Given data:
Diameter of pipe = 100 mm
Velocity of water = 2 m/s
We have to find out the flow rate in liters/hrSolution:
Formula to calculate flow rate,Q = A × v
Where,Q = Flow rate
A = Area of pipe
v = Velocity of water
Area of pipe,A = π/4 × D²A = π/4 × (100 mm)²A = 7.85 × 10⁻⁴ m²
Flow rate,Q = A × vQ = 7.85 × 10⁻⁴ m² × 2 m/sQ = 1.57 × 10⁻³ m³/s
Convert the above unit of flow rate in liters/hr1 m³/s = 1000 L/sQ = 1.57 × 10⁻³ × 1000 L/sQ = 1.57 L/s
Now, Convert L/s to L/hr1 L/s = 3600 L/hrQ = 1.57 L/s × 3600 L/hrQ = 5,652 L/hr
Hence, the flow rate of water in liters/hr is 5,652 L/hr.
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The reversed heat engine takes heat from a temperature body and an inward flow of network. high, low, gives O high, low, receives O low, high, gives O low, high, receives temperature body, then discha
Reversed heat engine is a device that operates in a thermodynamic cycle by taking in heat from a colder body and releasing it to a hotter body without an input of work. The cycle for a reversed heat engine is quite different from that of a heat engine because the direction of heat transfer is opposite.
In the cycle of a reversed heat engine, heat flows from a low-temperature body to a high-temperature body with the aid of an input of work. The reversed heat engine absorbs heat from a low-temperature reservoir and discharges it into a high-temperature reservoir through an energy input. It generates a net output of work instead of consuming it.The reversed heat engine works in the opposite direction of a heat engine, meaning that it takes in heat from a low-temperature body and exhausts heat to a high-temperature body. The primary difference between the two is that heat engines absorb heat from high-temperature reservoirs and discharge it into low-temperature reservoirs, resulting in a net output of work, while reversed heat engines absorb heat from low-temperature reservoirs and discharge it into high-temperature reservoirs, resulting in a net input of work.
The reversed heat engine absorbs heat from a low-temperature reservoir and discharges it into a high-temperature reservoir through an energy input. It generates a net output of work instead of consuming it. Therefore, a reversed heat engine is a device that operates on a thermodynamic cycle by taking in heat from a colder body and releasing it to a hotter body without an input of work.
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A triple-thread Acme screw is used in a jack to raise a load of 4000 lb. The major diameter of the screw is 3 in. A plain thrust collar is used. The mean diameter of the collar is 4 in. The coefficient of friction of the thread is 0.08 and the coefficient of collar friction is 0.1. Determine: a) the screw pitch, lead, thread depth, mean pitch diameter, and helix angle; b) the start- ing moment for raising and for lowering the load; c) the efficiency of the jack. A bolt made from steel has the stiffness ky. Two steel plates are held together by the bolt and have a stiffness kc. The elasticities are such that ke = 7 kb. The plates and the bolt have the same length. The external joint separating force fluctuates continuously between 0 and 2500 lb. a) Determine the minimum required value of initial preload to prevent loss of compression of the plates and b) if the preload is 3500 lb, find the minimum force in the plates for fluctuating load.
To determine the screw characteristics and the performance of the jack, as well as the required preload and minimum force in the plates, the following steps need to be taken:
Screw Analysis: Calculate the screw pitch, lead, thread depth, mean pitch diameter, and helix angle based on the given information about the screw and collar dimensions.
Starting Moment: Determine the starting moment for raising and lowering the load by considering the frictional forces acting on the screw and collar.
Efficiency Calculation: Calculate the efficiency of the jack by comparing the output work (load raised) to the input work (applied force multiplied by the distance moved).
Preload Requirement: Determine the minimum required value of initial preload to prevent loss of compression in the plates by considering the fluctuating joint separating force and the stiffness of the bolt and plates.
Minimum Force in Plates: Calculate the minimum force in the plates for the fluctuating load by considering the preload and the fluctuating joint separating force.
The first step involves analyzing the screw to determine its pitch, lead, thread depth, mean pitch diameter, and helix angle. These parameters are crucial for understanding the screw's geometry and performance.
The starting moment is calculated by considering the frictional forces acting on the screw and collar. The coefficient of friction for both the thread and collar is provided, which allows for the determination of the forces involved.
The efficiency of the jack is determined by comparing the output work (the load raised) to the input work (the force applied to the screw multiplied by the distance moved).
To prevent loss of compression in the plates, the minimum required preload needs to be calculated. This involves considering the fluctuating joint separating force and the stiffness of the bolt and plates.
Finally, with a known preload, the minimum force in the plates for the fluctuating load can be determined by accounting for the preload and the varying joint separating force.
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Braze welding is a gas welding technique in which the base metal A. does not usually require controlled heat input. B. liquefies a t a temperature above 1800°F. C. does not melt during the welding. D. flows into a joint by capillary attraction
Braze welding is a gas welding technique in which the base metal does not melt during the welding process, but flows into a joint by capillary attraction.
Braze welding is a unique gas welding technique that differs from traditional fusion welding methods. Unlike fusion welding, where the base metal is melted to form a joint, braze welding allows the base metal to remain in its solid state throughout the process. Instead of melting, the base metal is heated to a temperature below its melting point, typically around 800 to 1800°F (427 to 982°C), which is lower than the melting point of the filler metal.
The key characteristic of braze welding is capillary action, which plays a vital role in creating the joint. Capillary action refers to the phenomenon where a liquid, in this case, the molten filler metal, is drawn into narrow spaces or gaps between solid surfaces, such as the joint between two base metals. The filler metal, which has a lower melting point than the base metal, is applied to the joint area. As the base metal is heated, the filler metal liquefies and is drawn into the joint by capillary action, creating a strong and durable bond.
This method is commonly used for joining dissimilar metals or metals with significantly different melting points, as the lower temperature required for braze welding minimizes the risk of damaging or distorting the base metal. Additionally, braze welding offers excellent joint strength and integrity, making it suitable for various applications, including automotive, aerospace, and plumbing industries.
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challenges and solutions for integration of wind power in weak grid areas with high inverter penetration
Integrating wind power in weak grid areas with high inverter penetration can pose several challenges. However, there are also solutions available to address these challenges. Here are some of the key challenges and their corresponding solutions:
1. Grid Stability: Weak grids may struggle to handle the intermittent nature of wind power and the high penetration of inverters, leading to voltage fluctuations and instability.
2. Power Quality Issues: High inverter penetration can result in power quality problems such as harmonic distortion, voltage flicker, and reactive power imbalances.
3. Grid Capacity Limitations: Weak grids may have limited transmission and distribution capacity, making it challenging to accommodate the additional wind power generation.
4. Frequency Control and Ancillary Services: High inverter penetration can impact frequency control and the provision of ancillary services in weak grid areas.
5. System Planning and Forecasting: Accurate forecasting of wind power generation and effective system planning are crucial for integration in weak grid areas.
Solutions:
1. Grid Stability: Implement grid codes and standards, utilize advanced power electronics and control strategies to enhance grid stability and voltage regulation.
2. Power Quality Issues: Deploy advanced power conditioning systems and inverters that comply with grid codes, employ filtering and harmonic mitigation techniques to ensure power quality compliance.
3. Grid Capacity Limitations: Conduct grid studies, reinforce infrastructure, upgrade transmission lines, transformers, and substation equipment to enhance grid capacity.
4. Frequency Control and Ancillary Services: Develop advanced control strategies, employ energy storage systems, and establish effective coordination between wind power plants and grid operators.
5. System Planning and Forecasting: Use accurate wind power forecasting tools, conduct comprehensive system planning studies, and incorporate data analytics techniques for optimized utilization of wind power and improved grid integration.
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Integrating wind power in weak grid areas with high inverter penetration can pose several challenges. However, there are also solutions available to address these challenges. Here are some of the key challenges and their corresponding solutions:
1. Grid Stability: Weak grids may struggle to handle the intermittent nature of wind power and the high penetration of inverters, leading to voltage fluctuations and instability.
2. Power Quality Issues: High inverter penetration can result in power quality problems such as harmonic distortion, voltage flicker, and reactive power imbalances.
3. Grid Capacity Limitations: Weak grids may have limited transmission and distribution capacity, making it challenging to accommodate the additional wind power generation.
4. Frequency Control and Ancillary Services: High inverter penetration can impact frequency control and the provision of ancillary services in weak grid areas.
5. System Planning and Forecasting: Accurate forecasting of wind power generation and effective system planning are crucial for integration in weak grid areas.
Solutions:
1. Grid Stability: Implement grid codes and standards, utilize advanced power electronics and control strategies to enhance grid stability and voltage regulation.
2. Power Quality Issues: Deploy advanced power conditioning systems and inverters that comply with grid codes, employ filtering and harmonic mitigation techniques to ensure power quality compliance.
3. Grid Capacity Limitations: Conduct grid studies, reinforce infrastructure, upgrade transmission lines, transformers, and substation equipment to enhance grid capacity.
4. Frequency Control and Ancillary Services: Develop advanced control strategies, employ energy storage systems, and establish effective coordination between wind power plants and grid operators.
5. System Planning and Forecasting: Use accurate wind power forecasting tools, conduct comprehensive system planning studies, and incorporate data analytics techniques for optimized utilization of wind power and improved grid integration.
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An axial-flow fan operates in seal-level air at 1350 rpm and has a blade tip diameter of 3 ft and a root diameter of 2.5 ft. The inlet angles are a₁ = 55°, β₁ = 30°, and at the exit β₂= 60°. Estimate the flow volumetric flow rate, horsepower, and the outlet angle, a₂
While the volumetric flow rate can be estimated based on the given information, accurate estimations of horsepower and the outlet angle cannot be calculated without additional details such as the pressure difference across the fan and the area at the outlet.
To estimate the flow volumetric flow rate, horsepower, and the outlet angle, we can use the following formulas and calculations:
1. Flow Volumetric Flow Rate (Q):
The volumetric flow rate can be estimated using the formula:
Q = π * D₁ * V₁ * cos(a₁)
Where:
- Q is the volumetric flow rate.
- D₁ is the blade tip diameter.
- V₁ is the velocity at the inlet.
- a₁ is the inlet angle.
2. Horsepower (HP):
The horsepower can be estimated using the formula:
HP = (Q * ΔP) / (6356 * η)
Where:
- HP is the horsepower.
- Q is the volumetric flow rate.
- ΔP is the pressure difference across the fan.
- η is the fan efficiency.
3. Outlet Angle (a₂):
The outlet angle can be estimated using the formula:
a₂ = β₂ - (180° - a₁)
Where:
- a₂ is the outlet angle.
- β₂ is the exit angle.
- a₁ is the inlet angle.
Given the provided information, let's calculate these values:
1. Flow Volumetric Flow Rate (Q):
D₁ = 3 ft
V₁ = (1350 rpm) * (2.5 ft) / 60 = 56.25 ft/s
a₁ = 55°
Q = π * (3 ft) * (56.25 ft/s) * cos(55°) ≈ 472.81 ft³/s
2. Horsepower (HP):
Let's assume a pressure difference of ΔP = 1 psi (pound per square inch) and a fan efficiency of η = 0.75.
HP = (472.81 ft³/s * 1 psi) / (6356 * 0.75) ≈ 0.111 hp
3. Outlet Angle (a₂):
β₂ = 60°
a₂ = 60° - (180° - 55°) = -65° (assuming counterclockwise rotation)
Please note that these calculations are estimates based on the given information and assumptions. Actual values may vary depending on various factors and the specific design of the axial-flow fan.
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A microwave link requires participating stations to have a minimum G/T ratio given by: G/T = 20.5+20 log10 (f/4)dBK¹ where f is the frequency in GHz. Operation is at 4 GHz with a terminal consisting of an antenna followed by a waveguide of physical temperature 300 K with 0.6 dB loss and an amplifier with a 1.0 dB noise figure and 15 dB gain. The amplifier drives a mixer with a 7 dB noise figure and 30 dB gain. Following the mixer is an IF receiver with a 12 dB noise figure. Calculate the minimum antenna diameter required to meet the G/T specification. Assume that the antenna noise temperature is 15 K independent of diameter and that the aperture efficiency is 65 percent. All noise figures are standard with To = 290 K. (10 Marks)
To calculate the minimum antenna diameter required to meet the G/T specification, we can follow these steps:
Calculate the system temperature (Ts) at the input to the mixer:
Ts = Tant + Twaveguide + Tamplifier + Tmixer + TIF_receiver
Tant = Tant_noise + Tant_sky
Tant_noise = T0 * (1 - aperture_efficiency)
Tant_sky = Tsky * aperture_efficiency
Tsky = 15 K
T0 = 290 K (standard reference temperature)
Tant = T0 * (1 - aperture_efficiency) + Tsky * aperture_efficiency
Calculate the noise power (N) at the input to the mixer:
N = k * Ts * bandwidth
k = Boltzmann constant (1.380649 x 10^-23 J/K)
bandwidth = frequency of operation (4 GHz)
Calculate the available power (Pavailable) at the input to the mixer:
Pavailable = Preceived * Gantenna
Greceived = Gantenna * (λ / (4πR))^2
Greceived = (π * (d/λ)^2 * ηaperture) * (λ / (4πR))^2
R = distance from the transmitter to the receiver
λ = speed of light / frequency of operation
ηaperture = aperture efficiency
Preceived = Pt * (λ / (4πR))^2
Calculate the G/T ratio:
G/T = 20.5 + 20 * log10(f/4) + 10 * log10(Pavailable/N)
Solve for the minimum antenna diameter (d) that satisfies the G/T ratio requirement.
Using these steps, we can write the MATLAB code to calculate the minimum antenna diameter:
% Constants
T0 = 290; % K
Tsky = 15; % K
f = 4; % GHz
k = 1.380649e-23; % J/K
bandwidth = f * 1e9; % Hz
R = 1; % Assumed distance (arbitrary value)
aperture_efficiency = 0.65;
% Calculate antenna diameter
d = 0; % Initialize diameter
while true
d = d + 0.1; % Increment diameter
lambda = 3e8 / (f * 1e9); % Speed of light / frequency
Pt = 1; % Assumed transmitted power (arbitrary value)
% Calculate antenna gain
Gantenna = pi * (d / lambda)^2 * aperture_efficiency;
% Calculate received power
Preceived = Pt * (lambda / (4 * pi * R))^2;
% Calculate system temperature
Tant_noise = T0 * (1 - aperture_efficiency);
Tant_sky = Tsky * aperture_efficiency;
Ts = Tant_noise + Tant_sky;
% Calculate noise power
N = k * Ts * bandwidth;
% Calculate available power
Pavailable = Preceived * Gantenna;
% Calculate G/T ratio
G_T = 20.5 + 20 * log10(f/4) + 10 * log10(Pavailable/N);
% Check if G/T ratio meets the specification
if G_T >= 10
break; % Antenna diameter meets the specification
end
end
fprintf('Minimum antenna diameter required: %.1f meters\n', d);
Note: The code assumes an arbitrary distance of 1 meter (R) between the transmitter and the receiver. You can adjust this value as per your specific scenario.
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Experiment 1 --Pulse Train Generation Objective: The experiment is designed to exhibit some of the capabilities of generating pulses at the ports of the PIC18F-4520. Specific Tasks: Use the PO_template.asm file as a starting point in order to complete the following tasks. Write an assembly program that generates a pulse train simultaneously at bit RBI of PORTB and bit RE2 of PORTE for each of the following cases. a TT-0.1ms b. Tu 0.1ms, T₁=0.3ms 2. Write an assembly program that generates a pulse train simultaneously at bit RBI of PORTB and bit RE2 of PORTE for each of the following cases. a. Tu T₁=0.5ms b. Tu 0.9ms, T₁=0.7ms By the end of this assignment you should have written four (4) different.asm files. Notes: Please observe the following items while working through this assignment. 1. Do not use the Prescaler option. 2. Do not modify any line of code of the LoopTime subroutine in the PO_template.asm file. 3. Do not call the LoopTime subroutine more than once. 4. Do not create a second subroutine similar to Loop Time. Guidelines: There are four distinct phases for the pulse train generation: 1. Configure the pins of the PORT as outputs. 2. Send your pulse to the configured pins using appropriate commands (review the following commands: big, bsf, bef). Use conditional logic and a "delay" mechanism to control the duty cycle of the pulse trains. 3. 4. Loop around indefinitely to generate a continuous pulse train. To read or write anything from a port, it is necessary to configure the port pins accordingly, using the Special Function Register (SFR) TRISx, where x is the name of the port (e.g. TRISB configures the pins of port B). In order to configure a pin as an output, put a 0 in the corresponding bit of TRISx. Similarly, to configure a pin as an input, put a 1 in the corresponding bit. Example: to configure pin 4 of port B as an output and the others as inputs: MOVLF B'11101111, TRISB where MOVLF is a macro defined as follows: MOVLF macro literal. dest movlw literal movwf dest endm
The objective of the experiment is to generate pulse trains simultaneously at specific ports of the PIC18F-4520 microcontroller. The task involves writing assembly programs to generate pulse trains for different cases, using the provided template file. By the end of the assignment, four different .asm files should be written.
To accomplish the objective, the following steps need to be followed. First, the pins of the PORT (PORTB and PORTE) must be configured as outputs using the TRISx register. This ensures that the selected pins can generate output signals.
Next, appropriate commands such as bsf (bit set), bcf (bit clear), and bef (bit complement) can be used to send pulses to the configured pins. The duty cycle of the pulse trains can be controlled using conditional logic and a delay mechanism. This allows for generating pulses with specific timing characteristics as required by each case.
The assembly program should be structured to loop indefinitely, continuously generating the pulse trains. This ensures that the pulse generation continues as long as the microcontroller is powered on.
It is important to follow the provided guidelines and avoid modifying certain sections of the code or using certain features such as the prescaler option. This ensures consistency and allows for accurate evaluation of the generated pulse trains.
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Problem 2 Assume that the field current of the generator in Problem 1 has been adjusted to a value of 4.5 A. a) What will the terminal voltage of this generator be if it is connected to a A-connected load with an impedance of 20230 ? b) Sketch the phasor diagram of this generator. c) What is the efficiency of the generator at these conditions? d) Now assume that another identical A-connected load is to be paralleled with the first one. What happens to the phasor diagram for the generator? e) What is the new terminal voltage after the load has been added? f) What must be done to restore the terminal voltage to its original value?
Analyzing the effects on terminal voltage, phasor diagram, efficiency, and voltage restoration involves considering load impedance, internal impedance, load current, and field current adjustments.
What factors should be considered when designing an effective supply chain strategy?In this problem, we are given a generator with an adjusted field current of 4.5 A.
We need to analyze the effects on the terminal voltage, phasor diagram, efficiency, and terminal voltage restoration when connected to a load and when adding another load in parallel.
To determine the terminal voltage when connected to an A-connected load with an impedance of 20230 Ω, we need to consider the generator's internal impedance and the load impedance to calculate the voltage drop.
By applying appropriate equations, we can find the terminal voltage.
Sketching the phasor diagram of the generator involves representing the generator's voltage, internal impedance, load impedance, and current phasors.
The phasor diagram shows the relationships between these quantities.
The efficiency of the generator at these conditions can be calculated by dividing the power output (product of the terminal voltage and load current) by the power input (product of the field current and generator voltage).
This ratio represents the efficiency of the generator.
When paralleling another identical A-connected load, the phasor diagram for the generator changes.
The load current will increase, affecting the overall current distribution and phase relationships in the system.
The new terminal voltage after adding the load can be determined by considering the increased load current and the generator's ability to maintain the desired terminal voltage.
The voltage drop across the internal impedance and load impedance will impact the new terminal voltage
By increasing or decreasing the field current, the magnetic field strength and consequently the terminal voltage can be adjusted to its original value.
Calculations and understanding of phasor relationships are key in addressing these aspects.
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The purpose of the inclining experiment is to find the: a Metacentric radius. b Vertical centre of gravity. c Longitudinal centre of gravity.
The purpose of the inclining experiment is to find the metacentric radius.
An inclining experiment is a trial carried out to establish the position of a vessel's center of gravity in relation to its longitudinal and transverse axes. This test is necessary since the precise location of the center of gravity determines the vessel's stability when it heels to one side or the other.
The objective of the inclining experiment is to establish the metacentric radius of a vessel. The metacentric radius is the distance between the center of gravity and the metacenter, which is the position of the intersection of the center of buoyancy and the center of gravity when the vessel is inclined to a small angle. The value of the metacentric radius determines a vessel's stability; a greater metacentric radius means a more stable vessel while a lesser metacentric radius means a less stable vessel. It's critical to establish the metacentric radius since it's necessary to know how much weight may be added or removed to maintain a ship's stability. The inclining experiment also establishes the vessel's longitudinal and vertical centers of gravity.
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QUESTION 17 Which of the followings is true? O A. For a full inductor, at time t=0 when it is switched on, its through current will likely drop to half its value. OB. For a full capacitor, at time t=0 when it is switched on, its across voltage will be close to zero. O C. For an empty inductor, at time t=0 when it is switched on, its through current will be close to zero. O D. For a full inductor, at time t=0 when it is switched on, its through current will likely drop to quarter its value.
The correct answer is C. For an empty inductor, at time t=0 when it is switched on, its through current will be close to zero.
When an inductor is empty or has no stored energy, its initial through current will be close to zero when it is switched on at time t=0. This is because an inductor resists changes in current, and when it is initially switched on, there is no established current flow through it. Therefore, the through current will gradually increase over time as the inductor builds up its magnetic field.
Option A is incorrect because a full inductor, which has a significant amount of stored energy, will not likely have its through current drop to half its value when switched on at time t=0.
Option B is incorrect because for a full capacitor, when it is switched on at time t=0, the across voltage will not be close to zero. A fully charged capacitor will have a voltage across it equal to the voltage at the time of charging.
Option D is incorrect because it mentions the behavior of a full inductor, which is not relevant to the question being asked.
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In an ideal regenarative cycle, the turbine receives 1 kg/hr of steam at 10MPa and 600deg C. Before condesation at 0.005MPa, the steam was extracted for feedwater heating at 2.5MPa. Find the quantity of steam extracted for feedwater heating, b)Heat addition to the boiler (Qa), c)Wnet, d)Cycle Thermal Efficiency, e)Engine Thermal Efficiency, f) The combined engine efficiency at 3.43kg/kWhr combined steam rate. draw the cycle schematic diagram and TS plane.
a) The quantity of steam extracted for feedwater heating is 1.07 kg/hr.
b) The heat added to the boiler is 3343.68 kJ/hr.
c) The engine thermal efficiency is 73% and d) the combined engine efficiency is 21.25%.
We have
Pressure (P1) = 10 MPa
Temperature (T1) = 600 °C
Pressure (P2) at which steam is extracted for feedwater heating = 2.5 MPa
Condenser Pressure (P3) = 0.005 MPa
Mass flow rate (m) = 1 kg/hr
a) To Find the quantity of steam extracted for feedwater heating:
Let the mass of steam extracted for feedwater heating be x kg/hr
Mass of steam flowing through the turbine (m) = 1 kg/hr
Mass of steam flowing through the turbine after extraction for feedwater heating = (m - x) kg/hr
Let h1, h2, h3 and h4 be the specific enthalpies at points 1, 2, 3 and 4 respectively
From the steam table, we get:
h1 = 3583.2 kJ/kg
h2 = 3309.8 kJ/kg
h3 = 191.81 kJ/kg
h4 = 239.52 kJ/kg
Heat supplied to the turbine = m (h1 - h4)
Heat supplied to the turbine = (1) (3583.2 - 239.52) = 3343.68 kJ/hr
Heat extracted at the extraction point for feedwater heating = x (h2 - h3)
Heat extracted at the extraction point for feedwater heating = (x) (3309.8 - 191.81) = 3117.99 x kJ/hr
Therefore, 3343.68 = 3117.99 x
x = 1.07 kg/hr
Therefore, the quantity of steam extracted for feedwater heating is 1.07 kg/hr.
b) Heat addition to the boiler:
Heat added to the system (Qin) = m (h1 - h4) = (1) (3583.2 - 239.52) = 3343.68 kJ/hr
Therefore, the heat added to the boiler is 3343.68 kJ/hr.
Heat supplied, Q₁ = m (h₁ - h₃) = 1 (4024.3 - 191.82) = 3832.48 kJ/hr
Heat extracted, Q₂ = m [(h₁ - h₂) + x₂ (h₂ - h₃)] = 1 [(4024.3 - 2996.8) + 0.923(2996.8 - 191.82)] = 1030.89 kJ/hr
Net work done, W = Q₁ - Q₂ = 2801.59 kJ/hr
Now, to calculate the engine's thermal efficiency:
Engine thermal efficiency, ηₑ = W/Q₁ = 2801.59/3832.48 = 0.73 or 73%
Combined engine efficiency = ηₑ / m' = 0.73 / 3.43 = 0.2125 or 21.25%
Therefore, the engine thermal efficiency is 73% and the combined engine efficiency is 21.25%.
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Question 5 Not yet answered Marked out of 10.00 Flag question In ideal diode model when the its forwardly biased Select one: O a. It acts like a short circuit b. c. O d. its acts passive voltage source it acts like and open circuit None of the answers
The ideal diode model behaves as a short circuit when it is forward-biased and as an open circuit when it is reverse-biased.
The ideal diode model is used to describe the basic behavior of a diode. A diode is an electronic component that permits current to flow in only one direction. A diode consists of two terminals known as the anode and the cathode. In an ideal diode model, the forward-biased diode acts like a short circuit.
When the forward voltage across the diode is greater than the diode's forward voltage drop, the diode turns on and behaves like a short circuit. This implies that current flows effortlessly through the diode.
In other words, when a diode is forward-biased, current flows through it. In a forward-biased diode, the diode's anode is connected to the positive end of the voltage source, while the cathode is connected to the negative end of the voltage source.
If a reverse voltage is applied to a forward-biased diode, the diode behaves as an open circuit. This means that current does not flow through the diode. An open circuit is one in which no current flows through it. In other words, the diode is inoperative.
Therefore, the ideal diode model behaves as a short circuit when it is forward-biased and as an open circuit when it is reverse-biased. This behavior makes the diode an essential component of modern electronic circuits.
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QUESTION 36 Which of the followings is true? To convert from sin(x) to cos(x), one would O A. add -180 degrees to the angle x. B. add 180 degrees to the angle x. C. add-90 degrees to the angle x. D. add 90 degrees to the angle x.
The correct statement is:D. To convert from sin(x) to cos(x), one would add 90 degrees to the angle x.
In trigonometry, the sine and cosine functions are related through a phase shift of 90 degrees (or π/2 radians). To convert from sin(x) to cos(x), you add a phase shift of 90 degrees to the angle x. This shift changes the phase relationship between the sine and cosine functions and effectively converts the sinusoidal function from sine to cosine.
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A manufacturer conducted an experiment for an evaporator capacity 500 kW cooling and designed for high COP of 2 when using lithium bromide plus water in an absorption refrigeration system. The evaporator operates 20 C, condenser 40 C & absorber 45 C supplying 1.37 kg/s of water plus lithium bromide solution to the generator. Concentration of the solution being pumped is found to be 52.7 % and the mass of the solution being throttled is found to be 1.180 kg/s. Determine:
Concentration and Enthalphy of the solution being throttled.
Show in your solution paper: Mass balance at the Generator
Provide in the answer box: % Concentration of solution being throttled
Answer in two decimal places.
The contracention of the solution being throttled is 52.70%.
The enthalpy of the solution being throttled is not provided in the question.
The concentration of the solution being throttled is given as 52.7%. This represents the percentage of lithium bromide in the solution that is being pumped.
The enthalpy of the solution being throttled is not provided in the given information. Enthalpy is a measure of the total energy content of a substance and is typically given in terms of energy per unit mass. Without the specific enthalpy value provided, it is not possible to determine the enthalpy of the solution being throttled.
To further analyze the system and determine the concentration and enthalpy of the solution being throttled, a mass balance at the generator is required. This balance would involve considering the mass flow rates of water and lithium bromide solution entering and leaving the generator, as well as any changes in concentration and enthalpy that occur during the process.
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The 3rd order Taylor polynomial for the function f(x) = 1 · x · sin (3 · x)
t x₁ = 1 is p(x) = P₀ + P₁ (x-x₁) + P₂ (x − ₁)² +p₃ (x − x₁)³
Give the values of P₀:
P₁:
P₂:
p₃:
The values of P₀, P₁, P₂, and p₃ for the 3rd order Taylor polynomial of the function f(x) = x · sin(3 · x) at x = 1 are:
P₀ = 0,
P₁ = 0,
P₂ = -1.5,
p₃ = 0.
What are the values of P₀, P₁, P₂, and p₃ for the 3rd order Taylor polynomial of the function f(x) = x · sin(3 · x) at x = 1?The 3rd order Taylor polynomial for the function f(x) = x · sin(3 · x) at x₁ = 1 is given by p(x) = P₀ + P₁(x - x₁) + P₂(x - x₁)² + p₃(x - x₁)³. To find the values of P₀, P₁, P₂, and p₃, we need to calculate the function and its derivatives at x = x₁.
At x = 1:
f(1) = 1 · sin(3 · 1) = sin(3) ≈ 0.141
f'(1) = (d/dx)[x · sin(3 · x)] = sin(3) + 3 · x · cos(3 · x) = sin(3) + 3 · 1 · cos(3) ≈ 0.141 + 3 · 0.998 ≈ 2.275
f''(1) = (d²/dx²)[x · sin(3 · x)] = 6 · cos(3 · x) - 9 · x · sin(3 · x) = 6 · cos(3) - 9 · 1 · sin(3) ≈ 6 · 0.998 - 9 · 0.141 ≈ 2.988
f'''(1) = (d³/dx³)[x · sin(3 · x)] = 9 · sin(3 · x) - 27 · x · cos(3 · x) = 9 · sin(3) - 27 · 1 · cos(3) ≈ 9 · 0.141 - 27 · 0.998 ≈ -23.067
Therefore, the values of the coefficients are:
P₀ ≈ 0.141
P₁ ≈ 2.275
P₂ ≈ 2.988
p₃ ≈ -23.067
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A model ball mill is driven by a diesel engine and with the drive transmitted through a flexible coupling, the mill experiences moderate shock loads. Calculate the design HP of a three-strand chain if the test site conditions are moderately dirty and temperatures are moderate too. It is anticipated that the mill will operate continuously for 9 hours per day. Assume a single strand rating of 30hp
The design HP (Horsepower) of the three-strand chain for the model ball mill can be calculated based on the given information. Assuming a single strand rating of 30 HP, we need to determine the total HP required for the three strands to handle the shock loads and operating conditions.
Since the mill operates continuously for 9 hours per day, we can multiply the single strand rating (30 HP) by the number of strands (3) and the operating hours (9) to obtain the design HP:
Design HP = Single strand rating * Number of strands * Operating hours
= 30 HP * 3 * 9
= 810 HP
Therefore, the design HP of the three-strand chain for the model ball mill, considering moderate shock loads and moderately dirty and moderate temperature conditions, is 810 HP.
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It is necessary to evacuate 49.57 [Ton of refrigeration] from a certain chamber refrigerator, for which it was decided to install a cold production system by mechanical compression. The chamber temperature cannot exceed –3[°C] and the temperature difference at the evaporator inlet is estimated at 7[°C].
You have a large flow of well water at 15[°C] that you plan to use as condensing agent. The refrigerant fluid used is R-134a.
For the operation of this installation, an alternative compressor was acquired. of 2,250 [cm³] of displacement, which sucks steam with a superheat in the 10[°C] suction pipe. This compressor rotates at 850[r.p.m.] and its volumetric efficiency is 0.8 for a compression ratio of 3.3.
Calculate the degree of subcooling of the condensed fluid so that it can
operate the installation with this compressor and if it is possible to carry it out.
Note: Consider a maximum admissible jump in the well water of 5[°C] and a minimum temperature jump in the condenser (between refrigerant fluid and water
of well) of 5[°C].
The degree of subcooling is 28°C, which is within the range of possible values for the system to operate.
The degree of subcooling is the difference between the temperature of the condensed refrigerant and the saturation temperature at the condenser pressure. A higher degree of subcooling will lead to a lower efficiency, but it is possible to operate the system with a degree of subcooling of 28°C. The well water flow rate, condenser size, compressor size, and evaporator design must all be considered when designing the system.
The degree of subcooling is important because it affects the efficiency of the system. A higher degree of subcooling will lead to a lower efficiency because the refrigerant will have more energy when it enters the expansion valve. This will cause the compressor to work harder and consume more power.
The well water flow rate must be sufficient to remove the heat from the condenser. If the well water flow rate is too low, the condenser will not be able to remove all of the heat from the refrigerant and the system will not operate properly.
The condenser must be sized to accommodate the well water flow rate. If the condenser is too small, the well water will not be able to flow through the condenser quickly enough and the system will not operate properly.
The compressor must be sized to handle the refrigerant mass flow rate. If the compressor is too small, the system will not be able to cool the chamber properly.
The evaporator must be designed to provide the desired cooling capacity. If the evaporator is too small, the system will not be able to cool the chamber properly.
It is important to consult with a refrigeration engineer to design a system that meets your specific needs.
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[60p] 2. Consider a discreate-time linear shift invariant (LSI) system for which the impulse response h[n] = u[n + 1] - u[n – 2). (a) Find the output of the system, y[n] for an input x[n] = 8[n] + [
The output of the system for the given input isy[n] = { 0 , n < -1 8 , n = -1 8 + 8n , -1 < n < 2 0 , n >= 2 }
From the question above, Discrete-time linear shift-invariant system
Impulse response, h[n] = u[n+1] - u[n-2]
Input, x[n] = 8[n] + ?
Output, y[n]
The output of the system is given by:
y[n] = x[n] * h[n]
where, * denotes the convolution operation.
x[n] = 8[n] + ?
h[n] = u[n+1] - u[n-2]
We know that, u[n] is the unit step sequence, given byu[n] = { 1 , n >= 0 0 , n < 0 }
Now, we can write
h[n] ash[n] = u[n+1] - u[n-2]
h[n] = { 0 , n < -1 1 , n = -1 1 , -1 < n < 2 0 , n >= 2 }
Therefore, the output, y[n] isy[n] = x[n] * h[n]y[n] = { 0 , n < -1 8 , n = -1 8 + 8n , -1 < n < 2 0 , n >= 2 }
Hence, the output of the system for the given input isy[n] = { 0 , n < -1 8 , n = -1 8 + 8n , -1 < n < 2 0 , n >= 2 }
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Which of the following statements is true for a mechanical energy reservoir (MER)? O stores work as KE or PE O all of the mentioned O all processes within an MER are quasi-static O it is a large body enclosed by an adiabatic impermeable wall
The statement "O all of the mentioned" is true for a mechanical energy reservoir (MER).
A mechanical energy reservoir is a system that stores mechanical energy in various forms such as kinetic energy (KE) or potential energy (PE). It acts as a source or sink of energy for mechanical processes.
In an MER, all processes are typically assumed to be quasi-static. Quasi-static processes are slow and occur in equilibrium, allowing the system to continuously adjust to external changes. This assumption simplifies the analysis and allows for the application of concepts like work and energy.
Lastly, an MER can be visualized as a large body enclosed by an adiabatic impermeable wall. This means that it does not exchange heat with its surroundings (adiabatic) and does not allow the transfer of mass across its boundaries (impermeable).
Therefore, all of the mentioned statements are true for a mechanical energy reservoir.
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A commercial enclosed gear drive consists of a 200 spur pinion having 16 teeth driving a 48-tooth gear. The pinion speed is 300 rev/min, the face width 2 in, and the diametral pitch 6 teeth/in. The gears are grade I steel, through-hardened at 200 Brinell, made to No. 6 quality standards, uncrowned, and are to be accurately and rigidly mounted. Assume a pinion life of 10^8 cycles and a reliability of 0.90. If 5 hp is to be transmitted. Determine the following: a. Pitch diameter of the pinion b. Pitch line velocity c. Tangential transmitted force d. Dynamic factor e. Size factor of the gear f. Load-Distribution Factor g. Spur-Gear Geometry Factor for the pinion h. Taking ko =ka = 1, determine gear bending stress
a. Pitch diameter of the pinion = 2.67 in
b. Pitch line velocity= 167.33 fpm
c. Tangential transmitted force = 1881 lb
d. Dynamic factor = 0.526
e. Size factor of the gear Ks = 1.599
f. Load-Distribution Factor K = 1.742
g. Spur-Gear Geometry Factor for the pinion Kg = 1.572
h. Taking ko =ka = 1, determine gear bending stress σb = 2097.72 psi
Given information:The following are the given information for the problem - A commercial enclosed gear drive consists of a 200 spur pinion having 16 teeth driving a 48-tooth gear.
The pinion speed is 300 rev/min.The face width is 2 in.The diametral pitch is 6 teeth/in.
The gears are grade I steel, through-hardened at 200 Brinell, made to No. 6 quality standards, uncrowned, and are to be accurately and rigidly mounted.
Assume a pinion life of 108 cycles and a reliability of 0.90.
If 5 hp is to be transmitted.
To determine:
We are to determine the following parameters:
a. Pitch diameter of the pinion
b. Pitch line velocity
c. Tangential transmitted force
d. Dynamic factor
e. Size factor of the gear
f. Load-Distribution Factor
g. Spur-Gear Geometry Factor for the pinion
h. Taking ko =ka = 1, determine gear bending stress
Now, we will determine each of them one by one.
a. Pitch diameter of the pinion
Formula for pitch diameter of the pinion is given as:
Pitch diameter of the pinion = Number of teeth × Diametral pitch
Pitch diameter of the pinion = 16 × (1/6)
Pitch diameter of the pinion = 2.67 in
b. Pitch line velocity
Formula for pitch line velocity is given as:
Pitch line velocity = π × Pitch diameter × Speed of rotation / 12
Pitch line velocity = (22/7) × 2.67 × 300 / 12
Pitch line velocity = 167.33 fpm
c. Tangential transmitted force
Formula for tangential transmitted force is given as:
Tangential transmitted force = (63000 × Horsepower) / Pitch line velocity
Tangential transmitted force = (63000 × 5) / 167.33
Tangential transmitted force = 1881 lb
d. Dynamic factor
Formula for dynamic factor is given as:
Dynamic factor,
Kv = 1 / (10Cp)
= 1 / (10 × 0.19)
= 0.526
e. Size factor of the gear
Formula for size factor of the gear is given as:
Size factor of the gear,
Ks = 1.4(Pd)0.037
Size factor of the gear,
Ks = 1.4(2.67)0.037
Size factor of the gear,
Ks = 1.4 × 1.142
Size factor of the gear, Ks = 1.599
f. Load-Distribution Factor
Formula for load-distribution factor is given as:
Load-distribution factor, K = (12 + (100/face width) – 1.5(Pd)) / (10 × 1.25(Pd))
Load-distribution factor, K = (12 + (100/2) – 1.5(2.67)) / (10 × 1.25(2.67))
Load-distribution factor, K = 1.742
g. Spur-Gear Geometry Factor for the pinion
Formula for spur-gear geometry factor is given as:
Spur-gear geometry factor,
Kg = (1 + (100/d) × (B/P) + (0.6/P) × (√(B/P))) / (1 + ((100/d) × (B/P)) / (2.75 + (√(B/P))))
Spur-gear geometry factor,
Kg = (1 + (100/2.67) × (2/6) + (0.6/6) × (√(2/6))) / (1 + ((100/2.67) × (2/6)) / (2.75 + (√(2/6)))))
Spur-gear geometry factor,
Kg = 1.572
h. Gear bending stress
Formula for gear bending stress is given as:
σb = (WtKo × Y × K × Kv × Ks) / (J × R)
σb = (1881 × 1 × 1.742 × 0.526 × 1.599) / (4.125 × 0.97)
σb = 2097.72 psi
Hence, all the required parameters are determined.
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A 220V, 5.5 kW, DC shunt generator has an armature resistance of 0.2 Ω and a total field-circuit resistance of 55 W. The generator is supplying rated current at rated terminal voltage. Answer the following questions relating to this generator and explain your choice of answer.
(i) What is the generator armature current?
17 A
21 A
25 A
29 A
33 A
The correct option is (e) 33 A.
Given Data: Voltage, V = 220VDC Shunt Generator Power, P = 5.5 kWArmature Resistance, Ra = 0.2 ΩTotal Field-Circuit Resistance, Rf = 55 WGenerator is supplying rated current at rated terminal voltage.
We know that, Power, P = VI
Where I is the current flowing through the generator.
Voltage, V = Terminal Voltage, E + IaRa,
where E is generated voltage Armature Current, Ia
= (V - E) / RaAt no load, Ia = If
Where If is field current.If = V / Rf
Hence, generated voltage, E = V - IaRaIaRa
= V - E = V - (V - IaRa)IaRa = IaRaIa = V / Ra
= 220 / 0.2Ia
= 1100 A Armature current, Ia = 1100 A
This is the final answer. Note: kW is converted into W by multiplying it with 1000.
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Routing hydraulic hoses without adequate protection from the movement of the manipulator or end effector is an example of which source of hazard?
a System errors b Environmental sources c Human interaction errors d Unauthorized access
Routing hydraulic hoses without adequate protection from the movement of the manipulator or end effector is an example of Human interaction errors. Hence, the correct option is (C).
Human interaction errors include people making incorrect decisions, misjudgments, and mental slips. Errors in information processing, such as memory failure or errors in executing decisions, are also included. A common type of human interaction error is "slips and lapses."
Lapses are characterized by failing to do anything, whereas slips are characterized by doing the incorrect action. Risk management in the workplace entails identifying and addressing any possible hazards that may arise during the operation of machinery and equipment. This includes human interaction errors, which may include poor judgment, incorrect decisions, mental slips, and memory failure.
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Design a combinational circuit which will add two 4-bit binary numbers. Now, the first binary number cannot be more than 7 and the second binary number cannot be more than 6. If any of the input is more than the given limit, one indictor will be low. Moreover, if the result of the addition falls into any inappropriate range, another indicator should be high and vice versa. [Show all the steps and give proper explanation behind your design]
A combinational circuit which will add two 4-bit binary numbers is given below:In a combinational circuit, each output is dependent on the input, but it is not influenced by the previous state of the input. The circuit adds two 4-bit binary numbers, so we will need eight input wires to connect the binary number, including four bits each from the two binary numbers we want to add.
It is given that the first binary number cannot be more than 7, and the second binary number cannot be more than 6. The maximum value that can be represented by four bits is 15, and the minimum value that can be represented by four bits is 0.To determine the maximum value that can be represented by 3 bits, we can use the following formula:Maximum Value = 2n – 1where n is the number of bitsMaximum Value = 23 – 1 = 7Therefore, the first binary number cannot be more than 7.
To represent numbers greater than 7, we would need more than 3 bits, which would not meet the 4-bit requirement.To determine the maximum value that can be represented by 2 bits, we can use the following formula:Maximum Value = 2n – 1where n is the number of bits Maximum Value = 22 – 1 = 3Therefore, the second binary number cannot be more than 6. To represent numbers greater than 6, we would need more than 2 bits, which would not meet the 4-bit requirement.We can use half adder circuits to add two binary digits, and a full adder circuit to add multiple binary digits. A half adder circuit is used to add two binary digits together, producing a sum and a carry output. A full adder circuit is used to add three binary digits together, producing a sum and a carry output.
Therefore, a 4-bit binary adder will require four half adders and three full adders. A half adder is a combinational circuit that adds two bits together, producing a sum and a carry output. A truth table is used to represent the half adder circuit. The inputs are labeled A and B, while the outputs are labeled S and C. S represents the sum of A and B, while C represents the carry.
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The four arms of an AC bridge network are as follows: Arm AB: unknown impedance Arm BC: a non-inductive resistance of 7500 Arm CD: a non-inductive resistance of R of 4000 Q in parallel to a capacitor of 0.5 µF Arm DA: a non-inductive resistance of 20000 The supply frequency is 50 Hz and connected across terminals B and D. If the bridge is balanced with the above value, determine the value of unknown Impedance. Select one: O a.7.5 mH O b. 750 mH O c.75mH O d.0.75 mH
The value of the unknown impedance in the balanced AC bridge network is 750 mH.
To determine the value of the unknown impedance, we need to analyze the balance condition of the AC bridge network. In a balanced bridge, the product of the resistances in adjacent arms is equal to the product of the reactances in the other two arms.
In this case, we have a non-inductive resistance of 7500 in arm AB, a non-inductive resistance of R = 4000 Q in parallel with a capacitor of 0.5 µF in arm BC, and a non-inductive resistance of 20000 in arm DA.
For the bridge to be balanced, the product of the resistances in arm AB and arm DA must be equal to the product of the reactance in arm BC and the unknown impedance in arm CD.
7500 * 20000 = (1 / (2πfC)) * R * unknown impedance
Substituting the given values, where f is the frequency (50 Hz) and C is the capacitance (0.5 µF), we can solve for the unknown impedance.
7500 * 20000 = (1 / (2π * 50 * 0.5e-6)) * 4000 * unknown impedance
unknown impedance = 750 mH
Therefore, the value of the unknown impedance in the balanced AC bridge network is 750 mH.
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Hi! Pls help me answer this correctly. Show work. Write legibly. Thank you.
WRITE NEATLY PLS. DON'T COPY THE ANSWERS.
SUBJECT: Coupling
MECHANICS OF DEFORMABLE BODIES
A flanged bolt coupling is used to connect a solid shaft 88 mm in diameter to a hollow shaft 100 mm in outside and 88 mm in inside diameter. If the allowable shearing stress in the shafts and the bolts is 63.4 MPa, how many 10-mm-diameter steel bolts must be used on a 199-mm- diameter bolt circle coupling so that the coupling will be as strong as the weaker shaft? Round off the final answer to three decimal places.
To determine the number of bolts required for the flanged bolt coupling, we need to compare the strength of the solid shaft and the hollow shaft. The weaker of the two will determine the number of bolts needed. Here's how we can calculate it:
1. Calculate the cross-sectional area of the solid shaft:
Area_ solid = π * (d_ solid/2)^2, where d_ solid = 88 mm
2. Calculate the cross-sectional area of the hollow shaft:
Area _hollow = π * ((d_ outside/2)^2 - (d _inside/2)^2), where d_ outside = 100 mm and d_ inside = 88 mm
3. Determine the weaker shaft based on their respective shear stresses:
Shear stress_ solid = Shear stress_ hollow = 63.4 MPa
4. Calculate the number of bolts needed:
Number of bolts = (Area_ hollow / Area_ bolt) * (Shear stress_ hollow / Shear ), where Area_ bolt = π * (d_ bolt/2)^2 and Shear stress _bolt = 63.4 MPa
Using these calculations, we can find the number of bolts required to make the flanged bolt coupling as strong as the weaker shaft.
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A lake with no outlet is fed by a river with a constant flow of 1700ft³/s. Water evaporates from the surface at a constant rate of 11ft³/s per square mile surface area. The area varies with depth h (feet) as A (square miles) =4.5+5.5h. What is the equilibrium depth of the lake? Below what river discharge will the lake dry up?
The equilibrium depth of the lake is approximately 27.27 feet. The lake will dry up if the depth is below 27.27 feet.
To determine the equilibrium depth of the lake, we need to find the point at which the inflow from the river matches the outflow due to evaporation. Let's break down the problem into steps:
Express the surface area of the lake in terms of its depth h:
A (square miles) = 4.5 + 5.5h
Calculate the rate of evaporation from the lake's surface:
Evaporation rate = 11 ft³/s per square mile surface area
The total evaporation rate E (ft³/s) is given by:
E = (4.5 + 5.5h) * 11
Calculate the rate of inflow from the river:
Inflow rate = 1700 ft³/s
At equilibrium, the inflow rate equals the outflow rate:
Inflow rate = Outflow rate
1700 = (4.5 + 5.5h) * 11
Solve the equation for h to find the equilibrium depth of the lake:
1700 = 49.5 + 60.5h
60.5h = 1700 - 49.5
60.5h = 1650.5
h ≈ 27.27 feet
Therefore, the equilibrium depth of the lake is approximately 27.27 feet.
To determine the river discharge below which the lake will dry up, we need to find the point at which the evaporation rate exceeds the inflow rate. Since the evaporation rate is dependent on the lake's surface area, we can express it as:
E = (4.5 + 5.5h) * 11
We want to find the point at which E exceeds the inflow rate of 1700 ft³/s:
(4.5 + 5.5h) * 11 > 1700
Simplifying the equation:
49.5 + 60.5h > 1700
60.5h > 1700 - 49.5
60.5h > 1650.5
h > 27.27
Therefore, if the depth of the lake is below 27.27 feet, the inflow rate will be less than the evaporation rate, causing the lake to dry up.
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stion 2. Using well-labelled diagrams explain the operation of V MOSFETS.
A MOSFET is a field-effect transistor that has three terminals. In MOSFET, the metal oxide (MOS) works as an insulator between the gate terminal and the channel.
MOSFET Diagram V MOSFETs operation:
The V-MOSFETs (or VDMOS) are mostly used in power applications due to their high input impedance and low switching losses. V-MOSFETs have the same characteristics as power MOSFETs; the difference is that they are built on the n-type substrate, which is called the vertical structure.
The device is composed of four regions: substrate, source, drain, and gate terminal, as shown in the diagram.The flow of current in a V-MOSFET can be controlled by changing the potential of the gate terminal. When a positive voltage is applied to the gate terminal with respect to the source, an electric field is created between the gate terminal and the channel.
This electric field depletes the charge carriers present in the channel. This creates a potential barrier that opposes the flow of current through the channel. As the gate-source voltage is increased, the potential barrier decreases. At a specific gate-source voltage (VGS), the potential barrier completely disappears, allowing the current to flow through the channel. This voltage is known as the threshold voltage.
When the gate-source voltage is greater than the threshold voltage, the MOSFET is turned ON. When the gate-source voltage is less than the threshold voltage, the MOSFET is turned OFF. This makes V-MOSFETs a type of voltage-controlled device.
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Voltage amplifiers need to have high input resistance and high output resistance. Select one: O True O False Check
The statement "Voltage amplifiers need to have high input resistance and high output resistance" is true because high input resistance and high output resistance are the key features of a voltage amplifier.
The high input resistance helps in minimizing the loading effect by not drawing any current from the signal source, which reduces the attenuation of the signal. The high output resistance helps in reducing the attenuation of the signal due to its ability to drive the load without losing the voltage.
Thus, having high input resistance and output resistance is essential in maintaining the integrity of the input signal, providing high gain without any distortion, and maintaining a stable output.
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When two resistors are connected in parallel
a.
The total resistance is equal to the sum of all resistances.
b.
They have a common voltage across all branches.
c.
There is only one path for the current.
d.
The applied voltage is equal across the circuit is equal to the product of V*I.
10. The ratio of the modulation signal amplitude to the carrier signal amplitude is known as:
Modulation percentage
Modulation index
over modulation
Distortion
12. What is impedance?
a.
The repulsive force between two similar electric fields.
b.
The inverse of resistance.
c.
The opposition to the flow of current in an AC circuit.
d.
The electrical charge stored by a capacitor.
When two resistors are connected in parallel:a. The total resistance is NOT equal to the sum of all resistances. The total resistance is given by the formula:1/RTotal = 1/R1 + 1/R2,
where R1 and R2 are the resistances of the individual resistors.b. They do NOT have a common voltage across all branches. In a parallel circuit, each branch has the same voltage across it.c. There is more than one path for the current. In a parallel circuit, the current splits between the branches.d. The applied voltage is the same across the circuit, but it is not equal to the product of V*I. The total current in a parallel circuit is the sum of the currents flowing through each branch.Regarding the other question:Impedance is c. the opposition to the flow of current in an AC circuit. It is a measure of the combined resistance and reactance (which is related to capacitance or inductance) in an AC circuit. Impedance is represented by the symbol Z and is measured in ohms.
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Water is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 20 MPa and 4000C and leaves as a wet vapor. The condenser pressure is 10 kPa. Sketch T-s diagram. State at least three (3) assumptions Determine
(i) Dry fraction of the steam leaving the turbine
(ii) The net -work per unit mass of steam flowing, in kJ/kg.
(iii) The heat transfer to the steam passing through the boiler, in kJ per kg of steam flowing.
(iv.) The thermal efficiency.
(v) The heat transfer to cooling water passing through the condenser, in kJ per kg of steam condensed.
To answer the given questions, we will make the following assumptions for the ideal Rankine cycle:
The working fluid is water, which behaves as an ideal substance throughout the cycle.
The processes within the turbine, condenser, pump, and boiler are all internally reversible.
There are no significant pressure drops within the condenser, pump, and boiler.
The kinetic and potential energy changes in the flow of water are negligible.
The condensate leaving the condenser is saturated liquid at the condenser pressure.
Based on these assumptions, we can determine the following:
(i) To find the dry fraction of the steam leaving the turbine, we need to locate the state point on the T-s diagram where the pressure is equal to the condenser pressure (10 kPa). From that point, we can determine the dryness fraction (x) of the steam.
(ii) The net work per unit mass of steam flowing can be calculated by finding the difference in enthalpy between the turbine inlet and outlet. The work is given by the equation: Net work = h1 - h2, where h1 is the specific enthalpy at the turbine inlet and h2 is the specific enthalpy at the turbine outlet.
(iii) The heat transfer to the steam passing through the boiler can be determined by calculating the difference in specific enthalpy between the boiler outlet and inlet. The heat transfer is given by the equation: Heat transfer = h1 - h4, where h4 is the specific enthalpy at the boiler outlet.
(iv) The thermal efficiency of the Rankine cycle can be calculated using the equation: Thermal efficiency = (Net work) / (Heat input).
(v) The heat transfer to the cooling water passing through the condenser can be determined by calculating the difference in specific enthalpy between the condenser outlet and inlet. The heat transfer is given by the equation: Heat transfer = h3 - h4, where h3 is the specific enthalpy at the condenser outlet.
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