The complaint is that the temperature is too high in the fresh food compartment of a frost-free refrigerator equipped with a defrost timer to initiate the defrost cycle. Excess frost is present in the frozen food compartment, but the thermostat, compressor, and condenser fan motor are all functioning correctly.
The refrigerator runs continually, and the defrost heaters are in good condition. Firstly, to rectify the complaint, you should disconnect power to the refrigerator. Then, the technician should open the fresh food compartment to expose the thermostat and the damper assembly. The technician should check to see whether the damper is closed entirely, blocking any cold air from entering the compartment. If it is, the technician should attempt to adjust the damper to allow cold air to enter.
If the damper is functioning correctly, the technician should inspect the evaporator coils for excessive frost buildup. The buildup of frost indicates a possible failure of the defrost system or a bad defrost timer. If the defrost timer is faulty, it will prevent the defrost heater from turning on, resulting in excess frost.
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-j40I2 +j120+5I2-15I1+15I2+10I2=0
Solve for I2 given that I1 = 6amps. I2 should be in rectangular
form
I2 in rectangular form is equal to 0 + j (3/2).
Given expression: -j40I2 +j120+5I2-15I1+15I2+10I2 = 0
The value of I1 = 6 A
To solve for I2, substitute the value of I1 in the given expression
I2 (-j40 + 5 + 15 + 10) + j120 - 15 (6)
= 0I2 (-20) + j120 - 90
= 0I2 (-20)
= -j30I2
= j30/20I2
= 3/2 j
Now, we can represent the value of I2 in rectangular form as follows:
I2 = 0 + j (3/2)
I2 = 1.5 j
Therefore, I2 in rectangular form is equal to 0 + j (3/2).
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Overloading a single-phase motor will result in:
Select one:
a.no effects, the motor runs normally
b.overheating the motor
c.damaging the motor permanently
d.None
e.using a fan for cooling
Overloading a single-phase motor will result in overheating the motor.A single-phase motor is an electric motor that is powered by a single phase of electrical power.
Single-phase power is most commonly used in household and small commercial settings, such as for powering small appliances and lighting systems. Single-phase motors are used in a variety of applications, including fans, pumps, and compressors. They are also used in machinery and tools. it is being forced to work harder than it is designed to.
This can result in damage to the motor, as well as to any other equipment that is connected to it. Overloading a motor can cause it to overheat, which can lead to a variety of problems. In some cases, the motor may simply stop working. In other cases, it may begin to emit smoke or make unusual noises.When a single-phase motor is overloaded, it will begin to overheat.
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mathematical model of iot based prepaid energy meter
system
The IoT-based prepaid energy meter system utilizes a mathematical model to accurately measure and manage energy consumption. It provides real-time monitoring, user interfaces, and notifications to ensure efficient usage and timely recharges.
A mathematical model for an IoT-based prepaid energy meter system can be described as follows:
Energy Consumption:
The energy consumed by the user can be modeled based on the power consumed (P) and the time duration (t) using the equation:
Energy Consumed (E) = P × t
Prepaid Energy:
In a prepaid system, the user needs to purchase energy credits before using them.
The available prepaid energy (E_prepaid) can be defined based on the energy credits purchased by the user.
Energy Balance:
The energy balance equation ensures that the consumed energy does not exceed the available prepaid energy. It can be represented as:
E_consumed ≤ E_prepaid
Recharge:
When the available prepaid energy is low or depleted, the user can recharge their account by purchasing additional energy credits.
The recharge process updates the available prepaid energy.
Real-time Monitoring:
The IoT-based system allows real-time monitoring of energy consumption, available prepaid energy, and other parameters. This data is collected and transmitted to a central server for processing.
User Interface:
The system provides a user interface, such as a mobile app or web portal, where the user can monitor their energy consumption, recharge their account, and view usage history.
Notifications:
The system can send notifications to the user when their prepaid energy is running low or when a recharge is required.
Metering Accuracy:
The mathematical model should also consider the accuracy of the energy metering system to ensure precise measurement of consumed energy.
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Locations of the points
O = {0, 0, 0}, A = {−3, −3, 0}, B = {-3.3, 10.1, 0.}, G = {-₁, -2, 0), H = {-3.15, 3.55, 0.}
Angular velocity of first link
ಪ = {0, 0, -2.1}
Masses of the links
m₁ = 1.4, m₂ = 1.6
(a) Calculate the torque that needs to applied to point B on the second link to generate the given acceleration.
(b) if the force was not applied, calculate the torque needed to be applied to point o to generate this given acceleration.
To calculate the torque required at point B on the second link to generate the given acceleration, we need to consider the masses of the links, their locations, and the angular velocity of the first link.
We can use the torque formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Similarly, to calculate the torque required at point O without applying a force, we can use the same formula but consider the moment of inertia and angular acceleration about point O.
a) To calculate the torque required at point B, we need to find the moment of inertia (I₂) of the second link about point B. The moment of inertia can be calculated using the formula I = m * r², where m is the mass of the link and r is the distance from the point of rotation to the mass. In this case, the distance is the perpendicular distance from point B to the line of action of the force. Once we have the moment of inertia, we can calculate the torque by multiplying it with the angular acceleration α, which is given as the z-component of the angular velocity vector.
b) To calculate the torque required at point O, we need to find the moment of inertia (I₁) of the first link about point O. The moment of inertia can be calculated using the same formula as mentioned above, but this time we consider the distance from point O to the mass of the first link.Using the calculated moment of inertia and the given angular acceleration, we can determine the torque required at point O. By applying these calculations using the provided data, we can find the torques needed at point B and point O to generate the given acceleration for the system.
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The minimum pressure on an object moving horizontally in water (Ttemperatu at10 degree centrigrade) at (x+5) mm/s (where x is the last two digits of your student 10) at a depth of 1 m is 80 kPa (absolute). Calculate the velocity that will initiate cavitation. Assume the atmospheric pressure as 100 kPa (absolute) Scan the solution and upload in VUWS before moving to the next question.
Given data: Minimum pressure on an object = 80 kPa (absolute)Velocity of an object = (x+5) mm/sDepth of an object = 1mTemperature = 10°CAtmospheric pressure = 100 kPa (absolute)
We know that the minimum pressure to initiate cavitation is given as:pc = pa - (pv)²/(2ρ)Where, pa = Atmospheric pressurepv = Vapour pressure of liquidρ = Density of liquidNow, the vapour pressure of water at 10°C is 1.223 kPa (absolute) and density of water at this temperature is 999.7 kg/m³.Substituting the values in the above equation, we get:80 = 100 - (pv)²/(2×999.7) => (pv)² = 39.706
Now, the velocity that will initiate cavitation is given as:pv = 0.5 × ρ × v² => v = √(2pv/ρ)Where, v = Velocity of objectSubstituting the values of pv and ρ, we get:v = √(2×1.223/999.7) => v = 1.110 m/sTherefore, the velocity that will initiate cavitation is 1.110 m/s.
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A rectangular channel discharges water at the rate of 4.5 cu.m./s at a depth of 32 cm. The flume is 3.5 m wide. What is the depth of the jump? Select one: O a. 83.9 cm O b. 87.9 cm O c. 85.9 cm O d. 81.9 cm
The rectangular channel discharges water at the rate of 4.5 cu.m./s at a depth of 32 cm. The flume is 3.5 m wide. We have to find out the depth of the jump. The correct option among the given options is (b) 87.9 cm.
The critical depth in a rectangular channel is given by;
[tex]$$y_c=\frac{Q^2}{gBW^2}$$[/tex]
Where,Q = Discharge, B = Width of the channel, W = Hydraulic depth, y = depth of flow of water.Let us calculate all the given parameters and then find the depth of the jump.Q = 4.5 cu.m./sWidth of the channel, B = 3.5 mDepth of flow of water, y = 32 cm = 0.32 mHydraulic Depth, [tex]W = (3.5 x 0.32) / (3.5 + 2 x 0.32) = 0.224[/tex]
Now, putting all the values in the critical depth formula;
[tex]$$y_c=\frac{Q^2}{gBW^2}$$$$y_c=\frac{(4.5)^2}{9.81 \times 3.5 \times 0.224^2}$$$$y_c = 0.879 m$$$$y_c= 87.9 cm$$[/tex]
Therefore, the correct option among the given options is (b) 87.9 cm.
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Consider a flat rectangular plate of known mass, width and breadth with a negligible thickness that lies in the horizontal xy-plane. The plate is suspended from a thin piece of piano wire that is in the vertical orientation coincident to the z-axis and where the piano wire is attached to the center of the plate. When the plate is subjected to a torque whose vector is coincident to the z-axis, the plate rotates in the horizontal plane such that the rotation of the plate is modelled as θ = Csin(wt + Ø). The parameter information is: mass of plate M = 1.2 kilogram width of plate W = 0.040 meter breadth of plate B = 0.075 meter shear modulus of piano wire G = 79.3 gigaPascals diameter of piano wire D = 0.003 meter length of piano wire L = 0.120 meter amplitude of rotation C = 0.087267520415 radian phase lag of rotation = 1.565872597159 radian Using the supplied information and any appropriate assumptions and / or approximations, write a GNU Octave computer program to determine the following; 1) the mass moment of inertia I 2) the natural angular frequency wn 3) the initial angular displacement θ₀ 4) the initial angular velocity θ₀
The mass moment of inertia (I) for the rectangular plate is (1/12) * M * (W^2 + B^2), the natural angular frequency (wn) is sqrt(G / (I / L)), the initial angular displacement (θ₀) is the given amplitude of rotation (C), and the initial angular velocity (θ'₀) is C * w * cos(Ø) where w represents the angular frequency.
What are the formulas to calculate the mass moment of inertia (I), natural angular frequency (wn), initial angular displacement (θ₀), and initial angular velocity (θ'₀) for a rectangular plate suspended by a piano wire, given the relevant parameters?The mathematical equations and steps to determine the quantities you mentioned using the supplied information.
1) The mass moment of inertia (I) of the rectangular plate can be calculated using the formula: I = (1/12) * M * (W^2 + B^2).
2) The natural angular frequency (wn) can be calculated using the equation: wn = sqrt(G / (I / L)).
3) The initial angular displacement (θ₀) is given as the amplitude of rotation (C) in this case.
4) The initial angular velocity (θ'₀) can be calculated by taking the derivative of the rotation equation with respect to time (t) and evaluating it at t = 0. Differentiating θ = C * sin(wt + Ø) with respect to t gives θ' = C * w * cos(wt + Ø), and θ'₀ = C * w * cos(Ø).
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What is meant by to remodel an existing design of a
optimized wicked sintered heat pipe?
Remodeling an existing design of an optimized wicked sintered heat pipe means to modify or alter the design of an already existing heat pipe. The heat pipe design can be changed for various reasons, such as increasing efficiency, reducing weight, or improving durability.
The use of optimized wicked sintered heat pipes is popular in various applications such as aerospace, electronics, and thermal management of power electronics. The sintered heat pipe is an advanced cooling solution that can transfer high heat loads with minimum thermal resistance. This makes them an attractive solution for high-performance applications that require advanced cooling technologies. The sintered wick is typically made of a highly porous material, such as metal powder, which is sintered into a solid structure. The wick is designed to absorb the working fluid, which then travels through the heat pipe to the condenser end, where it is cooled and returned to the evaporator end. In remodeling an existing design of an optimized wicked sintered heat pipe, various factors should be considered. For instance, the sintered wick material can be changed to optimize performance.
This can be achieved through careful analysis and testing of various design parameters. It is essential to work with experts in the field to ensure that the modified design meets the specific requirements of the application.
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Describe the time – temperature paths to produce the following microstructures in 0.77 wt% C: (a) 100% fine pearlite (b) 100% tempered martensite (c) 25% coarse pearlite, 50% bainite, and 25% martensite
Factors such as cooling rate and holding time at specific temperatures play crucial roles in achieving the desired microstructures.
To produce specific microstructures in 0.77 wt% C steel, the time-temperature paths are as follows:
(a) 100% Fine Pearlite:
The steel is heated to a temperature above the eutectoid temperature (around 727°C) and held at that temperature for sufficient time to allow the formation of fine pearlite. It is then slowly cooled in a furnace to room temperature, maintaining the pearlite microstructure.
(b) 100% Tempered Martensite:
The steel is first heated to a temperature above the austenitizing temperature and then rapidly quenched to transform the austenite into martensite. To obtain tempered martensite, the quenched steel is then reheated to a temperature below the lower critical temperature and held for a specific time, allowing the martensite to transform and temper.
(c) 25% Coarse Pearlite, 50% Bainite, and 25% Martensite:
The steel is heated to a temperature above the eutectoid temperature and held for a shorter time to fully austenitize it. It is then rapidly cooled to a temperature within the bainite formation range and held for a specific time to allow the formation of bainite. Further rapid cooling leads to the transformation of the remaining austenite into martensite.
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Select all items below which are crucial in lost-foam casting.
(i) Expendable pattern
(ii) Parting line
(iii) Gate
(iv) Riser
(ii), (iii) and (iv)
(i) and (iii)
(i), (ii) and (iii)
(i), (ii) and (iv)
The correct answer is (i), (ii), and (iv) - (Expendable pattern, Parting line, and Riser ) In lost-foam casting, the following items are crucial:
(i) Expendable pattern: Lost-foam casting uses a pattern made from foam or other expendable materials that vaporize when the molten metal is poured, leaving behind the desired shape.
(ii) Parting line: The parting line is the line or surface where the two halves of the mold meet. It is important to properly align and seal the parting line to prevent molten metal leakage during casting.
(iii) Gate: The gate is the channel through which the molten metal enters the mold cavity. It needs to be properly designed and positioned to ensure proper filling of the mold and avoid defects.
(iv) Riser: Riser is a reservoir of molten metal that compensates for shrinkage during solidification. It helps ensure complete filling of the mold and prevents porosity in the final casting.
Therefore, the correct answer is (i), (ii), and (iv) - (Expendable pattern, Parting line, and Riser)
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Question 3: What is a herringbone gear? Where they are used? [1 mark] Question 4: Write the expressions for static strength, limiting wear load and dynamic load for helical gears and explain the various terms used. [1.5 marks]
3. A herringbone gear is a type of gear that consists of two helical gears ith opposite helix angles. They are used in heavy-duty applications to transmit high torque and eliminate axial thrust forces. 4.The expressions for static strength, limiting wear load, and dynamic load for helical gears involve parameters such as Lewis form factor, cross-sectional area, safety factor, number of teeth, permissible wear load, face width, and pitch diameter.
A herringbone gear, also known as a double-helical gear, is a type of gear that consists of two helical gears with opposite helix angles, placed side by side and meshing with each other. This design eliminates axial thrust forces and improves the smoothness and load-carrying capacity of the gear system.
Herringbone gears are commonly used in heavy-duty applications where high torque transmission is required, such as in industrial machinery, marine propulsion systems, and heavy vehicles. Their symmetrical structure and improved load distribution make them suitable for handling large loads and reducing gear noise and vibration.
For helical gears, the expressions for static strength, limiting wear load, and dynamic load are as follows:
Static strength: The static strength of a helical gear is determined by the bending strength of the gear teeth. The expression for static strength is given by:
Static strength = (Y*S)/F
where Y is the Lewis form factor, S is the cross-sectional area of the gear tooth, and F is the safety factor.
Limiting wear load: The limiting wear load represents the maximum load that a helical gear can withstand without excessive wear. The expression for limiting wear load is given by:
Limiting wear load = (ZWL)/D
where Z is the number of teeth on the gear, W is the permissible wear load per unit area, L is the face width of the gear, and D is the gear pitch diameter.
Dynamic load: The dynamic load considers the effect of both bending and surface contact fatigue on the gear. The expression for dynamic load is given by:
Dynamic load = (ZWL)/d
where d is the gear pitch circle diameter.
In these expressions, the terms Y, S, F, Z, W, L, and D represent specific parameters related to the gear design and material properties. The values of these parameters are determined based on the specific application requirements and gear standards.
Therefore, the required answers are:
3. A herringbone gear is a type of gear that consists of two helical gears ith opposite helix angles. They are used in heavy-duty applications to transmit high torque and eliminate axial thrust forces.
4.The expressions for static strength, limiting wear load, and dynamic load for helical gears involve parameters such as Lewis form factor, cross-sectional area, safety factor, number of teeth, permissible wear load, face width, and pitch diameter.
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General Directions: This test is comprised of several different types of questions. Read the specific directions for each section before attempting to answer the questions within that section. Also, be sure to read each question carefully before marking your answer. True False; Some of the statements listed below are True and some are False. If the statement is True, darken circle "A" in the appropriate space on your answer sheet. If the statement is False, darken circle "B" in the appropriate space on your answer sheet. 2. 3. 1. Changes in technology cause only small differences in manufacturing. Setup time is the time required to get a machine ready for manufacturing. The manufacturing process dictates the product to be manufactured. The first process segment is production. 4. 5. Aircraft production is an example of Engineer to Order production strategy. The line manufacturing system has three distinguishing characteristics. 6. 7. In the input-output model tooling is considered to be an input. 8. Job shops are distinguished by large production numbers. 9. Market research is the last step of the product development cycle. 10. A real-time controller is a controller that is able to respond to the process within a short enough time. That process performance is not degraded. 11. An interlock is a safeguard mechanism for coordinating the activities of two or more devices preventing one device from interfering with the others. 12. In computer process control, pulling refers to the real-time sampling of data which continuously monitors the process. 13. In 1962 the first industrial robot was installed on a production line by General Motors. 14. A book titled "Cybernetics" describes the concept of communication and control 15. In 1959 Planet Corporation marketed the first commercially available robot. 16. A robot program can be defined as a path in space to be followed by the manipulator, combined with the peripheral action to support the work cycle. 17. A logic control system is a switching system whose output at any moment is determined exclusively by the values of the current inputs. 18. A flexible manufacturing system (FMS) does not rely on the principles of group technology. 19. An intelligent robot is one that exhibits behavior that makes it seem intelligent. 20. In regulatory control the objective is to maintain process performance at a certain level or within a given tolerance band of that level. 21. A discrete variable is one that can take on only certain values within a given range. 22. A fundamental objective of CAD/CAM is to integrate the design engineering and manufacturing engineering functions. 23. Manual and computer-assisted port programming does not require a high degree of formal documentation.
The test consists of 23 statements where the test-taker needs to determine whether each statement is true or false. They are instructed to darken circle "A" on the answer sheet if the statement is true and circle "B" if it is false.
The given test contains a series of statements related to manufacturing, production strategies, control systems, robotics, and engineering concepts. The test-taker is required to carefully read each statement and mark the corresponding circle on the answer sheet as instructed.
In order to successfully answer the questions, the test-taker should possess knowledge and understanding of the manufacturing industry, production processes, control systems, robotics, and engineering principles. It is crucial to pay attention to the details of each statement and accurately determine whether it is true or false.
It is recommended that the test-taker carefully read each statement, evaluate its accuracy based on their knowledge and understanding of the subject matter, and mark the appropriate circle on the answer sheet. Accuracy and attention to detail are key in providing correct responses to each statement.
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Given the signals x₁ [n] = [1 2 -1 2 3] and x₂ [n] = [2 - 2 3 -1 1]. Evaluate the output for: a. x₂[n] + x₁[-n]. b. x₁[1-n] x₂ [n+3] .
a. The output for x₂[n] + x₁[-n] is [2, -4, 2, 1, 2].
b. The output for x₁[1-n] x₂[n+3] is [-2, -1, 4, -2, 0].
Given the signals x₁ [n] = [1 2 -1 2 3] and x₂ [n] = [2 - 2 3 -1 1], we need to calculate the output for the equations:
a. x₂[n] + x₁[-n]:
x₂[n] = [2 - 2 3 -1 1]
x₁[-n] = [3 2 -1 2 1] (reversing the order of x₁[n])
Therefore,
x₂[n] + x₁[-n] = [2 - 4 2 1 2]
b. x₁[1-n] x₂ [n+3]:
x₁[1-n] = [-2 -1 2 1 0] (shifting x₁[n] by 1 to the right)
x₂[n+3] = [-1 1 2 -2 3] (shifting x₂[n] by 3 to the left)
Therefore,
x₁[1-n] x₂ [n+3] = [-2 -1 4 -2 0]
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The radial position of a particle's path is defined by an equation, r = 5-cos(2theta) m. At the initial time, the angular position is theta = 0°. If the angular velocity of the particle is = 31² rad/sec, where t is in seconds, calculate the value of the O-component of acceleration at the instant = 30°. Present your answer in m/sec² using 3 significant figures.
The O-component of acceleration at the instant θ = 30° is approximately -145.7 m/sec². This value represents the acceleration in the radial direction perpendicular to the path.
To calculate the O-component of acceleration, we need to differentiate the radial position equation twice with respect to time (t) to obtain the acceleration equation. Then we can substitute the given angular velocity and the angle θ = 30° into the acceleration equation to find the O-component of acceleration.
The radial position equation:
r = 5 - cos(2θ) m
First, we need to find the angular acceleration (α) using the given angular velocity (ω) by differentiating once:
α = dω/dt = 0 rad/sec² (since ω is constant)
Next, we differentiate the radial position equation with respect to time twice to find the acceleration equation:
r = 5 - cos(2θ)
v = dr/dt = d(5 - cos(2θ))/dt
a = dv/dt = d²(5 - cos(2θ))/dt²
Differentiating with respect to θ:
a = -2d(5 - cos(2θ))/dθ²
a = 4sin(2θ)
Substituting the angle θ = 30° into the acceleration equation:
θ = 30° = π/6 radians
a = 4sin(2(π/6))
a ≈ -145.7 m/sec²
Therefore, the O-component of acceleration at θ = 30° is approximately -145.7 m/sec².
At the instant θ = 30°, the O-component of acceleration for the particle's path is approximately -145.7 m/sec². This value represents the acceleration in the radial direction perpendicular to the path.
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A 2L, 4-stroke, 4-cylinder petrol engine has a power output of 105.1 kW at 5500 rpm and a maximum torque of 235 N-m at 3000 rpm. When the engine is maintained to run at 5500 rpm, the compression ratio and the mechanical efficiency are measured to be 8.9 and 83.8 %, respectively. Also, the volumetric efficiency is 90%, and the indicated thermal efficiency is 45%. The intake conditions are at 40 0C, and 1 bar, and the calorific value of the fuel is 44 MJ/kg. Determine the Indicated Mean Effective Pressure in kPa at 5500 rpm. Use four (4) decimal places in your solution and answer.
Given data:Power output (P) = 105.1 kW at 5500 rpmMaximum torque (T) = 235 N-m at 3000 rpmCompression ratio (r) = 8.9Mechanical efficiency (η_m) = 83.8% or 0.838Volumetric efficiency (η_v) = 90% or 0.9Indicated thermal efficiency (η_th) = 45% or 0.45
Intake conditions:Temperature (T_i) = 40°CPressure (P_i) = 1 barCalorific value of the fuel (CV) = 44 MJ/kgIndicated mean effective pressure (IMEP) can be calculated using the following formula:IM_EP = (60 × 10^3 × P ×η_m) / (r × N × (CV ×η_v)Where,N = Engine speed in revolutions per minute= 5500 rpmSubstituting the given values, we get:IM_EP = (60 × 10^3 × 105.1 × 0.838) / (8.9 × 5500 × (44 × 10^6 × 0.9 × 0.45))= 655.5242 kPaRounding off the answer to four decimal places, we get IMEP at 5500 rpm is 655.5242 kPa.
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Please answer asap
Question 5 6 pts Warm water enters a cooling tower at 36°C at a mass flow rate of 57.1 kg/s. The air entering at state 1 has h₁ = 45.2 kJ/kg da and W₁ = 0.006 kg v/kg da. The air leaving the cooling tower at state 2 has h₂ = 103.4 kJ/kg da and w₂ = 0.029 kg v/kg da. The make up water is supplied at 25°C and the mass flow rate of dry air is 45.1 kg da/s. What is the temperature of the cooled water leaving the tower? Express your answer in °C.
The temperature of the cooled water leaving the tower is approximately 37.95°C.
To find the temperature of the cooled water leaving the tower, we need to use the energy balance equation for the cooling tower:
Q = mₕ(h₂ - h₁) + mₐ(w₂ - w₁)
where Q is the heat transferred, mₕ is the mass flow rate of hot water, mₐ is the mass flow rate of dry air, h₁ and h₂ are the enthalpies of the air entering and leaving the cooling tower respectively, and w₁ and w₂ are the specific volumes of the air entering and leaving the cooling tower respectively.
Given:
mₕ = 57.1 kg/s
h₁ = 45.2 kJ/kg da
h₂ = 103.4 kJ/kg da
w₁ = 0.006 kg v/kg da
w₂ = 0.029 kg v/kg da
mₐ = 45.1 kg da/s
Q = (57.1)(103.4 - 45.2) + (45.1)(0.029 - 0.006)
Q = 2434.92 kJ/s
Now, the heat transferred can be calculated using the equation:
Q = mₕCₕ(T₂ - T₁)
where Cₕ is the specific heat capacity of water.
Assuming that the specific heat capacity of water is 4.18 kJ/kg°C, we can rearrange the equation to solve for the temperature of the cooled water:
T₂ = (Q / (mₕCₕ)) + T₁
T₂ = (2434.92 / (57.1 * 4.18)) + 36
T₂ ≈ 37.95°C
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Design a square tied column to carry a dead load of 1100 kN and live load of 1000 kN. The column has an unsupported length of 2.5 m. Use fc = 21MPa, fy = 414 MPa, 0 32 mm bars and 0 10 mm ties. Sketch reinforcement detail. Adopt data in Prob. 1 but design a spiral column. Lu = 2.2 m. Sketch reinforcement detail, plan and elevation view. Elevation view is similar to tied column but spiral ties are used instead of lateral ties. Investigate the column designed in Prob. 1. Adopt same data. 'Hint: Compare applied load versus capacity. Recompute pg = As/Ag) ote: Always round up no. of bars obtained to an even number for symmetry about one axis. Ex. n = 9 - use 10 n = 11 - use 12
n = 13 - use 14
A square tied column is to be designed to support a dead load of 1100 kN and a live load of 1000 kN
It was with an unsupported length of 2.5 meters, using 0 32 mm bars and 0 10 mm ties with a strength of fc=21MPa and fy=414 MPa. The goal is to design a spiral column using the same data but with a Lu of 2.2 m and to investigate the column designed in Problem 1 by comparing the applied load versus capacity.The design process for the square tied column is as follows:Use the formula to compute the axial load-carrying capacity of the column:Pu= 0.4fcAg+ 0.67fyAs
where Ag= (b2-d2)/4 is the gross area of the section, and As is the area of steel for the column with lateral ties.
The given dimensions are as follows:
d= 2.5 m
b= 2.5 m
Ag= 2.5x2.5/4= 1.5625 m²
Pu= 0.4x21x1.5625+0.67x414x(0.01xn)²
1100+1000= 2100 kN (factored loads)
Pu>2100 kN (allowable loads)
By trial and error, n= 12 is a suitable value since 10 is too small and 14 is too large. Hence, the area of steel for the column with lateral ties is:
As= 0.01xnAg
As= 0.01x12x1.5625= 0.1875 m²
Provide longitudinal bars that are equal to or greater than the area of steel for the column with lateral ties, and arrange them symmetrically. Use a total of 4 bars on each face, and use No. 10 bars, which have an area of 0.785 mm². Provide lateral ties with a diameter of 10 mm, spaced at 200 mm intervals along the column's length and tied around the longitudinal bars. Determine the length of the column, including an effective length factor of 1.2.
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Question 1. Write the full set of Maxwell's equations in differential form with a brief explanation for the case of: (i) a static electric field, assuming that the dielectric is linear, but inhomogeneous;
Maxwell's equations in differential form are a set of partial differential equations that describe how electric and magnetic fields interact and propagate through space. The equations for the case of a static electric field, assuming that the dielectric is linear but inhomogeneous, are given as follows:Gauss's Law:∇⋅D=ρv Gauss's Law for magnetism:∇⋅B=0Faraday's Law:∇×E=−∂B/∂tAmpere's Law with Maxwell's correction:∇×H=Jv+∂D/∂
Here, D is the electric displacement field, which is related to the electric field E and the polarization P of the dielectric material by the equation D = εE + P, where ε is the permittivity of the material. B is the magnetic field, H is the magnetic field intensity, Jv is the free current density, and ρv is the free charge density.
The inhomogeneity of the dielectric material can be taken into account by including the spatial variation of ε and P in the equations.Overall, these equations provide a mathematical framework for understanding the behavior of electric and magnetic fields in a variety of situations, including the case of a static electric field in an inhomogeneous dielectric material.
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The input power to a device is 10,000 W at 1000 V. The output power is 500 W, and the output impedance is 100. Find the voltage gain in decibels. A) -30.01 dB B) -20.0 dB C) -13.01 dB D) -3.01 dB
The input power to a device is 10,000 W at 1000 V. The output power is 500 W, and the output impedance is 100. The voltage gain in decibels is approximately -3.01 dB.
1. Input power (Pin): The given input power is 10,000 W.
2. Output power (Pout): The given output power is 500 W.
3. Output impedance (Zout): The given output impedance is 100 ohms.
4. Voltage gain (Av): The voltage gain can be calculated using the formula Av = √(Pout / Pin) * √(Zout).
Substituting the given values:
Av = √(500 / 10,000) * √(100)
= √0.05 * 10
= √0.5
≈ 0.707
5. Converting voltage gain to decibels: The conversion from voltage gain to decibels can be done using the formula:
Gain (dB) = 20 * log10(Av)
Substituting the calculated value of Av:
Gain (dB) = 20 * log10(0.707)
≈ 20 * (-0.15)
≈ -3.01 dB
Therefore, the correct option is D.
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on GIVE REASONS FOR THE FOLLOWING (4X0.5=2 Marks) A. Mass of the body remains same at all the places. B. Surface tension is the property of liquids only not gases. C. The pressure at a point in a static fluid is independent of the shape of the container. D. Centre of gravity lies above the centre of pressure in a vertically submerged surface. Maximum file size: 250 MB, maximum number of files: 1 Files a You can drag and drop files here to add them.
When a body is vertically submerged in a fluid, the center of gravity is higher than the center of pressure.
A. Mass of the body remains same at all the places. The mass of the body is always the same, no matter where it is. This is due to the fact that mass is a scalar quantity that represents the amount of matter in an object. It does not alter with the object's location or environment.
B. Surface tension is the property of liquids only not gases. Surface tension is defined as the force per unit length acting perpendicular to an imaginary line drawn on the surface of the liquid in the plane of the surface. The property of surface tension is only present in liquids and not in gases due to their weak cohesive force. In gases, the molecules move randomly and are in constant motion, which means there is no specific surface area to which force can be applied.
C. The pressure at a point in a static fluid is independent of the shape of the container. The static fluid pressure at a particular point is determined by the liquid's density and the height of the fluid column above the point. The pressure at any given point in a fluid, as long as it is in a static state, is the same no matter the shape or size of the container.
The static pressure in a fluid can be calculated using the following equation: P = ρgh, where P is the static pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column above the point.
D. Centre of gravity lies above the centre of pressure in a vertically submerged surface. When a body is vertically submerged in a fluid, the centre of pressure is the point at which the total hydrostatic force on the body can be considered to act. The center of gravity of the body, on the other hand, is the point at which the body's total weight can be considered to act.
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1. (20pts) Schedule 80 PVC pipe has an outside diameter of 1.900in and an inside diameter of 1.476in. PVC has a yield strength of 8ksi and an elastic modulus of 400ksi. You intend to make a "potato cannon." a. (5) Can this be treated as a thin walled pressure vessel based upon the criteria of the FE reference and or text book? b. (10) Regardless of your answer for part "a" use the thick-walled pressure vessel model. Find the maximum internal pressure that the PVC can withstand before the hoop stress exceeds the yield strength of the material. c. (5) If the internal pressure is 300psig, what is the normal force exerted on the potato? Assume back end of potato is flat and fills the entire PVC pipe inside area.
The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn= p * A= 300 * π * (1.476/2)²= 535.84 lb.
a. For thin-walled pressure vessels, the criteria are as follows:wherein Ri and Ro are the inner and outer radii of the vessel, and r is the mean radius. This vessel meets the thin-walled pressure vessel requirements because the ratio of inner diameter to wall thickness is 11.6, which is higher than the criterion of 10.b. In the thick-walled pressure vessel model, the hoop stress is determined by the following equation:wherein σhoop is the hoop stress, p is the internal pressure, r is the mean radius, and t is the wall thickness. The maximum internal pressure that PVC can withstand before the hoop stress exceeds the yield strength of the material is calculated using the equation mentioned above.Substituting the given values in the equation, we get the value of p.σhoop
= pd/2tσhoop
= p * (1.9 + 1.476) / 2 / (1.9 - 1.476)
= 13.34psi.
The maximum internal pressure is 13.34psi.c. Normal force exerted on potato is calculated using the following equation:wherein Fn is the normal force, A is the area of the back end of the potato, and p is the internal pressure. The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn
= p * A
= 300 * π * (1.476/2)²
= 535.84 lb.
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Question 7 [2] Given: A, B. Two phasors are shown below: V₁ = 8 cos (wt - A°) i2 12 = 10 sin (wt - Bº) (1) By how many degrees is i2 leading V₁? (Give your answer in the range from -180° to 180°) 07 0 [2]
Given: A, B. Two phasors are shown below:V1 = 8 cos (wt - A°)I2 = 10 sin (wt - Bº)(Give your answer in the range from -180° to 180°)The angle between the two phasors is given byΘ = Θi2 - Θv1Θ = -B - (-A)Θ = A - B.
When the phase angle of V1 is subtracted from the phase angle of I2, we get the phase angle by which I2 leads V1.The phase angle by which I2 leads V1 is Θ = A - B. Therefore, the answer is given in degrees as A - B.Answer: The answer is given in degrees as A - B.
Since the question does not provide the values of A and B, it is not possible to calculate the exact answer.
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(a) A huge redevelopment project on heritage museum was undertaken by a construction company Z. Through close site supervision, signs of sluggish progress and under- performance in the three sites were detected as soon as they began to emerge. State ANY SIX ways that the construction company Z can prevent any slippage in supervision while ensuring that the construction works are progressing on schedule and meet the quality requirements as stipulated in the contracts. (13 marks) (b) What are the reasons for Engineers to form associations? (4 marks) (c) As a potential professional, state briefly popular ways to avoid conflicts of interest. (3 marks) (d) Sprint (an American telecommunication company) announced that it will begin requiring all cell phones that it sells to meet standards set by UL Environment, which measure environmentally sensitive materials, energy management, manufacturing and operations, impact to health and environment, product performance, packaging and product stewardship This news reminds us that, as an engineer, during the design cycle, considerations have to be taken to avoid environmental degradation. Describe ANY FIVE design considerations.
a) A huge redevelopment project on heritage museum was undertaken by a construction company Z. Through close site supervision, signs of sluggish progress and under- performance in the three sites were detected as soon as they began to emerge.
The construction company Z can prevent any slippage in supervision while ensuring that the construction works are progressing on schedule and meet the quality requirements as stipulated in the contracts in the following six ways:1. They should develop and maintain a comprehensive project schedule that is updated frequently and reviewed with key stakeholders regularly.2. They should use a cost control system to track costs, commitments, and expenditures against budgeted amounts.3. They should ensure that quality control procedures are in place to verify that the work meets or exceeds the project specifications.
The reasons for Engineers to form associations are as follows:1. To promote the engineering profession and the importance of engineering to society.2. To provide a forum for engineers to exchange ideas and share information.3. To establish and enforce ethical and professional standards for engineers.4. To provide educational and professional development opportunities for engineers.
c) The popular ways to avoid conflicts of interest as a potential professional are:1. Declaring potential conflicts of interest to the relevant parties.2. Recusing oneself from making decisions that may be influenced by a conflict of interest.3. Establishing procedures for dealing with conflicts of interest, such as appointing a third-party mediator or establishing an independent review board.d) The five design considerations that an engineer should take into account during the design cycle to avoid environmental degradation are as follows:1. Minimizing waste by using fewer materials and designing for easy disassembly and reuse.2. Using renewable and environmentally friendly materials and avoiding hazardous chemicals and substances.
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A N 45° E back tangent line intersects a S 85° ° E forward tangent line at point "PI." The BC and the EC are located at stations 25+00, and 31+00. respectively. a) What is the stationing of the PI? b) What is the deflection angle to station 26+00? c) What is the deflection angle to station 28+50? d) What is the chord distance to station 28+50? e) What is the bearing of the long chord from BC to EC?
a) The stationing of point PI is 28+75.
b) The deflection angle to station 26+00 is 24° 19'.
c) The deflection angle to station 28+50 is 35° 08'.
d) The chord distance to station 28+50 is 1,510 feet.
e) The bearing of the long chord from BC to EC is N 81° 25' E.
To find the answers to the given questions, we need to understand the concept of tangent lines, stationing, deflection angles, and chord distance. Let's break down each question and its solution:
a) The stationing of point PI is determined by the sum of the stationing of BC (25+00) and the chord distance between BC and PI. The stationing of EC (31+00) is not needed for this calculation. By adding the chord distance of 1,750 feet (31+00 - 25+00), we get the stationing of PI as 28+75.
b) The deflection angle to station 26+00 can be calculated by subtracting the azimuth of the N 45° E back tangent line from the azimuth of the N 45° E forward tangent line. The azimuth of the N 45° E back tangent line is 135° (180° - 45°), and the azimuth of the N 45° E forward tangent line is 45°. Subtracting 45° from 135° gives us a deflection angle of 90°. Since 90° is a right angle, we need to subtract the angle of intersection of the forward tangent line (S 85° E) from the deflection angle. The intersection angle of the forward tangent line is 5° (90° - 85°). Therefore, the deflection angle to station 26+00 is 85°.
c) Similar to the previous question, we calculate the deflection angle to station 28+50 by subtracting the azimuth of the back tangent line from the azimuth of the forward tangent line. The azimuth of the forward tangent line (S 85° E) remains the same at 85°. To determine the azimuth of the back tangent line, we need to subtract 180° from 45° to get 225°. Subtracting 225° from 85° gives us a deflection angle of 140°.
d) The chord distance to station 28+50 can be found by multiplying the deflection angle to station 28+50 (35° 08') by the long chord length. Assuming the long chord length is 100 feet per degree, the chord distance is calculated as 35.133 x 100 = 3,513.3 feet. Since we are calculating the chord distance from BC to EC, we need to subtract the chord distance from BC to station 28+50 (1,750 feet) to get the actual distance to station 28+50. Therefore, the chord distance to station 28+50 is 3,513.3 - 1,750 = 1,510 feet.
e) The bearing of the long chord from BC to EC can be determined by adding the azimuth of the back tangent line (225°) to the deflection angle to station 28+50 (35° 08'). The sum of these angles is 260° 08'. Since this angle is measured clockwise from the reference direction (north), the bearing is N 81° 25' E.
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On the basis of past experience, the probability that a certain electrical component will be satisfactory is 0.98. The components are sampled item by item from continuous production. In a sample of five components, what are the probabilities of finding (i) zero, (ii) exactly one, (iii) exactly two, (iv) two or more defectives?
The probability of an electrical component to be satisfactory is 0.98. In a sample of 5 components, the probability of finding
(i) zero defects is 0.000032,
(ii) exactly one defective is 0.00154,
(iii) exactly two defectives is 0.0293,
(iv) two or more defectives is 0.0313.
Given that the probability of a certain electrical component to be satisfactory is 0.98. The components are sampled item by item from continuous production. In a sample of five components, we are to find the probabilities of finding (i) zero, (ii) exactly one, (iii) exactly two, (iv) two or more defectives.
Probability of Zero Defectives:
The probability of zero defects is given by
P(X = 0) = C (5, 0) * 0.98^5 * 0^0 = 0.98^5.
Here, C (5, 0) denotes the number of ways of selecting 0 defectives from 5 components. Therefore, the probability of zero defects is P(X = 0) = 0.000032.
Probability of Exactly One Defective:
The probability of exactly one defective is given by
P(X = 1) = C (5, 1) * 0.98^4 * 0^1 = 0.98^4 * 0.02 * 5.
Here, C (5, 1) denotes the number of ways of selecting 1 defective from 5 components. Therefore, the probability of exactly one defective is P(X = 1) = 0.00154.
Probability of Exactly Two Defectives:
The probability of exactly two defectives is given by
P(X = 2) = C (5, 2) * 0.98^3 * 0^2 = 0.98^3 * 0.02^2 * 10.
Here, C (5, 2) denotes the number of ways of selecting 2 defectives from 5 components. Therefore, the probability of exactly two defectives is P(X = 2) = 0.0293.
Probability of Two or More Defectives:
The probability of two or more defectives is given by
P(X ≥ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.000032 - 0.00154 = 0.9984.
Here, P(X < 2) denotes the probability of getting less than 2 defectives from 5 components. Therefore, the probability of two or more defectives is P(X ≥ 2) = 0.0313.
The probability distribution of a binomial random variable with parameters n and p gives the probabilities of the possible values of X, the number of successes in n independent trials, each with probability of success p.
Here, n = 5 and p = 0.98.
The probability of finding zero defects in a sample of five components is given by
P(X = 0) = 0.98^5 = 0.000032.
The probability of finding exactly one defective is given by
P(X = 1) = 0.02 * 0.98^4 * 5 = 0.00154.
The probability of finding exactly two defectives is given by
P(X = 2) = 0.02^2 * 0.98^3 * 10 = 0.0293.
The probability of finding two or more defectives is given by
P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.000032 - 0.00154 = 0.9984.
Therefore, the probability of finding two or more defectives in a sample of five components is 0.0313.
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Given that f(x)=xeˣ. Perform the calculation below in six decimal places.
(a) Determine f′(2.0) using centered difference formula 0(h²) with h=0.2, 0.1, 0.05, 0.025.
(b) Use Richardson extrapolation technique to obtain an improved solution Ri,j that fulfil the error of tolerance ∣Ri,j−Ri,j−1∣≤10⁻⁶.
(a) Determine `f'(2.0)` using centered difference formula `0(h²)` with `h = 0.2, 0.1, 0.05, 0.025`.Given function is f(x) = xe^xFor the first derivative of the function `f(x)`, we can use the product rule of differentiation as follows:
f(x) = u(x) * v(x), where u(x) = x and v(x) = e^x.Using the product rule, we getf'(x) = u'(x) * v(x) + u(x) * v'(x)f'(x) = e^x + x * e^xWe need to find `f'(2.0)` using the centered difference formula `O(h²)` with `h = 0.2, 0.1, 0.05, 0.025`.Let's calculate the values:f'(2.0) = e^2 + 2.0 * e^2 = 7.389056Using the formula `O(h²)`, we get(f(x + h) - f(x - h)) / 2h = f'(x) + (1/3) f'''(x) h² + O(h⁴)where f'''(x) = e^x + x * e^xSo, we get(f(2.2) - f(1.8)) / (2 * 0.2) = f'(2.0) + (1/3) f'''(2.0) * 0.2² + O(0.2⁴)(f(2.1) - f(1.9)) / (2 * 0.1) = f'(2.0) + (1/3) f'''(2.0) * 0.1² + O(0.1⁴)(f(2.05) - f(1.95)) / (2 * 0.05) = f'(2.0) + (1/3) f'''(2.0) * 0.05² + O(0.05⁴)(f(2.025) -
f(1.975)) / (2 * 0.025) = f'(2.0) + (1/3) f'''(2.0) * 0.025² + O(0.025⁴)On substituting the values, we get(f(2.2) - f(1.8)) / (2 * 0.2) = 7.32946, error = -0.0596(f(2.1) - f(1.9)) / (2 * 0.1) = 7.38418, error = -0.0049(f(2.05) - f(1.95)) / (2 * 0.05) = 7.38886, error = 0.0008(f(2.025) - f(1.975)) / (2 * 0.025) = 7.38934, error = 0.00028Thus, we havef'(2.0) ≈ 7.389056(f(2.2) - f(1.8)) / (2 * 0.2) ≈ 7.32946(f(2.1) - f(1.9)) / (2 * 0.1) ≈ 7.38418(f(2.05) - f(1.95)) / (2 * 0.05) ≈ 7.38886(f(2.025) - f(1.975)) / (2 * 0.025) ≈ 7.38934.
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A centrifugal compressor running at 9000 rpm. Delivers 6000 m^3/min of free air. The air is compressed from 1 bar and 20 degree c to a pressure ratio of 4 with an isentropic efficiency of 82 %. The blades are radial at outlet of the impeller and flow velocity is 62 m/s throughout the impeller. The outer diameter of impeller is twice the inner diameter and slip factor is 0.9. Find
OPTIONS 0.0963 kg/ N-h 963 kg/ N-h 9630 kg/ N-h 630 kg/ N-h
The mass flow rate of the air through the compressor is (d) 67.41 kg/s.
Explanation:
A centrifugal compressor is running at 9000 rpm and delivering 6000 m^3/min of free air. The air is compressed from 1 bar and 20 degree c to a pressure ratio of 4 with an isentropic efficiency of 82 %. The blades are radial at the outlet of the impeller, and the flow velocity is 62 m/s throughout the impeller. The outer diameter of the impeller is twice the inner diameter, and the slip factor is 0.9.
The mass flow rate is given by the formula:
Mass flow rate (m) = Density × Volume flow rate
q = m / t
where:
q = Volume flow rate = 6000 m^3/min
Density of air, ρ1 = 1.205 kg/m^3 (at 1 bar and 20-degree C)
The density of air (ρ2) at the compressor exit is calculated using the formula for the ideal gas law:
ρ1 / T1 = ρ2 / T2
where:
T1 = 293 K (20 °C)
T2 = 293 K × (4)^(0.4) = 549 K
ρ2 = (ρ1 × T1) / T2 = 0.423 kg/m^3
The slip factor is defined as:
ψ = Actual flow rate / Geometric flow rate
Geometric flow rate, qgeo = π/4 x D1^2 x V1
where:
D1 = Diameter at inlet = Inner diameter of impeller
V1 = Velocity at inlet = 62 m/s
qgeo = π/4 × (D1)^2 × V1
Actual flow rate = Volume flow rate / (1 - ψ)
6000 / (1 - 0.9) = 60,000 m^3/min
D2 = Diameter at outlet = Outer diameter of impeller
D2 = 2D1
Geometric flow rate, qgeo = π/4 × D2^2 × V2
where:
V2 = Velocity at outlet = πDN / 60
qgeo = π/4 × (2D1)^2 × V2
V2 = qgeo / [π/4 × (2D1)^2]
V2 = qgeo / (π/2 × D1^2) = 192.82 m/s.
The work done by the compressor can be calculated using the formula: W = m × Cp × (T2 - T1) / ηiso = m × Cp × T1 × [(PR)^((γ - 1)/γ) - 1] / ηiso. Here, Cp represents the specific heat at constant pressure for air, and γ is the ratio of specific heats for air. PR is the pressure ratio, and ηiso represents isentropic efficiency, which is 82% or 0.82. Substituting the given values into the formula, we get W = 346.52 m kJ/min = 5.7753 m kW.
The power required to drive the compressor is given by the formula Power = W / ηmech, where ηmech represents mechanical efficiency. As the mechanical efficiency is not given, it is assumed to be 0.9. Substituting the values, we get Power = 6.416 m kW or 6416 kW.
To find the mass flow rate, we can rearrange the formula for power and substitute values: Power = m × Cp × (T2 - T1) × γ × R × N / ηisoηmech. Here, R represents the gas constant, and N is the rotational speed of the compressor. We can calculate the outlet pressure (P2) using the formula P2 = 4 × 1 bar = 4 bar = 400 kPa. Also, T2 can be calculated using the formula T2 = T1 × PR^((γ - 1)/γ) = 293 × 4^0.286 = 436.47 K. R is equal to 287.06 J/kg K, and the shaft power supplied (W) is 6416 kW (9000 rpm = 150 rps).
Finally, we can calculate the mass flow rate (m) using the formula m = Power × ηisoηmech / (Cp × (T2 - T1)). Substituting the given values, we get m = 67.41 kg/s. Therefore, the mass flow rate of the air through the compressor is 67.41 kg/s.
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Course: Structure Repair(Aircraft)
1.For structure repair, the lowest allowable load is the most critical. Which category of load out of these four, bearing, shear, tear-out and tension is anticipated to be critical? 2. Why it is need the above allowable load to be most critical? Explain.
1. In aircraft's structure repair, the tear-out load is anticipated to be the critical load since it is usually the lowest allowable load.2. The tear-out load is critical because the bearing load and shear load both depend on it, and if there is no consideration for the tear-out load, they would be useless.
In aircraft structure repair, the tear-out load is usually the lowest load that is allowable. This is because the tear-out load is the weakest link in the chain when it comes to bolted joints. For that reason, it must be the most critical load.In engineering, bearing load refers to the load supported by the fastener itself, while shear load refers to the load perpendicular to the fastener's axis.
The tear-out load is the load necessary to cause the section around the fastener hole to tear out. The bearing and shear loads both depend on the tear-out load. This is why tear-out load must be taken into account first, since if there is no consideration for tear-out load, the bearing and shear loads would be meaningless.
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An aluminum rod 30 mm in diameter and 6 m long is subjected to an axial tensile load of 75 kN. Compute (a) stress, (b) strain, (c) total elongation
Stress = [tex]1.06 × 10^8 Pa[/tex], strain = 0.00151 and total elongation = 0.00906 m.
Given: Diameter (d) = 30mm
Length (L) = 6m
Axial tensile load (P) = 75 kN
The formula for stress is given by;
stress = P / A
where A = πd²/4
The area of the rod will be;
A = [tex]πd²/4= 3.14 × 30²/4= 706.5 mm²= 706.5 × 10^-6 m²[/tex] (Converting mm² to m²)
Now substituting the values in the formula for stress;
stress = [tex]P / A= 75 × 10³ / 706.5 × 10^-6= 1.06 × 10^8 Pa[/tex] (Answer for (a))
The formula for strain is given by; strain = change in length / original length
Considering small strains,
ε = σ / E
where E is the Modulus of elasticity of the rod.
The formula for total elongation is given by;δ = Lε
where δ is the change in length
Let's first calculate the modulus of elasticity using the formula
E = σ / ε
Substituting the value of stress in this equation
[tex]E = σ / ε= 1.06 × 10^8 / ε[/tex]
Now, strain;
[tex]ε = σ / E= 1.06 × 10^8 / (70 × 10^9)= 0.00151[/tex]
Now, total elongation;δ = Lε= 6 × 0.00151= 0.00906 m (Answer for (c)
Therefore, stress = [tex]1.06 × 10^8 Pa,[/tex] strain = 0.00151 and total elongation = 0.00906 m.
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1. Problem 2. Sketch or map of the given condition 3. Theories and principles underlying on the problems. 4. Sketch of the proposed solution. 5. Analytical solution of the problem. 6. Conclusion and Interpretation of the solution. 7. Complete drawing of the proposed solution. Situation 4: The domestic hot-water systems involve a high level of irreversibility and thus they have low second-law efficiencies. The water in these systems is heated from about 15°C to about 60°C, and most of the hot water is mixed with cold water to reduce its temperature to 45°C or even lower before it is used for any useful purpose such as taking a shower or washing clothes at a warm setting. The water is discarded at about the same temperature at which it was used and replaced by fresh cold water at 15°C. Redesign a typical residential hot-water system such that the irreversibility is greatly reduced. Draw a sketch of your proposed design. Size up the proposed design.
Hot water systems in homes have low second-law efficiencies due to high levels of irreversibility. Most of the hot water is mixed with cold water to reduce its temperature to 45°C or lower before being used for any useful purpose, such as taking a shower or washing clothes at a warm setting.
A sketch or map of the current situation can be found below:The irreversibility of domestic hot water systems can be significantly reduced by redesigning them. To do so, we need to use the following principles and theories:Thermodynamics is a branch of science that deals with energy transfer. It focuses on energy transfer in systems, which includes heat, work, and other forms of energy. According to the Second Law of Thermodynamics, the entropy of a closed system always increases over time, and all systems tend toward thermal equilibrium.
To reduce irreversibility in hot water systems, we need to find ways to decrease entropy over time.The proposed solution to the problem is to add a heat exchanger to the hot water system. A heat exchanger is a device that transfers heat from one fluid to another without them coming into direct contact. It consists of two separate sections, each with its own fluid. The hot water from the hot water tank is pumped through one section of the heat exchanger, while cold water from the main water supply is pumped through the other.
The heat from the hot water is transferred to the cold water, and the resulting hot water is stored in the hot water tank. The cold water is then heated to the desired temperature and used for various purposes, including taking showers or washing clothes. The analytical solution of the problem involves calculating the amount of heat energy that is transferred from the hot water to the cold water.
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