The signal v = A cos(2πnfe) is a periodic signal, and its frequency spectrum consists of a single peak at the frequency fe. When transformed to approximate a square wave, the frequency spectrum of the resulting signal will contain the fundamental frequency and its odd harmonics.
How can the periodicity of the signal v = A cos(2πnfe) be proven, and what is the frequency spectrum of the signal? Additionally, how can this signal be transformed to approximate a square wave, and what is the resulting frequency spectrum?To prove that the signal v = A cos(2πnfe) is periodic, we need to show that it repeats itself after a certain interval.
To demonstrate the frequency spectrum of the signal, we can use Fourier analysis.
By applying the Fourier transform to the signal, we obtain its frequency components.
In this case, since v = A cos(2πnfe), the frequency spectrum will consist of a single peak at the frequency fe, representing the fundamental frequency of the cosine function.
To approximate a square wave using the given signal, we can use Fourier series expansion.
By adding multiple harmonics with appropriate amplitudes and frequencies, we can construct a square wave-like signal.
The Fourier series coefficients determine the amplitudes of the harmonics. The closer we get to an infinite number of harmonics, the closer the approximation will be to a perfect square wave.
By calculating the Fourier series coefficients and reconstructing the signal, we can visualize the transformation from the cosine signal to an approximate square wave.
The frequency spectrum of the approximate square wave will contain the fundamental frequency and its odd harmonics.
The amplitudes of the harmonics decrease as the harmonic number increases, following the characteristics of a square wave spectrum.
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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: Find the solution to the following linear programming problem using the simplex method Max (Z)=50x1+60x2 Subjected to: 2x1+x2 < 300 3x1+4x2 ≤ 509 4x1+7x2812 x1,x220
The simplex method is an approach to solve the linear programming problems. To solve the following linear programming problem using the simplex method: Max (Z)=50x1+60x2 Subjected to: 2x1+x2 < 3003x1+4x2 ≤ 5094x1+7x2 ≤ 812x1, x2
In this matrix, the last column represents the right-hand side of the constraints. The simplex method consists of the following - Identify the pivot element by selecting the most negative coefficient in the objective function row, which is -60 in our case. So, we will select x2 as the entering variable. Find the leaving variable by calculating the ratio of the RHS value to the coefficients of the entering variable in each constraint. The minimum non-negative ratio corresponds to the leaving variable.
From the first constraint, the ratio is 300/1 = 300, and from the third constraint, the ratio is 812/7 = 116. Therefore, we choose the first constraint for the leaving variable. So, s1 will leave the basis, and x2 will enter the basis. Perform elementary row operations to make the entering variable coefficient equal to 1 and all other coefficients in the entering column equal to 0. We can achieve this by dividing the first row by 1 and multiplying it by -1 and adding it to the second row.
Therefore, the solution to the following linear programming problem using the simplex method is x1 = 55, x2 = 85, and Z = 5750.
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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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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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A cantilever beam has length 24 in and a force of 2000 lbf at the free end. The material is A36/. For a factor of safety of 2, find the required cross section dimensions of the beam. The cross section can be assumed as square, rectangular, pipe or I-beam.
The formula for the shear stress in a cantilever beam subjected to a transverse force can be used to find the required cross-section dimensions for the beam.The formula is; τmax = VQ/ItWhere;V = the maximum force (2000 lbs.)Q = the first moment of the area around the neutral axis.
I = the moment of inertia.The maximum shear stress for A36 steel is 20,000 psi. For a factor of safety of 2, this value can be doubled to 40,000 psi.So,τmax = VQ/It = 40000 psi.The dimensions of the beam can be found using the shear stress equation and the bending moment equation.
Mmax = PL/4 = 2000 lbs. × 24 in./4 = 12000 in. lbs.τmax = Mmax*c/I = 40000 psiThe required cross-section dimensions of the beam can be found as follows;For a square beam;a = b ⇒ c = a / √6P = 12000 lbs.
[tex]Q = b × h × h / 2 = a × a × a / 2√3h = a/√3I = a^4/12c = I × τmax / b × h²a = (6 × P / (τmax × h²))^(1/4).[/tex]
For a rectangular beam;
[tex]a < b ⇒ c = a / √6P = 12000 lbs.Q = b × h × h / 2 = a × b × b / 2h = √(2a / 3)I = ab^3/12c = I × τmax / b × h²a = (6 × P / (τmax × h² × b))^(1/3) × b^2/3.[/tex]
For a pipe;a = b and D = 2rP = 12000 lbs.τavg = P/ (2A - a²) = 40000 psiThe diameter of the pipe can be found using the following equation;
[tex]r = (P/2τavg)(D² - d²)/D²d = D - 2ta = πr² - πr²/4A = πr²D = 2r(1 + (4a²/(πr^2))^(1/2)).[/tex]
For an I-beam;the required dimensions can be found by assuming that the beam is an equivalent rectangular beam and then using the above rectangular beam formula. In the equivalent rectangular beam, the width of the flanges is equal to the thickness of the web multiplied by a factor of 1.2 to 1.5. The thickness of the web is taken as the distance between the midpoints of the flanges.
From the above, we can conclude that the cross-section dimensions of a square beam, rectangular beam, pipe, and I-beam can be found.
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(a) American Standard Code for Information Interchange (ASCII) Code is use to transfer information between computers, between computers and printers, including for internal storage. Write the word of VictorY! using ASCII code in Decimal form and Hexadecimal form. Refer to Appendix 1 for the ASCII code table. Build a suitable table for each alphabets.
Therefore, the word “Victor Y” can be represented in decimal and hexadecimal forms using the ASCII code table, and a suitable table can be built for each alphabet.
The American Standard Code for Information Interchange (ASCII) Code is used to transfer information between computers, printers, and for internal storage. The ASCII code table is used for this purpose.
The word “Victor Y” can be written in decimal and hexadecimal forms using the ASCII code table. In decimal form, the word “Victor Y” can be written as:
86, 105, 99, 116, 111, 114, 89, 33. In hexadecimal form, it can be written as:
56, 69, 63, 74, 6F, 72, 59, 21.
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f₂ a b C 1 0 0 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 1 A. Predict Logical expression for the given truth table for the output function f2,if a,b,c. are the inputs.
B. Simplify expression a (write appropriate laws being used) C. Draw the logical diagram for the expression found in Question (B). D. Comment on the Number of gates required for implementing the original and reduced expression the Logical found in Question
To predict the logical expression for the given truth table for the output function F₂, we can analyze the combinations of inputs and outputs:
css
Copy code
a b c F₂
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
From the truth table, we can observe that F₂ is 1 when at least two of the inputs are 1. The logical expression for F₂ can be written as:
F₂ = (a AND b) OR (a AND c) OR (b AND c)
B. To simplify the expression, we can use Boolean algebra laws. Let's simplify the expression step by step:
F₂ = (a AND b) OR (a AND c) OR (b AND c)
Using the distributive law, we can factor out common terms:
F₂ = a AND (b OR c) OR b AND c
C. The logical diagram for the simplified expression can be represented using logic gates. In this case, we have two AND gates and one OR gate:
lua
Copy code
______
a ----| |
| AND |--- F₂
b ----|______|
______
c ----| |
| AND |
0 ----|______|
D. Comment on the number of gates required for implementing the original and reduced expression:
The original expression for F₂ required three AND gates and one OR gate. However, after simplification, the reduced expression only requires two AND gates and one OR gate.
Therefore, the reduced expression is more efficient in terms of the number of gates required for implementation.
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ie lbmol of pentane gas (C₅H₁₂) reacts with the theoretical amount of air in a closed, rigid tank. Initially, the reactants are at 77°F, 1 m. After complete combustion, the temperature in the tank is 1900°R. Assume air has a molar analysis of 21% O₂ and 79% N₂. Determine the heat transfer, in Btu. Q = i Btu
The heat transfer, Q, can be calculated using the equation:
Q = ΔHc + ΔHg. To determine the heat transfer in Btu for the given scenario, we need to calculate the heat released during the combustion of pentane and the subsequent increase in temperature of the gases in the tank.
Where ΔHc is the heat released during combustion and ΔHg is the heat gained by the gases in the tank due to the increase in temperature. To calculate ΔHc, we need to determine the moles of pentane reacted and the heat of combustion per mole of pentane. Since pentane reacts with air, we also need to consider the moles of oxygen available in the air. The heat of combustion of pentane can be obtained from reference sources. To calculate ΔHg, we can use the ideal gas law and the given initial and final temperatures, along with the molar analysis of air, to determine the change in enthalpy. By summing up ΔHc and ΔHg, we can obtain the total heat transfer, Q, in Btu. It's important to note that the actual calculations involve several steps and equations, including stoichiometry, enthalpy calculations, and gas laws. The specific values and formulas needed for the calculations are not provided in the question, so an exact numerical result cannot be determined without that information.
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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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a load absorbs 50 MVA at 0.6 pf leading at line to line voltage of 18 KV. find the perunit impedance of this load on a base of 100MVA and 20 KV. Select one: a. 3.888 +j 5.183 pu b. 3.888-j 5.183 pu c. 0.972 +j 1.295 pu N
d. one of these e. 0.972-j 1.295 pu
In order to determine the per unit impedance of a load on a base of 100 MVA and 20 kV, you need to calculate the total impedance of the load using the given information.
Load power, P = 50 MVA pf leading, cos(φ) = 0.6 Line to line voltage, V = 18 kV Base power, S = 100 MVA Base voltage, Vbase = 20 kVCalculation: Let's first convert the power to per unit value. For this we use the base power of 100 MVA and the base voltage of 20 kV. Per unit power, Ppu = P/S = 50/100 = 0.5 p u Now we can calculate the load current.
I using the given power and power factor. cos(φ) = P / (V x I)0.6 = 0.5 / (18 x I)I = 1.39 kA We can now calculate the load impedance in ohms using the formula: Z = V / IZ = 18 kV / 1.39 kA = 12973.38 ΩNow, we can convert this impedance value to per unit value.
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List the functions of a lubricant in a sliding contact
bearing
The following are the functions of a lubricant in a sliding contact bearing:
To reduce friction:
Friction generates heat in bearings, which can result in high temperatures and potential damage.
Lubricants are used to reduce friction in bearings by minimizing metal-to-metal contact and smoothing surfaces.
They function by developing an oil film that separates the two bearing surfaces and reduces friction.
To absorb heat:
Bearing lubrication also aids in the removal of heat generated by friction.
It absorbs heat, which it carries away from the bearing.
To prevent wear and tear:
Lubrication prevents wear by minimizing metal-to-metal contact between surfaces.
To prevent corrosion:
Lubricants help to minimize corrosion caused by exposure to moisture.
To provide stability:
It helps to maintain the shaft's stability while it is in motion.
To help cool down the system:
It helps to regulate the temperature in the system.
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3) 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. (10 marks) 10 marks: fully correct answer with correct description, interpretation with correct justification with appropriate NSPE Codes, discussion with appropriate ethical obligations 5-9: correct answer with missing interpretation with in correct correct justification with appropriate NSPE Codes, discussion with appropriate ethical obligations 0-4: incorrect/partial correct discussions with correct justification with appropriate NSPE Codes, discussion with appropriate ethical obligations
In this case, Company A, responsible for the design and development of a window cleaning system, neglected a major design requirement that could potentially lead to system failure.
Company A has an ethical responsibility to uphold the safety, health, and welfare of the public, as outlined in the National Society of Professional Engineers (NSPE) Code of Ethics. Specifically, section II.1.c of the NSPE code states that engineers must "hold paramount the safety, health, and welfare of the public." In this case, Company A should have recognized their negligence, acknowledged the suggestions provided by Party B, and taken appropriate action to rectify the design flaw. By ignoring the suggestions, Company A failed to fulfill their ethical obligations and jeopardized the safety of the window cleaning system.
On the other hand, Party B demonstrated a proactive approach and exhibited professional ethics by informing Company A about the design flaw. Their actions align with the NSPE code, particularly section II.4, which emphasizes the obligation of engineers to "act in professional matters for each employer or client as a faithful agent or trustee." Despite being a sub-contractor, Party B recognized their ethical duty to prioritize safety and welfare, showcasing integrity and responsibility.
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Vibrations of harmonic motion can be represented in a vectorial form. Analyze the values of displacement, velocity, and acceleration if the amplitude, A=2+Tm, angular velocity, ω=4+U radis and time, t=1 s. The values of T and U depend on the respective 5th and 6th digits of your matric number. For example, if your matric number is AD201414, it gives the value of T=1 and U=4.
Given that the amplitude A = 2+Tm, angular velocity [tex]ω = 4+U[/tex] radians and time t = 1 second. We need to find out the displacement, velocity, and acceleration values by using vectorial form of harmonic motion.
Vibrations of harmonic motion can be represented as a vectorial form i.e.,[tex]A sin (ωt + φ)[/tex]
The amplitude is denoted by 'A'Angular velocity is denoted by '[tex]ω[/tex]' time is denoted by 't'
The angle which the amplitude makes with the positive x-axis is denoted by 'φ' Displacement, Velocity, and acceleration values of a particle executing SHM at any time t
[tex]Displacement = A sin (ωt + φ)Velocity = Aω cos (ωt + φ)Acceleration = - Aω² sin (ωt + φ)Given A = 2+Tm, ω = 4+U and t = 1 s.[/tex]
Taking T = 1 and U = 4 from the given matric number.
Amplitude, A = 2+Tm = 2+1(m) = 2+m
Angular velocity, [tex]ω = 4+U = 4+4 = 8 rad/s[/tex]
Displacement, [tex]x = A sin(ωt + φ)[/tex]
Displacement = [tex](2 + m) sin(8(1) + φ)[/tex]......(1)
Velocity, [tex]v = Aω cos(ωt + φ)[/tex]
Velocity =[tex](2 + m)8 cos(8(1) + φ)[/tex]......(2)
Acceleration,[tex]a = -Aω² sin(ωt + φ)[/tex]
Acceleration =[tex]-(2 + m) 8² sin(8(1) + φ)[/tex]......(3)
Let us assume that the angle φ = 0.
Substituting [tex]φ = 0[/tex] in equation (1), (2) and (3)
Displacement, [tex]x = (2 + m) sin 8[/tex]
Velocity,[tex]v = (2 + m) 8 cos 8[/tex]
Acceleration,[tex]a = -(2 + m) 8² sin 8[/tex]
Therefore, Displacement is (2+m)sin8,
Velocity is (2+m)8cos8
Acceleration is -(2+m)64sin8.
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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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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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For the following unconventional manufacturing process, the initial cost is very high and the useful life of the flash lamp is short:
Answer Choice Group
a) EDM machining
b) plasma machining
c) laser beam machining
d) High pressure water jet machining
The unconventional manufacturing process whose initial cost is high and the useful life of the flash lamp is short is the laser beam machining. Laser beam machining (LBM) is an unconventional manufacturing process that employs a coherent, monochromatic, and high-energy laser beam to cut, machine, or otherwise modify materials with high accuracy and precision.LBM is classified as a thermal, non-contact, and high-speed machining method that offers a wide range of benefits over other machining methods, such as low heat-affected zone, no tool wear, high accuracy, and fine surface finish, among others.
The laser beam's energy is focused on the workpiece's surface, causing the material to melt, vaporize, or be ejected, depending on the laser power, pulse duration, and repetition rate.However, LBM has some drawbacks, such as high initial cost, limited beam divergence, small depth of cut, and short useful life of the flash lamp, among others. The initial cost of laser equipment is relatively high, which can make it difficult for small and medium-sized enterprises to adopt this technology.
The flash lamp, which is used as a pumping source for the laser, has a limited useful life, usually ranging from several hundred hours to a few thousand hours, depending on the flash lamp's type, size, and power density. Therefore, the replacement cost of the flash lamp should be considered when determining the overall cost of LBM.The other unconventional manufacturing processes, such as EDM machining, plasma machining, and high-pressure water jet machining, do not use flash lamps as pumping sources for energy.
They do not have a short useful life of the flash lamp as a disadvantage.
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6. A 2x4 made from southern pine is 10ft long supported at each end and laying flat. It is loaded in the center with 250 lbs. What is the max deflection? If the 2x4 is turned vertical, what will the deflection be?
A 10ft long 2x4 made from southern pine, supported at each end and loaded with 250 lbs in the center, will have a maximum deflection. If the 2x4 is turned vertical, the deflection will be different.
When a 2x4 made from southern pine is loaded at its center, it will experience a maximum deflection. The magnitude of this deflection can be calculated using beam deflection formulas, such as Euler-Bernoulli beam theory. However, the specific calculations depend on factors such as the material properties of southern pine and the dimensions of the 2x4.
If the 2x4 is turned vertically, its deflection will be influenced by different factors. The vertical orientation changes the beam's moment of inertia and the distribution of load along its length. These alterations can significantly affect the deflection characteristics of the beam.
It is important to note that without precise dimensions and material properties, it is challenging to provide an accurate numerical value for the maximum deflection in either case. To obtain a more precise result, it is recommended to consult a structural engineer or refer to relevant engineering handbooks and codes that provide deflection formulas and guidelines for specific beam configurations and materials.
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A 10ft long 2x4 made from southern pine, supported at each end and loaded with 250 lbs in the center, will have a maximum deflection. If the 2x4 is turned vertical, the deflection will be different.
When a 2x4 made from southern pine is loaded at its center, it will experience a maximum deflection. The magnitude of this deflection can be calculated using beam deflection formulas, such as Euler-Bernoulli beam theory.
However, the specific calculations depend on factors such as the material properties of southern pine and the dimensions of the 2x4.
If the 2x4 is turned vertically, its deflection will be influenced by different factors. The vertical orientation changes the beam's moment of inertia and the distribution of load along its length. These alterations can significantly affect the deflection characteristics of the beam.
It is important to note that without precise dimensions and material properties, it is challenging to provide an accurate numerical value for the maximum deflection in either case.
To obtain a more precise data , it is recommended to consult a structural engineer or refer to relevant engineering handbooks and codes that provide deflection formulas and guidelines for specific beam configurations and materials.
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A ship, travelling at 12 knots, has an autopilot system with a time and gain constants of 107 s and 0.185 s⁻¹, respectively. The autopilot moves the rudder heading linearly from 0 to 15 degrees over 1 minute. Determine the ships heading, in degrees, after 1 minute.
The ship's heading, in degrees, after 1 minute can be determined by considering the autopilot system's time and gain constants, as well as the rudder heading range. Using the given information and the rate of change in heading, we can calculate the ship's heading after 1 minute.
The autopilot system's time constant of 107 s represents the time it takes for the system's response to reach 63.2% of its final value. The gain constant of 0.185 s⁻¹ determines the rate at which the system responds to changes. Since the autopilot moves the rudder heading linearly from 0 to 15 degrees over 1 minute, we can calculate the ship's heading at the end of 1 minute. Given that the rudder heading changes linearly, we can divide the total change in heading (15 degrees) by the time taken (1 minute) to determine the rate of change in heading.
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The design of journal bearings usually involves two suitable combinations of variables: variables under control and dependent variables or performance factors. As such, a full journal bearing has a shaft journal diameter of 27 mm with a unilateral tolerance of 20.01 mm. The bushing bore has a diameter of 27.04 mm with a unilateral tolerance of 0.03 mm. The //d ratio is unity. The bushing load is 1.03 kN, and the journal rotates at 1153 rev/min. You are required to analyze the minimum clearance assembly if the average viscosity is 50 mPa.s to find the minimum oil film thickness, the power loss, and the percentage of side flow.
The variables include shaft journal bearings , bushing bore diameter, //d ratio, bushing load, and rotational speed, while the performance factors are minimum oil film thickness, power loss, and percentage of side flow.
What are the variables and performance factors involved in the design of journal bearings?
The paragraph describes the design of journal bearings and provides specific parameters for a full journal bearing assembly. The variables under control include the shaft journal diameter, bushing bore diameter, //d ratio, bushing load, and rotational speed. The dependent variables or performance factors to be analyzed are the minimum clearance assembly, minimum oil film thickness, power loss, and percentage of side flow.
To analyze the minimum clearance assembly, the given tolerances for the shaft journal and bushing bore diameters are considered. The minimum oil film thickness can be determined based on the average viscosity of the oil.
The power loss in the bearing can be calculated using appropriate formulas, considering factors such as speed, load, and oil viscosity. The percentage of side flow refers to the amount of oil escaping from the sides of the bearing.
Overall, the analysis aims to evaluate the performance and characteristics of the journal bearing assembly, taking into account various factors such as clearance, oil film thickness, power loss, and side flow.
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Explain in details how the processor can execute a couple of instructions, Given that the address of the first instruction in memory is AA2F.
The processor can execute a couple of instructions given that the address of the first instruction in memory is AA2F. The instruction set that the processor can execute depends on the architecture of the processor. Once an instruction is executed, the processor increments the memory address to the next instruction in the sequence. This process continues until the end of the program is reached.
Below are the details on how the processor executes instructions:
1. Fetching: The processor fetches the instruction from the memory location where it is stored. The address of the first instruction in memory is AA2F.
2. Decoding: The processor decodes the instruction to determine the operation that needs to be performed.
3. Executing: The processor executes the operation specified by the instruction.
4. Storing: The processor stores the result of the operation in a register or in memory.
5. Incrementing: The processor increments the memory address to the next instruction in the sequence.
The processor is designed to execute a large number of instructions. The instruction set that the processor can execute depends on the architecture of the processor. Some processors can execute more instructions than others. In general, the more complex the processor, the more instructions it can execute.
In conclusion, the processor can execute a couple of instructions given that the address of the first instruction in memory is AA2F. The processor fetches, decodes, executes, stores, and increments instructions in order to execute a program. The number of instructions that a processor can execute depends on the architecture of the processor.
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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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what is the properties(Mechanical,thermal and electrical) for Ultrahigh molecular weight Polyethylene (UHMWPE) and what is the application and uses of it?
What is all the forms that it can be on it (Like sheet) ?
Ultrahigh molecular weight polyethylene (UHMWPE) possesses several properties, including mechanical, thermal, and electrical characteristics. It finds applications in various fields. Additionally, UHMWPE can be available in different forms, such as sheets.
Ultrahigh molecular weight polyethylene (UHMWPE) is known for its exceptional mechanical properties, including high tensile strength, impact resistance, and abrasion resistance. It has a low coefficient of friction, making it self-lubricating and suitable for applications involving sliding or rubbing components. Thermally, UHMWPE has a high melting point, good heat resistance, and low thermal conductivity. In terms of electrical properties, UHMWPE exhibits excellent dielectric strength and insulation properties, making it suitable for electrical applications. Due to its unique combination of properties, UHMWPE finds wide applications. It is used in industries such as automotive, aerospace, medical, and defense.
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(a) (i) Determine and sketch the domain and range of the function f(x,y)=√√64-x² - y² . (5 Marks) (ii) Find the level curve of the function f(x, y) in part (i) and display this. (6 Marks) (b) (i) Find the rate of change of the temperature field T(x, y, z)=ze²+z+e" at the point P(1,0,2) in the direction of u = 2i-2j+lk. (8 Marks) (ii) In which direction does the temperature in part (i) decrease most rapidly at the point P? What is the minimum rate of change at that point? (3 Marks)
The domain and range of the function f(x, y) = √√(64 - x² - y²), we need to consider the restrictions on the square roots and the values that x and y can take.
Domain:
The square root function (√) requires its argument to be non-negative, so we must have 64 - x² - y² ≥ 0. This implies that x² + y² ≤ 64, which represents a disk centered at the origin with a radius of 8 units. Therefore, the domain of f(x, y) is the interior and boundary of this disk.
Domain: D = {(x, y) | x² + y² ≤ 64}
Range:
The range of the function depends on the values inside the square roots. The inner square root (√) requires its argument to be non-negative as well, so we need to consider √(64 - x² - y²). The outer square root (√) then requires this quantity to be non-negative too.
Since 64 - x² - y² can be at most 64, the inner square root (√) can take values from 0 to √64 = 8. The outer square root (√) can then take values from 0 to √8 = 2√2.
Range: R = [0, 2√2]
Sketch:
To sketch the function f(x, y) = √√(64 - x² - y²), we can plot points in the domain and indicate the corresponding values of f(x, y). Since the function is symmetric with respect to the x and y axes, we only need to consider one quadrant.
For example, when x = 0, the function simplifies to f(0, y) = √√(64 - y²). We can choose some values of y within the range of -8 to 8 and calculate the corresponding values of f(0, y). Similarly, we can calculate f(x, 0) for various values of x within the range of -8 to 8. Plotting these points will give us a portion of the graph of the function.
The level curve of a function represents the set of points where the function has a constant value. To find the level curve of the function f(x, y) = √√(64 - x² - y²), we need to set f(x, y) equal to a constant, say c, and solve for x and y.
√√(64 - x² - y²) = c
Squaring both sides twice, we can eliminate the square roots and obtain:
64 - x² - y² = c⁴
Now, rearranging the equation, we have:
x² + y² = 64 - c⁴
This equation represents a circle centered at the origin with a radius of √(64 - c⁴).
Therefore, the level curve of the function f(x, y) = √√(64 - x² - y²) is a family of circles centered at the origin, with each circle having a radius of √(64 - c⁴), where c is a constant.
find the rate of change of the temperature field T(x, y, z) = ze² + z + e^z at the point P(1, 0, 2) in the direction of u = 2i - 2j + lk, we can use the gradient of the function.
The gradient of T(x, y, z) is given by:
∇
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Differetiate between PI and pd controllers on the basis of
steady state error, overshoot and offset. Draw the hardware diagram
of each controler?
A controller is an electronic or mechanical device that regulates the system's physical parameters by operating on the signal it receives. A PD controller and PI controller are the two types of controllers. PD and PI are both closed-loop controllers.
PI and PD controllers are two types of proportional and derivative (PD) and proportional and integral (PI) controllers. Here's a detailed explanation of how they vary from one another:
PI Controller: On the basis of steady-state error, overshoot, and offset, here are some key features of the PI controller: Steady-state error: Since the PI controller includes an integral term, it can eliminate steady-state errors. If there is a constant disturbance, the integral term of the PI controller increases until it becomes equal to the disturbance's steady-state value.
Overshoot: Since the PI controller only includes a proportional term, there is no overshoot.
Offset: The PI controller is usually used to regulate systems that are difficult to model or that need fast action. Since there is no integral term in the PI controller, it is difficult to use for stable systems.
Therefore, there is no offset issue.
Hardware diagram: PD Controller: On the basis of steady-state error, overshoot, and offset, here are some key features of the PD controller:
Steady-state error: Since the PD controller only includes a derivative term, it cannot remove steady-state errors. This is because the steady-state error is generally proportional to the disturbance, and the PD controller's derivative term is zero for a constant disturbance.
Overshoot: Since the PD controller includes a derivative term, there is always an overshoot.
Offset: The PD controller is ideal for stable systems because there is no integral term. This implies that there is no offset.
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A centrifugal pump may be viewed as a vortex, where the 0.4m diameter impeller, rotates within a 1m diameter casing at a speed of 200 rpm.
Determine
The circumferential velocity, in m/s at a radius of 0.45 m
A centrifugal pump may be viewed as a vortex.
It consists of an impeller that rotates within a casing.
The impeller's diameter is 0.4m and rotates within a 1m diameter casing at a speed of 200rpm.
To determine the circumferential velocity, use the formula provided below:
Formula:
Circumferential velocity (v) = 2π x Radius (r) x Rotational Speed (N) / 60
Given:
Radius (r) = 0.45 m
Rotational speed
(N) = 200 rpm
Diameter of impeller = 0.4m
Diameter of casing = 1m
Solution:
Circumference of the impeller= π
diameter= π x 0.4 m
= 1.2566 m
Therefore,
Circumferential velocity (v) = 2π x Radius (r) x Rotational Speed (N) / 60
= (2 x π x 0.45 m x 200 rpm) / 60
= (0.1414 x 200) m/s
= 28.28 m/s
Therefore, the circumferential velocity at a radius of 0.45 m is 28.28 m/s.
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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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8. Newton's law for the shear stress is a relationship between a) Pressure, velocity and temperature b) Shear stress and velocity c) Shear stress and the shear strain rate d) Rate of shear strain and temperature 9. A liquid compressed in cylinder has an initial volume of 0.04 m² at 50 kg/cm' and a volume of 0.039 m² at 150 kg/em' after compression. The bulk modulus of elasticity of liquid is a) 4000 kg/cm² b) 400 kg/cm² c) 40 × 10³ kg/cm² d) 4 x 10 kg/cm² 10. In a static fluid a) Resistance to shear stress is small b) Fluid pressure is zero c) Linear deformation is small d) Only normal stresses can exist 11. Liquids transmit pressure equally in all the directions. This is according to a) Boyle's law b) Archimedes principle c) Pascal's law d) Newton's formula e) Chezy's equation 12. When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal b) Becomes curved c) Falls down on the front wall d) Falls down on the back wall 13. When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called a) Pascal's law b) Archimedes's principle c) Principle of flotation d) Bernoulli's theorem 14. An ideal liquid a) has constant viscosity b) has zero viscosity c) is compressible d) none of the above. 15. Units of surface tension are a) J/m² b) N/kg c) N/m² d) it is dimensionless 16. The correct formula for Euler's equation of hydrostatics is DE = a) a-gradp = 0 b) a-gradp = const c) à-gradp- Dt 17. The force acting on inclined submerged area is a) F = pgh,A b) F = pgh,A c) F = pgx,A d) F = pgx,A
The correct answers for the fluid mechanics problems are:
(c) Shear stress and the shear strain rate.
(a) 4000 kg/cm².
(b) Fluid pressure is zero.
(c) Pascal's law.
(a) Remains horizontal.
(b) Archimedes's principle.
b) has zero viscosity
(c) N/m².
∇·p = g
(b) F = pg[tex]h_{p}[/tex]A
How to interpret Fluid mechanics?8) Newton's law for the shear stress states that the shear stress is directly proportional to the velocity gradient.
Thus, Newton's law for the shear stress is a relationship between c) Shear stress and the shear strain rate .
9) Formula for Bulk modulus here is:
Bulk modulus =∆p/(∆v/v)
Thus:
∆p = 150 - 50 = 100 kg/m²
∆v = 0.040 - 0.039 = 0.001
Bulk modulus = 100/(0.001/0.040)
= 4000kg/cm²
10) In a static fluid, it means no motion as it is at rest and as such the fluid pressure is zero.
11) Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.
12) When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal
13) When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called b) Archimedes's principle
14) An ideal fluid is a fluid that is incompressible and no internal resistance to flow (zero viscosity)
15) Surface tension is also called Pressure or Force over the area. Thus:
The unit of surface tension is c) N/m²
16) The correct formula for Euler's equation of hydrostatics is:
∇p = ρg
17) The force acting on inclined submerged area is:
F = pg[tex]h_{p}[/tex]A
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A 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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