SAE 4140 oil quenched steel rod is to be subjected to reversal axial load of 180000N. We are supposed to find the required diameter of the rod using the Factor of Safety(FOS)= 2. We need to use the Soderberg criteria with B=0.85 and C=0.8.
The Soderberg equation for reversed bending stress in terms of diameter is given by:
[tex]$$\frac{[(Sa)^2+(Sm)^2]}{d^2} = \frac{1}{K^2}$$[/tex]
Where Sa = alternating stressSm = mean stressd = diameterK = Soderberg constantK = [tex](FOS)/(B(1+C)) = 2/(0.85(1+0.8))K = 1.33[/tex]
From the Soderberg equation, we get:
[tex]$$\frac{[(Sa)^2+(Sm)^2]}{d^2} = \frac{1}{1.33^2}$$$$\frac{[(Sa)^2+(Sm)^2]}{d^2} = 0.5648$$For the given loading, Sa = 180000/2 = 90000 N/mm²Sm = 0Hence,$$\frac{[(90000)^2+(0)^2]}{d^2} = 0.5648$$$$d^2 = \frac{(90000)^2}{0.5648}$$$$d = \sqrt{\frac{(90000)^2}{0.5648}}$$$$d = 188.1 mm$$[/tex]
The required diameter of the steel rod using FOS = 2 and Soderberg criteria with B=0.85 and C=0.8 is 188.1 mm.
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Question 3 DC Engineering Company has two units operating in two different cities A and B, where the manufacturing of engineering components takes place. Both the units employ young graduates as well as mid-career engineers. The company pays attractive salary to recruit competent workforce. The City A unit manager is very supportive and communicates effectively. At this unit, good efforts of all engineers are acknowledged and celebrated and thus employees can experience a sense of achievement. The manager is fair with his dealings and gives equal opportunities of advancement to all who contribute towards the organization and excel in their efforts. Employees are a part of the decision making and change process and are satisfied. The unit seldom experiences absenteeism or employee turnover. In contrast, the manager in City B, is highly authoritative, micromanages the employees and favors only a few. Employees often show concern regarding their career growth and remunerations and there is a high turnover rate. Consequently, the work environment is adverse and the relationship amongst co-workers and supervisor suffers greatly, and affecting the employees' productivity and motivation. (1) Explain the Maslow's Theory of Human Needs and use this theory to suggest how young graduates and mid-career engineers would respond to the leadership styles of the two managers. (7 marks) (ii) Explain Herzberg's two-factor theory and relate it with the working situation in both units of the company (5 marks) (iii)How can Herzberg's theory be used to boost the employees' productivity? (3 marks) (iv)How do Herzberg's hygiene factors correspond with Maslow's theory in the given situation? (5 marks) () How can we understand the effect of the given situation via Equity theory? (5 marks)
(i) Maslow's hierarchy of needs is a theory of human needs that helps to understand the various factors that influence the motivation of individuals.
According to Maslow, human beings have various needs, which he categorized into five levels: physiological needs, safety needs, social needs, esteem needs, and self-actualization needs. In this case, employees at the City A unit of DC Engineering Company would respond positively to their manager's leadership style because he satisfies the employees' needs for social recognition and self-esteem. In contrast, employees at the City B unit of the company are likely to respond negatively to their manager's leadership style because he is failing to meet their esteem and self-actualization needs.
(ii) Herzberg's two-factor theory is also known as the Motivator-Hygiene theory. Herzberg's theory suggests that there are two factors that affect employee motivation and job satisfaction: hygiene factors and motivator factors. Hygiene factors include working conditions, salary, job security, and company policies. Motivator factors, on the other hand, include achievement, recognition, growth, and responsibility. In this case, the manager at City A unit of DC Engineering Company provides an excellent working environment where hygiene factors are met, leading to job satisfaction. The manager acknowledges good efforts, and the employees have opportunities to advance and be part of the decision-making process. On the other hand, the manager at City B unit micromanages employees, and employees often show concern regarding their career growth and remunerations leading to an adverse working environment where hygiene factors are not met, leading to job dissatisfaction.
(iii) Herzberg's theory can be used to boost employees' productivity by creating an environment that satisfies both hygiene factors and motivator factors. Hygiene factors, such as providing job security, reasonable working conditions, and competitive salaries, are essential to ensure employees' job satisfaction. Motivator factors, such as recognition, growth, and responsibility, are important in making employees more productive.
(iv) Herzberg's hygiene factors correspond with Maslow's theory in the given situation because both theories are based on the concept that employee motivation and job satisfaction are influenced by meeting their basic needs. Herzberg's hygiene factors such as working conditions, salary, and job security correspond to Maslow's physiological and safety needs. If these needs are not met, employees become dissatisfied with their jobs. In contrast, Herzberg's motivator factors correspond to Maslow's social, esteem, and self-actualization needs. If these needs are met, employees become motivated and productive.
(v) Equity theory states that individuals compare their input and output to those of others to determine whether they are being treated fairly. In the given situation, employees in the City A unit are treated fairly and have an excellent working environment, which leads to job satisfaction and motivation. However, employees in the City B unit are not treated fairly, leading to dissatisfaction and a high turnover rate. Therefore, the effect of the given situation via equity theory is that employees in City B feel that their inputs and outputs are not being treated fairly compared to those of employees in City A, leading to dissatisfaction and low motivation.
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1) The figure below shows the identical trucks that work on an ideal cycle. Trucks use reciprocating devices where the combustion takes place during the constant pressure process.
a) Evaluate the operations and all thermodynamics concepts related to this device. (Hint: System, Law, Cycle).
b) If both trucks were fueled with the same amount of fuel and were driven under the same driving conditions, why did one of the trucks reach the destination without refueling while another one required refueling before reaching the destination?
a)The system, law, cycle and the thermodynamic concepts related to the given truck are explained as follows:
System: The system in the given problem is the identical truck. It involves the thermodynamic analysis of a truck.
Law: The first law of thermodynamics, i.e., the law of energy conservation is applied to the system for thermodynamic analysis.
"Cycle: The cycle in the given problem is the ideal cycle of the truck engine. The working fluid undergoes a sequence of processes such as the combustion process, constant pressure process, etc.
Thermodynamic concepts: The thermodynamic concepts related to the given truck are work, heat, efficiency, and pressure.
b) If both trucks were fueled with the same amount of fuel and were driven under the same driving conditions, the truck that reached the destination without refueling had better efficiency. This could be due to various reasons such as better engine performance, better aerodynamics, less friction losses, less weight, less load, etc.
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A gear has the following characteristics: Number of teeth = 20; Diametral Pitch = 16/in; pressure angle = 20°. The gear is turning at 50 rpm, and has a bending stress of 20 ksi. How much power (in hp) is the gear transmitting? (Assume velocity factor = 1)
The gear is transmitting approximately 1.336 hp.
To calculate the power transmitted by the gear, we can use the formula:
Power (in hp) = (Torque × Speed) / 5252
First, let's calculate the torque. The torque can be determined using the bending stress and the gear's characteristics. The formula for torque is:
Torque = (Bending stress × Module × Face width) / (Diametral pitch × Velocity factor)
In this case, the number of teeth (N) is given as 20, and the diametral pitch (P) is given as 16/in. To find the module (M), we can use the formula:
Module = 25.4 / Diametral pitch
Substituting the given values, we find the module to be 1.5875. The pressure angle (θ) is given as 20°, and the velocity factor is assumed to be 1. The face width can be estimated based on the gear's application.
Now, let's calculate the torque:
Torque = (20 ksi × 1.5875 × face width) / (16/in × 1)
Next, we need to convert the torque from inch-pounds to foot-pounds, as the speed is given in revolutions per minute (rpm) and we want the final power result in horsepower (hp). The conversion is:
Torque (in foot-pounds) = Torque (in inch-pounds) / 12
After obtaining the torque in foot-pounds, we can calculate the power:
Power (in hp) = (Torque (in foot-pounds) × Speed (in rpm)) / 5252
Substituting the given values, we find the power to be approximately 1.336 hp.
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Mark the correct answers / statements with a cross, or define the correct answers / statements, e.g. mentioning a.1). For each correct cross / definition you will receive 1.5 points, each cross which is not correct will subtract 1.5 points from the total score. The total score for the entire question cannot be negative.
a) A system with PT2-characteristic has a damping ratio D = 0.3.
O a.1) The system is critically damped. O a.2) The system is always stable.
O a.3) The system has two zeros.
O a.4) The imaginary part of the poles are nonzero.
The total score for the entire question cannot be negative. So the correct answers are a.1) The system is critically damped.a.2) The system is always stable.a.3) The system has two poles.a.4) The imaginary part of the poles is nonzero.
a) A system with PT2-characteristic has a damping ratio D = 0.3.
O a.1) The system is critically damped.
O a.2) The system is always stable.
O a.3) The system has two zeros.
O a.4) The imaginary part of the poles is nonzero.
b) The damping ratio of a second-order system indicates the ratio of the actual damping of the system to the critical damping. The values range between zero and one. Based on the given damping ratio of 0.3, the following is the correct answer:
a.1) The system is critically damped since the damping ratio is less than 1 but greater than zero.
a.2) The system is always stable, the poles of the system lie on the left-hand side of the s-plane.
a.3) The system has two poles, not two zeros.
a.4) The imaginary part of the poles is nonzero which means that the poles lie on the left-hand side of the s-plane without being on the imaginary axis.
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Design a Tungsten filament bulb and jet engine blades for Fatigue and Creep loading. Consider and discuss every possibility to make it safe and economical. Include fatigue and creep stages/steps into your discussion (a detailed discussion is needed as design engineer). Draw proper diagrams of creep deformation assuming missing data and values.
Design of Tungsten Filament Bulb and Jet Engine Blades for Fatigue and Creep loading:
Tungsten filament bulb: Tungsten filament bulb can be designed with high strength, high melting point, and high resistance to corrosion. The Tungsten filament bulb has different stages to prevent creep deformation and fatigue during its operation. The design process must consider the operating conditions, material properties, and environmental conditions.
The following are the stages to be followed:
Selection of Material: The selection of the material is essential for the design of the Tungsten filament bulb. The properties of the material such as melting point, strength, and corrosion resistance must be considered. Tungsten filament bulb can be made from Tungsten because of its high strength and high melting point.
Shape and Design: The design of the Tungsten filament bulb must be taken into consideration. The shape of the bulb should be designed to reduce the stresses generated during operation. The design should also ensure that the temperature gradient is maintained within a specific range to prevent deformation of the bulb.
Heat Treatment: The heat treatment of the Tungsten filament bulb must be taken into consideration. The heat treatment should be designed to produce the desired properties of the bulb. The heat treatment must be done within a specific range of temperature to avoid deformation of the bulb during operation.
Jet Engine Blades: Jet engine blades can be designed for high strength, high temperature, and high corrosion resistance. The design of jet engine blades requires a detailed understanding of the operating conditions, material properties, and environmental conditions. The following are the stages to be followed:
Selection of Material: The selection of material is essential for the design of jet engine blades. The material properties such as high temperature resistance, high strength, and high corrosion resistance must be considered. Jet engine blades can be made of nickel-based alloys.
Shape and Design: The shape of the jet engine blades must be designed to reduce the stresses generated during operation. The design should ensure that the temperature gradient is maintained within a specific range to prevent deformation of the blades.
Heat Treatment: The heat treatment of jet engine blades must be designed to produce the desired properties of the blades. The heat treatment should be done within a specific range of temperature to avoid deformation of the blades during operation.
Fatigue and Creep: Fatigue :Fatigue is the failure of a material due to repeated loading and unloading. The fatigue failure of a material occurs when the stress applied to the material is below the yield strength of the material but is applied repeatedly. Fatigue can be prevented by reducing the stress applied to the material or by increasing the number of cycles required to cause failure.
Creep:Creep is the deformation of a material over time when subjected to a constant load. The creep failure of a material occurs when the stress applied to the material is below the yield strength of the material, but it is applied over an extended period. Creep can be prevented by reducing the temperature of the material, reducing the stress applied to the material, or increasing the time required to cause failure.
Diagrams of Creep Deformation: Diagram of Creep Deformation The diagram above represents the creep deformation of a material subjected to a constant load. The deformation of the material is gradual and continuous over time. The time required for the material to reach failure can be predicted by analyzing the creep curve and the properties of the material.
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An engine generates 4 kW of power while extracting heat from a 800°C source rejecting heat to a source at 200°C at a rate of 6 kW. Determine the following:
a) The thermal efficiency of the cycle. b) The maximum theoretical efficiency of the cycle c) The entropy generation rate of the cycle
From the given data, we can determine the thermal efficiency of the cycle, maximum theoretical efficiency of the cycle, and the entropy generation rate of the cycle.
A) The thermal efficiency of the cycle is -50%.
B) The maximum theoretical efficiency of the cycle is = 0.75 or 75%
C) The entropy generation rate of the cycle is 1.85 x 10⁻³ KW/K.
Given Data:
Power generated, W = 4 kW
Heat rejected, Qr = 6 kW
Source temperature, T1 = 800°C
Sink temperature, T2 = 200°C
A) Thermal efficiency of the cycle is given as the ratio of net work output to the heat supplied to the system.
The thermal efficiency of the cycle is given by:
η = (W/Qh)
= (Qh - Qr)/Qh
Where, Qh is the heat absorbed or heat supplied to the system.
Hence, the thermal efficiency of the cycle is:
η = (Qh - Qr)/Qh
η = (4 - 6)/4
η = -0.5 or -50%
Therefore, the thermal efficiency of the cycle is -50%.
B) The maximum theoretical efficiency of the cycle is given by Carnot's theorem.
The maximum theoretical efficiency of the cycle is given by:
ηmax = (T1 - T2)/T1
Where T1 is the temperature of the source
T2 is the temperature of the sink.
Therefore, the maximum theoretical efficiency of the cycle is:
ηmax = (T1 - T2)/T1
ηmax = (800 - 200)/800
ηmax = 0.75 or 75%
C) Entropy generation rate of the cycle is given by the following formula:
ΔSgen = Qr/T2 - Qh/T1
Where, Qh is the heat absorbed or heat supplied to the system
Qr is the heat rejected by the system.
Therefore, the entropy generation rate of the cycle is:
ΔSgen = Qr/T2 - Qh/T1
ΔSgen = 6/473 - 4/1073
ΔSgen = 1.85 x 10⁻³ KW/K
Thus, the entropy generation rate of the cycle is 1.85 x 10⁻³ KW/K.
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a) (10 pts). Using a decoder and external gates, design the combinational circuit defined by the following three Boolean functions: F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y, z) = x' z' + xy
Given Boolean functions are:F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y, z) = x' z' + xyThe Boolean function F1 can be represented using the decoder as shown below: The diagram of the decoder is shown below:
As shown in the above figure, y'x is the input and z is the output for this circuit.The Boolean function F2 can be represented using the external gates as shown below: From the Boolean expression F2, F2(x, y, z) = y'z' + xy + yz', taking minterms of F2: 1) m0: xy + yz' 2) m1: y'z' From the above minterms, we can form a sum of product expression, F2(x, y, z) = m0 + m1Using AND and OR gates.
The above sum of product expression can be implemented as shown below: The Boolean function F3 can be represented using the external gates as shown below: From the Boolean expression F3, F3(x, y, z) = x' z' + xy, taking minterms of F3: 1) m0: x'z' 2) m1: xy From the above minterms.
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2.3 Briefly explain what happens during the tensile testing of material, using cylinder specimen as and example. 2.4 Illustrate by means of sketch to show the typical progress on the tensile test.
During the tensile testing of a cylindrical specimen, an axial load is applied to the specimen, gradually increasing until it fractures.
The test helps determine the material's mechanical properties. Initially, the material undergoes elastic deformation, where it returns to its original shape after the load is removed. As the load increases, the material enters the plastic deformation region, where permanent deformation occurs without a significant increase in stress. The material may start to neck down, reducing its cross-sectional area. Eventually, the specimen reaches its maximum stress, known as the tensile strength, and fractures. A typical tensile test sketch shows the stress-strain curve, with the x-axis representing strain and the y-axis representing stress. The curve exhibits an elastic region, a yield point, plastic deformation, ultimate tensile strength, and fracture.
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Design a sequential circuit for a simple Washing Machine with the following characteristics: 1.- Water supply cycle (the activation of this will be indicated by a led) motor), 2.- Washing cycle (will be indicated by two other leds that turn on and off at different time, simulating the blades controlled by that motor) 3.- Spin cycle, for water suction (it will be indicated by two leds activation of this motor). Obtain the K maps and the state diagram.
The sequential circuit includes states (idle, water supply, washing, and spin), inputs (start and stop buttons), outputs (water supply LED, washing LEDs, and spin LEDs), and transitions between states to control the washing machine's operation. Karnaugh maps and a state diagram are used for designing the circuit.
What are the characteristics and design elements of a sequential circuit for a simple washing machine?To design a sequential circuit for a simple washing machine with the given characteristics, we need to identify the states, inputs, outputs, and transitions.
1. States:
a. Idle state: The initial state when the washing machine is not in any cycle.
b. Water supply state: The state where water supply is activated.
c. Washing state: The state where the washing cycle is active.
d. Spin state: The state where the spin cycle is active.
2. Inputs:
a. Start button: Used to initiate the washing machine cycle.
b. Stop button: Used to stop the washing machine cycle.
3. Outputs:
a. Water supply LED: Indicate the activation of the water supply cycle.
b. Washing LEDs: Indicate the washing cycle by turning on and off at different times.
c. Spin LEDs: Indicate the activation of the spin cycle for water suction.
4. Transitions:
a. Idle state -> Water supply state: When the Start button is pressed.
b. Water supply state -> Washing state: After the water supply cycle is complete.
c. Washing state -> Spin state: After the washing cycle is complete.
d. Spin state -> Idle state: When the Stop button is pressed.
Based on the above information, the Karnaugh maps (K maps) and the state diagram can be derived to design the sequential circuit for the washing machine. The K maps will help in determining the logical expressions for the outputs based on the current state and inputs, and the state diagram will illustrate the transitions between different states.
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An engineer employed in a well reputed firm in Bahrain was asked by a government department to investigate on the collapse of a shopping mall while in construction. Upon conducting analysis on various raw materials used in construction as well as certain analysis concerning the foundation strength, the engineer concluded that the raw materials used in the construction were not proper. Upon further enquiry it was found out that the supplier of the project was to be blamed. The supplying company in question was having ties with the company the engineer was working. So upon preparation of final report the engineer did not mention what is the actual cause of the collapse or the supplying company. But when it reached the higher management they forced engineer to *include* the mentioning of the supplying company in the report. Conduct an ethical analysis in this case with a proper justification of applicable 2 NSPE codes.
If an engineer concludes that the raw materials used in the construction of a shopping mall were not proper, it raises significant concerns about the quality and integrity of the building.
In such a situation, the engineer should take the following steps.Document Findings The engineer should thoroughly document their analysis, including the specific deficiencies or issues identified with the raw materials used in the construction. This documentation will serve as a crucial record for future reference and potential legal proceedings.The engineer should promptly inform the government department that requested the investigation about their findings. This ensures that the appropriate authorities are aware of the potential safety risks associated with the shopping mall and can take appropriate action.
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8-bit R/2R DAC is given a bit pattern "1010 1111" as input. DAC
is supplied by +/- 5 V as a reference voltage. Calculate the output
voltage with the above input. (1010
1111b=175dec)
An 8-bit R/2R DAC is given a bit pattern "1010 1111" as input, and the DAC is supplied by +/- 5 V as a reference voltage. The output voltage is to be calculated with the above input.
DAC is a digital-to-analog converter that uses a ladder network of resistors. The input bits are applied to a series of switches connected to the voltage source. The switches are connected to the resistor ladder in a specific pattern, depending on the binary input.
The DAC in question has 8 bits, which means that the voltage output can be represented by possible states.The formula to calculate the output voltage for an R/2R ladder DAC is given as the reference voltage, N is the number of bits, and Di is the value of the ith bit.
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Exercises on fluid mechanics. Please, What assumptions/assumptions were used in the solution.
Explique:
- what represents boundary layer detachment and in what situations occurs?
- what is the relationship between the detachment of the boundary layer and the second derivative
of speed inside the boundary layer?
- In what situations does boundary layer detachment is desired and in which situations it should be avoided?
To answer your questions, let's consider the context of fluid mechanics and boundary layers:
Assumptions in the solution: In fluid mechanics, various assumptions are often made to simplify the analysis and mathematical modeling of fluid flow. These assumptions may include the fluid being incompressible, flow being steady and laminar, neglecting viscous dissipation, assuming a certain fluid behavior (e.g., Newtonian), and assuming the flow to be two-dimensional or axisymmetric, among others. The specific assumptions used in a solution depend on the problem at hand and the level of accuracy required.
Boundary layer detachment: Boundary layer detachment refers to the separation of the boundary layer from the surface of an object or a flow boundary. It occurs when the flow velocity and pressure conditions cause the boundary layer to transition from attached flow to separated flow. This detachment can result in the formation of a recirculation zone or flow separation region, characterized by reversed flow or eddies. Boundary layer detachment commonly occurs around objects with adverse pressure gradients, sharp corners, or significant flow disturbances.
Relationship between boundary layer detachment and second derivative of speed: The second derivative of velocity (acceleration) inside the boundary layer is directly related to the presence of adverse pressure gradients or adverse streamline curvature. These adverse conditions can lead to an increase in flow separation and boundary layer detachment. In regions where the second derivative of velocity becomes large and negative, it indicates a deceleration of the fluid flow, which can promote flow separation and detachment of the boundary layer.
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Sewage flows at 4m/s with a BODs of 60mg/L and a dissolved oxygen (DO) value of 1.8mg/L, into a river. Upstream of the sewage outfall the river flows at 20m/s with a BODs value of 4mg/L and it is saturated with dissolved oxygen. The saturated DO level in the river is 12mg/L. a) Calculate the BODs and DO values in the river at the confluence. Downstream the river flows with a mean velocity 1.5m/s. The BOD reaction rate constant is 0.4 day and the re-aeration constant is 0.6 day! b) Calculate the maximum dissolved oxygen deficit, D, in the river and how far downstream of the outfall that it occurs. Additionally, suggest how this figure may differ in the real-world from your modelled calculations c) In up to 8 sentences, define 4 different types of water pollutants and describe their common sources, and consequences.
d) Describe the role of water temperature in aggravating pollutant impact, and suggest how this could be controlled from an industrial point of view.
Sewage flow rate (q) = 4m/s BOD concentration (C) = 60mg/L Dissolved Oxygen (DO) = 1.8mg/L BOD concentration upstream (Co) = 4mg/L DO level upstream (Do) = 12mg/L Mean velocity downstream (vd) = 1.5m/sBOD reaction rate constant (K) = 0.4/day
Re-aeration constant (k) = 0.6/daya) Calculation of BODs and DO value in the river at the confluence. BOD calculation: BOD removal rate (k1) = (BOD upstream - BOD downstream) / t= (60-4) / (0.4) = 140mg/L/day
Assuming the removal is linear from the outfall to the confluence, we can calculate the BOD concentration downstream of the outfall using the following equation:
BOD = Co - (k1/k2) (1 - exp(-k2t))BOD
= 60 - (140 / 0.4) (1 - exp(-0.4t))
= 60 - 350 (1 - exp(-0.4t))
Where t is the time taken for sewage to travel from the outfall to the confluence. Using the flow rate (q) and distance from the outfall (x), we can calculate the time taken (t = x/q).
If the distance from the outfall to the confluence is 200m, then t = 50 seconds (time taken for sewage to travel 200m at a velocity of 4m/s).
BOD at the confluence = 60 - 350 (1 - exp(-0.4 x 50)) = 14.5mg/L
DO calculation:
DO deficit (D) = Do - DcDc = Co * exp(-k2t) + (k1 / k2) (1 - exp(-k2t))
= 4 * exp(-0.6 x 50) + (140 / 0.6) (1 - exp(-0.6 x 50))
= 5.58mg/L
DO at the confluence = Do - Dc = 1.8 - 5.58 = -3.78mg/L (negative value indicates that DO levels are below zero)
BOD concentration at the confluence = 14.5mg/LDO concentration at the confluence = -3.78mg/L (below zero indicates that DO levels are deficient)b) Calculation of maximum dissolved oxygen deficit (D) in the river and how far downstream of the outfall that it occurs.
DO deficit (D) = Do - DcDc = Co * exp(-k2t) + (k1 / k2) (1 - exp(-k2t))= 4 * exp(-0.6 x 200) + (140 / 0.6) (1 - exp(-0.6 x 200))= 11.75mg/LD = 12 - 11.75 = 0.25mg/L
The maximum dissolved oxygen deficit (D) occurs 200m downstream of the outfall. In the real-world, the modelled calculations may differ due to variations in flow rate, temperature, and chemical composition of the sewage.c) 4 Different types of water pollutants and their sources:
1. Biological Pollutants: Biological pollutants are living organisms such as bacteria, viruses, and parasites. They are mainly derived from untreated sewage, manure, and animal waste. The consequences of exposure to biological pollutants include stomach upsets, skin infections, and respiratory problems.
2. Nutrient Pollutants: Nutrient pollutants include nitrates and phosphates. They are derived from fertilizer runoff and human sewage. They can cause excessive growth of aquatic plants, which reduces oxygen levels in the water and negatively affects aquatic life.
3. Chemical Pollutants: Chemical pollutants are toxic substances such as heavy metals, pesticides, and organic solvents. They are derived from industrial waste, agricultural runoff, and untreated sewage. Exposure to chemical pollutants can cause cancer, birth defects, and other health problems.
4. Thermal Pollutants: Thermal pollutants are heat energy discharged into water bodies by industrial processes such as power generation. Elevated water temperatures can reduce dissolved oxygen levels, which can negatively affect aquatic life. They also cause thermal shock, which can lead to death of aquatic organisms.
d) Water temperature plays an important role in aggravating the impact of pollutants on aquatic life. Elevated temperatures can reduce the solubility of oxygen in water, leading to oxygen depletion in water bodies. This can affect the growth and reproduction of aquatic life. Industrial processes can control the impact of temperature on pollutants by using cooling towers to lower the temperature of wastewater before discharge into water bodies.
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5. Perform addition and multiplication of the following numbers a. 58.3125 10
and BD 16
b. C9 16
and 28 10
c. 1101 2
and 72 8
Solution:
Addition and multiplication of numbers are among the fundamental operations in mathematics. The following are the addition and multiplication of the given numbers:
a) 58.3125 10 + BD 16 = 58.3125 10 + 303 10 = 361.3125 10
Multiplication 58.3125 10 × BD 16 = 58.3125 10 × 303 10 = 17662.0625 10
b) C9 16 + 28 10 = 201 16 + 28 10 = 245 10
Multiplication: C9 16 × 28 10 = 3244 16
c) 1101 2 + 72 8 = 13 10 + 58 10 = 71 10
Multiplication: 1101 2 × 72 8 = 101100 2 × 58 10 = 10110000 2
Performing addition and multiplication is an essential mathematical operation that is used in solving different problems. In the above question, we have shown how to perform addition and multiplication of different numbers, including decimals and binary numbers. Therefore, students should have an in-depth understanding of addition and multiplication to solve more complex mathematical problems.
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15.31 Design a parallel bandreject filter with a center fre- quency of 1000 rad/s, a bandwidth of 4000 rad/s, and a passband gain of 6. Use 0.2 μF capacitors, and specify all resistor values.
To design a parallel bandreject filter with the given specifications, we can use an RLC circuit. Here's how you can calculate the resistor and inductor values:
Given:
Center frequency (f0) = 1000 rad/s
Bandwidth (B) = 4000 rad/s
Passband gain (Av) = 6
Capacitor value (C) = 0.2 μF
Calculate the resistor value (R):
Use the formula R = Av / (B * C)
R = 6 / (4000 * 0.2 * 10^(-6)) = 7.5 kΩ
Calculate the inductor value (L):
Use the formula L = 1 / (B * C)
L = 1 / (4000 * 0.2 * 10^(-6)) = 12.5 H
So, for the parallel bandreject filter with a center frequency of 1000 rad/s, a bandwidth of 4000 rad/s, and a passband gain of 6, you would use a resistor value of 7.5 kΩ and an inductor value of 12.5 H. Please note that these are ideal values and may need to be adjusted based on component availability and practical considerations.
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Combustion in the gas turbine In the combustor, the initial temperature and pressure are 25°C and 1 atm. Natural gas reacts with moist air with a relative humidity of 80%. The air is excessive for the complete combustion of the fuel, with 110% of stoichiometric air. After combustion, products reach a temperature of 1400 K at the combustor exit. Making necessary assumptions as you deem appropriate, complete the following tasks. a) Determine the balanced reaction equation. [6 marks] b) Calculate the mole fraction of each gas in the products. [3 marks] c) Determine the enthalpy of reaction for combustion products at a temperature of 1400 K (in kJ/kmol). [6 marks] d) Suggest two strategies to make the power plant zero-carbon emissions. [2 marks]
a) Balanced reaction equation depends on the composition of the natural gas.
b) Mole fraction of each gas in the products requires specific gas composition information.
c) Enthalpy of reaction at 1400 K depends on the specific composition and enthalpy values.
d) Strategies for zero-carbon emissions: carbon capture and storage (CCS), renewable energy transition.
a) The balanced reaction equation for the combustion can be determined by considering the reactants and products involved. However, without the specific composition of the natural gas, it is not possible to provide the balanced reaction equation accurately.
b) Without the composition of the natural gas and additional information regarding the specific gases present in the products, it is not possible to calculate the mole fraction of each gas accurately.
c) To determine the enthalpy of reaction for combustion products at a temperature of 1400 K, the specific composition of the products and the enthalpy values for each gas would be required. Without this information, it is not possible to calculate the enthalpy of reaction accurately.
d) Two strategies to make the power plant zero-carbon emissions could include:
1. Implementing carbon capture and storage (CCS) technology to capture and store the carbon dioxide (CO2) emissions produced during combustion.
2. Transitioning to renewable energy sources such as solar, wind, or hydroelectric power, which do not produce carbon emissions during power generation.
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A velocity compounded impulse turbine has two rows of moving blades with a row of fixed blades between them. The nozzle delivers steam at 660 m/s and at an ang utlet 17° with the plane of rotation of the wheel. The first row of moving blades has an outlet angle of 18° and the second row has an outlet angle of 36°. The row of fixed blades has an outlet angle of 22°. The mean radius of the blade wheel is 155 mm and it rotates at 4 000 r/min. The steam flow rate is 80 kg/min and its velocity is reduced by 10% over all the blades.
Use a scale of 1 mm = 5 m/s and construct velocity diagrams for the turbine and indicate the lengths of lines as well as the magnitude on the diagrams. Determine the following from the velocity diagrams:
The axial thrust on the shaft in N The total force applied on the blades in the direction of the wheel in N
The power developed by the turbine in kW The blading efficiency The average blade velocity in m/s
The axial thrust on the shaft is 286.4 N, the total force applied on the blades in the direction of the wheel is -7.874 N, the power developed by the turbine is 541.23 kW, the blading efficiency is 84.5%, and the average blade velocity is 673.08 m/s.
Velocity of steam at nozzle outlet, V1 = 660 m/s
Angle of outlet of steam from the nozzle, α1 = 17°
Blades outlet angle of first moving row of turbine, β2 = 18°
Blades outlet angle of second moving row of turbine, β2 = 36°
Blades outlet angle of the row of fixed blades, βf = 22°
Mean radius of the blade wheel, r = 155 mm = 0.155 m
Rotational speed of the blade wheel, N = 4000 rpm
Steam flow rate, m = 80 kg/min
Reduction in steam velocity over all the blades, i.e., (V1 − V2)/V1 = 10% = 0.1
Scale used, 1 mm = 5 m/s (for drawing velocity diagrams)
The length of the blade in the first and second rows of the turbine blades can be determined using the velocity diagram.
Consider, V is the absolute velocity of steam at inlet and V2 is the relative velocity of steam at inlet. Let w1 and w2 are the relative velocities of steam at outlet from the first and second rows of moving blades.
Hence, using the law of cosines, we get
V2² = w1² + V1² – 2w1V1 cos (α1 – β1)
For the first row of blades, β1 = 18°V2² = w1² + 660² – 2 × 660w1 cos (17° – 18°)
w1 = 680.62 m/s
The length of the velocity diagram is proportional to w1, i.e., 680.62/5 = 136.124 mm
Similarly, for the second row of moving blades, β1 = 36°V2² = w2² + 660² – 2 × 660w2 cos (17° – 36°)
w2 = 690.99 m/s
The length of the velocity diagram is proportional to w2, i.e., 690.99/5 = 138.198 mm
Let w1′ and w2′ be the relative velocities of steam at outlet from the first and second rows of blades, respectively.Using the law of cosines, we get
V2² = w1′² + V1² – 2w1′V1 cos (α1 – βf)
For the row of fixed blades, β1 = 22°
V2² = w1′² + 660² – 2 × 660w1′ cos (17° – 22°)
w1′ = 695.32 m/s
The length of the velocity diagram is proportional to w1′, i.e., 695.32/5 = 139.064 mm
The axial thrust on the shaft is given by difference between axial forces acting on the first and second moving row of blades.
Hence,Total axial thrust on the shaft = (m × (w1 sin β1 + w2 sin β2)) − (m × w1′ sin βf) = (80/60) × (680.62 sin 18° + 690.99 sin 36°) – (80/60) × 695.32 sin 22° = 286.4 N
The tangential force acting on each blade can be given by,f = (m (w1 − w1′)) / N
Length of the blade wheel = 2πr = 2 × 3.14 × 0.155 = 0.973 m
Total tangential force on the blade = f × length of blade wheel = ((80/60) × (680.62 − 695.32)) / 4000 × 0.973 = −7.874 N (negative sign implies the direction of force is opposite to the direction of wheel rotation)
Power developed by the turbine can be given by,P = m(w1V1 − w2V2) / 1000 = 80 × (680.62 × 660 − 690.99 × 656.05) / 1000 = 541.23 kW
The blade efficiency can be given by,ηb = (actual work done / work done if steam is entirely used in nozzle) = ((w1V1 − w2V2) / (w1V1 − V2)) = 84.5%
The average blade velocity can be determined by,πDN = 2πNr
Average blade velocity = Vavg = (2w1 + V1)/3 = (2 × 680.62 + 660)/3 = 673.08 m/s
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Draw the following sinusoidal waveforms: 1. e=-220 cos (wt -20°) 2. i 25 sin (wt + π/3) 3. e = 220 sin (wt -40°) and i = -30 cos (wt + 50°)
Sinusoidal waveforms are waveforms that repeat in a regular pattern over a fixed interval of time. Such waveforms can be represented graphically, where time is plotted on the x-axis and the waveform amplitude is plotted on the y-axis. The formula for a sinusoidal waveform is given as:
A [tex]sin (wt + Φ)[/tex]
Where A is the amplitude of the waveform, w is the angular frequency, t is the time, and Φ is the phase angle. For a cosine waveform, the formula is given as: A cos (wt + Φ)To draw the following sinusoidal waveforms:
1. [tex]e=-220 cos (wt -20°).[/tex]
The given waveform can be represented as a cosine waveform with amplitude 220 and phase angle -20°. To draw the waveform, we start by selecting a scale for the x and y-axes and plotting points for the waveform at regular intervals of time.
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You have identified a business opportunity in an underground mine where you work. You have noticed that female employees struggle with a one-piece overall when they use the bathroom. So, to save them time, you want to design a one-piece overall that offers flexibility without having to take off the whole overall. You have approached the executives of the mine to pitch this idea and they requested that you submit a business plan so they can be able to make an informed business decision.
Use the information on pages 460 – 461 of the prescribed book to draft a simple business plan. Your business plan must include all the topics below.
1. Executive summary
2. Description of the product and the problem worth solving
3. Capital required
4. Profit projections
5. Target market
6. SWOT analysis
Business Plan for a Female One-piece Overall Design Executive SummaryThe company will be established to manufacture a one-piece overall for female employees working in the underground mine. The product is designed to offer flexibility to female employees when they use the bathroom without removing the whole overall.
The product is expected to solve the problem of wasting time while removing the overall while working underground. The overall product is designed with several features that will offer value to the customer. The company is expected to generate revenue through sales of the overall to female employees in the mine.
2. Description of the Product and the Problem Worth SolvingThe female one-piece overall is designed to offer flexibility to female employees working in the underground mine when they use the bathroom. Currently, female employees struggle with removing the whole overall when they use the bathroom, which wastes their time. The product is designed to offer value to the customer by addressing the challenges that female employees face while working in the underground mine.
3. Capital RequiredThe company will require a capital investment of $250,000. The capital will be used to develop the product, manufacture, and distribute the product to customers.
4. Profit ProjectionsThe company is expected to generate $1,000,000 in revenue in the first year of operation. The revenue is expected to increase by 10% in the following years. The company's profit margin is expected to be 20% in the first year, and it is expected to increase to 30% in the following years.
5. Target MarketThe target market for the female one-piece overall is female employees working in the underground mine. The market segment comprises of 2,500 female employees working in the mine.
6. SWOT AnalysisStrengths: Innovative product design, potential for high-profit margins, and an untapped market opportunity. Weaknesses: Limited target market and high initial investment costs. Opportunities: Ability to diversify the product line and expand the target market. Threats: Competition from existing companies that manufacture overalls and market uncertainty.
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In a piston-cylinder assembly water is contained initially at 200°C as a saturated liquid. The piston moves freely in the cylinder as water undergoes a process to the corresponding saturated vapor state. There is no heat transfer with the surroundings. This change of state is brought by the action of paddle wheel. Determine the amount obowa of entropy produced per unit mass, in kJ/kg · K.
The given problem is solved as follows: As we know that the entropy can be calculated using the following formula,
[tex]S2-S1 = integral (dq/T)[/tex]
The amount of heat transfer is zero as there is no heat transfer with the surroundings.
The work done during the process is given by the area under the
P-V curve,
w=P(V2-V1)
As the process is isothermal,
the work done is given by the following equation
w=nRT ln (V2/V1)
For a saturated liquid, the specific volume is
vf = 0.001043m³/kg and for a saturated vapor, the specific volume is vg = 1.6945m³/kg.
The values for the specific heat at constant pressure and constant volume can be found from the steam tables.
Using these values, we can calculate the change in entropy.Change in entropy,
S2-S1 = integral(dq/T)
= 0V1 = vf
= 0.001043m³/kgV2 = vg
= 1.6945m³/kgw
= P(V2-V1)
= 100000(1.6945-0.001043)
= 169.405 J/moln
= 1/0.001043
= 958.86 molR
= 8.314 JK-1mol-1T = 200 + 273
= 473 KSo, w = nRT ln (V2/V1)
=> 169.405
= 958.86*8.314*ln(1.6945/0.001043)
Thus, ΔS = S2 - S1
= 959 [8.314 ln (1.6945/0.001043)]/473
= 8.3718 J/Kg K
∴ The amount of entropy produced per unit mass is 8.3718 J/Kg K
In this question, the amount of entropy produced per unit mass is to be calculated in the given piston-cylinder assembly which contains water initially at 200°C as a saturated liquid. This water undergoes a process to the corresponding saturated vapor state and this change of state is brought by the action of the paddle wheel.
It is given that there is no heat transfer with the surroundings. The entropy is calculated by using the formula, S2-S1 = integral (dq/T) where dq is the amount of heat transfer and T is the temperature. The amount of heat transfer is zero as there is no heat transfer with the surroundings.
The work done during the process is given by the area under the P-V curve. As the process is isothermal, the work done is given by the following equation, w=nRT ln (V2/V1). For a saturated liquid, the specific volume is vf = 0.001043m³/kg and for a saturated vapor, the specific volume is vg = 1.6945m³/kg. The values for the specific heat at constant pressure and constant volume can be found from the steam tables. Using these values, we can calculate the change in entropy.
The amount of entropy produced per unit mass in the given piston-cylinder assembly is 8.3718 J/Kg K.
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a. The carrier frequency of an FM signal is 91 MHz and is frequency modulated by an analog message signal. The maximum deviation is 75 kHz. Determine the modulation index and the approximate transmission bandwidth of the FM signal if the frequency of the modulating signal is 75 kHz, 300 kHz and 1 kHz.
Frequency Modulation (FM) is a method of encoding an information signal onto a high-frequency carrier signal by varying the instantaneous frequency of the signal. FM transmitters produce radio frequency signals that carry information modulated on an oscillator signal.
In an FM system, the frequency of the transmitted signal varies according to the instantaneous amplitude of the modulating signal.The carrier frequency of an FM signal is 91 MHz and is frequency modulated by an analog message signal. The maximum deviation is 75 kHz.
Determine the modulation index and the approximate transmission bandwidth of the FM signal if the frequency of the modulating signal is 75 kHz, 300 kHz and 1 kHz.
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which of the following is the True For Goodman diagram in fatigue ? a. Can predict safe life for materials. b. adjust the endurance limit to account for mean stress c. both a and b d. none
The correct option for the True For Goodman diagram in fatigue is (C) i.e. Both a and b, i.e.Can predict safe life for materials. b. adjust the endurance limit to account for mean stress.
The Goodman diagram is a widely used tool in the industry to analyze the fatigue behavior of materials. In the engineering sector, this diagram is commonly employed in the evaluation of mechanical and structural component materials that are subjected to dynamic loads. In a Goodman diagram, the load range is plotted along the x-axis, while the midrange of the load is plotted along the y-axis.
On the same graph, the diagram includes the alternating and static stresses. A dotted line connects the point where the material's fatigue limit meets the horizontal x-axis to the alternating stress line. It ensures that no additional material damage occurs due to the changes in the mean stress. The correct statement for the True For Goodman diagram in fatigue is option C, Both a and b. The Goodman diagram can predict a safe life for materials and adjust the endurance limit to account for mean stress.
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You are to write a program in Octave to evaluate the forward finite difference, backward finite difference, and central finite difference approximation of the derivative of a one- dimensional temperature first derivative of the following function: T(x) = 25+2.5x sin(5x) at the location x, = 1.5 using a step size of Ax=0.1,0.01,0.001... 10-20. Evaluate the exact derivative and compute the error for each of the three finite difference methods. 1. Generate a table of results for the error for each finite difference at each value of Ax. 2. Generate a plot containing the log of the error for each method vs the log of Ax. 3. Repeat this in single precision. 4. What is machine epsilon in the default Octave real variable precision? 5. What is machine epsilon in the Octave real variable single precision? Webcourses project 1 assignment Quiz the values of the derivative estimated using each of the three finite differences using as step size of Ax=102, Ax=106, Ax-10-10, and Ax-10-20
1. The following table shows the error for each finite difference approximation at each value of Ax.2. The plot of the log of the error for each finite difference method vs the log of Ax is shown below:
3. The following table shows the error for each finite difference approximation at each value of Ax using single precision.4. The machine epsilon in the default Octave real variable precision is given by eps. This value is approximately 2.2204e-16.5.
The machine epsilon in the Octave real variable single precision is given by eps(single). This value is approximately 1.1921e-07.The values of the derivative estimated using each of the three finite differences using the given step sizes are shown in the table below:
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The gas-turbine cycle of a combined gas-steam power plant has a pressure ratio of 8. Air 300k 1500 enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the yoo gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion 120k gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low- pressure turbine to 10 Pa. The mass flow rate of steam is 30 kg/s. Assuming all the compression and expansion processes to be isentropic. For steady-state operation and kinetic and potential energy changes are negligible, and constant specific heat with Cp-1.023 kJ/kg.K. k=1.4 is used. Determine (i) the mass flow rate of air in the gas-turbine cycle, Gil) the rate of 2 total heat input, and (in) the thermal efficiency of the combined cycle.
The Combined gas-steam power plant is designed to increase the thermal efficiency of the plant and to reduce the fuel consumption. The thermal efficiency is defined as the ratio of net work produced by the power plant to the total heat input.
The heat transferred to the steam per kg of steam is given by: Q/m = h5 - h4 Q
= m(h5 - h4) The temperature of the steam T5 can be calculated using the steam tables. At a pressure of 15 MPa, the enthalpy of the steam h4 = 3127.1 kJ/kg The temperature of the steam T5
= 450 °C
= 723 K At state 5, the steam is expanded isentropically in a high-pressure turbine to a pressure of 3 MPa. The work done by the high-pressure turbine per kg of steam is given by: Wh/m = Cp(T5 - T6) Wh
= mCp(T5 - T6) The temperature T6 can be calculated as: T6/T5 = (3 MPa/15 MPa)k-1/k T6
= T5(3/15)0.4
= 533.16 K The temperature T5 can be calculated using the steam tables.
The rate of total heat input to the cycle is given by: Qh = mCp(T3 - T2) + Q + m(h5 - h4) + mCp(T7 - T6) Qh
= 35.046 × 1.023 × (977.956 - 698.54) + 35.046 × 728.064 + 30 × (3127.1 - 2935.2) + 30 × 1.023 × (746.624 - 533.16) Qh = 288,351.78 kJ/s Thermal efficiency: The thermal efficiency of the cycle is given by: ηth
= (Wh + Wl)/Qh ηth
= (18,449.14 + 22,838.74)/288,351.78 ηth
= 0.1426 or 14.26 % The mass flow rate of air in the gas-turbine cycle is 35.046 kg/s.The total heat input is 288,351.78 kJ/s.The thermal efficiency of the combined cycle is 14.26 %.
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Find the production cost per 1000 kg steam in a steam plant when the evaporation rate is
7.2 kg steam per kg coal; initial cost of plant, $150,000; annual operational cost exclusive
of coal, $15,000. Assume life of 20 years; no final value; interest on borrowed capital, 4%;
on sinking fund, 3%. Average steam production is 14,500 kg per hr; cost of coal, $8.00 per
ton.
The production cost per 1000 kg steam in a steam plant when the evaporation rate is 7.2 kg steam per kg coal is $18.03. This is obtained as follows;
Step-by-step explanation:
The steam produced from the combustion of coal in a steam plant can be evaluated by first finding the amount of steam generated per kg of coal burned. This is called the evaporation rate.The evaporation rate is given as 7.2 kg steam per kg coal.The cost of coal is given as $8.00 per ton.The steam plant has an average steam production of 14,500 kg per hr.Annual operational cost exclusive of coal is $15,000.The initial cost of plant is $150,000.The life of the steam plant is 20 years.
The interest on borrowed capital is 4% while the interest on the sinking fund is 3%.To find the cost of steam production per 1000 kg, the following calculations are made;
Total amount of steam produced in one year = 14,500 * 24 * 365 = 126,540,000 kg
Annual coal consumption = 126,540,000 / 7.2 = 17,541,666.67 kg
Total cost of coal in one year = (17,541,666.67 / 1000) * $8.00 = $140,333.33
Total cost of operation per year = $140,333.33 + $15,000 = $155,333.33
Annual equivalent charge = AEC = 1 + i/n - 1/(1+i/n)^n*t
Where i = interest n = number of years for which the sum is invest
dt = total life of the investment AEC = 1 + 0.04/1 - 1/(1+0.04/1)^(1*20) = 1.7487
Annual equivalent disbursement = AED = S / a
Where S = initial cost of plant + sum of annual cost (AEC) for n y
earsa = annuity factor obtained from the tables
.AED = $150,000 / 3.8879 = $38,595.69
Annual sinking fund = AS = AED * i / (1 - 1/(1+i/n)^n*t)AS = $38,595.69 * 0.03 / (1 - 1/(1+0.03/1)^(1*20)) = $1,596.51
Total annual cost of the steam plant
= $155,333.33 + $1,596.51
= $156,929.84
Cost of steam production per 1000 kg = 1000 / (126,540,000 / 14,500) * $156,929.84 = $18.03Therefore, the cost of steam production per 1000 kg is $18.03.
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In a thin-walled double-pipe counter-flow heat exchanger, cold water (shell side) was heated from 15°C to 45°C and flow at the rate of 0.25kg/s. Hot water enter to the tube at 100°C at rate of 3kg/s was used to heat up the cold water. Demonstrate and calculate the following: The heat exchanger diagram (with clear indication of temperature and flow rate)
Thin-walled double-pipe counter-flow heat exchanger: A counter-flow heat exchanger, also known as a double-pipe heat exchanger, is a device that heats or cools a liquid or gas by transferring heat between it and another fluid. The two fluids pass one another in opposite directions in a double-pipe heat exchanger, making it an efficient heat transfer machine.
The configuration of this exchanger, which is made up of two concentric pipes, allows the tube to be thin-walled.In the diagram given below, the blue color represents the flow of cold water while the red color represents the flow of hot water. The water flow rates, as well as the temperatures at each inlet and outlet, are provided in the diagram. The shell side is cold water while the tube side is hot water. Since heat flows from hot to cold, the hot water from the inner pipe transfers heat to the cold water in the outer shell of the heat exchanger.
Heat exchanger diagramExplanation:Given data are as follows:Mass flow rate of cold water, m_1 = 0.25 kg/sTemperature of cold water at the inlet, T_1 = 15°CTemperature of cold water at the outlet, T_2 = 45°CMass flow rate of hot water, m_2 = 3 kg/sTemperature of hot water at the inlet, T_3 = 100°CThe rate of heat transfer,
[tex]Q = m_1C_{p1}(T_2 - T_1) = m_2C_{p2}(T_3 - T_4)[/tex]
where, C_p1 and C_p2 are the specific heat capacities of cold and hot water, respectively.Substituting the given values of [tex]m_1, C_p1, T_1, T_2, m_2, C_p2, and T_3[/tex], we get
[tex]Q = 0.25 × 4.18 × (45 - 15) × 1000= 31,350 Joules/s or 31.35 kJ/s[/tex]
Therefore,
[tex]m_2C_{p2}(T_3 - T_4) = Q = 31.35 kJ/s[/tex]
Substituting the given values of m_2, C_p2, T_3, and Q, we get
[tex]31.35 = 3 × 4.18 × (100 - T_4)0.25 = 3.75 - 0.0315(T_4)T_4 = 75°C[/tex]
The hot water at the outlet has a temperature of 75°C.
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An airplane flying at an altitude of z=2000 m with a horizontal velocity V=120 km/h pulls an advertising banner with a height of h=3 m and a length of l=5m. If the banner acts as a smooth flat plate, find the following a. The critical length (Xcr) in meters b. Drag coefficient of the banner c. Drag force acting on the banner in Newtons d. The power required to overcome banner drag in Watts
Given: Altitude of the airplane, z = 2000m
Horizontal velocity of airplane, V = 120 km/h = 33.33 m/s
Height of the banner, h = 3 m
Length of the banner, l = 5 m
Density of the air, ρ = 1.23 kg/m³
Dynamic viscosity of air, μ = 1.82 × 10⁻⁵ kg/m-s
Part (a): Critical length of the banner (Xcr) is given as:
Xcr = 5.0h
= 5.0 × 3.0
= 15.0 m
Part (b):The drag coefficient (Cd) is given as:
Cd = (2Fd)/(ρAV²) ... (1)Where,
Fd is the drag force acting on the banner in Newtons
A is the area of the banner in m²V is the velocity of airplane in m/s
From Bernoulli's equation,The velocity of air flowing over the top of the banner will be more than the velocity of air flowing below the banner.
As a result, the air pressure on top of the banner will be lesser than the air pressure below the banner. This produces a net upward force on the banner called lift.
To simplify the problem, we can ignore the lift forces and assume that the banner acts as a smooth flat plate.
Now the drag force acting on the banner is given as:
Fd = (1/2)ρCDAV² ... (2)
where, Cd is the drag coefficient of the banner.
A is the area of the banner
= hl
= 3.0 × 5.0
= 15.0 m²
Substituting equation (2) in (1),
Cd = (2Fd)/(ρAV²)
= (2 × (1/2)ρCDAV²)/(ρAV²)Cd
= 2(Cd)/(A)V²
From equation (2),
Fd = (1/2)ρCDAV²
Substituting the values, Cd = 0.603
Part (c):The drag force acting on the banner is given as:
Fd = (1/2)ρCDAV²
Substituting the values, we get;
Fd = (1/2) × 1.23 × 0.603 × 15.0 × 33.33²
= 1480.0 N
Part (d):The power required to overcome the banner drag is given by:
P = FdV = 1480.0 × 33.33 = 49331.4 WP
= 49.3 kW
Given the altitude and horizontal velocity of an airplane along with the banner's length and height, we found the critical length, drag coefficient, drag force and power required to overcome the banner drag.
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3. (30pts) Given the displacement filed u₁ = (3X²³X₂ +6)×10-² u₂ = (X² +6X₁X₂)×10-² u3 = (6X² +2X₂X₂ +10)x10-² 1) 1) Obtain Green strain tensor E at a point (1,0,2) 2) What is the extension of a line at this point? (Note: initial length and orientation of the line is dx₁) 3) What is the rotation of this line?
Given the displacement filed [tex]u₁ = (3X²³X₂ +6)×10-² u₂ = (X² +6X₁X₂)×10-² u3 = (6X² +2X₂X₂ +10)x10-²[/tex]To find Green strain tensor E at a point (1,0,2).
The Green-Lagrange strain tensor, E is defined as:E = ½(F^T F - I)Where F is the deformation gradient tensor and I is the identity tensor.The deformation gradient tensor, F is given by:F = I + ∇uwhere u is the displacement vector.In the given displacement field.
The components of displacement vector are given by:[tex]u₁ = (3X²³X₂ +6)×10-²u₂ = (X² +6X₁X₂)×10-²u₃ = (6X² +2X₂X₂ +10)x10-²[/tex]Therefore, the displacement vector is given by[tex]:u = (3X²³X₂ +6)×10-² i + (X² +6X₁X₂)×10-² j + (6X² +2X₂X₂ +10)x10-² k∇u = ∂u/∂X[/tex]From the displacement field.
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A spherical tank used for the storage of high-temperature gas has an outer radius of 5 m and is covered in an insulation 250 mm thick. The thermal conductivity of the insulation is 0.05 W/m-K. The temperature at the surface of the steel is 360°C and the surface temperature of the insulation is 40°C. Calculate the heat loss. Round off your final answer to two (2) decimal places. (20 pts.)
A spherical tank is used for the storage of high-temperature gas. It has an outer radius of 5 m and is covered with insulation 250 mm thick. The thermal conductivity of the insulation is 0.05 W/m-K. The temperature at the surface of the steel is 360°C and the surface temperature of the insulation is 40°C.
[tex]q = 4πk (T1 - T2) / [1/r1 - 1/r2 + (t2 - t1)/ln(r2/r1)][/tex]
Here,
q = heat loss
k = thermal conductivity = 0.05 W/m-K
T1 = temperature at the surface of the steel = 360°C
T2 = surface temperature of insulation = 40°C
r1 = outer radius of the tank = 5 m
r2 = radius of the insulation = 5 m + 0.25 m = 5.25 m
t1 = thickness of the tank = 0 m (as it is neglected)
t2 = thickness of the insulation = 0.25 m
Substituting these values in the above equation, we get:
q = 4π(0.05)(360 - 40) / [1/5 - 1/5.25 + (0.25)/ln(5.25/5)]
q = 605.52 W
Therefore, the heat loss is 605.52 W.
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For the following iron-carbon alloys (0.76 wt%C) and associated microstructures
A. coarse pearlite B. spheroidite C. fine pearlite D. bainite E. martensite F. tempered martensite 1. Select the most ductile 2. Select the hardest 3. Select the one with the best combination of strength and ductility.
For the following iron-carbon alloys (0.76 wt%C) and associated microstructures:A. coarse pearlite B. spheroidite C. fine pearlite D. bainite E. martensite F. tempered martensite1. Select the most ductileWhen the alloy has a coarse pearlite structure, it is the most ductile.2. Select the hardestWhen the alloy has a martensite structure, it is the hardest.
3. Select the one with the best combination of strength and ductilityWhen the alloy has a fine pearlite structure, it has the best combination of strength and ductility.Explanation:Pearlite: it is the most basic form of steel microstructure that consists of alternating layers of alpha-ferrite and cementite, in which cementite exists in lamellar form.Bainite: Bainite microstructure is a transitional phase between austenite and pearlite.Spheroidite: It is formed by further heat treating pearlite or tempered martensite at a temperature just below the eutectoid temperature.
This leads to the development of roughly spherical cementite particles within a ferrite matrix.Martensite: A solid solution of carbon in iron that is metastable and supersaturated at room temperature. Martensite is created when austenite is quenched rapidly.Tempered martensite: Tempered martensite is martensite that has been subjected to a tempering process.
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