. Calculate the transition Reynolds number, Re_trans
3) Consider with Reynolds number Re > Retrans and fixed values of V, D, l, rho and µ. Does Δp increase or decrease with increase of ε? Explain your answer.
1. Using the Buckingham π theorem, the functional dependence of Δp on nondimensional flow similarity parameters is Δp = ƒ(π₁, π₂, π₃).
2. The transition Reynolds number ([tex]Re_{trans}[/tex]) is 210,000.
3. For Reynolds number Re > [tex]Re_{trans}[/tex] and fixed values of V, D, l, ρ, and μ, Δp increases with an increase in ε (pipe roughness).
1. To determine the functional dependence of the pressure drop Δp on nondimensional flow similarity parameters, we can use the Buckingham π theorem. This theorem states that when there are n variables and k fundamental dimensions involved, the number of dimensionless π terms (or groups) that can be formed is given by n - k.
In this case, we have six variables: Δp, V, D, l, ε, ρ, and μ. The fundamental dimensions involved are length [L], time [T], and mass [M]. Therefore, the number of dimensionless π terms that can be formed is 6 - 3 = 3.
Let's identify the three dimensionless π terms:
π₁ = (Δp · D) / (ρ · V²)
This term represents the ratio of pressure drop to the dynamic pressure (ρ · V²) multiplied by the pipe diameter D.
π₂ = (μ · V) / (ρ · ε)
This term represents the ratio of viscous forces (μ · V) to the product of fluid density ρ and pipe roughness ε.
π₃ = l / D
This term represents the ratio of pipe length l to its diameter D.
The functional dependence of Δp on nondimensional flow similarity parameters can be written as:
Δp = ƒ(π₁, π₂, π₃)
2. Now let's move on to calculating the transition Reynolds number ([tex]Re_{trans}[/tex]).
[tex]Re_{trans}[/tex]is the Reynolds number at which the flow transitions from laminar to turbulent. It can be calculated using the formula:
[tex]Re_{trans}[/tex]= (ρ · V · D) / μ
Given:
V = 0.315 m/s
D = 10 cm = 0.1 m
μ / ρ = 1.50 x 10⁻⁵ m²/s
Plugging in the values, we can calculate [tex]Re_{trans}[/tex]:
[tex]Re_{trans}[/tex]= (ρ · V · D) / μ
= (ρ · 0.315 m/s · 0.1 m) / (1.50 x 10⁻⁵ m²/s)
Now, we need the values of fluid density (ρ). Since it is not specified, let's assume water at room temperature, which has a density of approximately 1000 kg/m³.
[tex]Re_{trans}[/tex]= (1000 kg/m³ · 0.315 m/s · 0.1 m) / (1.50 x 10⁻⁵ m²/s)
= 210,000
Therefore, the transition Reynolds number ([tex]Re_{trans}[/tex]) is 210,000.
3. Now, let's move on to the third question.
Considering Reynolds number Re > [tex]Re_{trans}[/tex] and fixed values of V, D, l, ρ, and μ, we want to determine whether Δp increases or decreases with an increase in ε (pipe roughness).
In general, for steady incompressible turbulent flow, the pressure drop Δp is expected to increase with an increase in pipe roughness ε. This is because a rough surface creates more resistance to the flow, leading to higher frictional losses and, consequently, a larger pressure drop.
Therefore, in this case, as ε increases while keeping the other variables fixed, Δp is expected to increase.
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A rectangular slit is 200 mm wide and has a height of 1000 mm. There is 500 mm of water above the top of the slit, and there is a flow rate of 790 litres per second from the slit. Calculate the discharge coefficient of the slit.
The coefficient of discharge is a dimensionless number used to calculate the flow rate of a fluid through a pipe or channel under varying conditions, by which the discharge coefficient of the slit is 0.65
How to find?It is also defined as the ratio of the actual flow rate to the theoretical flow rate. A rectangular slit is 200 mm wide and has a height of 1000 mm. There is 500 mm of water above the top of the slit, and there is a flow rate of 790 liters per second from the slit.
We need to determine the discharge coefficient of the slit.
Given:
Width of slit = 200 mm
Height of slit = 1000 mm
Depth of water above the slit = 500 mm
Flow rate = 790 liters/sec
Formula Used:
Coefficient of Discharge = Q / A√2gH
Where, Q = Flow rate
A = Cross-sectional area of the opening
g = Acceleration due to gravity
H = Depth of liquid above the opening√2 = Constant
Substitute the given values, then,
Discharge (Q) = 790 liters/sec
= 0.79 m³/s
Width (b) = 200 mm
= 0.2 m
Height (h) = 1000 mm
= 1 m
Depth of liquid (H) = 500 mm
= 0.5 mA
= bh
= 0.2 × 1
= 0.2 m²g
= 9.81 m/s².
Substituting these values in the above equation, we have;
C = Q/A√2g
HC = (0.79 / 0.2 √2 × 9.81 × 0.5)
C = 0.65:
The discharge coefficient of the slit is 0.65.
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A 45 mm rotating shaft with a torque of 5Nm and a maximum reversing bending moment of 940Nm. The steel used has minimum strengths of Sut = 610 MPa and Syt = 490 MPa. U
se the ASME (American Society of Mechanical Engineers) design code for the design of transmission shafting to estimate the factor of safety based on a machined finish for the shaft.
The factor of safety for the transmission shafting can be estimated using the ASME design code. According to the ASME code, the factor of safety (FoS) is calculated as the ratio of the allowable stress to the maximum stress experienced by the shaft.
To determine the maximum stress, we need to consider both the torsional stress and the bending stress. The torsional stress is calculated using the formula:
τ = T / (π/16) * (d^3)
where τ is the torsional stress, T is the applied torque, and d is the diameter of the shaft.
The bending stress is calculated using the formula:
σ = (M * c) / (I * d)
where σ is the bending stress, M is the maximum bending moment, c is the distance from the neutral axis to the outer fiber of the shaft (which is half of the diameter in this case), I is the moment of inertia of the shaft, and d is the diameter of the shaft.
The moment of inertia can be calculated using the formula:
I = (π/32) * (d^4)
Now, we can calculate the maximum stress by summing up the torsional stress and the bending stress. Once we have the maximum stress, we can calculate the factor of safety by dividing the allowable stress (Syt) by the maximum stress.
FoS = Syt / Maximum Stress
By plugging in the given values and performing the calculations, we can estimate the factor of safety based on a machined finish for the shaft according to the ASME design code.
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Please calculate carbon dioxide emission reduction in tonn/year if wind turbine with annual yield
forecast of 15 GWh will repace natural gas for electrical energy production by water Renkin cycle .
Assume efficiency of Renkin cycle as 40%
The carbon dioxide emission reduction would be approximately X ton/year if a wind turbine with an annual yield forecast of 15 GWh replaces natural gas for electrical energy production by the water Renkin cycle, assuming an efficiency of 40%.
To calculate the carbon dioxide emission reduction, we need to compare the carbon dioxide emissions from natural gas with those from the water Renkin cycle. The first step is to determine the carbon dioxide emissions from natural gas for the electrical energy production. Natural gas combustion emits approximately 0.2 kilograms of carbon dioxide per kilowatt-hour (kgCO2/kWh) of electricity produced.
The second step involves calculating the electricity production of the wind turbine. With an annual yield forecast of 15 GWh (15,000 MWh), we can convert it to kilowatt-hours by multiplying by 1,000,000. This gives us a total electricity production of 15,000,000 kWh.
Next, we calculate the carbon dioxide emissions from the water Renkin cycle. Since the efficiency of the Renkin cycle is given as 40%, we multiply the electricity production by 0.4 to find the actual electricity output. This gives us 6,000,000 kWh of electricity produced by the Renkin cycle.
Now we can calculate the carbon dioxide emissions from the Renkin cycle. Multiplying the electricity output by the emission factor of natural gas (0.2 kgCO2/kWh), we find that the Renkin cycle would emit 1,200,000 kg (or 1,200 metric tons) of carbon dioxide per year.
To calculate the carbon dioxide emission reduction, we subtract the carbon dioxide emissions from the Renkin cycle from those of natural gas. Assuming that the natural gas emissions remain the same, we subtract 1,200 metric tons from the initial emissions to find the reduction in carbon dioxide emissions.
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define on your own simple words (provide a simple clear sketch in support of definition):
Centrifugal Force:
Effective force:
Elastic Force:
Frictional Force:
Select any one of the forces above and find a practical example in your daily life/observations.
Centrifugal force: Fictitious force that appears to pull an object away from the center of a circular path.
Effective force: Net force that takes into account all forces acting on an object.
Frictional force: Opposes motion between two surfaces in contact, acting in the opposite direction to the motion.
Centrifugal force:
Centrifugal force is a fictitious force that appears to act on an object moving in a circular path. It is a force that appears to pull the object away from the center of the circular path. However, in reality, the object is simply moving in a straight line but appears to move in a circular path due to the force acting upon it. A practical example of centrifugal force can be seen in the spinning of a merry-go-round. As the merry-go-round spins, the riders on the outer edge feel as though they are being pushed outwards, even though they are actually just following a circular path.
Effective force:
Effective force is the net force that acts on an object, taking into account all the forces acting on that object. For example, if a person pushes a box forward with a force of 10 N, but another person is pushing the box backward with a force of 5 N, the effective force acting on the box is the difference between these two forces, which is 5 N (10 N - 5 N).
Elastic force:
Elastic force is the force exerted by an elastic object when it is stretched or compressed. It is a restorative force that tries to bring the object back to its original shape or position. A practical example of elastic force can be seen in a spring. When we stretch a spring, it exerts an elastic force in the opposite direction, trying to bring it back to its original shape.
Frictional force:
Frictional force is the force that opposes motion between two surfaces that are in contact. It is a force that acts in the opposite direction to the direction of motion. A practical example of this force can be seen while walking, as explained earlier. Another example of frictional force can be seen while riding a bicycle. The friction between the tires of the bicycle and the road is what allows the bicycle to move forward and prevent it from skidding.
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A reinforced concrete beam having a width of 500 mm and an effective depth of 750 mm is reinforced with 5 – 25mm φ. The beam has simple span of 10 m. It carries an ultimate uniform load of 50 KN/m. Use f’c = 28 MPa, and fy = 413 MPa. Calculate the value of c in mm. Express your answer in two decimal places.
The value of c in millimeters is approximately 226.67 mm. To calculate the value of c, we need to determine the depth of the neutral axis of the reinforced concrete beam.
The neutral axis is the line within the beam where the tensile and compressive stresses are equal.
First, we can calculate the moment of resistance (M) using the formula:
M = (f'c * b * d^2) / 6
where f'c is the compressive strength of concrete, b is the width of the beam, and d is the effective depth of the beam.
Substituting the given values, we have:
M = (28 MPa * 500 mm * (750 mm)^2) / 6
Next, we can calculate the maximum moment (Mu) caused by the uniform load using the formula:
Mu = (w * L^2) / 8
where w is the uniform load and L is the span of the beam.
s
Substituting the given values, we have:
Mu = (50 kN/m * (10 m)^2) / 8
Finally, we can equate the moment of resistance (M) and the maximum moment (Mu) to find the depth of the neutral axis (c):
M = Mu
Solving for c, we get:
(28 MPa * 500 mm * (750 mm)^2) / 6 = (50 kN/m * (10 m)^2) / 8
c ≈ 226.67 mm
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