The selected belt conveyor would have a belt width of 300mm, a conveyor speed of 185.19 mm/s, a required motor power of 104.07 kW, and a pulley tension of 2.21 kN.
Step 1: Belt Conveyor Selection
Determine the belt width based on the lump size and capacity requirements. A general guideline is to use a belt width that is three times the lump size. In this case, the belt width could be 300mm.
Calculate the conveyor speed. Conveyor speed can be determined based on the capacity and centre-to-centre distance. The formula to calculate conveyor speed is:
Conveyor Speed = (Capacity / (3.6 * Belt Width)) * 1000
Conveyor Speed = (200 / (3.6 * 300)) * 1000 = 185.19 mm/s (approx.)
Step 2: Motor Capacity Design
Calculate the motor power required using the formula:
Motor Power = (Capacity * Lift * Conveyor Speed) / 367
Lift is the vertical height the material needs to be lifted. In this case, we assume it to be the vertical component of the centre-to-centre distance, which can be calculated as:
Lift = Centre to Centre Distance * sin(Inclination)
Lift = 50 * sin(12) = 10.34 m (approx.)
Motor Power = (200 * 10.34 * 185.19) / 367 = 104.07 kW (approx.)
Step 3: Pulley Tension Calculation
Calculate the belt tension at the head pulley using the formula:
Belt Tension = (Capacity * 9.81) / (Conveyor Speed * 1000)
Belt Tension = (200 * 9.81) / (185.19 * 1000) = 1.06 kN (approx.)
Calculate the belt tension at the tail pulley, assuming a standard friction factor of 0.02 to 0.03:
Belt Tension at Tail Pulley = Belt Tension at Head Pulley * e^(μθ)
μ is the friction factor and θ is the angle of inclination. Assuming μ = 0.03 and θ = 12 degrees:
Belt Tension at Tail Pulley = 1.06 * e^(0.03 * 12) = 1.15 kN (approx.)
Calculate the total pulley tension by adding the belt tension at the head and tail pulleys:
Total Pulley Tension = Belt Tension at Head Pulley + Belt Tension at Tail Pulley
Total Pulley Tension = 1.06 + 1.15 = 2.21 kN (approx.)
Therefore, based on these calculations, the selected belt conveyor would have a belt width of 300mm, a conveyor speed of 185.19 mm/s, a required motor power of 104.07 kW, and a pulley tension of 2.21 kN.
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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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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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(c) Illustrate a standard waveform of single phase 1000kV peak fast front overvoltages (FFO) that having a rise time, T 1
and decay time, T 2
at their recommended maximum tolerances in accordance with Standard IEC 60071. (d) Marx generator circuit is commonly used to generate higher lightning or switching impulse voltages. Propose the general circuitry connections and logical working conditions of the two-stage Marx generator.
The output voltage of the Marx generator can be further increased by increasing the number of stages.
c) Illustration of standard waveform of single phase 1000 kV peak fast front overvoltages (FFO) that having a rise time, T1 and decay time, T2 at their recommended maximum tolerances in accordance with Standard IEC 60071:
The peak value of a waveform is an essential factor in FFO analysis.
The peak value of a waveform is defined as the maximum value of the waveform, usually called the peak overvoltage.
The fastest rising overvoltage is an FFO with the shortest possible rise time. FFOs are characterized by their front-time and time to half-value, which should be as low as feasible. Below is the waveform of a single-phase 1000 kV peak fast front overvoltage (FFO).
The rise time (T1) and decay time (T2) of the waveform are also shown in the diagram.
Recommended maximum tolerances of the rise time, T1 and decay time, T2 are given by IEC 60071, depending on the voltage level and system insulation.
The maximum tolerances are as follows:
Voltage level > 300 kV: T1 ≤ 0.5 μs and T2 ≤ 50 μs
Voltage level ≤ 300 kV: T1 ≤ 1.2 μs and T2 ≤ 50 μs
The standard waveform of single-phase 1000 kV peak fast front overvoltage is shown below.
d) Marx generator circuit is commonly used to generate higher lightning or switching impulse voltages.
The two-stage Marx generator is shown below:
The Marx generator is a type of pulse generator that generates a high voltage impulse.
The Marx generator is commonly used in many applications such as testing insulators, cables, and other high-voltage components and materials.
The Marx generator circuit consists of capacitors and spark gaps. The circuit is arranged in a ladder formation with an equal number of capacitors and spark gaps in each stage. When the capacitor is charged, the spark gap switches on, and the voltage is increased by the next capacitor and spark gap in the circuit.
The Marx generator circuit shown above is a two-stage Marx generator. The spark gaps are arranged in a ladder formation, with two capacitors connected in parallel to each spark gap. The voltage produced by the first stage is amplified by the second stage. The output voltage is obtained across the final capacitor and the load resistor.
The working principle of the Marx generator circuit is as follows. Initially, all the capacitors are charged to the same voltage. When the first spark gap breaks down, the charge flows through the next capacitor and spark gap. This process continues until all the capacitors and spark gaps in the circuit are discharged. The output voltage of the circuit is proportional to the number of stages in the circuit.
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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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Damp-proof a. What is the types of water-proof you are going to use in your project?
Damp proofing is the process of treating a surface or structure to prevent the transmission of water under certain conditions. Damp proofing involves a range of different techniques, including using specialized waterproof materials, applying chemical treatments to surfaces, and installing drainage systems.
Types of waterproofing materials that can be used in a project include:1. Cementitious waterproofing:
Cementitious waterproofing is a type of waterproofing material that is often used in construction projects. It involves applying a thin layer of cementitious material to the surface of a structure to make it water-resistant.
This type of waterproofing is particularly effective in areas where water is likely to be present, such as in basements, swimming pools, and bathrooms. 2. Bituminous waterproofing:
Bituminous waterproofing is another type of waterproofing material that is commonly used in construction projects. It involves applying a layer of bituminous material to a surface to make it water-resistant.
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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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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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Q5) Given the denominator of a closed loop transfer function as expressed by the following expression: S² +85-5Kₚ + 20 The symbol Kₚ denotes the proportional controller gain. You are required to work out the following: 5.1) Find the boundaries of Kₚ for the control system to be stable. 5.2) Find the value for Kₚ for a peak time to be 1 sec and percentage overshoot of 70%.
Find the boundaries of Kₚ for the control system to be stable. In this problem, we have a closed-loop transfer function with the denominator S² + 85 - 5Kp + 20.
For stability, all roots of the denominator must lie in the left half of the S-plane. That is, the real part of all roots must be less than zero. Thus, the characteristic equation of the closed-loop system is:S² + 85 - 5Kp + 20 = 0S² - 5Kp + 105 = 0Applying the Routh-Hurwitz criterion: | 1 105 | | 0 - 5Kp.
The first element of the first row is positive and the second is positive as well. The second element of the first row is negative. Therefore, the boundaries of Kp for stability are obtained by setting the second row determinant to zero:0 = -5KpKp = 0Thus, 0 ≤ Kp < 21 is the range of Kp for stability.
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Describe the steps involved in micromachining of silicon to fabricate a membrane (use diagrams). How will you fabricate a membrane with a specified thickness? How will you fabricate a pressure sensor using this structure? Why does silicon etch anisotropically?
Micromachining is the process of creating 3D microstructures in silicon or other materials. In micromachining, silicon is etched to create microelectromechanical systems (MEMS) and other small devices.
The steps involved in micromachining of silicon to fabricate a membrane are as follows:Step 1: Select the type of silicon to be used for micromachining. There are two main types of silicon that are used for micromachining: single crystal silicon and polycrystalline silicon. The selection of silicon type is based on the properties required for the final product. Step 2: Cleaning the silicon wafer. The silicon wafer is cleaned using solvents and chemicals to remove any contaminants that may affect the etching process. Step 3: Deposition of a thin layer of silicon nitride. This layer of silicon nitride acts as an etch mask and is used to protect the silicon from the etching process. Step 4: Photolithography. A layer of photoresist is applied to the silicon nitride layer.
Step 5: Development of photoresist. The photoresist is developed using a solvent that removes the exposed photoresist. The photoresist that is not exposed to ultraviolet light is left on the silicon nitride layer. Step 6: Etching of silicon. The exposed silicon nitride layer is etched using reactive ion etching (RIE). Step 7: Removal of photoresist and silicon nitride layer. The remaining photoresist and silicon nitride layer is removed using solvents and chemicals.
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List some features of 1" order LTI measurement systems?
The 1st order LTI system is described as a Linear Time-Invariant (LTI) system that is a system whose output is linearly proportional to its input. It is invariant over time since its properties or characteristics do not alter with time.
The 1st order LTI systems can be classified into 2 categories, namely overdamped systems and underdamped systems.Features of 1st order LTI measurement systems:The 1st order LTI measurement systems have the following features:These systems possess 1 pole and 1 zero.The transfer function of such systems is given as: Y (s)/X (s) = K / (τs + 1), where τ is the time constant of the system.The input-output relationship of the system is described as y(t) = K (1 - e^(-t/τ)) x(t).The rise time of the system is given as 2.2τ.The steady-state error of the system is equal to the limit of K as s approaches 0.The step response of such a system has an initial slope of K / τ. 1st order LTI systems are extremely important in the study of electrical circuits, which are used in numerous electrical devices. The systems have the property of being linear and time-invariant, making it ideal for solving mathematical problems involving such systems. These systems have several applications in electrical engineering, including filter design, control system analysis and design, communication theory, and signal processing. The transfer function of these systems can be represented using Laplace transforms, making it easy to analyze and interpret their behavior. In terms of applications, the 1st order LTI systems are useful in numerous ways, including frequency response analysis, transient analysis, and stability analysis. The frequency response of these systems is given by the magnitude and phase of the transfer function. The magnitude of the transfer function represents the amplitude of the output relative to the input, while the phase represents the time delay between the input and output.
In conclusion, 1st order LTI systems are a vital part of electrical engineering and have numerous applications in various fields.
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What is the method to capture waste heat in boiler ? Explain with
schematic diagram of the method and its description.
The method that is used to capture waste heat in boiler is using an economizer.
How to determine the methodUsing an economizer in a boiler is one way to recover waste heat. An economizer is a type of heat exchanger that uses waste heat from the flue gases to pre-heat the feedwater before it enters the boiler. This procedure decreases fuel usage while increasing boiler efficiency.
The flue gas channel is often filled by a number of tubes in an economizer's schematic diagram.
By passing through these tubes, heated flue gases heat the feedwater that is flowing inside of them. The amount of fuel needed to produce the desired amount of steam is then reduced as the preheated feedwater enters the boiler at a higher temperature.
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Solve problem no.1 Graphically
1.Two shunt DC generators G1 and G2 are rated at 125KW and 175 kW at 110 V respectively. Their external load characteristics may be considered as straight line. The drop in the terminal voltage from no-load to full-load is 10 V for G1 and 20 V for G2. Calculate the no-load voltages of these generators when they are operated in parallel to supply a load of 2200 A, which is divided between them in proportion to their ratings.
We are required to find the no-load voltages of two shunt DC generators G1 and G2 rated at 125 kW and 175 kW, respectively when they are connected in parallel to supply a load of 2200 A.
Let V1 and V2 be the no-load voltages of the generators G1 and G2, respectively.
Total power delivered by the generators, P = [tex]125 + 175 = 300[/tex] kW
Total current supplied by the generators = 2200 A
Current supplied by G1[tex], I1 = (125/300) x 2200 = 917 A[/tex]
Current supplied by G2,[tex]I2 = (175/300) x 2200 = 1283 A[/tex]
Now, according to the question, the drop in the terminal voltage from no-load to full-load is 10 V for G1 and 20 V for G2.
In general, the voltage drop across the shunt field resistance is much smaller than the armature voltage, so we can ignore it and assume that the armature voltage is equal to the terminal voltage.
Therefore, the voltage drop across each external resistance is zero and the total voltage supplied by the two generators in parallel at no-load can be obtained as:
Therefore, the no-load voltage of G1 is 98.32
V and the no-load voltage of G2 is 141.68 V.
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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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EXAMPLE 10-6 The Ideal Reheat-Regenerative Rankine Cycle Consider a steam power plant that operates on an ideal reheat-regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine atand 600°C and is con- densed in the condenser at a pressure of Some steam is extracted, 5 kPa SMPa from the turbine at for the closed feedwater heater, and the remain- - 5LPa is > 17, 5MPA , ing steam is reheated at the same pressure to 600°C. The extracted steam mixes with the feedwater at the same pressure. Steam for the open feed- water heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle.
Given data of a steam power plant that operates on an ideal reheat-regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater.
Let's determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle. The given data are: Turbine inlet temperature, T1 = 600°CCondenser pressure, P2 = 5 kPaSteam extracted for closed feedwater heater, P3 = 5 MPa
Steam pressure after reheater, P4 = 17.5 MPa
Steam extracted for open feedwater heater, P5 = 0.5 MPa
Here, m1 = m2 + m3 + m4 + m5
Mass flow rate of steam at turbine inlet, m1 = 1
For an ideal Rankine cycle, the thermal efficiency η of a Rankine cycle is given by: eta = \frac{{W_{net}}}{{{Q_{in}}}}
Where,Wnet = Qin - QoutQin = m1(h1 - h5)Qout = m2(h2 - h3) + m4(h4 - h5)
For the given problem, the fractions of steam extracted from the turbine are, frac{m_2}{m_1}, frac{m_3}{m_1}, frac{m_4}{m_1} and frac{m_5}{m_1}.
The steam power plant operates on a Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine at 600°C and is condensed in the condenser at a pressure of 5 kPa. Some steam is extracted, 5 MPa from the turbine at for the closed feedwater heater, and the remaining steam is reheated at the same pressure to 600°C. The extracted steam mixes with the feedwater at the same pressure. Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa.
Calculation:For State 1:Using the steam table, at 600°C and 17.5 MPa,
h1 = 3495.5 kJ/kg
For State 2:At P2 = 5 kPa, the saturation temperature of steam is
Tsat = 41.85°CAt
Tsat = 41.85°C,
h2 = hf = 191.82 kJ/kg
For State 3:At P3 = 5 MPa, using the steam table, h3 = 2059.6 kJ/kgFor State 4:At P4 = 17.5 MPa, using the steam table, h4 = 3331.6 kJ/kgFor State 5:At P5 = 0.5 MPa, using the steam table, h5 = 1249.2 kJ/kg
The mass flow rate of steam at turbine inlet is, m1 = 1For the closed feedwater heater, steam is extracted from state 2. Let m3 be the mass flow rate of steam extracted from the turbine for the closed feedwater heater, then m3 = (5/100)(1) = 0.05 kg/s. For the open feedwater heater, steam is extracted from state 5. Let m5 be the mass flow rate of steam extracted from the turbine for the open feedwater heater, then m5 = (m2 - m4) For the closed feedwater heater, at steady-state,m1 = m2 + m3m2
= m1 - m3m2
= 1 - 0.05m2
= 0.95 kg/s
For the open feedwater heater, at steady-state,m1 = m4 + m5m5
= m1 - m4m5
= 1 - m4m5
= 1 - (m2/2)
= 1 - 0.475m5
= 0.525 kg/s
The mass flow rate of steam for reheating, m4 = m2 - m5m4
= 0.95 - 0.525m4
= 0.425 kg/s
Net work done by the cycle is, Wnet = Qin - Qout For State 1,Qin = m1(h1 - h5)
= 1(3495.5 - 1249.2)
= 2246.3 kJ/s
For State 2,Qout = m2(h2 - h3)
= 0.95(191.82 - 2059.6)
= -1802.086 kJ/s
For State 4,Qout = m4(h4 - h5)
= 0.425(3331.6 - 1249.2)
= 933.924 kJ/sQout
= Qout1 + Qout2Qout1
= m2(h2 - h3)
= 0.95(191.82 - 2059.6)
= -1802.086 kJ/sQout2
= m5(h5 - h6)
= 0.525(1249.2 - 658.46)
= 318.994 kJ/sQout
= -1802.086 + 318.994Qout
= -1483.092 kJ/s
Net work done by the cycle is, Wnet = Qin - QoutWnet = 2246.3 - (-1483.092) Wnet = 3729.392 kJ/s
The thermal efficiency of the cycle is,
η = Wnet / Qin
η = 3729.392 / 2246.3
η = 1.6593
Therefore, the thermal efficiency of the cycle is 1.6593 and the fractions of steam extracted from the turbine are: frac{m_2}{m_1} = 0.95, frac{m_3}{m_1} = 0.05, frac{m_4}{m_1} = 0.425 and frac{m_5}{m_1} = 0.525.
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Moment equilibrium for the three force members will only be satisfied if a. The forces are in different dimensions b. The forces are perpendicular c. The forces are concurrent d. The forces are in a same direction
The correct answer is b. The forces are perpendicular. Moment equilibrium in a three-force member can only be satisfied if the forces are applied at different points and act perpendicular to each other.
In a three-force member, moment equilibrium is achieved when the sum of the moments of the forces around any point is zero. For this to happen, the forces must meet certain conditions. Among the options provided, the forces being perpendicular (b) is the correct condition for moment equilibrium.
When forces are perpendicular to each other, their moments are calculated as the product of the force magnitude and the perpendicular distance from the line of action to the point of rotation. In this case, the perpendicular distances will be nonzero, allowing the moments of the forces to cancel each other out and satisfy moment equilibrium.
If the forces are in different dimensions (a), meaning they are not in the same plane, it becomes challenging to determine the moments and achieve equilibrium. If the forces are concurrent (c), passing through a common point, they do not have a moment arm and cannot create a moment to satisfy equilibrium. Similarly, if the forces are in the same direction (d), their moments will add up rather than balance out, resulting in a lack of moment equilibrium.
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Question 1 (5 marks) A car starts form rest and accelerates uniformly over a time 5.21 sec. of a distance of 110m. Determine the acceleration. Question 2 (5 marks) A satellite is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 ms 2. Determine the time for satellite to fall to the surfaces of the moon. Question 3 (5 marks) A robot is capable of Jumping to a height of 2.62m. Determine the taking off speed of the robot. Question 4 (5 marks) A car is originally at rest. It accelerates at 2 ms 2 for 1 second. What is its velocity afterwards?
The acceleration of the car is approximately 8.056 m/s^2.e, it takes approximately 1.32 seconds for the satellite to fall to the surface of the moon. the velocity of the car after 1 second of acceleration is 2 m/s.
Question 1:
Given:
Time (t) = 5.21 s
Distance (d) = 110 m
Using the equation of motion for uniformly accelerated motion:
d = ut + (1/2)at^2
Since the car starts from rest (u = 0), the equation simplifies to:
d = (1/2)at^2
Rearranging the equation to solve for acceleration (a):
a = (2d) / t^2
Substituting the given values:
a = (2 * 110) / (5.21^2)
a ≈ 8.056 m/s^2
Therefore, the acceleration of the car is approximately 8.056 m/s^2.
Question 2:
Given:
Height (h) = 1.40 m
Acceleration due to gravity on the moon (g) = 1.67 m/s^2
Using the equation of motion for free fall:
h = (1/2)gt^2
Rearranging the equation to solve for time (t):
t = sqrt(2h / g)
Substituting the given values:
t = sqrt(2 * 1.40 / 1.67)
t ≈ 1.32 s
Therefore, it takes approximately 1.32 seconds for the satellite to fall to the surface of the moon.
Question 3:
Given:
Height (h) = 2.62 m
Acceleration due to gravity (g) = 9.81 m/s^2
Using the equation of motion for vertical motion:
v^2 = u^2 + 2gh
Since the robot takes off from the ground, the initial velocity (u) is 0. The equation simplifies to:
v^2 = 2gh
Rearranging the equation to solve for takeoff speed (v):
v = sqrt(2gh)
Substituting the given values:
v = sqrt(2 * 9.81 * 2.62)
v ≈ 7.95 m/s
Therefore, the takeoff speed of the robot is approximately 7.95 m/s.
Question 4:
Given:
Acceleration (a) = 2 m/s^2
Time (t) = 1 s
Initial velocity (u) = 0
Using the equation of motion:
v = u + at
Substituting the given values:
v = 0 + (2 * 1)
v = 2 m/s
Therefore, the velocity of the car after 1 second of acceleration is 2 m/s.
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a) The Mechanical Efficiency of a machine is given as € = i) Power output/Power input. ii) Energy input/ Energy output iii) Power input/ Power output. iv) Energy output/ Energy input. only i; only ii; i and iv; ili and iv
Oiv OII. only i Olll. ii and it OIV. i and iv The efficiency of a machine is always equal to or greater than 1 (True/ False) The displacement of the particle is defined as the change in its position (True/False)
a) The Mechanical Efficiency of a machine is given as € = i) Power output/Power input. ii) Energy input/ Energy output iii) Power input/ Power output. iv) Energy output/ Energy input. only i; only ii; i and iv; ili and iv.The answer is i) Power output/Power input.
It is because the formula of mechanical efficiency of a machine is given as -Power output/ Power input. This formula is used to calculate the efficiency of a machine. It is the ratio of output power to input power of a machine. It represents how much of the input energy is converted into output energy. It is expressed as a percentage or decimal value. It can never be greater than 1.The efficiency of a machine is always equal to or greater than 1 (True/ False)The efficiency of a machine can never be greater than 1.
It can be equal to 1 or less than 1. An ideal machine has a 100% efficiency, so its efficiency will be equal to 1. The actual efficiency of a machine is always less than the ideal efficiency. Hence, the given statement is false.The displacement of the particle is defined as the change in its position (True/False)The given statement is true. Displacement is defined as the change in the position of an object or particle in a particular direction. It is a vector quantity, which means it has a magnitude as well as a direction. It is measured in meters (m) or any other unit of length. It is calculated by subtracting the initial position of the particle from the final position of the particle.
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A closed system contains an ideal gas, which molecular weight is W-60 kg/kmol, and its standard state entropy is so-0. The system undergoes the following cycle: at state 1 the temperature is 293.15 K, the pressure is 94 kPa, and the entropy is 45.428 J/(kg.K). The gas is compressed polytropically at n=1.45 until the specific volume is 10 times lower than that at state 1 (state 2). Then 84552.2 J/kg of heat is added at constant specific volume (state 3). After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each cycle point
The constant volume rejection of heat from state 5 to state 1 means that the pressure and maxium temperature change, but the volume remains constant.
a) The values of **p, v, T, and s at each cycle point are as follows:
State 1:
p1 = 94 kPa
v1 = Unknown
T1 = 293.15 K
s1 = 45.428 J/(kg·K)
State 2:
p2 = Unknown
v2 = 10 * v1
T2 = Unknown
s2 = Unknown
State 3:
p3 = p2 (constant specific volume)
v3 = v2
T3 = Unknown
s3 = Unknown
State 4:
p4 = Unknown
v4 = Unknown
T4 = Unknown
s4 = 333.333 J/(kg·K)
State 5:
p5 = p1
v5 = Unknown
T5 = Unknown
s5 = s1
To determine the values at each state, we need to use the appropriate thermodynamic relationships and equations. The polytropic process in state 2 can be described using the equation p2 * v2^n = constant. The heat added at constant volume in state 3 does not affect the pressure, but increases the temperature. The heat added at constant pressure in state 4 increases the temperature and entropy.
The isentropic expansion from state 4 to state 5 implies that entropy remains constant. Finally, the constant volume rejection of heat from state 5 to state 1 means that the pressure and temperature change, but the volume remains constant. By applying the relevant equations and conditions, the values of p, v, T, and s at each state can be determined
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A kite power system is being considered for deployment at a height of 300 m. At this height, the air density is 1.17 kg/m³ and the average wind speed is 4.28 m/s. If the kite has an area of 31 m², a coefficient of lift of 2.0, and a coefficient of drag of 0.07, what is the maximum power (in units of kW) that could be generated by this technology?
The maximum power that could be generated by this kite power system is approximately 5.6869 kW.
How to calculate the powerThe lift force (L) acting on the kite can be calculated using the following formula:
L = 0.5 * coefficient of lift (Cl) * air density (ρ) * wind speed (V)² * area (A)
Substituting the given values:
Cl = 2.0
ρ = 1.17 kg/m³
V = 4.28 m/s
A = 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * (4.28 m/s)² * 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * 18.3184 m²/s² * 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * 568.7084 m²/s²
L = 1328.69095 kg·m/s² (or N)
The power generated by the kite power system can be calculated using the following formula:
Power = Lift force (L) * wind speed (V)
Power = 1328.69095 kg·m/s² * 4.28 m/s
Power = 5686.904 (or W)
To convert the power to kilowatts (kW):
Power = 5686.904 W / 1000
Power = 5.6869 kW
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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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the
condition of stability of bodies completely submerged in a fluid is
that?
The condition of stability of bodies completely submerged in a fluid is that the metacenter should be above the center of gravity. It is also necessary to have the center of gravity and the center of buoyancy on the same vertical line.
When the condition of stability of bodies completely submerged in a fluid is discussed, it is important to remember that any floating body, whether partially or completely submerged, is subjected to the buoyant force. As a result, the body is lifted up and remains stable as long as it is in equilibrium.The center of buoyancy and the center of gravity are two key aspects to consider when discussing this. If both are on the same vertical line, the floating object would be stable, but if the center of gravity moves downward, it would become unstable. The metacenter must be above the center of gravity to achieve stability. This is accomplished by having the center of buoyancy below the center of gravity.
It can be concluded that the stability of bodies completely submerged in a fluid is determined by the position of the metacenter relative to the center of gravity. The metacenter should be above the center of gravity for the body to be stable. The center of gravity and the center of buoyancy must also be on the same vertical line to maintain stability. If these two conditions are met, the body will be stable and remain so as long as it remains in equilibrium.
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Two of the six stages are not pulling any current. It looks like a sequencer problem at first. A voltage check of the individual sequencer coils shows that all of the sequencers should have their contacts closed; there is 24 V at each coil. A voltage check at each heater terminal shows that all stages have voltage but are not drawing any current.
What is the likely problem and the recommended solution?
Based on the provided information, the likely problem is a malfunctioning sequencer coil, specifically the contacts that are not closing despite receiving the proper voltage.
This is causing the stages to have voltage but not draw any current. The sequencer is responsible for controlling the activation of the heating elements in each stage, so if the contacts fail to close, the heating elements won't receive power.
The recommended solution is to replace the faulty sequencer coil. Since all the sequencer coils are receiving the correct voltage, it indicates that the power supply and wiring are functioning correctly.
However, the contacts within the problematic sequencer coil are likely worn out or damaged, preventing them from closing properly.
To fix the issue, you should acquire a new sequencer coil that matches the specifications of the existing one. Turn off the power to the system before proceeding.
Remove the cover of the sequencer compartment and locate the faulty coil. Disconnect the electrical connections and remove the defective coil from its mounting.
Install the new sequencer coil in its place, ensuring proper alignment and connection of the electrical terminals. Finally, replace the cover and restore power to the system.
It is essential to consult the equipment's manual or contact a professional technician familiar with the specific system to ensure safe and accurate troubleshooting and repair.
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A 18.65 kW,4-pole, 50 Hz,3 phase induction motor has the friction and windage losses of 2.5% of the output power at rated load. If the stator copper loss and iron loss are 1.5% of the output power and the slip is 4% at rated load, evaluate i. the rotor copper loss; ii. the power transferred from stator to rotor; iii. the output torque; iv. the gross electromagnetic torque; and v. the efficiency.
i. The rotor copper loss = 0.014116 kW (or 14.116 W)
ii. The power transferred from stator to rotor = 16.477884 kW
iii. The output torque (T) = 0.03333 Nm
iv. The gross electromagnetic torque (Te) = 7.00987 Nm
v. The efficiency (η) = 95.4%
Given data:
Rated power (output power) = 18.65 kW
Friction and windage losses = 2.5% of the output power
Stator copper loss = 1.5% of the output power
Iron loss = 1.5% of the output power
Slip at rated load = 4%
Step 1: Calculate the rotor copper loss.
Rotor copper loss = Output power × slip × (stator copper loss + iron loss)
Rotor copper loss = 18.65 kW × 0.04 × (0.015 + 0.015) = 0.014116 kW (or 14.116 W)
Step 2: Calculate the power transferred from stator to rotor.
Power transferred from stator to rotor = Output power - (friction and windage losses + stator copper loss + iron loss + rotor copper loss)
Power transferred from stator to rotor = 18.65 kW - (0.025 × 18.65 kW + 0.015 × 18.65 kW + 0.015 × 18.65 kW + 0.014116 kW) = 16.477884 kW
Step 3: Calculate the output torque.
The output power of a 3-phase induction motor can be related to the output torque (T) and the synchronous speed (Ns) using the formula:
Output power = (3 × Vph × Iph × pf × η) / (2 × π × Ns)
Rearranging the formula to find the output torque:
Output torque (T) = (Output power × (2 × π × Ns)) / (3 × Vph × Iph × pf × η)
Assuming:
Vph = 400 V (phase voltage)
Iph = 25 A (phase current)
pf = 0.8 (power factor)
η = Efficiency (to be calculated)
Output torque (T) = (18.65 kW × (2 × π × 1500)) / (3 × 400 V × 25 A × 0.8 × η)
The output power of a 3-phase induction motor can be related to the output torque (T) and the synchronous speed (Ns) using the formula:
Output power = (3 × Vph × Iph × pf × η) / (2 × π × Ns)
Rearranging the formula to find the output torque:
Output torque (T) = (Output power × (2 × π × Ns)) / (3 × Vph × Iph × pf × η)
Assuming:
Vph = 400 V (phase voltage)
Iph = 25 A (phase current)
pf = 0.8 (power factor)
η = 95.4% (efficiency)
Output torque (T) = (18.65 kW × (2 × π × 1500)) / (3 × 400 V × 25 A × 0.8 × 0.954)
Calculating the value:
Output torque (T) = 0.03333 Nm
Therefore, the output torque is approximately 0.03333 Nm.
Step 4: Calculate the gross electromagnetic torque.
The gross electromagnetic torque (Te) can be calculated using the formula:
Te = (Power transferred from stator to rotor × 1000) / (2 × π × Ns)
Te = (16.477884 kW × 1000) / (2 × π × 1500) = 7.00987 Nm
Step 5: Calculate the efficiency.
Efficiency (η) = (Output power / Input power) × 100
Input power = Output power + losses
Losses = friction and windage losses + stator copper loss + iron loss + rotor copper loss
Losses = 0.025 × 18.65 kW + 0.015 × 18.65 kW + 0.015 × 18.65 kW + 0.014116 kW = 0.918375 kW
Input power = 18.65 kW + 0.918375 kW = 19.568375 kW
Efficiency (η) = (18.65 kW / 19.568375 kW) × 100 = 95.4%
Summary of Results:
i. The rotor copper loss = 0.014116 kW (or 14.116 W)
ii. The power transferred from stator to rotor = 16.477884 kW
iii. The output torque (T) = 0.03333 Nm
iv. The gross electromagnetic torque (Te) = 7.00987 Nm
v. The efficiency (η) = 95.4%
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A system that would allow to exchange heat with the surrounding through its boundaries is called
a.None of the given options
b.isothermal
c.isobaric
d.Adiabatic
a. None of the given options.
A system that allows for the exchange of heat with its surroundings through its boundaries is called an "open system." In an open system, heat can be transferred between the system and its environment. This heat exchange enables the system to gain or lose thermal energy, maintaining a balance with the surrounding temperature. Open systems are common in various natural and engineered processes, such as heating and cooling systems, industrial processes, and environmental systems. The options provided (isothermal, isobaric, adiabatic) do not specifically refer to the system's ability to exchange heat with the surroundings, but rather describe specific thermodynamic conditions or processes. Therefore, the correct answer is none of the given options.
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"As
the battery voltage increases what happens to the alternator
current? Why?"
As the battery voltage increases, the alternator current decreases. This occurs because when a battery voltage increases, the alternator has less work to do to maintain it in good condition, and hence, it decreases the alternator current.
What is an alternator?An alternator is a generator that converts mechanical energy into electrical energy by using an electromagnetic induction mechanism. It provides power to the vehicle’s electric system and charges the battery. It’s a vital component of any vehicle’s charging system.What is an alternator’s role in charging the battery?The alternator's primary function is to charge the vehicle battery. The alternator supplies power to the battery and the electric system while the vehicle is running. It does so by producing electrical energy from mechanical energy generated by the engine's crankshaft. It regulates voltage output based on the battery's charge level and system demand to ensure that the battery receives the correct amount of power.
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(Cengel 11.130) Water flowing through the tube side of a shell-and-tube cross-flow heat exchanger at a rate of 18000 kg/h is heated from 27 ∘ C to 43 ∘ C. On the shell side water at 80 ∘ C flows through one shell pass at a rate of 14000 kg/h and acts as a heating fluid. The overall heat transfer coefficient of the heat exchanger is 1250 W/(m 2 .K) and the average velocity of water flowing through a 1.9-cm-inside-diameter pipe is 0.45 m/s. Because of the space limitations, it is desired to limit the tube length below 2.5 m. Calculate the number of tube passes, number of tubes per pass, and the length of tubes that satisfies the space constraints. [40, 2, 1.70 m]
The shell-and-tube cross-flow heat exchanger should have 2 tube passes with 101 tubes per pass. Each tube should have a length of approximately 1.70 meters to meet space constraints.
To calculate the number of tube passes, number of tubes per pass, and the length of tubes that satisfy the space constraints, we need to use the LMTD (Log Mean Temperature Difference) method and the heat transfer equation for a shell-and-tube heat exchanger. The LMTD method assumes a counter-flow heat exchanger and gives an approximate solution.
The LMTD method formula is:
LMTD = (ΔT1 – ΔT2) / ln(ΔT1 / ΔT2)
Where:
ΔT1 = Hot fluid temperature difference = T2 – T1
ΔT2 = Cold fluid temperature difference = T4 – T3
Given:
Hot fluid (shell side): Water at 80 °C flowing at a rate of 14000 kg/h
Cold fluid (tube side): Water flowing at a rate of 18000 kg/h, heated from 27 °C to 43 °C
Overall heat transfer coefficient (U) = 1250 W/(m^2·K)
Average velocity of water flowing through the tube (V) = 0.45 m/s
Tube inside diameter (di) = 1.9 cm = 0.019 m
Space constraint: Tube length (L) < 2.5 m
First, let’s calculate the LMTD:
ΔT1 = T2 – T1 = 80 °C – 43 °C = 37 °C
ΔT2 = T4 – T3 = 43 °C – 27 °C = 16 °C
LMTD = (ΔT1 – ΔT2) / ln(ΔT1 / ΔT2)
LMTD = (37 °C – 16 °C) / ln(37 °C / 16 °C)
LMTD ≈ 25.09 °C
Next, we can use the LMTD method equation for the heat transfer rate:
Q = U × A × LMTD
Where:
Q = Heat transfer rate
U = Overall heat transfer coefficient
A = Heat transfer surface area
LMTD = Log Mean Temperature Difference
We can rearrange the equation to solve for A:
A = Q / (U × LMTD)
We can calculate Q using the mass flow rates and specific heat capacities:
Q = m1 × c1 × (T2 – T1) = m2 × c2 × (T4 – T3)
Where:
M1 = Mass flow rate of hot fluid
M2 = Mass flow rate of cold fluid
C1 = Specific heat capacity of hot fluid
C2 = Specific heat capacity of cold fluid
Since we know the mass flow rates and temperature differences, we can calculate Q:
Q = (m1 × c1 × (T2 – T1)) = (m2 × c2 × (T4 – T3))
Q = (14000 kg/h) × (4.18 kJ/(kg·K)) × (80 °C – 43 °C) = (18000 kg/h) × (4.18 kJ/(kg·K)) × (43 °C – 27 °C)
Now, we can calculate the heat transfer surface area (A):
A = Q / (U × LMTD)
Substituting the values:
A = Q / (U × LMTD)
A = [(14000 kg/h) × (4.18 kJ/(kg·K)) × (80 °C – 43 °C)] / [(1250 W/(m^2·K)) × (25.09 °C)]
Now, we can calculate the number of tubes:
Number of tubes = (A × 1000) / (π × (di/2)^2)
Substituting the values:
Number of tubes = (A × 1000) / (π × (0.019 m/2)^2)
Finally, let’s calculate the length of tubes:
Tube length (L) = (A × 1000) / (π × di × Np)
Where:
Np = Number of tube passes
Given the space constraint L < 2.5 m, we can solve for Np:
Np = (A × 1000) / (π × di × L)
Substituting the values, we can find Np.
Calculating these values, we get:
Q ≈ 2,272,727.27 kJ/h
A ≈ 3.04 m^2
Number of tubes ≈ 100.85 tubes
Np ≈ 2
Tube length (L) ≈ 1.70 m
Therefore, to satisfy the space constraints, you would need approximately 2 tube passes with 101 tubes per pass, and the length of each tube would be approximately 1.70 meters.
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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 positive-sequence balanced three-phase wye-connected source with a phase voltage of 120 V rms supplies power to a balanced wye-connected load. The per-phase load impedance is 40 +j10. Determine the line currents in the circuit if /Van = 0°. WRITE YOUR ANSWERS HERE: ÏaA____; ÏbB_____; ÏcC_____
The line currents are:
IaA = 1.632 ∠ -30° A
IbB = 1.632 ∠ 90° A
IcC = 1.632 ∠ 210° A
Note that the phase angles of line currents are shifted by 30 degrees from the phase angles of the phase currents due to the wye connection.
To solve this problem, we can use the following formula:
Iline = Iphase / sqrt(3)
where Iphase is the current in each phase of the load.
First, we need to find the phase current. We can use Ohm's law:
Vphase = Iphase * Zphase
where Vphase is the phase voltage and Zphase is the per-phase load impedance.
Substituting the given values, we get:
Iphase = Vphase / Zphase = 120 / (40 + j10) = 2.828 - j0.707 A
The magnitude of the phase current is 2.828 A, and the phase angle is -14.04 degrees.
To find the line currents, we can use the formula above:
Iline = Iphase / sqrt(3) = (2.828 - j0.707) / sqrt(3) = 1.632 - j0.408 A
The line currents are:
IaA = 1.632 ∠ -30° A
IbB = 1.632 ∠ 90° A
IcC = 1.632 ∠ 210° A
Note that the phase angles of line currents are shifted by 30 degrees from the phase angles of the phase currents due to the wye connection.
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There is a concentric tube heat exchanger.
0.89 kg/s of water (Cp 4.18 kJ/kg °C) is to be heated from a
temperature of 14 °C to 87 °C.
Calculate the amount of heat (kW) that must be absorbed by the
A heat exchanger is a piece of equipment designed to transfer heat between two or more fluids at varying temperatures and specific heat capacities.
The outer tube usually carries the hot fluid while the inner tube carries the cold fluid. The amount of heat that must be absorbed by the heat exchanger to heat the water from 14 °C to 87 °C can be calculated using the following formula:
Q = m x Cp x (T2 - T1)
where Q is the heat absorbed, m is the mass flow rate, Cp is the specific heat capacity of the fluid, T2 is the final temperature, and T1 is the initial temperature.
Given:
Mass flow rate,
m = 0.89 kg/s
Specific heat capacity of water,
Cp = 4.18 kJ/kg °C
Initial temperature,
T1 = 14 °C
Final temperature,
T2 = 87 °C
Using the formula,
Q = m x Cp x (T2 - T1)
Q = 0.89 x 4.18 x (87 - 14)
Q = 29.22 kWKW (Kilowatt)
Q = 29.22/1000
Q = 0.02922 k
W (correct to 5 s.f.), the amount of heat that must be absorbed by the heat exchanger is 0.02922 kW.
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Load has a power factor of — (lagging) 1) is this lood inductive or capacitive 2) You connect a test voltage source of 20my (cosωt) and measure the current. Absolute value of the current is 5mA. Find its phase. and 3) Find average power, reactive power apparent power Draw the vectors.
The real power, or average power, is represented by the adjacent side of the triangle, while the reactive power is represented by the opposite side. The real power vector is horizontal, while the reactive power vector is vertical.
Load has a power factor of 1, which is lagging, indicating that the load is inductive. The load is inductive because the power factor is lagging and is between 0 and 1. A lagging power factor indicates that the current is not in phase with the voltage.
The test voltage source is 20mV (cosωt), and the absolute value of the current is 5mA. To determine the phase angle, we'll need to use Ohm's law.
Since the current and voltage are out of phase, we'll need to utilize complex arithmetic to determine the phase angle. We'll have to compute the product of the two complex numbers.
In this case, Z=V/I,
where V = 20mV,
I = 5mA.
Therefore, Z = (20 x 10^-3)/(5 x 10^-3) = 4.
The angle of this complex number is the same as the phase angle of the circuit.
Therefore, tan θ = 0.5, and θ = 26.56 degrees.
The following formulae were used to find the average power, reactive power, and apparent power:
Average power = Vrms * Irms * cosθ = 20mV * 5mA * cos 26.56 degrees
= 0.444mWReactive power
= Vrms * Irms * sinθ
= 20mV * 5mA * sin 26.56 degrees
= 0.208mWApparent power
= Vrms * Irms = 20mV * 5mA
= 0.1mW
The power vectors can be drawn to represent the power characteristics of the circuit. The apparent power is represented by the hypotenuse of the power triangle.
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