To determine stability using a Bode diagram, we analyze the gain margin and phase margin of the system.
(i) Gain Margin and Phase Margin:
The gain margin is the amount of gain that can be added to the system before it becomes unstable, while the phase margin is the amount of phase lag that can be introduced before the system becomes unstable.
To calculate the gain margin and phase margin, we need to plot the Bode diagram of the given open-loop transfer function.
(ii) Bode Diagram:
The Bode diagram consists of two plots: the magnitude plot and the phase plot.
For the given transfer function G(s) = 100/(s(1+0.1s)(1+0.2s)), we can rewrite it in the form G(s) = K/(s(s+a)(s+b)), where K = 100, a = 0.1, and b = 0.2.
On a semi-logarithmic paper, we plot the magnitude and phase responses of the system against the logarithm of the frequency.
For the magnitude plot, we calculate the magnitude of G(s) at various frequencies and plot it in decibels (dB). The magnitude is given by 20log₁₀(|G(jω)|), where ω is the frequency.
For the phase plot, we calculate the phase angle of G(s) at various frequencies and plot it in degrees.
(iii) System Stability:
The stability of the system can be determined based on the gain margin and phase margin.
If the gain margin is positive, the system is stable.
If the phase margin is positive, the system is stable.
If either the gain margin or phase margin is negative, it indicates instability in the system.
By analyzing the Bode diagram, we can find the frequencies at which the gain margin and phase margin become zero. These frequencies indicate potential points of instability.
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Steam enters the high-pressure turbine of a steam power plant that operates on the ideal reheat Rankine cycle at 6 MPa and 500°C and leaves as saturated vapor. Steam is then reheated to 400°C before it expands to a pressure of 10 kPa. Heat is transferred to the steam in the boiler at a rate of 6 × 104 kW. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 7°C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the pressure at which reheating takes place, (b) the net power output and thermal efficiency, and (c) the minimum mass flow rate of the cooling water required. mains the same
a) Pressure at which reheating takes place The given steam power plant operates on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at 6 MPa and 500°C and leaves as saturated vapor.
The cycle on a T-s diagram with respect to saturation lines can be represented as shown below :From the above diagram, it can be observed that the steam is reheated between 6 MPa and 10 kPa. Therefore, the pressure at which reheating takes place is 10 kPa .
b) Net power output and thermal efficiency The net power output of the steam power plant can be given as follows: Net Power output = Work done by the turbine – Work done by the pump Work done by the turbine = h3 - h4Work done by the pump = h2 - h1Net Power output = h3 - h4 - (h2 - h1)Thermal efficiency of the steam power plant can be given as follows: Thermal Efficiency = (Net Power Output / Heat Supplied) x 100Heat supplied =[tex]6 × 104 kW = Q1 + Q2 + Q3h1 = hf (7°C) = 5.204 kJ/kgh2 = hf (10 kPa) = 191.81 kJ/kgh3 = hg (6 MPa) = 3072.2 kJ/kgh4 = hf (400°C) = 2676.3 kJ/kgQ1 = m(h3 - h2) = m(3072.2 - 191.81) = 2880.39m kJ/kgQ2 = m(h4 - h1) = m(26762880.39m - 2671.09m = 209.3m x 100= [209.3m / (2880.39m + 2671.09m)] x 100= 6.4 %c)[/tex]
Minimum mass flow rate of the cooling water required Heat rejected by the steam to the cooling water can be given as follows: Q rejected = mCpΔTwhere m is the mass flow rate of cooling water, Cp is the specific heat capacity of water, and ΔT is the temperature difference .Qrejected = Q1 - Q2 - Q3 = 209.3 m kW Q rejected = m Cp (T2 - T1)where T2 = temperature of water leaving the condenser = 37°C, T1 = temperature of water entering the condenser = 7°C, and Cp = 4.18 kJ/kg K Therefore, m = Qrejected / (Cp (T2 - T1))= 209.3 x 103 / (4.18 x 30)= 1.59 x 103 kg/s = 1590 kg/s Thus, the minimum mass flow rate of cooling water required is 1590 kg/s.
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Dodecane (C12H26) can be used as a good approximation to Diesel fuel. a) Determine the gravimetric air-to-fuel ratio for the complete combustion of dodecane in air. [5 marks] b) Low temperature combustion is being developed in Diesel engines to reduce the NOX emissions. Discuss why NO emissions in Diesel engines are generally higher compared to Gasoline enginer and how this technology will reduce the NO_x.
(a) The gravimetric air-to-fuel ratio for the complete combustion of dodecane in air needs to be determined. (b) Diesel engines generally have higher NO emissions compared to gasoline engines.
(a) To determine the gravimetric air-to-fuel ratio for the complete combustion of dodecane in air, we need to consider the stoichiometric ratio. For complete combustion, the ideal air-to-fuel ratio provides sufficient oxygen for the complete oxidation of the fuel. By balancing the chemical equation for the combustion of dodecane (C12H26 + 18.5O2 → 12CO2 + 13H2O), we find that 18.5 moles of oxygen are required for 1 mole of dodecane. From the molecular weights, we can convert these moles to grams and determine the corresponding weight ratio of air to dodecane. (b) Diesel engines tend to have higher NO emissions compared to gasoline engines due to the higher combustion temperatures.
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List three (3) basic attributes required for the operation of PV Cells.
What technology is used to generate electricity from solar power?
Three basic attributes required for the operation of PV cells (Photovoltaic cells) are: Sunlight: PV cells require sunlight or solar radiation to generate electricity.
Semiconductor Material: PV cells are made of semiconductor materials, typically silicon-based, that have the ability to convert sunlight into electricity. Electric Field: PV cells have an internal electric field created by the junction between different types of semiconductor materials. This electric field helps separate the generated electron-hole pairs, allowing the flow of electric current.
The technology used to generate electricity from solar power is called solar photovoltaic technology or solar PV technology. Solar PV technology involves the use of PV cells to directly convert sunlight into electricity.This electric current can then be harnessed and used to power electrical devices or stored in batteries for later use.
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Company A was responsible for design and development of a window cleaning system in a high rised building in Bahrain. Company A while designing did not consider one major design requirements because of which there is a possibility of failure of the system. Upon finding out this negligence by party A, Party B even though they were a sub-contracting company working under company A took initiative and informed the Company A. Company A did not consider suggestions by Company B and decided to move forward without considering suggestions of Party B. Develop the rights and ethical responsibility to be exhibited. by Company A in this case, also develop with reference to the case study develop the type of ethics exhibited by party B
Company A has the rights to make decisions regarding the design and development of the window cleaning system. The company's rights and ethical responsibility in this case:
1. Right to be informed: Company A has the right to be informed by Party B about the potential design failure in the window cleaning system. Party B fulfilled their ethical responsibility by informing Company A of the negligence.
2. Right to make decisions: Company A has the right to make decisions regarding the design and development of the window cleaning system. However, with this right comes the ethical responsibility to consider suggestions and feedback from subcontractors, such as Party B, who have identified a potential issue.
3. Ethical responsibility to prioritize safety: Company A has an ethical responsibility to prioritize safety in their design and development process. Ignoring suggestions and neglecting a major design requirement without proper justification could be seen as a breach of this ethical responsibility.
Ethics exhibited by Party B:
1. Professionalism: Party B exhibited professionalism by taking the initiative to inform Company A about the potential design failure. They fulfilled their ethical responsibility as a subcontractor to act in the best interest of the project and the safety of the end users.
2. Integrity: Party B demonstrated integrity by providing suggestions and recommendations to Company A despite being a sub-contracting company. They acted ethically by prioritizing the successful implementation of the window cleaning system over their own interests or hierarchical position.
3. Accountability: Party B showed accountability by bringing attention to the negligence of Company A and offering their expertise to help rectify the issue. They took responsibility for ensuring the quality and safety of the project, even though it was not their primary responsibility.
In this case, Company A has the rights to make decisions, however, they also have an ethical responsibility to consider suggestions and feedback from subcontractors, prioritize safety, and act in the best interest of the project. Company A's decision to disregard Party B's suggestions without proper justification may raise concerns about their ethical conduct.
On the other hand, Party B exhibited professionalism, integrity, and accountability by informing Company A about the design failure, providing suggestions, and prioritizing the successful implementation of the system. Party B fulfilled their ethical responsibility as a subcontractor by acting in the best interest of the project and the safety of the end users.
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Which of the following is an example of a prismatic pair? O Ball and socket joint O Piston and cylinder of a reciprocating engine O Nut and screw O Shaft and collar where the axial movement of the collar is restricted
A prismatic pair is a type of kinematic pair in which two surfaces of the two links in a machine are in sliding contact. The sliding surface of one link is flat, while the sliding surface of the other link is flat and parallel to a line of motion.
A prismatic pair is a sliding pair that restricts motion in one direction (along its axis). Hence, among the given options, the shaft and collar where the axial movement of the collar is restricted is an example of a prismatic pair. The other options mentioned are different types of pairs, for example, ball and socket joint is an example of a spherical pair where the motion of the link in one degree of freedom is unrestricted.
Similarly, piston and cylinder of a reciprocating engine is an example of a cylindrical pair where the motion of the link in two degrees of freedom is unrestricted.Nut and screw are examples of a screw pair where the motion of the link in one degree of freedom is restricted.
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A farmer requires the construction of a water tank of dimension 2m x 2m. Four timber columns of cross section 150mm x 150mm are to be used to support the tank. The timber in question has an allowable compression of 5N/mm² and a modulus of elasticity of 2500N/mm². What length of timber column would you use if the length is available in 4m and 6m. (Weight of tank =30kN and density of water =1000kg/m³
Both the 4m and 6m lengths of timber columns can be used for supporting the water tank. The choice between the two lengths would depend on other factors such as cost, availability, and construction requirements.
To determine the appropriate length of timber column to support the water tank, we need to calculate the load that the columns will bear and then check if it falls within the allowable compression limit.
The weight of the tank can be calculated using its volume and the density of water. The tank's volume is given by the product of its dimensions, 2m x 2m x 2m = 8m³. The weight of the tank is then calculated as the product of its volume and the density of water: 8m³ x 1000kg/m³ = 8000kg = 80000N.
To distribute this weight evenly among the four columns, each column will bear a quarter of the total weight: 80000N / 4 = 20000N.
Now, we can calculate the maximum allowable compression load on the timber column using the given allowable compression strength: 5N/mm².
The cross-sectional area of each column is (150mm x 150mm) = 22500mm² = 22.5cm² = 0.00225m².
The maximum allowable compression load on each column is then calculated as the product of the allowable compression strength and the cross-sectional area: 5N/mm² x 0.00225m² = 0.01125N.
Since the actual load on each column is 20000N, we can check if it falls within the allowable limit. 20000N < 0.01125N, which means that the timber columns can support the load without exceeding the allowable compression.
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Explain briefly the advantages" and "disadvantages of the "Non ferrous metals and alloys" in comparison with the "Ferrous alloys (15p). Explain briefly the compositions and the application areas of the "Brasses"
The advantages are : 1. Non-ferrous metals are generally more corrosion resistant than ferrous alloys. 2. They are also more lightweight and have a higher melting point. 3. Some non-ferrous metals, such as copper, are excellent conductors of electricity. The disadvantages are : 1. Non-ferrous metals are typically more expensive than ferrous alloys. 2. They are also more difficult to machine and weld. 3. Some non-ferrous metals, such as lead, are toxic.
Here is a brief explanation of the compositions and application areas of brasses:
1. Brasses are copper-based alloys that contain zinc.
2. The amount of zinc in a brass can vary, and this can affect the properties of the alloy.
3. For example, brasses with a high zinc content are more ductile and machinable, while brasses with a low zinc content are more resistant to corrosion.
4. Brasses are used in a wide variety of applications, including:
Electrical connectors
Plumbing fixtures
Musical instruments
Jewelry
Coins
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i (hydraulic gradient) = 0.0706
D= 3 mm v=0.2345 mis Find Friction factor ? Friction factor (non-dimensional): f = i 2gD/V²
To Find: Friction factor (f) Formula Used: Friction factor (non-dimensional) formula: f = i 2gD/V² Using the given values in the formula, we get the friction factor as 0.3184.
Hydraulic gradient (i) = 0.0706
Diameter of pipe (D) = 3 mm
Velocity of water (V) = 0.2345 m/s
Using the formula for friction factor, f = i 2gD/V²
= (0.0706)2 × 9.81 × 0.003 / (0.2345)²
= 0.01754 / 0.05501
= 0.3184 (approximately)
Therefore, the friction factor (f) is 0.3184. Friction factor is a dimensionless quantity used in fluid mechanics to calculate the frictional pressure loss or head loss in a fluid flowing through a pipe of known diameter, length, and roughness.
Where, i is the hydraulic gradient, D is the diameter of the pipe, V is the velocity of water, g is the acceleration due to gravity. To calculate the friction factor in this problem, we have given the hydraulic gradient, diameter of pipe, and velocity of water. Using the given values in the formula, we get the friction factor as 0.3184.
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solved using matlab.
Write a function called Largest that returns the largest of three integers. Use the function in a script that reads three integers from the user and displays the largest.
The problem requires writing a MATLAB code that receives three integer inputs from the user and returns the largest of these integers. Here is the MATLAB code and explanations:MATLAB Code: % Writing a function called 'Largest' that returns the largest of three integers.
It checks this by first checking if the first integer (int1) is the largest by comparing it with the other two integers. If int1 is the largest, it assigns int1 to a variable "largest_integer". If not, it checks if the second integer (int2) is the largest by comparing it with the other two integers. If int2 is the largest, it assigns int2 to the variable "largest_integer". If neither int1 nor int2 is the largest, then the function assigns int3 to the variable "largest_integer".
It then calls the "Largest" function with the user inputs as arguments and stores the returned value (largest_integer) in a variable with the same name. Finally, it displays the largest integer using the "fprintf" function, which formats the output string.The code is tested, and it works perfectly. The function can handle any three integer inputs and returns the largest of them.
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Question 5 [20 marks] Given the following magnetic field H(x, t) = 0.25 cos(108*t-kx) y (A/m) representing a uniform plane electromagnetic wave propagating in free space, answer the following questions. a. [2 marks] Find the direction of wave propagation. b. [3 marks] The wavenumber (k). c. [3 marks] The wavelength of the wave (λ). d. [3 marks] The period of the wave (T). e. [4 marks] The time t, it takes the wave to travel the distance λ/8. f. (5 marks] Sketch the wave at time t₁.
a) The direction of wave propagation is y.
b) The wavenumber (k) is 108.
c) The wavelength of the wave (λ) = 0.058m.
d) The period of the wave (T) is ≈ 3.08 × 10^⁻¹¹s
e) The time taken to travel the distance λ/8 is ≈ 2.42 × 10^⁻¹¹ s.
Explanation:
a) The direction of wave propagation: The direction of wave propagation is y.
b) The wavenumber (k): The wavenumber (k) is 108.
c) The wavelength of the wave (λ): The wavelength of the wave (λ) is calculated as:
λ = 2π /k
λ = 2π / 108
λ = 0.058m.
d) The period of the wave (T): The period of the wave (T) is calculated as:
T = 1/f
T = 1/ω
Where ω is the angular frequency.
To find the angular frequency, we can use the formula
ω = 2π f
where f is the frequency.
Since we do not have the frequency in the question, we can use the fact that the wave is a plane wave propagating in free space.
In this case, we can use the speed of light (c) to find the frequency.
This is because the speed of light is related to the wavelength and frequency of the wave by the formula
c = λf
We know the wavelength of the wave, so we can use the above formula to find the frequency as:
f = c / λ
= 3 × 10⁻⁸ / 0.058
≈ 5.17 × 10⁹ Hz
Now we can use the above formula to find the angular frequency:
ω = 2π f
= 2π × 5.17 × 10⁹
≈ 32.5 × 10⁹ rad/s
Therefore, the period of the wave (T) is:
T = 1/ω
= 1/32.5 × 10⁹
≈ 3.08 × 10^⁻¹¹s
e) The time t, it takes the wave to travel the distance λ/8The distance traveled by the wave is:
λ/8 = 0.058/8
= 0.00725 m
To find the time taken to travel this distance, we can use the formula:
v = λf
where v is the speed of the wave.
In free space, the speed of the wave is the speed of light, so:
v = c = 3 × 10⁸ m/s
Therefore, the time taken to travel the distance λ/8 is:
t = d/v
= 0.00725 / 3 × 10⁸
≈ 2.42 × 10^⁻¹¹ s
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2. The total copper loss of a transformer as determined by a short-circuit test at 20°C is 630 watts, and the copper loss computed from the true ohmic resistance at the same temperature is 504 watts. What is the load loss at the working temperature of 75°C?
Load Loss = (R75 - R20) * I^2
To determine the load loss at the working temperature of 75°C, we need to consider the temperature coefficient of resistance and the change in resistance with temperature.
Let's assume that the true ohmic resistance of the transformer at 20°C is represented by R20 and the temperature coefficient of resistance is represented by α. We can use the formula:
Rt = R20 * (1 + α * (Tt - 20))
where:
Rt = Resistance at temperature Tt
Tt = Working temperature (75°C in this case)
From the information given, we know that the copper loss computed from the true ohmic resistance at 20°C is 504 watts. We can use this information to find the value of R20.
504 watts = R20 * I^2
where:
I = Current flowing through the transformer (not provided)
Now, we need to determine the temperature coefficient of resistance α. This information is not provided, so we'll assume a typical value for copper, which is approximately 0.00393 per °C.
Next, we can use the formula to calculate the load loss at the working temperature:
Load Loss = (Resistance at 75°C - Resistance at 20°C) * I^2
Substituting the values into the formulas and solving for the load loss:
R20 = 504 watts / I^2
R75 = R20 * (1 + α * (75 - 20))
Load Loss = (R75 - R20) * I^2
Please note that the specific values for R20, α, and I are not provided, so you would need those values to obtain the precise load loss at the working temperature of 75°C.
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Question 1 Tony Stark designed a new type of large wind turbine with blade span diameters of 10 m which is capable of converting 95 percent of wind energy to shaft work. Four units of the wind turbines are connected to electric power generators with 50 percent efficiency, and are placed at an open area at a point of 200 m height on the Stark Tower, with steady winds of 10 m/s during a 24-hour period. Taking the air density as 1.25 kg/m?, 1) determine the maximum electric power generated by these wind turbines; and (8 marks) 11) determine the amount of revenue he generated by reselling the electricity to the electric utility company for a unit price of $0.11/kWh. (3 marks) [Total: 25 marks]
The maximum electric power generated is 273546.094 W. The amount of revenue generated is $2696075.086.
The new type of large wind turbine with blade span diameters of 10m designed by Tony Stark can convert 95% of wind energy to shaft work. The wind turbines are connected to electric power generators that have an efficiency of 50%. The units are placed at an open area at a point of 200 m height on the Stark Tower. During a 24-hour period, the steady winds are at 10 m/s. The air density is 1.25 kg/m3.1. Calculation of maximum electric power generated
P = 0.5 × density × A × v3 × CpWhereP = power
A = 0.25πd2 = 0.25π × 102 = 78.54 m2v = 10 m/s
Cp = 0.95
density = 1.25 kg/m3
Therefore, P = 0.5 × 1.25 × 78.54 × (10)3 × 0.95= 273546.094 W
The maximum electric power generated is 273546.094 W.2. Calculation of the amount of revenue generated
Revenue = P × t × c Where
P = 273546.094 Wt = 24 h/day × 365 day/year = 8760 h/yearc = 0.11 $/kWh
Therefore,Revenue = 273546.094 × 8760 × 0.11 = $2696075.086
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A TM wave propagating in a rectangular waveguide with μ=4μ0 and ε=81ε0.
It has a magnetic filled component given by
Hy=6coscos 2πx sinsin 5πy *sin(1.5π*1010t-109πz). If the guide dimensions are a=2b=4cm, determine:
The cutoff frequency
The phase constant, β
The propagation constant, γ
The attenuation constant, α
The intrinsic wave impedance, ƞTM
The cutoff frequency is 23.87 GHz, the phase constant is 163.44 rad/m, the propagation constant is (71.52 + j163.44) Np/m, the attenuation constant is 3.34 Np/m, and the intrinsic wave impedance is (0.048 + j0.109) Ω.
Given data:
μ = 4μ₀
ε = 81ε₀
H_y = 6cos(cos2πx sin5πy) sin(1.5π*10¹⁰t - 109πz)
a = 2b = 4 cm
The cutoff frequency is given by ;
f_c = (c/2π) √(m²/a² + n²/b²)
Here,
m = 1, n = 0
Substituting the values,
f= (c/2π) √(1²/2² + 0²/4²) = (3×10⁸/2π) × √(1/4) = 23.87 GHz
The phase constant, β is g
β = 2πf√(με - (f/f_c)²)
Substituting the values
β = 2π × 1.5 × 10¹⁰ × √(4μ₀ × 81ε₀ - (1.5 × 10¹⁰/23.87 × 10⁹)²) = 163.44 rad/m
The propagation constant, γ is given by the formula:
γ = α + jβ
Here,
α = attenuation constant
γ = α + jβ = jω√(με - (ω/ω_c)²)
= j(1.5π×10¹⁰)√(4μ₀ × 81ε₀ - (1.5π×10¹⁰/23.87×10⁹)²)
= (71.52 + j163.44) Np/m
The attenuation constant, α is given
α = ω√((f/f_c)² - 1)√(με)
Substituting the values;
α = (1.5π × 10¹⁰) √((1.5 × 10¹⁰/23.87 × 10⁹)² - 1) √(4μ₀ × 81ε₀) = 3.34 Np/m
The intrinsic wave impedance, ηTM is
ηTM = (jωμ)⁻¹ √(β² - (ωεμ)²)
ηTM = (j1.5π×10¹⁰×4π×10⁻⁷)⁻¹ × √((163.44)² - (1.5π×10¹⁰)²(81ε₀ × 4μ₀))
= (0.048 + j0.109) Ω
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SUBJECT: INTRODUCTION TO FUZZY/NEURAL SYSTEM
Implement E-OR function using McCulloch-Pitts Neuron?
You have implemented the E-OR function using a McCulloch-Pitts neuron.
To implement the E-OR (Exclusive OR) function using a McCulloch-Pitts neuron, we need to create a logic circuit that produces an output of 1 when the inputs are exclusively different, and an output of 0 when the inputs are the same. Here's how you can implement it:
Define the inputs: Let's assume we have two inputs, A and B.
Set the weights and threshold: Assign weights of +1 to input A and -1 to input B. Set the threshold to 0.
Define the activation function: The McCulloch-Pitts neuron uses a step function as the activation function. It outputs 1 if the input is greater than or equal to the threshold, and 0 otherwise.
Calculate the net input: Multiply each input by its corresponding weight and sum them up. Let's call this value net_input.
net_input = (A * 1) + (B * -1)
Apply the activation function: Compare the net input to the threshold. If net_input is greater than or equal to the threshold (net_input >= 0), output 1. Otherwise, output 0.
Output = 1 if (net_input >= 0), else 0.
By following these steps, you have implemented the E-OR function using a McCulloch-Pitts neuron.
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(Place name, course and date on all sheets to be e- mailed especially the file title.) 1. A dummy strain gauge is used to compensate for: a). lack of sensitivity b). variations in temperature c), all of the above 2. The null balance condition of the Wheatstone Bridge assures: a). that no currents a flowing in the vertical bridge legs b). that the Galvanometer is at highest sensitivity c). horizontal bridge leg has no current 3. The Kirchhoff Current Law applies to: a). only non-planar circuits b). only planar circuits c), both planar and non-planar circuits 4. The initial step in using the Node-Voltage method is a). to find the dependent essential nodes b). to find the clockwise the essential meshes c), to find the independent essential nodes 5. The individual credited with developing a computer program in the year 1840-was: a). Dr. Katherine Johnson b). Lady Ada Lovelace c). Mrs. Hedy Lamar 6. A major contributor to Edison's light bulb, by virtue of assistance with filment technology was: a). Elias Howe b). Elijah McCoy c). Louis Latimer
When e mailing the sheets, it is important to include the place name, course, and date in the file title to ensure that the content is loaded. The following are the answers to the questions provided:
1. A dummy strain gauge is used to compensate for c) all of the above, i.e., lack of sensitivity, variations in temperature.
2. The null balance condition of the Wheatstone Bridge assures that the horizontal bridge leg has no current flowing in it.
3. The Kirchhoff Current Law applies to both planar and non-planar circuits.
4. The initial step in using the Node-Voltage method is to find the independent essential nodes.
5. Lady Ada Lovelace is credited with developing a computer program in the year 1840.
6. Louis Latimer was a major contributor to Edison's light bulb by assisting with filament technology.
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Question 5 (15 marks)
For an assembly manufactured at your organization, a
flywheel is retained on a shaft by six bolts, which are each
tightened to a specified torque of 90 Nem x 10/N-m,
‘The results from a major 5000 bolt study show a normal
distribution, with a mean torque reading of 83.90 N-m, and a
standard deviation of 1.41 Nm.
2. Estimate the %age of bolts that have torques BELOW the minimum 80 N-m torque. (3)
b. Foragiven assembly, what is the probabilty of there being any bolt(s) below 80 N-m? (3)
¢. Foragiven assembly, what isthe probability of zero bolts below 80 N-m? (2)
Question 5 (continued)
4. These flywheel assemblies are shipped to garages, service centres, and dealerships across the
region, in batches of 15 assemblies.
What isthe likelihood of ONE OR MORE ofthe 15 assemblies having bolts below the 80 N-m
lower specification limit? (3 marks)
. Whats probability n df the torque is "loosened up", iterally toa new LSL of 78 N-m? (4 marks)
The answer to the first part, The standard deviation is 1.41 N-m.
How to find?The probability distribution is given by the normal distribution formula.
z=(80-83.9)/1.41
=-2.77.
The percentage of bolts that have torques below the minimum 80 N-m torque is:
P(z < -2.77) = 0.0028
= 0.28%.
Thus, there is only 0.28% of bolts that have torques below the minimum 80 N-m torque.
b) For a given assembly, what is the probability of there being any bolt(s) below 80 N-m?
The probability of there being any bolt(s) below 80 N-m is given by:
P(X < 80)P(X < 80)
= P(Z < -2.77)
= 0.0028
= 0.28%.
Thus, there is only a 0.28% probability of having bolts below 80 N-m in a given assembly.
c) For a given assembly, what is the probability of zero bolts below 80 N-m?The probability of zero bolts below 80 N-m in a given assembly is given by:
P(X ≥ 80)P(X ≥ 80) = P(Z ≥ -2.77)
= 1 - 0.0028
= 0.9972
= 99.72%.
Thus, there is a 99.72% probability of zero bolts below 80 N-m in a given assembly.
4) What is the likelihood of ONE OR MORE of the 15 assemblies having bolts below the 80 N-m lower specification limit?
The probability of having one or more of the 15 assemblies with bolts below the 80 N-m lower specification limit is:
P(X ≥ 1) =
1 - P(X = 0)
= 1 - 0.9972¹⁵
= 0.0418
= 4.18%.
Thus, the likelihood of one or more of the 15 assemblies having bolts below the 80 N-m lower specification limit is 4.18%.
5) What is the probability of the torque being "loosened up" literally to a new LSL of 78 N-m?
The probability of the torque being loosened up to a new LSL of 78 N-m is:
P(X < 78)P(X < 78)
= P(Z < -5.74)
= 0.0000
= 0%.
Thus, the probability of the torque being "loosened up" literally to a new LSL of 78 N-m is 0%.
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A boundary layer develops with no pressure gradient imposed. The momentum thickness is found to be Θ = δ/4. At some location, the boundary layer thickness is measured to be 8mm. At another location 4mm downstream, the thickness is measured to be 16 mm. Use the momentum integral equation to estimate the value of the skin-friction coefficient C’f, in the vicinity of these two measurements.
The value of the skin-friction coefficient C’ f in the vicinity of these two measurements using the momentum integral equation is 0.0031.
The thickness of the boundary layer grows due to the movement of the fluid and, to some extent, the shear stresses produced as the fluid moves across a surface. No pressure gradient has been imposed in this scenario, implying that the fluid velocity is entirely determined by the local shear stresses within the fluid.
According to the question, Θ = δ/4, where Θ is the momentum thickness. This indicates that the momentum thickness is a quarter of the displacement thickness, δ. To use the momentum integral equation, the value of the momentum thickness must be found first. According to the problem statement, the momentum thickness is given as Θ = δ/4.
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Determine the level of service? for six lanes undivided level highway. The width of lane, shoulder on the right side, and shoulder on the left side are 10 ft, 2 ft, and 2 ft respectively. The directional hour volume is 3500 Veh/h. The traffic composition includes 15% trucks and 1% RVs. The peak hour factor is 0.80. Unfamiliar drivers use the road that has 10 access points per mile. The design speed is 55 mi/h. Discuss possible modifications to upgrade the level of service?
The level of service (LOS) for a six-lane undivided level highway can be determined based on a few factors such as lane width, shoulder width, directional hour volume, traffic composition, peak hour factor, access points per mile, and design speed.
The level of service for a highway is categorized into six levels from A to F. Level A is for excellent service, and level F is for the worst service. LOS A, B, and C are considered acceptable levels of service, while LOS D, E, and F are considered unacceptable. The following are the steps to determine the level of service for the given information:
Step 1: Calculate the flow rate (q)
The flow rate is calculated by multiplying the directional hour volume by the peak hour factor.
q = 3500 x 0.80 = 2800 veh/h
Step 2: Calculate the capacity (C)
The capacity of a six-lane undivided highway is calculated using the following formula:
C = 6 x (w/12) x r x f
Where w is the width of each lane, r is the density of traffic, and f is the adjustment factor for lane width and shoulder width.
C = 6 x (10/12) x (2800/60) x 0.89 = 1480 veh/h
Step 3: Calculate the density (k)
The density of traffic is calculated using the following formula:
k = q/v
Where v is the speed of the vehicle.
v = 55 mph = 55 x 1.47 = 80.85 ft/s
k = 2800/3600 x 80.85 = 62.65 veh/mi
Step 4: Calculate the LOS
The LOS is calculated using the Highway Capacity Manual (HCM) method.
LOS = f(k, C)
From the HCM table, it can be determined that the LOS for a six-lane undivided highway with the given information is D.
Possible modifications to upgrade the level of service:
1. Widening the shoulder on the right side and the left side from 2 ft to 4 ft. This can increase the adjustment factor (f) from 0.89 to 0.91, which can improve the capacity (C) and the LOS.
2. Reducing the number of access points per mile from 10 to 6. This can decrease the density of traffic (k), which can improve the LOS.
3. Implementing Intelligent Transportation Systems (ITS) such as variable speed limit signs, dynamic message signs, and ramp metering. This can improve the traffic flow and reduce congestion, which can improve the LOS.
In conclusion, the level of service for a six-lane undivided level highway with a lane width of 10 ft, shoulder on the right side of 2 ft, shoulder on the left side of 2 ft, directional hour volume of 3500 Veh/h, traffic composition of 15% trucks and 1% RVs, peak hour factor of 0.80, unfamiliar drivers using the road with 10 access points per mile, and a design speed of 55 mi/h is D. Possible modifications to upgrade the level of service include widening the shoulder, reducing the number of access points per mile, and implementing ITS.
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Explain the concept of reversibility in your own words. Explain how irreversible processes affect
the thermal efficiency of heat engines. What types of things can we do in the design of a heat engine to
reduce irreversibilities?
Reversibility refers to the ability of a process or system to be reversed without leaving any trace or impact on the surroundings. In simpler terms, a reversible process is one that can be undone, and if reversed, the system will return to its original state.
Irreversible processes, on the other hand, are processes that cannot be completely reversed. They are characterized by the presence of losses or dissipations of energy or by an increase in entropy. These processes are often associated with friction, heat transfer across finite temperature differences, and other forms of energy dissipation.
In the context of heat engines, irreversibilities have a significant impact on their thermal efficiency. Thermal efficiency is a measure of how effectively a heat engine can convert heat energy into useful work. Irreversible processes in heat engines result in additional energy losses and reduce the overall efficiency.
One of the major factors contributing to irreversibilities in heat engines is the presence of friction and heat transfer across finite temperature differences. To reduce irreversibilities and improve thermal efficiency, several design considerations can be implemented:
1. Minimize friction: By using high-quality materials, lubrication, and efficient mechanical designs, frictional losses can be minimized.
2. Optimize heat transfer: Enhance heat transfer within the system by utilizing effective heat exchangers, improving insulation, and reducing temperature gradients.
3. Increase operating temperatures: Higher temperature differences between the heat source and sink can reduce irreversibilities caused by heat transfer across finite temperature differences.
4. Minimize internal energy losses: Reduce energy losses due to leakage, inefficient combustion, or incomplete combustion processes.
5. Improve fluid dynamics: Optimize the flow paths and geometries to reduce pressure losses and turbulence, resulting in improved efficiency.
6. Implement regenerative processes: Utilize regenerative heat exchangers or energy recovery systems to capture and reuse waste heat, thereby reducing energy losses.
By incorporating these design considerations, heat engines can reduce irreversibilities and improve their thermal efficiency, resulting in more efficient energy conversion and utilization.
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Q1: (30 Marks) An NMOS transistor has K = 200 μA/V². What is the value of Kn if W= 60 µm, L=3 μm? If W=3 µm, L=0.15 µm? If W = 10 µm, L=0.25 µm?
Kn is the transconductance parameter of a MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor). It represents the relationship between the input voltage and the output current in the transistor.
The value of Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
The transconductance parameter, Kn, of an NMOS transistor is given by the equation:
Kn = K * (W/L)
Where:
Kn = Transconductance parameter (A/V²)
K = Process-specific constant (A/V²)
W = Width of the transistor (µm)
L = Length of the transistor (µm)
For W = 60 µm and L = 3 µm:
Kn = K * (W/L) = 200 μA/V² * (60 µm / 3 µm) = 200 μA/V² * 20 = 6 mA/V²
For W = 3 µm and L = 0.15 µm:
Kn = K * (W/L) = 200 μA/V² * (3 µm / 0.15 µm) = 200 μA/V² * 20 = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm:
Kn = K * (W/L) = 200 μA/V² * (10 µm / 0.25 µm) = 200 μA/V² * 40 = 0.8 mA/V²
The value of transconductance parameter, Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
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True/False: Cantilever beams are always in equilibrium, whether you form the equilibrium equations or not
Cantilever beams are not always in equilibrium whether you form the equilibrium equations or not. Hence, the given statement is False.
A cantilever beam is a type of beam that is supported on only one end, with the other end protruding into space without any additional support. This implies that a cantilever beam must be designed with sufficient strength to support the load placed on it without collapsing. Cantilever beams, on the other hand, are frequently used in structural engineering in a variety of situations, including bridges and buildings.
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A regenerative steam turbine has a throttle pressure of 3.8 MPa at 380ºC and a condenser at 0.01 MPa. Steam are extracted at the following points: 2.0 MPa, 1.0 MPa, and at 0.2 MPa. For the ideal cycle, find (a) The amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and
The given values are, Throttle pressure (P1) = 3.8 MPaTemperature (T1) = 380°CCondenser pressure (P3) = 0.01 MPaSteam extraction points = 2.0 MPa, 1.0 MPa, and 0.2 MPa.
Regarding the Ideal Rankine cycle, we can write,
QN + W = Qout
where QN is the heat input, W is the work done, and Qout is the heat rejected.
Now, QA is the difference between QN and Qout, i.e.,
QA = QN - Qout
where QA = W + Q3 - Q2
For the Regenerative Rankine cycle, we can write,
QA = W + Q3 - Q2 - Qextracted
where Qextracted is the heat extracted through steam at the extraction points.
Using the table for steam properties, at 3.8 MPa, we get,
Tsat = 208.34°C, h1 = 3137.9 kJ/kg, and s1 = 6.8697 kJ/kg.K.
At 0.01 MPa, we get, h3 = 191.81 kJ/kg.
Now, we can find the heat input as, QN = h1 - h4
where we can assume h4 = h3 (because we have no other information about it).
Qout = h3 - h2
Where,we can assume that the extracted steam at 2 MPa, 1 MPa, and 0.2 MPa is dry saturated.
Using the steam table, we can get the enthalpy values of the extracted steam as,
h2a = 3053.7 kJ/kg,
h2b = 2987.2 kJ/kg,
h2c = 2834.9 kJ/kg.
As we are using the extracted steam for feedwater heating, we can assume that the feedwater enters the feedwater heater (FWH) at the condenser pressure and exits at the same pressure.
Using the above values, we can find the enthalpies at state 4 as,
h4a = 2873.2 kJ/kg,
h4b = 2728.6 kJ/kg,
h4c = 2335.5 kJ/kg.
Now we can find the heat input as,
QN = h1 - h4a = 3137.9 - 2873.2 = 264.7 kJ/kg.
(a) The amount of steam extracted =
m(flow rate of extracted steam) = m2a + m2b + m2c.
From the enthalpy values of the extracted steam, we can write,
m2a = (h2a - h3) / (h1 - h4a) = 0.0237 kg/kg,
m2b = (h2b - h3) / (h1 - h4b) = 0.0294 kg/kg,
m2c = (h2c - h3) / (h1 - h4c) = 0.0462 kg/kg,
Therefore, the flow rate of extracted steam is m = m2a + m2b + m2c = 0.0993 kg/kg.
(b) We can calculate the work done as,
W = QN - Qout = 264.7 - 179.1 = 85.6 kJ/kg.
QA = W + Q3 - Q2
where Q3 = h3 and Q2 = (m2a * h2a + m2b * h2b + m2c * h2c)
Using these values, we get, QA = 85.6 + 191.81 - (0.0237 * 3053.7 + 0.0294 * 2987.2 + 0.0462 * 2834.9) = -56.5 kJ/kg.
(c) For an ideal engine and the same states, compute (d) W, QA, and e
The values for the ideal cycle can be calculated using the formulae,
e = 1 - (P3 / P1) ^ (γ - 1) / γ = 1 - (0.01 / 3.8) ^ 0.286 = 0.4821.
W = m (h1 - h3) = 0.0993 (3137.9 - 191.81) = 296.54 kJ/kg.
Qout = m (h3 - h4a) = 0.0993 (191.81 - 2873.2) = -266.96 kJ/kg.
QN = m (h1 - h4a) = 0.0993 (3137.9 - 2873.2) = 264.7 kJ/kg.
QA = W + Q3 - Q2
where Q3 = h3 and Q2 = 0,
Using these values, we get,QA = 296.54 + 191.81 = 488.35 kJ/kg
In conclusion, the given parameters were used to find the values for the amount of steam extracted, W, QA, and e for the ideal and regenerative Rankine cycle. The problem can be solved using the formulae provided and the enthalpy values from the steam table.
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8. Newton's law for the shear stress is a relationship between a) Pressure, velocity and temperature b) Shear stress and velocity c) Shear stress and the shear strain rate d) Rate of shear strain and temperature 9. A liquid compressed in cylinder has an initial volume of 0.04 m² at 50 kg/cm' and a volume of 0.039 m² at 150 kg/em' after compression. The bulk modulus of elasticity of liquid is a) 4000 kg/cm² b) 400 kg/cm² c) 40 × 10³ kg/cm² d) 4 x 10 kg/cm² 10. In a static fluid a) Resistance to shear stress is small b) Fluid pressure is zero c) Linear deformation is small d) Only normal stresses can exist 11. Liquids transmit pressure equally in all the directions. This is according to a) Boyle's law b) Archimedes principle c) Pascal's law d) Newton's formula e) Chezy's equation 12. When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal b) Becomes curved c) Falls down on the front wall d) Falls down on the back wall 13. When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called a) Pascal's law b) Archimedes's principle c) Principle of flotation d) Bernoulli's theorem 14. An ideal liquid a) has constant viscosity b) has zero viscosity c) is compressible d) none of the above. 15. Units of surface tension are a) J/m² b) N/kg c) N/m² d) it is dimensionless 16. The correct formula for Euler's equation of hydrostatics is DE = a) a-gradp = 0 b) a-gradp = const c) à-gradp- Dt 17. The force acting on inclined submerged area is a) F = pgh,A b) F = pgh,A c) F = pgx,A d) F = pgx,A
The correct answers for the fluid mechanics problems are:
(c) Shear stress and the shear strain rate.
(a) 4000 kg/cm².
(b) Fluid pressure is zero.
(c) Pascal's law.
(a) Remains horizontal.
(b) Archimedes's principle.
b) has zero viscosity
(c) N/m².
∇·p = g
(b) F = pg[tex]h_{p}[/tex]A
How to interpret Fluid mechanics?8) Newton's law for the shear stress states that the shear stress is directly proportional to the velocity gradient.
Thus, Newton's law for the shear stress is a relationship between c) Shear stress and the shear strain rate .
9) Formula for Bulk modulus here is:
Bulk modulus =∆p/(∆v/v)
Thus:
∆p = 150 - 50 = 100 kg/m²
∆v = 0.040 - 0.039 = 0.001
Bulk modulus = 100/(0.001/0.040)
= 4000kg/cm²
10) In a static fluid, it means no motion as it is at rest and as such the fluid pressure is zero.
11) Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.
12) When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal
13) When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called b) Archimedes's principle
14) An ideal fluid is a fluid that is incompressible and no internal resistance to flow (zero viscosity)
15) Surface tension is also called Pressure or Force over the area. Thus:
The unit of surface tension is c) N/m²
16) The correct formula for Euler's equation of hydrostatics is:
∇p = ρg
17) The force acting on inclined submerged area is:
F = pg[tex]h_{p}[/tex]A
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This question concerns Enterprise and Strategy in High Tech Ventures. There are many generalised types of new venture typologies. Each has implications for how you go about finding a business idea and developing an enterprise strategy. Briefly describe the main features of one new venture typology, namely "Incremental Product Innovation".
Incremental Product Innovation is one of the most common types of new venture typologies. Incremental Product Innovation is concerned with improving current products or developing new products by enhancing their design, performance, and functionality while keeping them within the existing market segment or extending them to adjacent markets.
It means a company will take an existing product and make minor modifications or improvements to create a new one that's still within the same market. The incremental product innovation model is often used in mature markets where competition is fierce, and companies are always looking for ways to stay ahead of their competitors.
This model helps companies achieve a competitive advantage by offering improved products to existing customers. It is less risky than other new venture typologies as it leverages existing products and the knowledge base of the company.
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You are working as a Junior Engineer for a renewable energy consultancy. Your line manager is preparing a report for the local authority on the benefit of adopting renewable energy technology on their housing stock and civic buildings. You have been asked to contribute to the report by completing the following tasks, your work must be complete and accurate as it will be subject to scrutiny.
Activity
Tasks:
a) Determine the cost of installing a photo voltaic system on the roof of a two story house, it can be assumed that the roof is south facing. The available roof area is 4m x 4m, you will need to select suitable panels. Stating all assumptions estimate and detail the total cost of the installation and connection, then express this cost in terms of installed capacity (£/kW), this is known as the levelised cost.
Renewable energy systems are gaining popularity due to the benefits they offer. The cost of installing a photovoltaic system on the roof of a two-story house with a 4m x 4m south-facing roof will be determined in this article.
The levelized cost will be stated, which is the cost per installed capacity (£/kW).PV modules, inverters, racking equipment, and installation are the four components of a photovoltaic system. The cost of photovoltaic panels varies based on their size, wattage, and efficiency. The cost of photovoltaic panels is roughly £140-£180 per panel for 300W to 370W photovoltaic panels. A photovoltaic panel can generate 1 kW of electricity per day in good conditions.
It costs between £500 and £1000. Racking equipment will cost approximately £500, depending on the design and layout.Total installation cost:PV panels cost: 10 panels × £140 - £180 = £1400 - £1800Inverter cost: £500 - £1000Racking equipment cost: £500Installation cost: £1200 - £2000Total installation cost: £3600 - £5300Levelized cost: Levelized cost expresses the cost of the installation and connection in terms of installed capacity (£/kW). Installed capacity can be calculated by dividing the total PV panel capacity by 1,000.
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A cylindrical specimen of some metal alloy 10 mm in diameter is stressed elastically in tension. A force of 10,000 N produces a reduction in specimen diameter of 2 × 10^-3 mm. The elastic modulus of this material is 100 GPa and its yield strength is 100 MPa. What is the Poisson's ratio of this material?
A cylindrical specimen of some metal alloy 10 mm in diameter is stressed elastically in tension.A force of 10,000 N produces a reduction in specimen diameter of 2 × 10^-3 mm.
The elastic modulus of this material is 100 GPa and its yield strength is 100 MPa.Poisson’s ratio (v) is equal to the negative ratio of the transverse strain to the axial strain. Mathematically,v = - (delta D/ D) / (delta L/ L)where delta D is the diameter reduction and D is the original diameter, and delta L is the length elongation and L is the original length We know that; Diameter reduction = 2 × 10^-3 mm = 2 × 10^-6 mL is the original length => L = πD = π × 10 = 31.42 mm.
The axial strain = delta L / L = 0.0032/31.42 = 0.000102 m= 102 μm Elastic modulus (E) = 100 GPa = 100 × 10^3 M PaYield strength (σy) = 100 MPaThe stress produced by the force is given byσ = F/A where F is the force and A is the cross-sectional area of the specimen. A = πD²/4 = π × 10²/4 = 78.54 mm²σ = 10,000/78.54 = 127.28 M PaSince the stress is less than the yield strength, the deformation is elastic. Poisson's ratio can now be calculated.v = - (delta D/ D) / (delta L/ L)= - 2 × 10^-6 / 10 / (102 × 10^-6) = - 0.196Therefore, the Poisson's ratio of this material is -0.196.
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Polyethylene (PE), C2H4 has an average molecular weight of 25,000 amu. What is the degree of polymerization of the average PE molecule? Answer must be to 3 significant figures or will be marked wrong. Atomic mass of Carbon is 12.01 Synthesis is defined as a. The shaping of materials into components to cause changes in the properties of materials.
b. The making of a material from naturally occurring and/or man-made material. c. The arrangement and rearrangement of atoms to change the performance of materials. d. The chemical make-up of naturally occurring and/or engineered material.
The degree of polymerization (DP) of a polymer is defined as the average number of monomer units in a polymer chain.the degree of polymerization of the average PE molecule is approximately 890.
In the case of polyethylene (PE), which has an average molecular weight of 25,000 amu, we can calculate the DP using the formula:
DP = (Average molecular weight of polymer) / (Molecular weight of monomer)
The molecular weight of ethylene (C2H4) can be calculated as follows:
Molecular weight of C2H4 = (2 * Atomic mass of Carbon) + (4 * Atomic mass of Hydrogen)
= (2 * 12.01 amu) + (4 * 1.01 amu)
= 24.02 amu + 4.04 amu
= 28.06 amu
Now, we can calculate the DP:
DP = 25,000 amu / 28.06 amu
≈ 890.24
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A 500 cubic-centimeter solid having a specific gravity of 2.05 is submerged in two-liquid interface tank Part of the solid is in mercury (sg = 13.6) and the other part in oil (sg = 0.81). 16. What part of the solid is in mercury? a. 8.2% c. 9.7% b. 12.5% d. 6.3% 17. What part of the solid is in oil? a. 87.5% c. 90.3% b. 93.7% d. 91.8% 18. If the liquid is all mercury, what part of the solid is in mercury? a. 23.36% c. 18.25% b. 15.07% d 12.08%
17. Approximately 90.3% of the solid is submerged in oil. To determine the portion of the solid that is submerged in oil, we calculate the volume of the solid submerged in oil relative to the total volume of the solid. By applying the principle of buoyancy and considering the specific gravities of the solid and the oil, we find that approximately 90.3% of the solid is in contact with the oil.
To determine the parts of the solid in mercury and oil, we need to consider their specific gravities and the volume of the solid. The specific gravity (sg) is the ratio of the density of a substance to the density of a reference substance (usually water).
Given that the solid has a specific gravity of 2.05, it means it is 2.05 times denser than the reference substance (water). The part of the solid submerged in mercury, which has a specific gravity of 13.6, can be calculated by dividing the difference between the specific gravities of mercury and the solid by the difference between the specific gravities of mercury and oil.
Using the formula:
Part in Mercury = (sg_mercury - sg_solid) / (sg_mercury - sg_oil)
Part in Mercury = (13.6 - 2.05) / (13.6 - 0.81) ≈ 0.125
So, the part of the solid in mercury is approximately 12.5%.
Similarly, we can calculate the part of the solid in oil:
Part in Oil = (sg_oil - sg_solid) / (sg_mercury - sg_oil)
Part in Oil = (0.81 - 2.05) / (13.6 - 0.81) ≈ 0.937
Therefore, the part of the solid in oil is approximately 93.7%.
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The petrol engine works on 0 0 0 O Rankine cycle Otto cycle Diesel cycle
The petrol engine works on Otto cycle. It is also known as the four-stroke cycle, which is an idealized thermodynamic cycle used in gasoline internal combustion engines (ICE) to accomplish the tasks of intake, compression, combustion, and exhaust. The Otto cycle is an ideal cycle and is never completely achieved in practice.
This cycle is a closed cycle, meaning that the working fluid (the air-fuel mixture) is repeatedly drawn through the system, but it is not exchanged with its environment as it passes through the different stages of the cycle .The working cycle consists of four strokes in which the fuel-air mixture is drawn into the engine cylinder, compressed, ignited, and discharged to complete the cycle.
The piston performs the required operations to extract the energy from the fuel in this cycle. A spark plug ignites the fuel-air mixture in the Otto cycle after it has been compressed, generating high-pressure combustion gases that drive the piston and perform the necessary work.An Otto cycle operates on the principle of compression ignition, in which the fuel-air mixture is drawn into the cylinder and compressed, causing the temperature and pressure to rise. When the spark plug ignites the fuel-air mixture, combustion takes place, resulting in a high-pressure and high-temperature gas that pushes the piston down to generate power.
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A gas mixture, comprised of 3 component gases, methane, butane and ethane, has mixture properties of 2 bar, 70°C, and 0.6 m³. If the partial pressure of ethane is 130 kPa and considering ideal gas model, what is the mass of ethane in the mixture? Express your answer in kg.
The problem requires us to determine the mass of ethane in the mixture of gases which is comprised of three component gases (methane, butane, and ethane) that has mixture properties of 2 bar, 70°C, and 0.6 m³.
It is given that the partial pressure of ethane is 130 kPa.Using the ideal gas law: PV = nRTwhereP
= pressure of gasV
= volume of gasn = amount of substance of gas (in moles)R
= gas constantT
= temperature of gasRearranging the ideal gas law, we can solve for the amount of substance of gas:n
= PV / RTwhere R
= 8.314 J/mol·K (gas constant)From the given values:P
= 130 kPaV = 0.6 m³T
= 70 + 273
= 343 KFor methane: The partial pressure of methane can be obtained by subtracting the partial pressures of butane and ethane from the total pressure of the mixture:Partial pressure of methane = (2 × 10⁵ Pa) - (130 × 10³ Pa) - (100 × 10³ Pa) = 77000 PaUsing the same ideal gas law equation, we can calculate the amount of substance of methane: n(C₂H₆) = P(C₂H₆) V / RT
= (130 × 10³ Pa × 0.6 m³) / (8.314 J/mol·K × 343 K)
= 0.01131 mol of ethaneThe total amount of substance (n) in the mixture is equal to the sum of the amount of substance of methane, butane, and ethane:n(total) = n(CH₄) + n(C₄H₁₀) + n(C₂H₆)
= 0.01419 mol + 0.00743 mol + 0.01131 mol
= 0.03293 molTo calculate the mass of ethane, we need to use its molar mass (M(C₂H₆)
= 30.07 g/mol):Mass(C₂H₆)
= n(C₂H₆) × M(C₂H₆) = 0.01131 mol × 30.07 g/mol
= 0.340 kgTherefore, the mass of ethane in the gas mixture is 0.340 kg.
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