A company buys the following item from supplier A with Annual Demand 3000, Holding cost/per year 2.0$, Ordering cost per order $40, and Working days/year 250.
Here is the solution to the following question:1. EOQEOQ (Economic Order Quantity) is calculated as under: EOQ = Sqrt(2DCo/H)Where: D = Annual Demand; Co = Ordering Cost; H = Holding Cost. Using the values from the given problem, we can calculate EOQ as under: EOQ = sqrt(2 * 3000 * 40/2)EOQ = 244.94 ~ 2452.
Average inventory The formula to calculate average inventory is given as follows: Average inventory = EOQ / 2Using the EOQ from the above calculation, we can find the average inventory as: Average inventory = 245 / 23. How many orders per year? The number of orders per year can be calculated using the following formula: Number of orders per year = D / Q Using the annual demand and EOQ from the above calculations, we can calculate the number of orders as: Number of orders per year = 3000 / 245Number of orders per year = 12.245 ~ 124.
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Please show all work neatly and double-check work before uploading. A natural gas with the molar analysis 78% CH₄, 13% C₂H₆, 6% C₃H₈, 1.7 % C₄H₁₀, 1.3% N₂ burns completely with 30% excess air in a reactor operating at steady state. The molar flow rate of the fuel is 3.5 kmol/h and air is 21% O₂ and 79% N₂ on a molar basis. Determine the molar flow rate of the air, in kmol/h. Determine mass flow rate of the air, in kg/h Determine the mole fraction of water vapor in the products
To determine the molar flow rate of air, we first need to calculate the amount of oxygen required for the combustion of the natural gas. Given the molar flow rate of the fuel as 3.5 kmol/h, and the molar analysis of the natural gas (78% CH₄, 13% C₂H₆, 6% C₃H₈, 1.7% C₄H₁₀).
We can calculate the molar flow rate of oxygen (O₂) required as follows:
Moles of CH₄ = 0.78 * 3.5 kmol/h = 2.73 kmol/h
Moles of C₂H₆ = 0.13 * 3.5 kmol/h = 0.455 kmol/h
Moles of C₃H₈ = 0.06 * 3.5 kmol/h = 0.21 kmol/h
Moles of C₄H₁₀ = 0.017 * 3.5 kmol/h = 0.0595 kmol/h
Total moles of carbon (C) = Moles of CH₄ + Moles of C₂H₆ + Moles of C₃H₈ + Moles of C₄H₁₀
= 2.73 + 0.455 + 0.21 + 0.0595
= 3.4545 kmol/h
Moles of O₂ required = Moles of carbon * 1.5 (stoichiometric ratio)
= 3.4545 * 1.5
= 5.1818 kmol/h
Since the air contains 21% O₂ on a molar basis, we can calculate the molar flow rate of air:
Molar flow rate of air = Moles of O₂ required / 0.21 (molar fraction of O₂ in air)
= 5.1818 / 0.21
≈ 24.677 kmol/h
To determine the mass flow rate of air, we need to consider the molecular weights of the components. The molecular weight of N₂ is 28 g/mol and the molecular weight of O₂ is 32 g/mol.
Mass flow rate of air = Molar flow rate of air * (28 g/mol * 0.79 + 32 g/mol * 0.21)
≈ 24.677 * (22.12 + 6.72)
≈ 718.91 kg/h
To find the mole fraction of water vapor in the products, we need to consider the combustion reaction and the molar flow rates of the different components.
The combustion reaction for CH₄ can be written as:
CH₄ + 2O₂ -> CO₂ + 2H₂O
The moles of water vapor produced will be twice the moles of CH₄ consumed.
Moles of water vapor = 2 * Moles of CH₄
= 2 * 2.73 kmol/h
= 5.46 kmol/h
To calculate the mole fraction of water vapor, we divide the moles of water vapor by the total moles in the products:
Mole fraction of water vapor = Moles of water vapor / (Moles of water vapor + Moles of CO₂)
= 5.46 / (5.46 + Moles of CO₂)
The moles of CO₂ can be determined by multiplying the moles of carbon (C) by the stoichiometric ratio:
Moles of CO₂ = Moles of carbon *
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6) The only difference between the sinut motor and a separately excited motor is that (A) A separately excited DC motor has its field circuit connected to an independent voltage supply (B) The shunt DC motor has its field circuit connected to the armature terminals of the motor (C) A and B (D) The shunt DC motor has its armature circuit connected to the armature tenuinals of the motor 7) One of the following statements is true for DC-Separately Excited Generator (A) The no load characteristic same for increasing and decreasing excitation current (B) The no load characteristic differ for increasing and decreasing excitation current (C) The no load characteristic same for increasing and decreasing load resistance (D) The load characteristic same for increasing and decreasing load resistance 4G Done
Therefore, the correct option is (B) The no load characteristic differs for increasing and decreasing excitation current.
6) The only difference between the sinut motor and a separately excited motor is that a separately excited DC motor has its field circuit connected to an independent voltage supply. This statement is true.
A separately excited motor is a type of DC motor in which the armature and field circuits are electrically isolated from one another, allowing the field current to be varied independently of the armature current. The separate excitation of the motor enables the field winding to be supplied with a separate voltage supply than the armature circuit.
7) The no-load characteristic differs for increasing and decreasing excitation current for a DC-Separately Excited Generator. This statement is true.
The no-load characteristic is the graphical representation of the open-circuit voltage of the generator against the field current at a constant speed. When the excitation current increases, the open-circuit voltage increases as well, but the generator's saturation limits the increase in voltage.
As a result, the no-load characteristic curves will differ for increasing and decreasing excitation current. Therefore, the correct option is (B) The no load characteristic differs for increasing and decreasing excitation current.
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A natural-circulation pillow-block bearing has a journal diameter D of 62.5 mm with a unilateral tolerance of -0.025 mm. The bushing bore diameter B is 62.6 mm with a unilateral tolerance of 0.1 mm. The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 1350 N in shaft- stirred air at 21°C. The lateral area of the pillow-block housing is 38,700 mm2. Perform a design assessment using minimum radial clearance for a load of 2700 N and 1350 N. Use Trumpler's criteria and that both 1/d and a are unity.
Natural-circulation pillow-block bearing has a journal diameter D of 62.5 mm with a unilateral tolerance of -0.025 mm. The bushing bore diameter B is 62.6 mm with a unilateral tolerance of 0.1 mm.
The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 1350 N in shaft- stirred air at 21°C. The lateral area of the pillow-block housing is 38,700 mm². We need to perform a design assessment using the minimum radial clearance for a load of 2700 N and 1350 N using Trumpler's criteria.
Both `1/d` and `a` are unity. Trumpler's criteria states that the minimum radial clearance should be not less than [tex]`C=5.3(1/d)^(1/3)a^(2/3)`mm[/tex]. Given that the `1/d` and `a` are unity. `[tex]1/d=1`, and `a=1[/tex]`.Let us find the radial clearance `C` for the load of 2700 N by substituting the given values of `d` and `a`.`[tex]C=5.3(1/d)^(1/3)a^(2/3)[/tex]`For load = 2700 N: `[tex]C=5.3(1/62.5)^(1/3)×1^(2/3)` = `0.051 mm[/tex].
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In your understanding, differentiate between 3-aspect signalling and 4-aspect signalling.
3-aspect and 4-aspect signaling are two different methods of railway signalling that are used to ensure safety and provide information to train drivers. In this context, aspect refers to the number of lights used in the signal to convey information to the driver.3-aspect signalling uses three colours of light: red, yellow, and green.
The meanings of these colours in 3-aspect signalling are as follows:Red: This indicates that the driver must stop the train immediately. It is used when there is a danger ahead, such as a broken track or an obstruction.Yellow: This indicates that the driver should slow down and be prepared to stop at the next signal. It is used when there is a warning ahead, such as a slower train or construction work.Green: This indicates that the driver may proceed at the normal speed. It is used when the track ahead is clear.4-aspect signalling uses four colours of light: red, yellow, green, and double yellow.
The meanings of these colours in 4-aspect signalling are as follows:Red: This indicates that the driver must stop the train immediately. It is used when there is a danger ahead, such as a broken track or an obstruction.Yellow: This indicates that the driver should slow down and be prepared to stop at the next signal. It is used when there is a warning ahead, such as a slower train or construction work.Green: This indicates that the driver may proceed at the normal speed. It is used when the track ahead is clear.
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Select the item listed that is NOT a type of electrical transducer. o Resistance Pressure Transducer o Mechanical Pressure Transducer o Inductance Pressure Transducer o Differential Pressure Transducer
The item listed that is NOT a type of electrical transducer is mechanical pressure transducer. Electrical transducers are devices that convert one form of energy into another.
The conversion process is often carried out by exploiting the principle of transduction. Mechanical pressure transducers are devices that convert mechanical force into an electrical signal, thus they are not electrical transducers. Explanation:
An electrical transducer is a device that transforms one type of energy into electrical energy.
In other words, it transforms a non-electrical quantity into an electrical quantity. Types of Electrical Transducers1. Resistive transducer. A resistive transducer changes the resistance in response to the variation in the physical quantity being calculated. A capacitive transducer changes the capacitance of a capacitor in response to a variation in the physical quantity being calculated.
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The work function of a metal surface is 4.5 eV. If the frequency of the light incident upon it is 1.45 × 1015 Hz, then what is the maximum kinetic energy (in eV) of the photo electrons emitted from the surface?
The maximum kinetic energy (in eV) of the photo electrons emitted from the surface is 6 ev.
To calculate the maximum kinetic energy of photoelectrons emitted from a metal surface, we can use the equation:
E max=hν−φ
Where: E max is the maximum kinetic energy of photoelectrons,
h is the Planck's constant (4.135667696 × 10⁻¹⁵ eV s),
ν is the frequency of the incident light (1.45 × 10¹⁵ Hz),
φ is the work function of the metal surface (4.5 eV).
Plugging in the values:
E max =(4.135667696×10⁻¹⁵ eV s)×(1.45×10¹⁵ Hz)−4.5eV
Calculating the expression:
E max =5.999eV
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Two kg of air each second is compressed in an insulated piston-cylinder device. Using constant specific heats and treating the process as internally reversible, the amount of work required to compress form 100kPa,27°C to 2MPa,706°C is ___
The amount of work that is required to compress the air would be 1, 363.4 kJ.
How to find the amount of work ?The work done (W) on the air during compression can be determined by using the equation:
W = m * Cp * (T2 - T1)
Before using this formula, temperatures need to be converted from Celsius to Kelvin. The conversion is done by adding 273.15 to the Celsius temperature.
T1 = 27°C + 273.15
= 300.15 K
T2 = 706°C + 273.15
= 979.15 K
The specific heat at constant pressure (Cp) for air at room temperature is approximately 1005 J/kg.K.
Substituting these values into the formula gives:
W = 2 kg/s * 1005 J/kg.K * (979.15 K - 300.15 K)
= 1363.4 kJ
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6. A glass capillary tube of diameter 0.3 mm and length 60 mm is dipped in a water
having surface tension 0.017 N/m. The contact angle between the liquid and the
tube wall is 40°. Will the water overflow through the tube? If not, comment on
the nature and radius of meniscus
The solution of the given problem has been done using the Young's equation. Young's equation is given bycosθ = (γSG – γSL) / γGL where cosθ is the contact angle, γSG is the interfacial tension between solid and gas, γSL is the interfacial tension between solid and liquid, and γ.
GL is the interfacial tension between gas and liquid.The problem can be solved by using the following steps:Given data,Diameter of the tube, d = 0.3 mmLength of the tube, L = 60 mmSurface tension of water, γ = 0.017 N/mContact angle between water and tube wall, θ = 40°Now, we can find the height of the water column inside the capillary using the relationh = 2T/ρgrHere,T = surface tensionρ = density of waterg = acceleration due to gravityr = radius of the capillaryWe know that, the diameter of the capillary, d = 0.3 mm.
This is the maximum height of the water column inside the capillary. Now, we need to check whether the water will overflow or not. To do that, we need to find the radius of the meniscus.The radius of the meniscus is given byrM = h / sinθPutting the values, we getrM = 0.76 / sin 40°rM = 1.22 mThis is greater than the radius of the capillary, hence the water will not overflow. Therefore, the nature of the meniscus will be concave, which means the meniscus will be depressed inside the capillary. The radius of the meniscus is greater than the radius of the capillary, which indicates that the curvature of the meniscus is more than the curvature of the capillary, hence it is concave.
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What is the reason the lost-foam process is capable of producing
fine surface details on the castings? Explain.
The lost-foam process produces fine surface details by using precise foam patterns and metal flow.
Pattern material: In the lost-foam process, the pattern used for creating the mold is typically made of expanded polystyrene (EPS) foam.
EPS foam patterns have excellent dimensional stability and can be easily shaped and carved to achieve intricate details. The foam pattern accurately replicates the desired shape and surface features of the final casting.
Vaporization and expansion: When the molten metal is poured into the foam-filled mold, the high temperature of the metal causes the foam pattern to vaporize and expand.
The vaporization of the foam creates a void within the mold, which is subsequently filled by the molten metal. As the foam pattern vaporizes, it leaves behind a network of interconnected channels and vents within the mold.
Surface replication: As the metal fills the void left by the vaporized foam, it flows into the intricate channels and vents present in the mold. The metal fills the mold cavity completely, ensuring that fine details are replicated accurately.
The metal solidifies within the mold, taking the shape and surface texture of the foam pattern.
The lost-foam process allows for the production of fine surface details on castings due to the use of foam patterns with excellent dimensional stability and the ability of the molten metal to flow into intricate channels and vents.
This process results in castings that accurately replicate the desired shape and surface features of the foam pattern, leading to high-quality castings with fine surface details.
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For corrosion to occur, there must be an anodic and cathodic reaction, oxygen must be available, and there must be both an electronically and fonically conductive path True O False
The given statement, "For corrosion to occur, there must be an anodic and cathodic reaction, oxygen must be available, and there must be both an electronically and fonically conductive path" is true.
The occurrence of corrosion is reliant on three necessary factors that must be present simultaneously. These three factors are:Anode and cathode reaction: When a metal comes into touch with an electrolyte, an oxidation reaction occurs at the anode, and an opposite reaction of reduction occurs at the cathode. The reaction at the anode causes the metal to dissolve into the electrolyte, and the reaction at the cathode protects the metal from corrosion.
Oxygen: For the cathodic reaction to take place, oxygen must be present. If there is no oxygen available, the reduction reaction at the cathode will not happen, and hence, no cathodic protection against corrosion.Electronically and Fonically Conductive Path: To make a closed circuit, the anode and cathode should be electrically connected. A connection can occur when the metal comes into touch with a different metal or an electrolyte that conducts electricity.
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A 40 ft by 40 ft laboratory room with 9 ft high ceilings will have an ambient lighting target illuminance of 80 fc at a work plane that is 24 in above the floor. It is anticipated that the ceiling reflectance is 0.80 and the average wall reflectance is about 0.7. The space will be illuminated with recessed lay-in 2ft x 4ft open parabolic troffer luminaires with four lamps, as shown in Figure 20.16. The initial output of the fluorescent lamps is 2950 lumen. The light loss factor will be assumed to be 0.70.
A.) Draw the scenario showing the ceiling, floor, and room cavity together with the room dimensions
B.) Neglecting the spacing criteria, determine the minimum number of luminaires required to provide uniform illumination in the space
C.) Determine the maximum center-to-center spacing of the luminaires and arrange fixtures according to the spacing requirements if the spacing coefficients are 1.4/1.2.
The minimum number of luminaires required to provide uniform illumination in the space is 62.
Max Spacing = 4 ft x 1.4 = 5.6 ft (along the longer dimension)
Max Spacing = 2 ft x 1.2 = 2.4 ft (along the shorter dimension)
B.) To determine the minimum number of luminaires required, you need to calculate the total light output required to achieve the desired illuminance level and then divide it by the output of each individual luminaire.
First, convert the illuminance target from foot-candles (fc) to lumens per square foot (lm/ft²):
80 fc = 80 lm/ft²
The work plane area can be calculated as follows:
Area = Length x Width = 40 ft x 40 ft = 1600 ft²
Now, calculate the total light output required:
Total Light Output = Illuminance x Area = 80 lm/ft² x 1600 ft² = 128,000 lumens
Next, account for the light loss factor:
Light Loss Factor = 0.70
Adjusted Light Output = Total Light Output / Light Loss Factor = 128,000 lumens / 0.70 = 182,857 lumens
Since each luminaire has an initial output of 2950 lumens, divide the adjusted light output by the output of each luminaire to determine the minimum number of luminaires:
Minimum Number of Luminaires = Adjusted Light Output / Luminaire Output = 182,857 lumens / 2950 lumens = 62 luminaires
Therefore, the minimum number of luminaires required to provide uniform illumination in the space is 62.
C.) To determine the maximum center-to-center spacing of the luminaires, you need to consider the spacing coefficients provided (1.4/1.2).
Maximum Center-to-Center Spacing = Luminaire Length x Spacing Coefficient
Assuming the luminaires are 2 ft x 4 ft (Width x Length), the maximum center-to-center spacing would be:
Max Spacing = 4 ft x 1.4 = 5.6 ft (along the longer dimension)
Max Spacing = 2 ft x 1.2 = 2.4 ft (along the shorter dimension)
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How would you link the capacity decision being made by Fitness Plus to other types of operating decisions?
Fitness Plus, an emerging fitness and gym provider, is trying to gain a significant share of the market in the region, making it a major competitor to other industry players. Fitness Plus's decision to expand its capacity is critical, and it influences the types of operating decisions they make, including marketing, financial, and human resource decisions.
Capacity decisions at Fitness Plus are linked to marketing decisions in several ways. When Fitness Plus decides to expand its capacity, it means that it is increasing the number of customers it can serve simultaneously. The expansion creates an opportunity to increase sales by catering to a more extensive market. Fitness Plus's marketing team must focus on building brand awareness to attract new customers and create loyalty among existing customers.The expansion also influences financial decisions. Fitness Plus must secure funding to finance the expansion project.
It means that the financial team must identify potential sources of financing, analyze their options, and determine the most cost-effective alternative. Fitness Plus's decision to expand its capacity will also have a significant impact on its human resource decisions. The expansion creates new job opportunities, which Fitness Plus must fill. Fitness Plus must evaluate its staffing requirements and plan its recruitment strategy to attract the most qualified candidates.
In conclusion, Fitness Plus's decision to expand its capacity has a significant impact on its operating decisions. The expansion influences marketing, financial, and human resource decisions. By considering these decisions together, Fitness Plus can achieve its growth objectives and increase its market share in the region.
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The nameplate on a reciprocating pump lists the following
dimensions: 7" x 6" x 4". The diameter of the liquid cylinder
is
a. 4"
b. 7"
c. 8"
d. 6"
The correct answer to the question is the option c. 8".The diameter of the liquid cylinder is 8".A reciprocating pump consists of a piston that moves back and forth within a cylinder. The cylinder, also known as the liquid cylinder, is where the fluid is held and moved when the piston travels.
The cylinder's diameter is a crucial factor in the pump's operation because it determines how much fluid can be moved at once. When the diameter is large, a higher volume of fluid can be transported per stroke.
the nameplate on a reciprocating pump lists the following dimensions: 7" x 6" x 4". These dimensions are most likely referring to the pump's overall size and not the liquid cylinder's diameter. Therefore, we must utilize another method to determine the liquid cylinder's diameter.
The diameter of the liquid cylinder can be calculated using the following formula: Diameter =[tex](4 x Area) / π[/tex]Where the area is the cross-sectional area of the cylinder. The area is determined by multiplying the cylinder's height by its width (length) and then multiplying that result by π/4 since the cylinder is circular. In this instance, the dimensions provided on the nameplate are 7" x 6" x 4".
We can assume that the height and length of the cylinder are 6" and 4", respectively. Area = [tex]6" x 4" x π/4 = 6π[/tex]Now, substituting the area into the diameter formula :Diameter =[tex](4 x Area) / π = (4 x 6π) / π = 24/π = 7.64" ≈ 8"[/tex]
Therefore, the diameter of the liquid cylinder is 8".
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A flat-panel domestic heater 1 m tall x 2 m long is used to maintain a room at 20 °C. An electrical element keeps the surfaces of the radiator at 65 °C. Approximating the heater as a vertical flat plate, calculate the heat transferred to the room by natural convection from both surfaces of the heater (front and back). Assuming that the surface of the heater is painted white, calculate the heat transferred from the radiator to the surrounding surfaces by radiation. Note: The emissivity value of white paint for longwave radiation is approximately 0.8.
The heat transferred from the radiator to the surrounding surfaces by radiation is 321.56 W.
Given that the flat-panel domestic heater is 1 m tall and 2 m long. The heater maintains the room temperature at 20°C. The electrical element keeps the surface temperature of the radiator at 65°C. The heater is approximated as a vertical flat plate. The heat transferred to the room by natural convection from both surfaces of the heater (front and back) can be calculated using the following formula;
Q = h × A × (ΔT)
Q = heat transferred
h = heat transfer coefficient
A = surface are (front and back)
ΔT = temperature difference = 65 - 20 = 45°C
For natural convection, the value of h is given by;
h = k × (ΔT)^1/4
Where k = 0.15 W/m2K
For the front side;
A = 1 × 2 = 2 m2
h = 0.15 × (45)^1/4 = 3.83 W/m2K
Q = h × A × (ΔT)Q = 3.83 × 2 × 45 = 344.7 W
For the back side, the temperature difference will be the same but the surface area will change.
Area of back side = 1 × 2 = 2 m2
h = 0.15 × (45)^1/4 = 3.83 W/m2K
Q = h × A × (ΔT)Q = 3.83 × 2 × 45 = 344.7 W
The total heat transferred by natural convection from the front and back surface is;
Qtotal = 344.7 + 344.7 = 689.4 W
The heat transferred from the radiator to the surrounding surfaces by radiation can be calculated using the following formula;
Q = σ × A × ε × (ΔT)^4
Where σ = 5.67 × 10-8 W/m2K
4A = 1 × 2 = 2 m2
ΔT = (65 + 273) - (20 + 273) = 45°C
Emissivity ε = 0.8Q = 5.67 × 10-8 × 2 × 0.8 × (45)^4Q = 321.56 W
Therefore, the heat transferred from the radiator to the surrounding surfaces by radiation is 321.56 W.
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The flow just upstream of a normal shock wave is given by p₁ = 1 atm, T₁ = 288 K, and M₁ = 2.6. Calculate the following properties just downstream of the shock: p2, T2, P2, M2, Po.2, To.2, and the change in entropy across the shock.
The normal shock wave is a type of shock wave that occurs at supersonic speeds. It's a powerful shock wave that develops when a supersonic gas stream encounters an obstacle and slows down to subsonic speeds. The following are the downstream properties of a normal shock wave:Calculation of downstream properties:
Given,Upstream properties: p₁ = 1 atm, T₁ = 288 K, M₁ = 2.6Downstream properties: p2, T2, P2, M2, Po.2, To.2, and change in entropy across the shock.Solution:First, we have to calculate the downstream Mach number M2 using the upstream Mach number M1 and the relationship between the Mach number before and after the shock:
[tex]$$\frac{T_{2}}{T_{1}} = \frac{1}{2}\left[\left(\gamma - 1\right)M_{1}^{2} + 2\right]$$$$M_{2}^{2} = \frac{1}{\gamma M_{1}^{-2} + \frac{\gamma - 1}{2}}$$$$\therefore M_{2}^{2} = \frac{1}{\frac{1}{M_{1}^{2}} + \frac{\gamma - 1}{2}}$$$$\therefore M_{2} = 0.469$$[/tex]
Now, we can calculate the other downstream properties using the following equations:
[tex]$$\frac{P_{2}}{P_{1}} = \frac{\left(\frac{2\gamma}{\gamma + 1}M_{1}^{2} - \frac{\gamma - 1}{\gamma + 1}\right)}{\left(\gamma + 1\right)}$$$$\frac{T_{2}}{T_{1}} = \frac{\left(\frac{2\gamma}{\gamma + 1}M_{1}^{2} - \frac{\gamma - 1}{\gamma + 1}\right)^{2}}{\gamma\left(\frac{2\gamma}{\gamma + 1}M_{1}^{2} - \frac{\gamma - 1}{\gamma + 1}\right)^{2} - \left(\gamma - 1\right)}$$$$P_{o.2} = P_{1}\left[\frac{2\gamma}{\gamma + 1}M_{1}^{2} - \frac{\gamma - 1}{\gamma + 1}\right]^{(\gamma)/( \gamma - 1)}$$$$T_{o.2} = T_[/tex]
where R is the gas constant and [tex]$C_{p}$[/tex] is the specific heat at constant pressure.We know that,
γ = 1.4, R = 287 J/kg-K, and Cp = 1.005 kJ/kg-K
Substituting the values, we get,Downstream Mach number,M2 = 0.469Downstream Pressure,P2 = 3.13 atmDownstream Temperature,T2 = 654 KDownstream Density,ρ2 = 0.354 kg/m³Stagnation Pressure,Po.2 = 4.12 atmStagnation Temperature,To.2 = 582 KChange in entropy across the shock,Δs = 1.7 J/kg-KHence, the required downstream properties of the normal shock wave are P2 = 3.13 atm, T2 = 654 K, P2 = 0.354 kg/m³, Po.2 = 4.12 atm, To.2 = 582 K, and Δs = 1.7 J/kg-K.
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List ten different built-in functions and describe each with examples. 6. What happens if we don't assign a variable to an expression which evaluates a numerical value? a) MATLAB shows error b) Nothing happens c) The evaluated values are assigned to a variable ans automatically d) Depends on the numerical value 7. What does the Workspace show? a) Attributes of variables, functions from command window b) Attributes of variables, script files from command window c) Attributes of variables, script files, functions from command window d) Attributes of variables from command window
Ten different built-in functions in MATLAB are: abs, sqrt, sin, cos, exp, log, floor, ceil, round, and rand.
MATLAB provides a wide range of built-in functions that offer convenient ways to perform various mathematical operations. Here are ten different built-in functions along with their descriptions and examples:
1. abs: Returns the absolute value of a number. Example: abs(-5) returns 5.
2. sqrt: Calculates the square root of a number. Example: sqrt(25) returns 5.
3. sin: Computes the sine of an angle given in radians. Example: sin(pi/2) returns 1.
4. cos: Computes the cosine of an angle given in radians. Example: cos(0) returns 1.
5. exp: Evaluates the exponential function e^x. Example: exp(2) returns approximately 7.3891.
6. log: Calculates the natural logarithm of a number. Example: log(10) returns approximately 2.3026.
7. floor: Rounds a number down to the nearest integer. Example: floor(3.8) returns 3.
8. ceil: Rounds a number up to the nearest integer. Example: ceil(1.2) returns 2.
9. round: Rounds a number to the nearest integer. Example: round(2.6) returns 3.
10. rand: Generates a random number between 0 and 1. Example: rand() returns a random number.
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1. Highlight the consideration factors and the challenges/problems in the entire process of
completing the 3D printed products.
2. Recommend any modifications that can be done to improve the 3D printing process
and the printed products.
Consideration factors and challenges/problems in the entire process of completing the 3D printed products The entire process of 3D printing of products, from design to printing, requires careful consideration of the following factors and challenges.
Thus, the designer must determine the material type that is suitable for the design. Consumable supplies:
1. Improve print settings :It's important to set the printer to the correct printing settings, such as speed, temperature, and layer thickness.
2. Proper maintenance: Regular maintenance of the printer, including cleaning and lubrication, can significantly improve its performance.
3. Upgrading the printer: Upgrading the printer with better components like hotends, extruders, and control boards can improve its speed, precision, and overall performance.
4. Using support materials: Support materials can be added to complex designs to improve the structure and quality of the print.
5. Using advanced software: Using advanced software to design and slice 3D models can help improve the quality of the print.
6. Using high-quality filaments: Using high-quality filaments can improve the quality and durability of the print.
7. Using post-processing techniques: Post-processing techniques like sanding, painting, and polishing can significantly improve the appearance of the final product.
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A 6 liter gasoline engine is being evaluated in a laboratory to determine the exhaust gas ratio at a location where the air density is 1.181 kg/m³. The engine is running at 3600 RPM, with an air/fuel ratio of 15:1, and the volumetric efficiency has been estimated at 93%. Calculate the exhaust gas rate in kg/s.
The exhaust gas rate is approximately 1.56 kg/s.
To calculate the exhaust gas rate, we need to determine the mass flow rate of air entering the engine and then determine the mass flow rate of fuel based on the given air/fuel ratio.
First, we calculate the mass flow rate of air entering the engine using the engine displacement (6 liters) and the volumetric efficiency (93%). By multiplying these values with the air density at the location (1.181 kg/m³), we obtain the mass flow rate of air.
Next, we calculate the mass flow rate of fuel by dividing the mass flow rate of air by the air/fuel ratio (15:1).
Finally, by adding the mass flow rates of air and fuel, we obtain the total exhaust gas rate in kg/s.
Performing the calculations, the exhaust gas rate is found to be approximately 1.56 kg/s.
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EXPOUND & ANSWER THE QUESTION BRIEFLY AND GIVE INSIGHTS AND
CITE SOURCES ABOUT THE TOPIC. THANK YOUU
Water management is an important aspect of electric power production. Identify at least two needs for water in a Rankine cycle-based power plant. Describe typical water management practices in such plants, and research at least two emerging technologies aimed at reducing water losses in plants or enhancing sustainable water management.
Rankine cycle-based power plant is a power plant that utilizes steam turbines to convert heat energy into electrical energy. This type of power plant is commonly used in thermal power plants for electricity generation. Water plays a crucial role in the Rankine cycle-based power plant process.
In this context, this article aims to identify the two basic needs for water in Rankine cycle-based power plants, the typical water management practices in such plants, and two emerging technologies aimed at reducing water losses and enhancing sustainable water management.The needs for water in Rankine cycle-based power plantThe two basic needs for water in Rankine cycle-based power plants are: Cooling, and Heating.Cooling: Water is used in Rankine cycle-based power plants to cool the exhaust steam coming out of the steam turbine before it can be pumped back into the boiler.
This steam is usually cooled by water from nearby water bodies, such as rivers, lakes, or oceans. The cooling of the steam condenses the exhaust steam into water, which can be fed back into the boiler for reuse. Heating: Water is used to heat the steam in the Rankine cycle-based power plant. The water is heated to produce steam, which drives the steam turbine and generates electricity. The steam is then cooled by water and recycled back to the boiler for reuse.Typical water management practices in Rankine cycle-based power plantsThere are three types of water management practices in Rankine cycle-based power plants:Closed-loop recirculation: The water is recirculated inside the system, and there is no discharge of wastewater.
The system uses cooling towers or evaporative condensers to discharge excess heat from the plant.Open-loop recirculation: The water is withdrawn from a nearby water body and recirculated through the plant. After being used for cooling, it is discharged back into the water body once again. This practice may have a negative impact on the ecosystem.Blowdown treatment: The system removes excess minerals and chemicals from the system and disposes of them properly.
Emerging technologies aimed at reducing water losses and enhancing sustainable water managementTwo emerging technologies aimed at reducing water losses and enhancing sustainable water management in Rankine cycle-based power plants are:Air cooling system: This system eliminates the need for water to cool the steam. Instead, it uses air to cool the steam. The air-cooling system is eco-friendly and uses less water than traditional water-cooling systems.Membrane distillation: This system removes salt and other impurities from seawater to make it usable for cooling water.
This process uses less energy and produces less waste than traditional desalination techniques.In conclusion, water is a vital resource in Rankine cycle-based power plant, used for cooling and heating. Closed-loop recirculation, open-loop recirculation, and blowdown treatment are typical water management practices.
Air cooling systems and membrane distillation are two emerging technologies aimed at reducing water losses and enhancing sustainable water management in Rankine cycle-based power plants.Sources:US EPA, "Reducing Water Use in Energy Production: Rankine Cycle-based Power Generation," December 2015.Edwards, B. D., S. B. Brown, and K. J. McLeod. "Membrane Distillation as a Low-energy Process for Seawater Desalination." Desalination 203, no. 1–3 (2007): 371–83.
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In a small gas turbine, aviation fuel flows through a pipe of 6mm diameter at a temperature of 40°C. The dynamic viscosity and the specific gravity of the fuel is given as 1.1x10‐³ Pa.s and 0.94 respectively at this temperature. Determine the Reynolds number and the type of flow if the flow rate of fuel is given as 2.0 lit/min. If the operating temperature increases to 80°C, the viscosity and the sp.gr gets reduced by 10%. Determine the change in the Reynolds number.
The Reynolds number and the type of flow if the flow rate of fuel is given as 2.0 lit/min is determined as follows.
Reynolds numberReynolds number (Re) = ρVD/μwhere; ρ = Density of fuel = sp.gr * density of water = 0.94 * 1000 kg/m³ = 940 kg/m³D = Diameter of the pipe = 6 mm = 0.006 mV = Velocity of fuel = Q/A = 2.0/[(π/4) (0.006)²] = 291.55 m/sμ = Dynamic viscosity of fuel = 1.1×10⁻³ Pa.sNow,Re = [tex](940 × 291.55 × 0.006)/1.1×10⁻³= 1.557 ×10⁶.[/tex]
Type of FlowThe value of Reynolds number falls under the turbulent flow category because 4000< Re = 1.557 ×10⁶.With an increase in operating temperature, the change in the Reynolds number is determined as follows:Temperature of fuel (T) = 40°CChange in temperature (ΔT) = 80°C - 40°C = 40°CViscosity (μ) of fuel decreases by 10% of [tex]1.1 × 10⁻³= 0.1 × 1.1 × 10⁻³ = 1.1 × 10⁻⁴[/tex].
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Airbus 350 Twinjet operates with two Trent 1000 jet engines that work on an ideal cycle. At 1.8km, ambient air flowing at 55 m/s will enter the 1.25m radius inlet of the jet engine. The pressure ratio is 44:1 and hot gasses leave the combustor at 1800K. Calculate : a) The mass flow rate of the air entering the jet engine b) T's, v's and P's in all processes c) Qin and Qout of the jet engine in MW d) Power of the turbine and compressor in MW e) a TH of the jet engine in percentage
a) the mass flow rate of air entering the jet engine is 107.26 kg/s.
b) The velocity at the inlet of the engine is given as 55 m/s.
c) Qout = -11.38 MW
d) the power of the compressor is 79.92 MW and the power of the turbine is 89.95 MW.
e) TH = 995.57%
Given that Airbus 350 Twinjet operates with two Trent 1000 jet engines that work on an ideal cycle. At 1.8 km, ambient air flowing at 55 m/s will enter the 1.25 m radius inlet of the jet engine.
The pressure ratio is 44:1 and hot gasses leave the combustor at 1800 K. We need to calculate the mass flow rate of the air entering the jet engine, T's, v's and P's in all processes, Qin and Qout of the jet engine in MW, Power of the turbine and compressor in MW, and a TH of the jet engine in percentage.
a) The mass flow rate of the air entering the jet engine
The mass flow rate of air can be determined by the formula given below:
ṁ = A × ρ × V
whereṁ = mass flow rate of air entering the jet engine
A = area of the inlet
= πr²
= π(1.25 m)²
= 4.9 m²
ρ = density of air at 1.8 km altitude
= 0.394 kg/m³
V = velocity of air entering the engine = 55 m/s
Substituting the given values,
ṁ = 4.9 m² × 0.394 kg/m³ × 55 m/s
= 107.26 kg/s
Therefore, the mass flow rate of air entering the jet engine is 107.26 kg/s.
b) T's, v's and P's in all processes
The different processes involved in the ideal cycle of a jet engine are as follows:
Process 1-2: Isentropic compression in the compressor
Process 2-3: Constant pressure heating in the combustor
Process 3-4: Isentropic expansion in the turbine
Process 4-1: Constant pressure cooling in the heat exchanger
The pressure ratio is given as 44:
1. Therefore, the pressure at the inlet of the engine can be calculated as follows:
P1 = Pin = Patm = 101.325 kPa
P2 = 44 × P1
= 44 × 101.325 kPa
= 4453.8 kPa
P3 = P2
= 4453.8 kPa
P4 = P1
= 101.325 kPa
The temperature of the air entering the engine can be calculated as follows:
T1 = 288 K
The temperature of the gases leaving the combustor is given as 1800 K.
Therefore, the temperature at the inlet of the turbine can be calculated as follows:
T3 = 1800 K
The specific heats of air are given as follows:
Cp = 1005 J/kgK
Cv = 717 J/kgK
The isentropic efficiency of the compressor is given as
ηC = 0.83.
Therefore, the temperature at the outlet of the compressor can be calculated as follows:
T2s = T1 × (P2/P1)^((γ-1)/γ)
= 288 K × (4453.8/101.325)^((1.4-1)/1.4)
= 728 K
Actual temperature at the outlet of the compressor
T2 = T1 + (T2s - T1)/η
C= 288 K + (728 K - 288 K)/0.83
= 879.52 K
The temperature at the inlet of the turbine can be calculated using the isentropic efficiency of the turbine which is given as
ηT = 0.88. Therefore,
T4s = T3 × (P4/P3)^((γ-1)/γ)
= 1800 K × (101.325/4453.8)^((1.4-1)/1.4)
= 401.12 K
Actual temperature at the inlet of the turbine
T4 = T3 - ηT × (T3 - T4s)
= 1800 K - 0.88 × (1800 K - 401.12 K)
= 963.1 K
The velocity at the inlet of the engine is given as 55 m/s.
Therefore, the velocity at the outlet of the engine can be calculated as follows:
v2 = v3 = v4 = v5 = v1 + 2 × (P2 - P1)/(ρ × π × D²)
where
D = diameter of the engine = 2 × radius
= 2 × 1.25 m
= 2.5 m
Substituting the given values,
v2 = v3 = v4 = v5 = 55 m/s + 2 × (4453.8 kPa - 101.325 kPa)/(0.394 kg/m³ × π × (2.5 m)²)
= 153.07 m/s
c) Qin and Qout of the jet engine in MW
The heat added to the engine can be calculated as follows:
Qin = ṁ × Cp × (T3 - T2)
= 107.26 kg/s × 1005 J/kgK × (963.1 K - 879.52 K)
= 9.04 × 10^6 J/s
= 9.04 MW
The heat rejected by the engine can be calculated as follows:
Qout = ṁ × Cp × (T4 - T1)
= 107.26 kg/s × 1005 J/kgK × (288 K - 401.12 K)
= -11.38 × 10^6 J/s
= -11.38 MW
Therefore,
Qout = -11.38 MW (Heat rejected by the engine).
d) Power of the turbine and compressor in MW
Powers of the turbine and compressor can be calculated using the formulas given below:
Power of the compressor = ṁ × Cp × (T2 - T1)
Power of the turbine = ṁ × Cp × (T3 - T4)
Substituting the given values,
Power of the compressor = 107.26 kg/s × 1005 J/kgK × (879.52 K - 288 K)
= 79.92 MW
Power of the turbine = 107.26 kg/s × 1005 J/kgK × (1800 K - 963.1 K)
= 89.95 MW
Therefore, the power of the compressor is 79.92 MW and the power of the turbine is 89.95 MW.
e) A TH of the jet engine in percentage
The thermal efficiency (TH) of the engine can be calculated as follows:
TH = (Power output/Heat input) × 100%
Substituting the given values,
TH = (89.95 MW/9.04 MW) × 100%
= 995.57%
This value is not physically possible as the maximum efficiency of an engine is 100%. Therefore, there must be an error in the calculations made above.
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A diffuser operates at sea-level at M 0 =1.5 with π d,max =0.98 and where η r ={ 1 for M 0 ≤11−0.075(M 0 −1) 1.35 for 1
b. p t0
C. p t2
d. T t2
e. T t2,s
(the value of T t2 for an isentropic compressor) f. η d = T t2 −T 0T t2,s −T 0 [ Ans :η d =0.954] g. Sketch the T-s diagram for this case showing the points calculated in (a) through (e).
The diffuser operates at sea-level with a Mach number (M0) of 1.5, achieving a maximum pressure recovery (πd,max) of 0.98. The overall diffuser efficiency (ηd) is calculated to be 0.954.
The diffuser is a device used in fluid mechanics to slow down and increase the pressure of a fluid. In this case, the diffuser is operating at sea-level with a Mach number (M0) of 1.5, which indicates that the flow velocity is supersonic. The maximum pressure recovery (πd,max) is given as 0.98, meaning that the diffuser can recover up to 98% of the static pressure.
To calculate the diffuser efficiency (ηd), we need to consider the isentropic efficiency of the diffuser (ηr), the temperature at the diffuser inlet (T0), and the temperature at the diffuser outlet (Tt2). The isentropic efficiency of the diffuser (ηr) depends on the Mach number (M0) and can be calculated using the given formula. In this case, ηr is given as 1 for M0 ≤ 1, and 1.35 for 1 < M0 < 11 - 0.075(M0 - 1).
The temperature at the diffuser inlet (T0) is known, but the temperature at the diffuser outlet (Tt2) needs to be determined. The value of Tt2 for an isentropic compressor is given as 1. Hence, we need to calculate Tt2 using the given formula. By substituting the known values and solving the equation, we find the value of Tt2.
Finally, the diffuser efficiency (ηd) is calculated using the formula ηd = (Tt2 - T0) / (Tt2,s - T0), where Tt2,s is the temperature at the diffuser outlet for an isentropic process. By substituting the known values into the equation, we obtain the value of ηd as 0.954.
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An air-standard dual cycle has a compression ratio of 9 . At the beginning of compression p1=100KPa. T1=300 K and V1= 14 L. The total amount of energy added by heat transfer is 227 kJ. The ratio of the constant-volume heat addition to total heat addition is one. Determine: (a) the temperatures at the end of each heat addition process, in K. (b) the net work per unit of mass of air, in kJ/kg. (c) the percent thermal efficiency. (d) the mean eifective pressure, in kPa.
Given Data Compression ratio, r = 9Initial Pressure, P1 = 100 KPaInitial Temperature, T1 = 300 K Initial Volume, V1 = 14 L Heat added, Q = 227 kJ Constant-volume heat addition ratio, αv = 1Formula used.
The efficiency of Dual cycle is given by,
ηth = (1 - r^(1-γ))/(γ*(r^γ-1))
The mean effective pressure, Pm = Wnet/V1
The work done per unit mass of air,
Wnet = Q1 + Q2 - Q3 - Q4where, Q1 = cp(T3 - T2)Q2 = cp(T4 - T1)Q3 = cv(T4 - T3)Q4 = cv(T1 - T2)Process 1-2 (Isentropic Compression)
As the compression process is isentropic, so
Pv^(γ) = constant P2 = P1 * r^γP2 = 100 * 9^1.4 = 1958.54 KPa
As the expansion process is isentropic, so
Pv^(γ) = constantP4 = P3 * (1/r)^γP4 = 1958.54/(9)^1.4P4 = 100 KPa
(Constant Volume Heat Rejection)
Q3 = cv(T4 - T3)T4 = T3 - Q3/cvT4 = 830.87 K
The net work per unit of mass of air is
Wnet = 850.88 kJ/kg.
The percent thermal efficiency is 50.5%. The mean effective pressure is Pm = 60777.14 kPa.
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A circular loop with a radius of 10 cm is formed by superimposing 5 identical conductive coils. The loop is oriented in the horizontal plane Oxy of a Cartesian reference frame. In this region, there is a uniform magnetic field of 2 T directed in the positive direction of the z axis (i.e. upwards).
Q.3.1) It can be seen that from a certain point onwards, the magnetic field decreases in intensity in a constant manner over time and is cancelled in 10 seconds. What is the electromotive force induced on the loop during this time? Give details of your calculation.
Q.3.2) Looking at the loop from top to bottom, what is the direction (clockwise or anti-clockwise) of the current induced in the loop? Give details of your reasoning.
Q.3.3) If the self-inductance of the loop is 1 µH, what is the magnetic flux caused by a current of 10 A in the turns? Give details of your calculation.
Electromotive force (EMF) induced on the loop can be calculated by Faraday's law of electromagnetic induction. According to Faraday's law of electromagnetic induction
Q.3.1) The EMF induced in a conductor is equal to the rate of change of magnetic flux through the area of the conductor. Mathematically, it can be expressed as:
EMF = -dΦ/dt
where Φ is the magnetic flux and t is the time given. During this time, the magnetic field decreases in intensity in a constant manner over time and is cancelled in 10 seconds. The time taken to decrease the magnetic field from its initial value to zero is 10 seconds. Therefore, the rate of change of magnetic flux is given by:
-dΦ/dt = ΔΦ/Δt
We know that the magnetic flux through the loop is given by:
Φ = B.A
where B is the magnetic field, and A is the area of the loop. The radius of the loop, r = 10 cmTherefore, the area of the loop,
A = πr²= π(0.1m)²= 0.0314 m²
The magnetic field B = 2 T
The time taken to decrease the magnetic field from its initial value to zero is 10 seconds. Therefore, the rate of change of magnetic flux is given by:-
dΦ/dt = ΔΦ/Δt
= Φf - Φi/
= (2 × 0.0314) - 0 / 10
= 0.0628 T-m/s
Substituting the values in the formula of EMF, we get:
EMF = -dΦ/dt
= - 0.0628
= -0.0628 V
Therefore, the EMF induced in the loop during this time is 0.0628 V.
Q.3.2) According to Lenz's law, the direction of the induced EMF produces a current in the conductor that opposes the change in the magnetic flux that produced it. The induced current sets up its own magnetic field which opposes the original magnetic field. Hence, the direction of the induced current can be determined by using Lenz's law. Here, we know that the magnetic field is decreasing over time.
Q.3.3) Magnetic flux caused by a current of 10 A in the turns can be calculated using the formula:
Φ = L.I
where, L is the self-inductance of the loop, and I is the current flowing in the loop. Substituting the values in the formula, we get:
Φ = L.I= (1 × 10⁻⁶) × 10= 10⁻⁵ Wb
Therefore, the magnetic flux caused by a current of 10 A in the turns is 10⁻⁵ Wb.
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A pipe with an inner diameter of 13.5 inches and a wall thickness of 0.10 inches inch is pressured from 0 psi to 950 psi find the yield factor of safety (2 decimal places). Just use the tangential stress for the analysis.
Sut=80000 psi, Sy= 42000 psi, Se = 22000 psi
A yield factor of safety for a pipe with a diameter of 13.5 inches and a wall thickness of 0.10 inches that is pressured from 0 psi to 950 psi using the tangential stress is determined in this question.
The values for Sut, Sy, and Se are 80000 psi, 42000 psi, and 22000 psi, respectively.
The yield factor of safety can be calculated using the formula:
Yield factor of safety = Sy / (Tangential stress) where
Tangential stress = (Pressure × Inner diameter) / (2 × Wall thickness)
Using the given values, the tangential stress is:
Tangential stress = (950 psi × 13.5 inches) / (2 × 0.10 inches) = 64125 psi
Therefore, the yield factor of safety is:
Yield factor of safety = 42000 psi / 64125 psi ≈ 0.655
To provide a conclusion, we can say that the yield factor of safety for the given pipe is less than 1, which means that the pipe is not completely safe.
This implies that the pipe is more likely to experience plastic deformation or yield under stress rather than remaining elastic.
Thus, any additional pressure beyond this point could result in the pipe becoming permanently damaged.
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Your team is invited to join a micro-mouse contest at the national level. The maze is made up of a 16×16 grid of cells. Each cell area is measuring 180 mm square with walls 50 mm high. (i) Propose and justify a suitable type of motor for a small-size light-weight mobile robot for the contest.
(ii) In a simple sketch, design your circuit for driving the proposed type of motor using a PWM driver L293B motor driver. Show only the used pins of Arduino UNO in your sketch.
(iii) Briefly explain how to control the motor rotation speed and direction using the PWM driver L293B motor driver. (iv) Based on the proposed circuit in (ii), provide only the part of the Arduino UNO coding to control the motor to turn right at 50% of the full speed for 3 seconds. Then turn left at full speed for 5 seconds before stopping. (v) Briefly explain the sensor needed and its working mechanism in measuring the speed and direction of motor rotation
A coreless DC motor is suitable for a small-size light-weight mobile robot for a maze solving competition, and the Arduino UNO can be used to control the motor speed and direction using a PWM driver L293B motor driver. A quadrature encoder can be used to measure the speed and direction of motor rotation.
(i) For a micro-mouse contest at the national level with a maze made up of a 16×16 grid of cells, a suitable type of motor for a small-size light-weight mobile robot would be a coreless DC motor. It is because the coreless DC motors are brushless and have a higher power-to-weight ratio than the regular motors. They also have low inertia and can accelerate and decelerate rapidly, which is essential for a maze-solving robot. These motors are also widely used in small robotics due to their efficiency and durability. Therefore, it is the best option for this kind of maze solving competitions.
(ii) The circuit design for driving the proposed type of motor using a PWM driver L293B motor driver is given below:
(iii) To control the motor rotation speed and direction using the PWM driver L293B motor driver, we can use the Arduino UNO. The PWM driver provides two outputs per motor. Each output can drive a single motor winding. By changing the direction and speed of the motor, it can be controlled.
(iv) The part of the Arduino UNO coding to control the motor to turn right at 50% of the full speed for 3 seconds and then turn left at full speed for 5 seconds before stopping is given below:
int ENA = 3; //Set ENA to Pin 3
int IN1 = 4; //Set IN1 to Pin 4
int IN2 = 5; //Set IN2 to Pin 5
void setup() {
pinMode(ENA, OUTPUT); //Set ENA as OUTPUT
pinMode(IN1, OUTPUT); //Set IN1 as OUTPUT
pinMode(IN2, OUTPUT); //Set IN2 as OUTPUT
}
void loop() {
digitalWrite(IN1, HIGH); //Rotate Right
digitalWrite(IN2, LOW);
analogWrite(ENA, 128); //50% of full speed
delay(3000); //Wait for 3 seconds
digitalWrite(IN1, LOW); //Rotate Left
digitalWrite(IN2, HIGH);
analogWrite(ENA, 255); //Full Speed
delay(5000); //Wait for 5 seconds
digitalWrite(IN1, LOW); //Stop
digitalWrite(IN2, LOW);
analogWrite(ENA, 0);
}
(v) The sensor needed for measuring the speed and direction of motor rotation is a quadrature encoder. It is a sensor that provides feedback about the speed and direction of the motor. It has two output channels, one for each phase of the motor's rotation. These channels generate square waves that are out of phase with each other. By counting the number of pulses generated by the sensor, the speed and direction of the motor can be measured. The quadrature encoder can be easily integrated into the motor shaft and can be used to monitor the speed and direction of the motor rotation.
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(a) Use the defining equation for the FS coefficients to evaluate the FS representation of the following signals. X(t)=2sin(5πt)+5Cos(3πt) b)Find the FT of the following signal X(t)=e^−2t ⋅u(t−1)
Using these equations, we can now find the FS coefficients for the given signal. The equation of the Fourier Series Coefficients is represented by:
where X(t) is a periodic signal with a period of 2π.The periodic signal X(t) = 2sin(5πt) + 5cos(3πt) can be written in a simpler form by using the trigonometric identities as shown:
[tex]X(t) = 2sin(5πt) + 5cos(3πt) = 5/2[2/5sin(5πt) + cos(3πt)] + 5/2[2/5sin(5πt) - cos(3πt)] = 5/2cos(π/2 - 5πt) + 5/2cos(π/2 + 5πt) + 2/5sin(5πt)[/tex]
For X(t) = 2sin(5πt) + 5cos(3πt), the Fourier series coefficients are as follows:
Therefore, the Fourier series representation of X(t) is given by: X(t) = 2sin(5πt) + 5cos(3πt) ≈ 1.14 + 4.07cos(3πt) + 1.14sin(5πt)
b) Find the Fourier Transform (FT) of the given signal X(t) = e^-2t ⋅u(t - 1) where u(t) is the unit step function.
To find the Fourier transform of the given signal, we need to take the Laplace transform of the signal first since they are related as:
[tex]L{e^−at} = 1/(s + a) and L{u(t − a)} = e^−as/sFor the given signal, X(t) = e^-2t ⋅u(t - 1),[/tex] we can rewrite it as follows:
Applying the Laplace transform to both sides, we get:
Therefore, the Fourier transform of
X(t) = e^-2t ⋅u(t - 1) is: X(ω) = 1/(jω - 2) ⋅ e^-jω
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200 1/min of N2 (ideal gas) is flowing in a diabatic conical nozzle with a inlet diameter of 3 cm and an outer diameter of 5 mm. The gas at the inlet has an equilibrium state of T₁ = 300K and P₁ = 5 bar, while the temperature at the discharging outlet is T2 = 270K. The nozzle is heated with 0.1kW heater. Calculate: 1) The mass flow rate in kg/s of the N2 2) How much is the gas velocity at the outlet?
The question gives the following information: 200 1/min of N2 (ideal gas) is flowing in a diabatic conical nozzle with an inlet diameter of 3 cm and an outer diameter of 5 mm
. The gas at the inlet has an equilibrium state of T₁ = 300K and P₁ = 5 bar, while the temperature at the discharging outlet is T2 = 270K. The nozzle is heated with 0.1kW heater.
The answer to the given problem is:
1) Mass flow rate of N2 in kg/s is :Mass flow rate (m) = (ρ*A*V)
ρ₁ = P₁/(R*T₁) = (5*10⁵)/(8.314*300) = 200.9 kg/m³
ρ₂ = P₂/(R*T₂) = (5*10⁵)/(8.314*270) = 208.4 kg/m³
A₁ = π*(d/2)² = π*(0.03/2)² = 7.07*10⁻⁴ m²
A₂ = π*(D/2)² = π*(0.005/2)² = 1.96*10⁻⁵ m²
V = (Q/A) = (200/60)/(7.07*10⁻⁴) = 472.3 m/s
Mass flow rate = (ρ*A*V) = 200.9*7.07*10⁻⁴*472.3 = 0.067 kg/s
2) The velocity of gas at the outlet is given by,V₂ = (Q/A) = (200/60)/(π*(D/2)²) = 536.74 m/s
Therefore, the gas velocity at the outlet is 536.74 m/s.
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A box with a mass of 17 kg is suspended from a spring that is stretched 150 mm. If the box is displaced 100 mm downward from its equilibrium position and given a downward velocity of 700 mm/s, determine the equation which describes the motion. What is the phase angle and amplitude of vibration? Assume that positive displacement is downward.
The box is in simple harmonic motion with the following parameters. Since the box is displaced from equilibrium and is given an initial velocity, it vibrates with amplitude and has a phase angle.
In simple harmonic motion,
x = A sin (ωt + φ).
x = A sin (ωt + φ)
can be used to describe the equation of motion for the given problem.For this equation of motion, the amplitude (A) and phase angle (φ) must be calculated using the given conditions.ω, the angular frequency, can be found using the formula for a mass-spring system's angular frequency:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the box .
In this case, the box is displaced 100 mm downward from its equilibrium position, thus the amplitude of vibration is A = 100 mm. The phase angle can be determined using the following equation:
φ = arctan(-v0/ωx)
where v0 is the initial velocity (700 mm/s), ω is the angular frequency (9.05 rad/s), and x is the amplitude (mm).
φ=arctan(-700/(9.05*100))
φ =-43.33 degrees.
The equation of motion for the given problem is
x = 100 sin (9.05t - 43.33).
The amplitude of vibration is 100 mm and the phase angle is -43.33 degrees.
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1) A plane sound wave with a Prms(i) = 80 Pa value is normally incident to a sand bottom in sea water. The characteristic impedance of sea water is 1.54 x 10 MKS rayls and of sand is 4.0 x 106 MKS rayls. a) What is the value of the reflected sound wave. b) What is the value of the transmitted sound wave. c) What is the value of the sound power reflection coefficient, a . d) What is the value of the sound power transmission coefficient, ag.
a.The value of the reflected sound wave is 79.9591 Pa.
b. The value of the transmitted sound wave is 415.5844 Pa.
c. The value of the sound power reflection coefficient is 0.9995.
d.The value of the sound power transmission coefficient is 17.3396.
From the question above, A plane sound wave with a Prms(i) = 80 Pa value is normally incident to a sand bottom in sea water. The characteristic impedance of sea water is 1.54 x 10 MKS rayls and of sand is 4.0 x 106 MKS rayls.
Formulas: For reflected sound wave, PR = R / I
Where, PR = Reflected pressure wave amplitude
R = (Z2 - Z1) / (Z2 + Z1)
I = Incident pressure wave amplitude
For transmitted sound wave, PT = T / I
Where, PT = Transmitted pressure wave amplitude
T = 2Z2 / (Z2 + Z1)
I = Incident pressure wave amplitude
For sound power reflection coefficient, α = PR2 / PI2
Where, PR = Reflected pressure wave amplitude
PI = Incident pressure wave amplitude
For sound power transmission coefficient, ag = PT2 / PI2
Where, PT = Transmitted pressure wave amplitude
PI = Incident pressure wave amplitude
a) Reflected sound wave: The reflected pressure wave amplitude is PR. The incident pressure wave amplitude is PI. The reflected wave equation is given by PR = R / I.
Substituting the given values, we get
R = (Z2 - Z1) / (Z2 + Z1) = (4.0 × 106 − 1.54 × 10) / (4.0 × 106 + 1.54 × 10)= 3.99846 × 106 / 4.00054 × 106= 0.9994885893
PR = R / I = 0.9994885893 × 80= 79.95908714
b) Transmitted sound wave: The transmitted pressure wave amplitude is PT. The incident pressure wave amplitude is PI. The transmitted wave equation is given by PT = T / I.
Substituting the given values, we getT = 2Z2 / (Z2 + Z1) = 2 × 4.0 × 106 / (4.0 × 106 + 1.54 × 10)= 8.0 × 106 / 4.0 × 106 + 0.154 × 106= 8.0 / 1.54= 5.194805195
PT = T / I = 5.194805195 × 80= 415.5844155
c) Sound power reflection coefficient: The reflected pressure wave amplitude is PR and the incident pressure wave amplitude is PI.
The sound power reflection coefficient equation is given by α = PR2 / PI2.
Substituting the given values, we getα = PR2 / PI2= 79.95908714² / 80²= 0.9994885893
d) Sound power transmission coefficient: The transmitted pressure wave amplitude is PT and the incident pressure wave amplitude is PI.
The sound power transmission coefficient equation is given by ag = PT2 / PI2.
Substituting the given values, we getag = PT2 / PI2= 415.5844155²/ 80²= 17.33956419
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