5.Hence, option (a) is correct.
1. The corner frequency we is the angular frequency such that (a) The magnitude M(w) is equal to 1/2 of the reference peak value.
2. Concatenating a lowpass filter with wew
LP in series with a high pass filter with we = WHP will
(a) Generate a bandpass filter if WLP < WHP.
3. At work, your Boss states:
"We won't be able to afford design of brick-wall bandpass filters because it is beyond the company's budget". This statement is indeed
(b) Not true due to the causality constraint.
4. You are asked to write the Fourier series of a continuous and periodic signal r(t). You plot the series representation of the signal with 500 terms.
Do you expect to see the Gibbs phenomenon?
(a) Yes, irrespective of the number of terms.
5. The power of an AM modulated signal
(A+ cos (27 fmt)) cos(2π fet) depends on
(a) The DC amplitude A and the frequency fm.
1. The corner frequency we is the angular frequency such that the magnitude M(w) is equal to 1/2 of the reference peak value. Hence, option (a) is correct.
2. Concatenating a lowpass filter with wew
LP in series with a high pass filter with we = WHP will generate a bandpass filter
if WLP < WHP. Hence, option (a) is correct.
3. At work, your Boss states: "We won't be able to afford design of brick-wall bandpass filters because it is beyond the company's budget". This statement is not true due to the causality constraint. Hence, option (b) is correct.
4. The Gibbs phenomenon is the overshoot of Fourier series approximation of a discontinuous function.
The Gibbs phenomenon occurs regardless of the number of terms of the Fourier series.
Hence, option (a) is correct.
5. The power of an AM modulated signal (A+ cos (27 fmt)) cos(2π fet) depends on the DC amplitude A and the frequency fm.
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Solve the following problems: 1. A reciprocating compressor draws in 500ft 3/min. of air whose density is 0.079lb/ft 3 and discharges it with a density of 0.304lb/ft 3. At the suction, p1=15psia; at discharge, p2 = 80 psia. The increase in the specific internal energy is 33.8Btu/lb, and the heat transferred from the air by cooling is 13Btu/lb. Determine the horsepower (hp) required to compress (or do work "on") the air. Neglect change in kinetic energy. 2. The velocities of the water at the entrance and at the exit of a hydraulic turbine are 10 m/sec and 3 m/sec, respectively. The change in enthalpy of the water is negligible. The entrance is 5 m above the exit. If the flow rate of water is 18,000 m3
/hr, determine the power developed by the turbine. 3. A rotary compressor draws 6000 kg/hr of atmospheric air and delivers it at a higher pressure. The specific enthalpy of air at the compressor inlet is 300 kJ/kg and that at the exit is 509 kJ/kg. The heat loss from the compressor casing is 5000 watts. Neglecting the changes in kinetic and potential energy, determine the power required to drive the compressor.
1.The horsepower required to compress the air is 0.338 hp
2.The power developed by the turbine is 2,235,450 W.
3. The power required to drive the compressor is 349.03 kW.
1. The calculation of horsepower required to compress the air is shown below:Mass flow rate, m = density × volume flow rate= 0.079 lb/ft³ × 500 ft³/min = 39.5 lb/min.
The energy added to the air, q = increase in internal energy + heat transferred from the air by cooling.= 33.8 Btu/lb × 39.5 lb/min + 13 Btu/lb × 39.5 lb/min= 1340.3 Btu/min.
To determine the horsepower required to compress the air, use the following relation:
Horsepower = q/3960 = 1340.3 Btu/min ÷ 3960 = 0.338 hp.
.2. The calculation of the power developed by the turbine is shown below:
Volume flow rate, Q = 18,000 m³/hr ÷ 3600 s/hr = 5 m³/s
.The mass flow rate, m = ρQ = 1000 kg/m³ × 5 m³/s = 5000 kg/s.
The difference in kinetic energy, Δv²/2g = (10² − 3²)/2g = 43.5 m
. The velocity head is, hv = Δv²/2g = 43.5 m.
The potential energy difference, Δz = 5 m.
Power developed, P = m(gΔz + hv) = 5000 kg/s(9.81 m/s² × 5 m + 43.5 m) = 2,235,450 W.
3. The calculation of power required to drive the compressor is shown below:
Mass flow rate, m = 6000 kg/hr ÷ 3600 s/hr = 1.67 kg/s.
The energy added to the air, q = change in specific enthalpy of the air= (509 − 300) kJ/kg = 209 kJ/kg.
Power input, P = m × q + heat loss from the compressor casing.= 1.67 kg/s × 209 kJ/kg + 5000 W = 349.03 kW.
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Develop a project with simulation data of a DC-DC converter: Buck Boost a) 12V output and output current between (1.5 A-3A) b) Load will be two 12 V lamps in parallel/Other equivalent loads correction criteria c) Simulation: Waveforms (input, conversion, output) of voltage and current in general. Empty and with load. d) Converter efficiency: no-load and with load e) Frequency must be specified f) Development of the high frequency transformer, if necessary g) Smallest size and smallest possible mass. Reduce the use of large transformers. >>> Simulation can be done in Multisim or in another software of your choice.
Project Description:In this project, we will simulate a DC-DC converter known as a Buck-Boost converter. The objective is to design a converter that produces a 12V output with an output current ranging between 1.5A and 3A.
The load for the converter will consist of two 12V lamps connected in parallel or other equivalent loads as per the correction criteria.
The simulation will involve analyzing the waveforms of the input voltage and current, conversion voltage and current, and output voltage and current. The simulation will be conducted for both empty (no-load) conditions and with the specified load.
Efficiency analysis will be performed to determine the converter's efficiency under both no-load and loaded conditions. The efficiency will be calculated as the ratio of the output power to the input power.
The frequency of operation for the converter needs to be specified. Generally, a high-frequency operation is preferred to reduce the size and mass of the components. The specific frequency will depend on the requirements and constraints of the project.
If necessary, the design will involve the development of a high-frequency transformer. The transformer will be designed to meet the size and mass requirements while ensuring efficient power transfer.
The main objective of the project is to achieve the smallest possible size and mass for the converter while reducing the reliance on large transformers. The design will prioritize compactness and efficiency.
Simulation software such as Multisim or any other suitable software of your choice can be used to perform the simulation and analysis of the DC-DC converter.
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Consider a pulsed Doppler system, which operates at a center frequency of 7.5 MHz. This system is used to image blood flow in a vein located at a distance of 5 cm from the transducer. The angle between the beam and blood flow is 60°. Assume that the minimum velocity that this instrument can measure is 2 cm/s (FYI, this limit is typically imposed by tissue movement, but this information is not needed to solve the problem). If needed, please assume c = 1540 m/s. Please find: i) i the maximum velocity that this instrument can measure; ii) the minimum Doppler frequency; iii) the spectral broadening: iv) the pulse repetition frequency.
Center frequency of 7.5 MHz, Distance of 5 cm, angle of 60°, minimum velocity of 2 cm/s, c= 1540 m/s.The relationship between the Doppler shift, the angle between the ultrasound beam and blood flow, the velocity of the blood, and the ultrasound frequency can be calculated as:
ƒ_D = (2ƒ_0v cos θ) / cwhere ƒ_D is the Doppler frequency shift, ƒ_0 is the ultrasound frequency, v is the velocity of the blood, θ is the angle between the blood flow and the ultrasound beam, and c is the speed of sound in tissue.
The maximum frequency shift is obtained when the angle between the ultrasound beam and the blood flow is 0. This is due to the fact that cos (0) = 1. The minimum detectable velocity is 2 cm/s.The maximum velocity, therefore, is:
[tex]v_max = cƒ_D / (2ƒ_0cos θ)Where cos θ = cos (60°) = 1/2v_max = cƒ_D / (2ƒ_0 cos θ)= (1540 x 7.5 x 10^6) / (2 x 7.5 x 10^6 x 1/2)= 1540 m/s.[/tex]
Therefore, the maximum velocity that this system can detect is 1540 m/s.The Doppler frequency shift for the minimum detectable velocity can be calculated using the equation above with v = 2 cm/s and θ = 60°.
[tex]ƒ_D,min = (2ƒ_0v min cos θ) / cƒ_D,min = (2 x 7.5 x 10^6 x 2 x 10^-2 x 1/2) / 1540= 0.0245 MHz[/tex]
The minimum detectable frequency shift is 0.0245 MHz.
Spectral broadening is the result of the flow rate being non-uniform across the sample volume. The spectral broadening of the Doppler signal is a measure of the degree of spectral overlap. This can be calculated using the following equation:β = (2kv max) / cwhere β is the spectral broadening, k is a constant that depends on the particular type of flow, and v_max is the maximum velocity.
The spectral broadening is calculated as follows:
[tex]β = (2k v max) / c= (2 x v max) / c= (2 x 1540) / 1540= 2.[/tex]
The spectral broadening is 2.Pulse repetition frequency (PRF) is determined by the depth of the sample volume and the time required for each pulse to travel to the target and return.
The PRF is calculated using the following formula:PRF = (c/2) x d_maxwhere PRF is the pulse repetition frequency, c is the speed of sound in tissue, and d_max is the maximum distance that the pulse can travel in one-half cycle of the PRF. The maximum distance is calculated using the Pythagorean theorem:
[tex]d_max = (5^2 + (sin 60° x 5)^2)1/2= 5.77 cmPRF = (c/2) x d_max= (1540 x 5.77) / (2 x 10^-2)= 2.22 x 10^5 Hz.[/tex]
In a pulsed Doppler system, the maximum velocity that can be measured is calculated using the formula:
v_max = cƒ_D / (2ƒ_0cos θ)where c is the speed of sound in tissue, ƒ_D is the Doppler frequency shift, ƒ_0 is the ultrasound frequency, and θ is the angle between the blood flow and the ultrasound beam. The maximum Doppler frequency shift occurs when the angle between the blood flow and the ultrasound beam is 0. The maximum velocity that can be detected in this system is 1540 m/s.
The minimum detectable velocity is 2 cm/s, and the minimum Doppler frequency shift is 0.0245 MHz. The spectral broadening is 2. The pulse repetition frequency (PRF) is calculated using the formula PRF = (c/2) x d_max, where d_max is the maximum distance that the pulse can travel in one-half cycle of the PRF. The PRF for this system is 2.22 x 10^5 Hz.
In summary, a pulsed Doppler system with a center frequency of 7.5 MHz, located at a distance of 5 cm from a vein, with an angle of 60° between the blood flow and the ultrasound beam, and a minimum detectable velocity of 2 cm/s can detect a maximum velocity of 1540 m/s, with a minimum detectable Doppler frequency shift of 0.0245 MHz. The spectral broadening is 2. The PRF for this system is 2.22 x 10^5 Hz.
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A single phase half-wave controlled rectifier is used to control a power of 230V, 1500W, DC heater. To get 100W of heating power output from so called heater, find the firing angle of the SCR, if the system is powered by a 230V, 50Hz power supply.(Assume the heater efficiency is 100%)
A single-phase half-wave controlled rectifier is used to control a power of 230V, 1500W, DC heater. The power can be calculated by using the formula P = VI, where P is power, V is voltage and I is current.
Therefore, the current is I = P/V which equals I = 1500/230 = 6.52Amps. Hence, to get 100W of heating power output, the power delivered to the heater can be calculated as 100W = VI. Therefore, the voltage required can be calculated as V = 100/6.52 = 15.33V.
The remaining voltage is 230 - 15.33 = 214.67V. To calculate the firing angle of the SCR, the formula is α = cos-1(Po/Pi) where Po is the power output and Pi is the input power. Therefore, the firing angle is α = cos-1(100/1500) = 82.32°.Therefore, the firing angle of the SCR to get 100W of heating power output from the heater in a single-phase half-wave controlled rectifier is 82.32° when the system is powered by a 230V, 50Hz power supply.
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Write a Matlab code to plot the continuous time domain signal for the following spectrum:
X (jω) = 2sin(ω)/ω
Here is a MATLAB code to plot the continuous-time domain signal for the given spectrum: X(jω) = 2sin(ω)/ω.
% Define the frequency range
w = -10*pi:0.01*pi:10*pi;
% Compute the spectrum X(jω)
X = 2*sin(w)./w;
% Plot the signal in the time domain
plot(w, X)
xlabel('Frequency (rad)')
ylabel('Amplitude')
title('Continuous-Time Domain Signal')
grid on
The MATLAB code provided above allows us to plot the continuous-time domain signal for the given spectrum X(jω) = 2sin(ω)/ω.
First, we define the frequency range 'w' over which we want to evaluate the spectrum. In this case, we use a range of -10π to 10π with a step size of 0.01π.
Next, we compute the values of the spectrum X(jω) using the element-wise division operator './'. We calculate 2*sin(w)./w to obtain the values of X for each frequency 'w'.
Finally, we plot the signal in the time domain using the 'plot' function. The 'xlabel', 'ylabel', and 'title' functions are used to label the axes and title of the plot. The 'grid on' command adds a grid to the plot for better visualization.
By running this MATLAB code, we can obtain a plot that represents the continuous-time domain signal corresponding to the given spectrum.
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A piston-cylinder device contains 0.8 lbm of Helium, initially at 30 psia and 100 oF. The gas is then heated, at constant pressure, using a 400-watt electric heater to a final temperature of 450°F.
a) Calculate the initial and final volumes
b) Calculate the net amount of energy transferred (Btu) to the gas
c) Calculate the amount of time the heater is operated
a) Calculation of the initial and final volumes of the given piston-cylinder device: Given data, Pressure, P1 = 30 psia Temperature, T1 = 100 °F Molar mass of helium, M = 4.0026 l bm/lbm-mol Specific heat of helium, Cp = 3.117 Btu/lbm-°FR = 53.35 ft. lbf/lbm-°R Using the ideal gas law.
PV = m R TInitial volume, V1 can be calculated as;V1 = (mRT1) /[tex](P1) = (0.8 × 53.35 × (100 + 460)) / (30) = 8.30 ft3Now, using the Gay-Lussac's law, (p1 / T1) = (p2 / T2)The final pressure P2 can be found as, P2 = (P1 × T2) / T1 = (30 × 910) / (100 + 460) = 35.9 psia Final volume, V2 can be found asV2 = (mRT2) / (P2) = (0.8 × 53.35 × (450 + 460)) / (35.9) = 17.06 ft3Therefore, the initial volume, V1 = 8.30 ft3 and the final volume, V2 = 17.06 ft3.[/tex]
b) Calculation of the net amount of energy transferred (Btu) to the gas The net amount of energy transferred can be calculated as [tex];W = Q - ΔE,where, ΔE = U2 - U1 as,ΔE = mCpΔT,where,ΔT = T2 - T1 = 450 - 100 = 350 °FΔE = 0.8 × 3.117 × 350 = 868.68 Btu The heat added to the gas, Q is given by; Q = W + ΔE = PΔV + ΔEHere,ΔV = V2 - V1 = 17.06 - 8.30 = 8.76 ft3Thus,Q = 30 × 8.76 + 868.68 = 1154.08 1154.08[/tex]
c) Calculation of the time the heater is operated The rate of energy supplied by the heater, E = 400 watts = 400 J/s The time for which the heater operates, t can be calculated as[tex]; t = Q / E = 1154.08 / 400 = 2.885[/tex] s Therefore, the amount of time the heater is operated is 2.885 seconds.
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An air standard Otto cycle has the following characteristics; 1. It draws air from the environment at 98 kPa and 14°C. 2. The cycle has a compression ratio of 9.5: 1. 3. Heat (990 kJ/kg) is added to the compressed gases at constant volume. The working fluid is air, a perfect gas with 4. ratio of specific heats y = 1.4 and gas constant R = 287 J/kgK. Follow the instructions below: a) Describe each of the four processes from the thermodynamic point of view. [4 marks] b) Sketch the P-v and T-S plots for this cycle add labels starting as air intake at (1). [2 marks] c) Calculate the peak in cylinder pressure. [2 marks] d) Calculate the thermal efficiency of the cycle. [1 mark] Evaluate the Break Mean Effective Pressure. [1 mark] Q2 (Unseen Part) f) During the Diesel combustion process, work is extracted giving constant pressure. This process results in lower peak temperatures than the equivalent constant volume combustion process. However it is reported that Diesel engines produce less CO2 in their exhausts compared to Otto cycle engines for the amount of work supplied. Explain in detail why this is so. [5 marks] g) In recent years Diesel powered motor cars have become much less popular in spite of their superior efficiency. Describe why this is so, identify both important mechanisms and clearly explain how these problems influence human health. [5 marks]
Description of the four processes of Otto cycle from a thermodynamic point of view:Process 1-2 is Isentropic compression: During this process, the gas is compressed isentropically from point 1 to point 2. The compression ratio is given as 9.5: 1, which means that the volume at point 2 is 1/9.5 times the volume at point 1.Process 2-3 is Constant volume heat addition: Heat is added to the compressed air at a constant volume.
This process is represented by a vertical line on the P-v diagram. During this process, the temperature increases, and the pressure also increases. The specific heat of the air is given as 990 kJ/kg.Process 3-4 is Isentropic expansion: The air is expanded isentropically from point 3 to point 4. During this process, the temperature and pressure of the air decrease, and the volume increases.
Process 4-1 is Constant volume heat rejection: The air is cooled at a constant volume from point 4 to point 1. This process is represented by a vertical line on the P-v diagram. During this process, the temperature and pressure of the air decrease, and the specific heat of the air is rejected. Sketch the P-v and T-S diagrams for the cycle The P-v and T-S diagrams for the cycle
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a The AC power transmission and distribution system has several important advantages over a DC system. However, there would still be advantages for a DC power system. What are those? Note: Assume the same voltage and current ratings for DC as for AC. e a) The design of circuit breakers and transformers would be much simplified for DC. b) The voltage drop across the transmission lines would be reduced. c) The losses in a DC transformer are lower than in an AC transformer. Why do outdoor insulators often have disks? a) To reduce the magnetic field. b) To reduce the electric field. c) To increase the creepage distance. Who was the biggest proponent for the development of early alternating current power system? a) Thomas A. Edison b) Antonio Pacinotti c) Nikola Tesla A complex load of 3+j4 ohms is connected to 120V. What is the power factor? a) 53.1 deg b) 0.6 lagging c) 0.6 leading How can the power factor be corrected for the load in the previous question? How can the power factor be corrected for the load in the previous question? a) An inductor in parallel to the load. b) A capacitor in series to the load. c) A capacitor in parallel to the loa
Advantages of DC power system over AC system:There are several advantages of a DC power system over an AC power lines such as:Circuit breakers and transformers would be much simplified for DC.The voltage drop across the transmission lines would be reduced.
The losses in a DC transformer are lower than in an AC transformer.Disk-shaped insulators:To increase the creepage distance, outdoor insulators often have disks.Proponent for the development of early alternating current power system:The biggest proponent for the development of early alternating current power systems was Nikola Tesla. The Serbian American inventor, electrical engineer, mechanical engineer, and futurist is best known for his contributions to the design of the modern alternating current (AC) electricity supply system.
Complex load power factor:Given a complex load of 3+j4 ohms connected to 120V, the power factor is 0.6 lagging.Power factor correction:To correct the power factor of a load, a capacitor should be added in parallel with the load. The capacitor, which is essentially a reactive component, produces a current that lags behind the voltage across it. In this manner, the load's reactive power demand is balanced out by the capacitor's reactive power supply.
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For a 3-bus power system, the real and reactive powers are specified at all buses except the swing bus. The Newton Rephson method is chosen to solve the lood flow problem 1- What is the order of the Jacobian matrix ? 2- Determine the element in the Jacobson matrix, representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2 3- Determine the element in the Jacobian matrix, representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2
1. The order of the Jacobian matrix is equal to the number of unknowns in the power flow problem. In a 3-bus power system, the unknowns typically include the voltage magnitudes and voltage angles at each bus except the swing bus. Therefore, the order of the Jacobian matrix would be (2n - 1), where n is the number of buses in the system. In this case, since there are three buses, the order of the Jacobian matrix would be (2 * 3 - 1) = 5.
2. To determine the element in the Jacobian matrix representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2, we need to compute the partial derivative of the real power at bus 2 with respect to the voltage magnitude at bus 2 (∂P2/∂|V2|).
The Jacobian matrix for the power flow problem consists of partial derivatives of the power injections at each bus with respect to the voltage magnitudes and voltage angles at all buses. Let's denote the Jacobian matrix as J.
The element representing ∂P2/∂|V2| in the Jacobian matrix can be denoted as J(2, 2), indicating the second row and second column of the matrix.
To determine the element in the Jacobian matrix representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2, we need to compute the partial derivative of the reactive power at bus 3 with respect to the voltage angle at bus 2 (∂Q3/∂θ2).
Similarly to the previous question, the element representing ∂Q3/∂θ2 in the Jacobian matrix can be denoted as J(3, 2), indicating the third row and second column of the matrix.
1. The order of the Jacobian matrix for a 3-bus power system is 5.
2. The element in the Jacobian matrix representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2 is J(2, 2).
3. The element in the Jacobian matrix representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2 is J(3, 2).
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Merits and Demerits between HRC/Drop-Out Fuses and other
different types of
fuses
HRC and drop-out fuses have both merits and demerits when compared to other types of fuses. It is up to the user to decide which type of fuse is best suited for their specific needs.
HRC (High Rupturing Capacity) and drop-out fuses are some of the types of fuses that have both merits and demerits as compared to other types of fuses.
The demerits and merits of each type of fuse are discussed in detail as follows:
Demerits of HRC and Drop-Out Fuses:
The following are the demerits of the HRC and drop-out fuses:
They are more expensive than other types of fuses. Due to their complexity, they require more maintenance, which adds to their cost.
They are unsuitable for low voltages because they require a lot of current to trigger, which can be dangerous.
They have a higher tripping time than other types of fuses, which can cause damage to equipment.
Merits of HRC and Drop-Out Fuses:
The following are the merits of the HRC and drop-out fuses:
They can handle a larger amount of current than other types of fuses, which means they can protect larger electrical systems.
They have a higher breaking capacity, which means they can handle large current surges without breaking down.
They have a longer lifespan than other types of fuses, which makes them more reliable.
They are safer because they have a lower risk of causing a fire or explosion due to their design. Other types of fuses have a higher risk of failure due to their design, which can lead to a fire or explosion.
Overall, HRC and drop-out fuses have both merits and demerits when compared to other types of fuses. It is up to the user to decide which type of fuse is best suited for their specific needs.
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Q5) Given the denominator of a closed loop transfer function as expressed by the following expression: S²+85-5Kₚ + 20 The symbol Kₚ denotes the proportional controller gain. You are required to work out the following: 5.1) Find the boundaries of Kₚ for the control system to be stable.
5.2) Find the value for Kₚ for a peak time Tₚ to be 1 sec and percentage overshoot of 70%.
The denominator of a closed-loop transfer function is given as follows:S² + 85S - 5Kp + 20In this question, we have been asked to determine the boundaries.
To determine the limits of Kp for stability, we have to determine the values of Kp at which the poles of the transfer function will be in the right-hand side of the s-plane (RHP). This is also referred to as the instability criterion. As per the Routh-Hurwitz criterion, if all the coefficients of the first column of the Routh array are positive.
So let us form the Routh array for the given transfer function. Routh array:S² 1 -5Kp85 20The first column of the Routh array is [1, 85]. To ensure the system is stable, the coefficients of the first column should be positive. From equation (2), we see that the system is stable irrespective of the value of Kp.
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At inlet, in a steady flow process, 1.7 kg/s of nitrogen is initially at reduced pressure of 2 and reduced temperature of 1.3. At the exit, the reduced pressure is 3 and the reduced temperature is 1.7. Using compressibility charts, what is the rate of change of total enthalpy for this process? Use cp-1.039 kJ/kg K. Express your answer in kW.
The rate of change of total enthalpy for this process is 84.35 kW.Processes can be classified as steady or unsteady. In a steady flow process, the flow properties (temperature, pressure.
The energy or mass entering a system is equal to the energy or mass leaving the system. Given the information provided in the question, it is a steady flow process.As per the given data,Mass flow rate = 1.7 kg/sReduced pressure at inlet (P1) = 2Reduced temperature at inlet Reduced temperature at outlet (T2) = 1.7The compressibility factor (Z) can be obtained from the compressibility chart
The compressibility factor at the inlet and outlet can be found as follows:Compressibility factor at inlet, Z1:From the chart .Compressibility factor at outlet, Z2:From the chart, for P2 = 3 and T2 = 1.7, Z2 = 0.97.The specific heat of nitrogen at constant pressure .The rate of change of total enthalpy for this process can be calculated as follows Therefore, the rate of of total enthalpy for this process.
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21. A(n) ____. is a material that has a very high resistance and resists the flow of electrons a. Circuit breaker b. insulator c. fuse d. conductor e. none of the above 22. The process by which general contractors and electrical contractors obey during construction for safety purposes around electrical equipment is referred to as: a. Saf-T-tag b. Keep out watch out c. Lock out tag out d. Suns out guns out 23. Explain the difference between 12-2 and 10-3 Romex: 24. Which type of light bulb currently used in construction draws the least amount of power? 25. (A) What does GFCI stand for? (B) What does a GFCI do, and where does it belong?
21 A(n) insulator. is a material that has a very high resistance and resists the flow of electrons
b. insulatorWhat contractors and electrical contractors must adhere to22. During construction, general contractors and electrical contractors must adhere to the lock out tag out process for safety purposes around electrical equipment.
c. Lock out tag out23. The numbers in 12-2 and 10-3 Romex refer to the gauge of the wire and the number of conductors.
12-2 Romex has a 12-gauge wire, which is thicker than 10-gauge wire. It contains two conductors, typically a black (hot) wire and a white (neutral) wire.
10-3 Romex has a 10-gauge wire, which is thicker than 12-gauge wire. It contains three conductors, typically a black (hot) wire, a red (hot) wire, and a white (neutral) wire.
The difference in gauge affects the current-carrying capacity of the wire, with lower gauge numbers being able to handle higher currents.
24. LED (Light Emitting Diode) light bulbs currently used in construction draw the least amount of power compared to traditional incandescent or fluorescent bulbs. LEDs are highly efficient and provide significant energy savings.
25. (A) GFCI stands for Ground Fault Circuit Interrupter.
(B) A GFCI is a safety device designed to protect against electrical shocks caused by ground faults. It constantly monitors the electrical current flowing through a circuit and quickly shuts off power if it detects any imbalance between the hot and neutral wires. It helps prevent electric shock hazards, particularly in areas with water such as bathrooms, kitchens, or outdoor outlets. GFCIs are typically installed in electrical outlets or incorporated into circuit breakers.
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(a) Explain a model for the angle y(t)=q(t). The input u(t) is given by
Y(s) = G(s)U(s) = 1.35/s(0.1s+1)U(s)
(b) This is given:
E(s)0.1s+1/0.1s+1-1.35KPR(s) - 1.35/0.1s+1-1.35KPV(s)
Show and explain how the error signal with a reference is given by this information.
The error signal with a reference in the given model is represented by the equation E(s) = (0.1s + 1)/(0.1s + 1 - 1.35KP)R(s) - 1.35/(0.1s + 1 - 1.35KP)V(s).
In the given model, the error signal E(s) represents the difference between the reference signal R(s) and the output of the system represented by V(s). The term (0.1s + 1)/(0.1s + 1 - 1.35KP) represents the transfer function of the proportional controller, while 1.35/(0.1s + 1 - 1.35KP) represents the transfer function of the velocity controller.
The error signal E(s) is calculated by multiplying the reference signal R(s) with the proportional controller transfer function, subtracting the output signal V(s) multiplied by the velocity controller transfer function, and dividing it by the difference between the proportional controller transfer function and 1.35KP.
The given equation provides a mathematical representation of the error signal in terms of the reference signal and the output of the system. It takes into account the proportional controller and velocity controller transfer functions to calculate the error signal. Understanding and analyzing this equation allows for better understanding and control of the system's behavior.
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(a) Prepare a schematic diagram to show the provision and distribution of fire hydrants and hose reels on all residential floors based on the Code of Practice for Minimum Fire Services Installations and Equipment, Fire Service Department, HKSAR (2012).
(b) Each flat has the following water draw-off points: I washbasin, 1 WC-cistern, 1 shower head, I kitchen sink and I washing machine. Find the total loading unit and the diversified flow rate for a typical residential floor based on relevant data in BS EN 806-3:2006. Find also the external pipe diameter of the main stack serving all residential floors. It is assumed that the plumbing facilities are supplied by hot-dip galvanized steel pipes.
The schematic diagram that shows the provision and distribution of fire hydrants and hose reels on all residential floors based on the Code of Practice for Minimum Fire Services Installations and Equipment, Fire Service Department, HKSAR (2012) is shown below.
The total loading unit and the diversified flow rate for a typical residential floor based on relevant data in BS EN 806-3:2006 is given as follows;I washbasin - 1 WCI WC-cistern - 2 WCI shower head - 1 WCI kitchen sink - 1 WCI washing machine - 2 WCI
Total Loading Unit = 1+2+1+1+2= 7 WCI
Diversified Flow Rate = Total Loading Unit x 0.114
= 7 x 0.114
= 0.798 l/s.
The external pipe diameter of the main stack serving all residential floors is given by Therefore, the external pipe diameter of the main stack serving all residential floors is 399 mm.
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When filled to capacity, the unpressurized storage tank contains water to a height of h = 34 ft. The outside diameter of the tank is 7.3 ft and the wall thickness is 0.646 in. Determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base. (Weight density of water = 62.4 lb/ft3.)
The given data:Height of the storage tank, h = 34 ftOutside diameter of the tank, D = 7.3 ftWall thickness, t = 0.646 inWeight density of water, w = 62.4 lb/ft³.
We need to determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base.So, the following formulae are used:Volume of the tank = [tex]πD²h/4 = π(7.3)²(34)/4 = 1988.29 ft³.[/tex]
Weight of the water = Volume of the tank × weight density of water = 1988.29 × 62.4 = 124236.1 lb.
The water in the tank is trying to expand and the tank is resisting this expansion. Thus, there will be a radial stress on the tank at the bottom.The maximum normal stress at the base of the tank,
σmax = wH/2t + P/4t
Where P = Weight of the water in the tank = 124236.1 lbH = Height of the water in the tank = 34 ft
[tex]σmax = (62.4 × 34)/(2 × 0.646) + 124236.1/(4 × 0.646) = 23618.2 + 48325.6 = 71943.8 lb/ft²= 71943.8/144 = 499.6 psi[/tex].
The absolute maximum shear stress on the outer surface of the tank at its base, τmax = P/2At the base, the direction of the normal stress is radial and the direction of the shear stress is tangential.
Therefore, τmax = 124236.1/2 = 62118.05 lb/ft²= 62118.05/144 = 431.4 psi
In this question, the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base is to be determined. The formulae used to solve this problem are as follows:
The maximum normal stress at the base of the tank, σmax = wH/2t + P/4tThe absolute maximum shear stress on the outer surface of the tank at its base, τmax = P/2When the water is filled in the tank, it tries to expand and the tank resists this expansion.
Therefore, there is a radial stress on the tank at the bottom. The maximum normal stress at the base of the tank is calculated by using the formula σmax = wH/2t + P/4t. Here, w is the weight density of water, H is the height of the water in the tank, t is the thickness of the wall, and P is the weight of the water in the tank.
Substituting the given values, we get
[tex]σmax = (62.4 × 34)/(2 × 0.646) + 124236.1/(4 × 0.646) = 23618.2 + 48325.6 = 71943.8 lb/ft².[/tex]
The absolute maximum shear stress on the outer surface of the tank at its base is calculated by using the formula τmax = P/2. Here, P is the weight of the water in the tank. Substituting the given values, we get
τmax = 124236.1/2 = 62118.05 lb/ft².
Therefore, the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base are 499.6 psi and 431.4 psi, respectively.
Thus, we can conclude that the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base are 499.6 psi and 431.4 psi, respectively.
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Volume charge density is located as follows: p = 0 for p < 1 mm and for p> 2mm, p = 4pµC/m³ for 1 < p < 2 mm. (a) Calculate the total charge in the region, 0 < p < 0₁, 0 < z Previous question
The total charge in the given region is 7.8548 × 10⁻⁷ C
Given that, Volume charge density (p) is located as follows:
p = 0 for p < 1 mm and for p> 2mm,
p = 4pµC/m³ for 1 < p < 2 mm.
To calculate the total charge in the region, 0 < p < 0₁, 0 < z, we need to use integration.
Let's see the calculation in detail below:
Formula used:
Total charge = ∫∫∫ρdτ
where ρ is the volume charge density, and dτ is the volume element.
To calculate the total charge in the region, we integrate the volume charge density with respect to the volume element.
Here, we have to consider the cylindrical coordinates. So, the volume element is given asdτ = r dr dθ dz Where r is the radius, θ is the angle, and z is the height.
So, Total charge, Q = ∫∫∫ρdτ= ∫∫∫ρr dr dθ dz Bounds:0 < r < 0₁0 < θ < 2π0 < z
Let's calculate the total charge in three parts
Part 1: For 0 < p < 1 mm Given that, p = 0 for p < 1 mm Bounds: 0 < r < 0₁0 < θ < 2π0 < z < 0.001∫∫∫ρr dr dθ dz= ∫∫∫(0) r dr dθ dz= 0
Part 2: For 1 < p < 2 mm Given that, p = 4pµC/m³ Bounds: 0 < r < 0₁0 < θ < 2π0.001 < z < 0.002∫∫∫ρr dr dθ dz= ∫∫∫(4 × 10⁻⁶) r dr dθ dz= (4 × 10⁻⁶) ∫∫∫r dr dθ dz= (4 × 10⁻⁶) × (π/4) (0₁²) (0.002 - 0.001)= (10⁻⁶) (0.25 π) (0₁²)
Part 3: For 2 < p Given that, p = 0 for p> 2mm Bounds: 0 < r < 0₁0 < θ < 2π0.002 < z∫∫∫ρr dr dθ dz= ∫∫∫(0) r dr dθ dz= 0
Therefore, Total charge, Q = (10⁻⁶) (0.25 π) (0₁²)= 7.8548 × 10⁻⁷ C
Hence, the total charge in the given region is 7.8548 × 10⁻⁷ C.
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a=6
Use Kaiser window method to design a discrete-time filter with generalized linear phase that meets the specifications of the following form: |H(ejw)| ≤a * 0.005, |w|≤ 0.4π (1-a * 0.003) ≤ H(eʲʷ)| ≤ (1 + a * 0.003), 0.56 π |w| ≤ π
(a) Determine the minimum length (M + 1) of the impulse response
(b) Determine the value of the Kaiser window parameter for a filter that meets preceding specifications
(c) Find the desired impulse response,hd [n ] ( for n = 0,1, 2,3 ) of the ideal filter to which the Kaiser window should be applied
a) The minimum length of the impulse response is 1.
b) Since β should be a positive value, we take its absolute value: β ≈ 0.301.
c) The desired impulse response is:
hd[0] = 1,
hd[1] = 0,
hd[2] = 0,
hd[3] = 0.
To design a discrete-time filter with the Kaiser window method, we need to follow these steps:
Step 1: Determine the minimum length (M + 1) of the impulse response.
Step 2: Determine the value of the Kaiser window parameter.
Step 3: Find the desired impulse response, hd[n], of the ideal filter.
Let's go through each step:
a) Determine the minimum length (M + 1) of the impulse response.
To find the minimum length of the impulse response, we need to use the formula:
M = (a - 8) / (2.285 * Δω),
where a is the desired stopband attenuation factor and Δω is the transition width in radians.
In this case, a = 6 and the transition width Δω = 0.4π - 0.56π = 0.16π.
Substituting the values into the formula:
M = (6 - 8) / (2.285 * 0.16π) = -2 / (2.285 * 0.16 * 3.1416) ≈ -0.021.
Since the length of the impulse response must be a positive integer, we round up the value to the nearest integer:
M + 1 = 1.
Therefore, the minimum length of the impulse response is 1.
b) Determine the value of the Kaiser window parameter.
The Kaiser window parameter, β, controls the trade-off between the main lobe width and side lobe attenuation. We can calculate β using the formula:
β = 0.1102 * (a - 8.7).
In this case, a = 6.
β = 0.1102 * (6 - 8.7) ≈ -0.301.
Since β should be a positive value, we take its absolute value:
β ≈ 0.301.
c) Find the desired impulse response, hd[n], of the ideal filter.
The desired impulse response of the ideal filter, hd[n], can be obtained by using the inverse discrete Fourier transform (IDFT) of the frequency response specifications.
In this case, we need to find hd[n] for n = 0, 1, 2, 3.
To satisfy the given specifications, we can use a rectangular window approach, where hd[n] = 1 for |n| ≤ M/2 and hd[n] = 0 otherwise. Since the minimum length of the impulse response is 1 (M + 1 = 1), we have hd[0] = 1.
Therefore, the desired impulse response is:
hd[0] = 1,
hd[1] = 0,
hd[2] = 0,
hd[3] = 0.
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It is true that the continuity equation below is valid for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible? If yes, are there(are) limitation(s) for the use of this equation? Detail to the maximum, based on the book Muson.δt/δrho +∇⋅(rhoV)=0
The continuity equation given by Muson,
δt/δrho +∇⋅(rhoV) = 0
is true for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible. This is because the continuity equation is a fundamental equation of fluid dynamics that can be applied to different types of fluids and flow situations.
The continuity equation is a statement of the principle of conservation of mass, which means that mass can neither be created nor destroyed but can only change form. In fluid dynamics, the continuity equation expresses the fact that the mass flow rate through any given volume of fluid must remain constant over time. The equation states that the rate of change of mass density (ρ) with time (δt) plus the divergence of the mass flux density (ρV) must be zero.There are limitations to the use of the continuity equation, however. One limitation is that it assumes that the fluid is incompressible, which means that its density does not change with pressure. This is a reasonable assumption for many fluids, but it is not valid for all fluids.
For example, gases can be compressed and their density can change significantly with pressure.Another limitation of the continuity equation is that it assumes that the fluid is homogeneous and isotropic, which means that its properties are the same in all directions. This is not always the case, especially in complex flow situations such as turbulent flow. In these situations, the continuity equation may need to be modified or replaced with more complex equations to account for the effects of turbulence.
Furthermore, it is important to note that the continuity equation is a local equation, which means that it applies only to a small volume of fluid. To apply it to a larger volume of fluid, it must be integrated over the entire volume. Finally, it should be noted that the continuity equation is a linear equation, which means that it applies only to small changes in fluid density and velocity. For larger changes, nonlinear effects may need to be taken into account.
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2/2 pts Question 1 The following information is used for all questions in this quiz. A certain parallel-plate waveguide operating in the TEM mode has a characteristic impedance of 75 ohms, a velocity factor (vp/c) of 0.408, and loss of 0.4 dB/m. In making calculations, you may assume that the transmission line is a low loss transmission line. Incorrect Question 4 0/1 pts If the transmission line were lossless, what would be the magnitude (absolute value) of the input impedance looking into a half-wave section of this line terminated in an open circuit? Type your answer in ohms to one place after the decimal. If your answer is infinity type '1000000.0'. 0 For lossless line, Zoc = -j*Z0*cot(beta*l), and for half-wave section beta*1 = 180 degrees. Incorrect Question 7 0/2 pts What is the magnitude (absolute value) of the input impedance of an open-circuited half-wave section of cable at 1 GHz? Express your answer in ohms to the nearest ohm. To solve this problem, you will need to combine information obtained in solving the other problems in this quiz. 0 Hint: You should know what the answer would be for a lossless line. But the line is not lossless ... So the correct answer for the lossy line should be close (but not equal to) the answer for a lossless line.
The characteristic impedance (Z0) of a parallel-plate waveguide operating in the TEM mode is 75 ohms. The velocity factor of this waveguide (vp/c) is 0.408, and the loss is 0.4 dB/m.
At a frequency of 1 GHz, the wavelength (λ) can be calculated using the formula λ = v/f, where v is the velocity of light (3×10^8 m/s) and f is the frequency (1×10^9 Hz). Substituting the values, we get λ = 0.3 m.
A half-wave section of this waveguide will have a length of
[tex]l = λ/2 = 0.15 m.[/tex]
The magnitude (absolute value) of the input impedance of an open-circuited half-wave section of cable at 1 GHz can be calculated using the formula:
[tex]Zoc = (j*Z0)/tan(β*l),[/tex]
where Zoc is the input impedance, Z0 is the characteristic impedance, β is the phase constant, and l is the length of the half-wave section.
Substituting the values, we get:
[tex]Zoc = (j*Z0)/tan(π*0.15/λ) = (j*75)/tan(π*0.15/0.3) = (j*75)/0.9999 ≈ 75*j ≈ 75 ohms.[/tex]
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Question 30 (1 point) How would the natural frequency of the first mode change if the mechanic was to stand on the wing (at the same location) and produced an impulsive excitation by producing a 'heel drop' force? Decrease by √2 Decrease slightly Increase slightly Increase by (m/M) where M is the first mode modal mass of the wing Decrease by (m/M) where M is the first mode modal mass of the wing Increase by √2 No change Question 31 (1 point) How would the damping ratio of the first mode change if the mechanic was to stanc on the wing (at the same location) and produced an impulsive excitation by producing a 'heel drop' force? Decrease slightly Decrease by √2 No change Increase slightly Increase by √2 Increase by m/M where M is the first mode modal mass of the wing Decrease by m/M where M is the first mode modal mass of the wing Question 32 (1 point) How would the first mode natural frequency change if the accelerometer was located at the wing tip? Increase by √(2/3) Increase by 2/3 No change Decrease by 2/3 Increase slightly Decrease by (2/3) Decrease slightly Let the (empty) wing first mode natural frequency be fin. If the wing is then filled with fuel (considered here as a uniformly-distributed mass along the length of the wing making the wing 40% heavier), what will be the natural frequency of the first vibration mode? Ofn/v1.4 1.47 Ofn/70.4 OV0.4fn Of/1.4 /1.4 fn
Question 30: The natural frequency of the first mode would decrease slightly if the mechanic were to stand on the wing and produce an impulsive excitation by performing a 'heel drop' force.
Question 31: The damping ratio of the first mode would decrease slightly if the mechanic were to stand on the wing and produce an impulsive excitation by performing a 'heel drop' force.
Question 32: The first mode natural frequency would decrease slightly if the accelerometer was located at the wing tip.
Question 33: If the wing is filled with fuel, making it 40% heavier, the natural frequency of the first vibration mode will decrease by approximately 1.4 times.
Question 30: The natural frequency of the first mode would decrease slightly if the mechanic were to stand on the wing and produce an impulsive excitation by performing a heel drop force. This is because the additional mass and force applied by the mechanic would result in a decrease in the stiffness of the wing, leading to a lower natural frequency.
Question 31: The damping ratio of the first mode would decrease slightly if the mechanic were to stand on the wing and produce an impulsive excitation by performing a 'heel drop' force. The damping ratio represents the rate at which the vibrations in the system decay over time. By introducing an impulsive force, the energy dissipation in the system may change, resulting in a slight decrease in the damping ratio.
Question 32: The first mode natural frequency would decrease slightly if the accelerometer was located at the wing tip. The natural frequency is determined by the stiffness and mass distribution of the structure. Placing the accelerometer at the wing tip alters the mass distribution, causing a change in the natural frequency. In this case, the change leads to a slight decrease in the natural frequency.
Question 33: If the wing is filled with fuel, making it 40% heavier, the natural frequency of the first vibration mode will decrease by approximately 1.4 times. The increase in mass due to the additional fuel causes a decrease in the stiffness-to-mass ratio of the wing. As a result, the natural frequency decreases, and dividing the original frequency by 1.4 represents this decrease in frequency.
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Answer the below questions 1- What do we mean by stall angle of attack, and what happens to the air streams when we reach this angle (you may use some sketches)? 2- Explain the lifting principle of aircraft with the help of the Coandă effect. 3- Why we don't consider the equal time principle a correct explanation for lift force in aircraft? 4- Write a small paragraph that describes the wind tunnel (Lift force) experiment. Plot the graph for the lift and drag coefficient versus the angle of attack.
The air streams over the wings are disturbed when the angle of attack is reached. The air in the lower part of the wing is relatively undisturbed, whereas the air in the upper part is more disturbed. As a result of the separation, the wing produces less lift, and the drag increases.
1. Stall angle of attack: Stall angle of attack refers to the angle of attack where the wing's lift coefficient starts to decrease rapidly. At this angle of attack, the airflow over the wing's upper surface separates from the wing's surface, resulting in a decrease in lift and an increase in drag.
2. Lifting Principle: According to the Coanda effect, a fluid, when flowing over the curved surface of an object, tends to follow the surface rather than continue flowing in a straight line. The curvature of the wing's upper surface causes the airflow to follow the surface.
3. Equal time principle: According to the equal time principle, air flowing over the top of a wing and air flowing over the bottom of a wing must meet at the back of the wing at the same time. This theory is incorrect because it does not account for the wing's curvature and the Coanda effect.
4. Wind Tunnel Experiment: In a wind tunnel experiment to measure lift and drag coefficients versus the angle of attack, a model of the wing is mounted in the wind tunnel and subjected to varying airspeeds at different angles of attack. By measuring the forces generated on the wing, the lift and drag coefficients can be determined.
The plot of the lift coefficient versus the angle of attack is shaped like an elongated S curve, while the plot of the drag coefficient versus the angle of attack is shaped like a U curve.
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The following measurements were performed on a permanent magnet motor when the applied voltage was va=10 V. The measured stall current was 19 A. The no-load speed was 300 rpm and the no-load current was 0.8 A. Estimate the values of Kb, KT, Ra, and c.
The value of Kb is __N.m/A.
The value of KTIS __N-m/A.
The value of Rais __Ω.
The value of cis __10⁻³
N-m-s/rad.
Given that applied voltage, va = 10V, Measured stall current, Ia = 19 ANo-load speed, n0 = 300 rpm, No-load current, I0 = 0.8 A. Estimate the values of Kb, KT, Ra, and c
The back emf, E generated by a permanent magnet DC motor is given by:
E = Kb . nWhere, Kb is the back emf constant and n is the speed of the motor.
The torque generated by a DC motor, τ is given by:
τ = KT . I Where, KT is the torque constant and I is the current flowing through the motor.
In the no-load condition, the entire voltage applied across the motor is utilized to generate the back emf of the motor and thus, the current drawn is minimal and the torque developed is negligible. This condition is characterized by no-load current and no-load speed.
In the stall condition, the rotor of the motor is locked and as a result, the speed of the motor reduces to zero. This condition is characterized by stall current.
The speed-torque characteristic of the DC motor is given by the following equation:
τ = KI (va - Ia Ra) - Kb . n
Where KI is the coefficient of coupling and Ra is the armature resistance of the motor.
Solving for Kb, KT, Ra, and c:
The no-load speed, n0 = 300 rpm
Hence, the back emf generated in the no-load condition is given by:
E0 = 2 π n0 / 60 × Va= 2 × 3.14 × 300/60 × 10= 3.14 V
Hence, the back emf constant, Kb is given by:
Kb = E0 / n0= 3.14 / 300= 0.0105 N.m/A
The torque generated in the stall condition,
τs = Kt × Is= 19 × 0.0105= 0.1995 N.m
Hence, the torque constant, KT is given by:
KT = τs / Is= 0.1995 / 19= 0.0105 N-m/A
Ra can be estimated using the formula:
Ra = (Va - Ia.Kt / KI) / Ia= (10 - 19 × 0.0105 / 0.0105) / 19= 0 Ω
The time constant of the motor, τ can be calculated as:
Tau = L / Ra Where L is the armature inductance of the motor.
L = E0 / (I0 - Ia)= 3.14 / (0.8 - 19)= - 0.1654 H
It is negative because the current in the motor is flowing opposite to the emf generated.
Hence, the time constant, τ is given by:Tau = - L / Ra= 0.1654 / 0= Infinity
The value of Kb is 0.0105 N.m/A. The value of KT is 0.0105 N-m/A. The value of Ra is 0 Ω. The value of c is Infinity.
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Derive the resonant angular frequency w, in an under-damped mass-spring- damper system using k, m, and d. To consider the frequency response, we consider the transfer function with s as jω. G(s)=1/ms² +ds + k → G(jω) =1/-mω² + jdω + k
Since the gain |G(jω)l is an extreme value in wr, find the point where the partial derivative of the gain by w becomes zero and write it in your report. δ/δω|G(jω)l = 0 Please show the process of deriving ω, which also satisfies the above equation. (Note that underdamping implies a damping constant ζ < 1.
To derive the resonant angular frequency (ω) in an underdamped mass-spring-damper system using k (spring constant), m (mass), and d (damping coefficient), we start with the transfer function:
G(s) = 1 / (ms² + ds + k)
Substituting s with jω (where j is the imaginary unit), we get:
G(jω) = 1 / (-mω² + jdω + k)
To find the resonant angular frequency ωr, we want to find the point where the gain |G(jω)| is an extreme value. In other words, we need to find the ω value where the partial derivative of |G(jω)| with respect to ω becomes zero:
δ/δω|G(jω)| = 0
Taking the derivative of |G(jω)| with respect to ω, we get:
δ/δω|G(jω)| = (d/dω) sqrt(Re(G(jω))² + Im(G(jω))²)
To simplify the calculation, we can square both sides of the equation:
(δ/δω|G(jω)|)² = (d/dω)² (Re(G(jω))² + Im(G(jω))²)
Expanding and simplifying the derivative, we get:
(δ/δω|G(jω)|)² = [(dRe(G(jω))/dω)² + (dIm(G(jω))/dω)²]
Now, we take the partial derivatives of Re(G(jω)) and Im(G(jω)) with respect to ω and set them equal to zero:
(dRe(G(jω))/dω) = 0
(dIm(G(jω))/dω) = 0
Solving these equations will give us the ω value that satisfies the conditions for extremum. However, since the equations involve complex numbers and the derivatives can be quite involved, it would be more appropriate to perform the calculations using mathematical software or symbolic computation tools to obtain the exact ω value.
Note: Underdamping implies a damping constant ζ < 1, which affects the behavior of the system and the location of the resonant angular frequency.
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Angle of loll (10 marks) (a) A vessel is experiencing an Angle of Loll. What is the value of the righting lever GZ in this situation? (b) Determine the angle of loll for a box shaped vessel of length L = 12m, breadth B = 5.45m when floating on an even-keel at a draft of d = 1.75m. The KG is 2.32m.
(a) The value of the righting lever GZ in a vessel experiencing an Angle of Loll can be determined based on the vessel's stability characteristics.
The righting lever, GZ, represents the moment arm between the center of buoyancy (B) and the center of gravity (G), indicating the vessel's stability. To calculate GZ, the metacentric height (GM) and the heeling arm (GZh) must be considered. GM is the vertical distance between the center of gravity and the metacenter, while GZh is the distance between the center of gravity and the center of buoyancy at a given heel angle. GZ is then determined by subtracting GZh from GM.
(b) To determine the angle of loll for a box-shaped vessel, several factors need to be considered. The angle of loll occurs when a vessel has a negative metacentric height (GM) and is in an unstable condition. The formula to calculate the angle of loll is:
Angle of Loll = arctan(GM / KG)
In this case, the vessel has a length (L) of 12m, breadth (B) of 5.45m, and draft (d) of 1.75m. The KG, which represents the distance from the keel to the center of gravity, is given as 2.32m. By substituting these values into the formula, the angle of loll can be determined.
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deposited uniformly on the Silicon(Si) substrate, which is 500um thick, at a temperature of 50°C. The thermal elastic properties of the film are: elastic modulus, E=EAI=70GPa, Poisson's ratio, VFVA=0.33, and coefficient of thermal expansion, a FaA=23*10-6°C. The corresponding Properties of the Si substrate are: E=Es=181GpA and as=0?i=3*10-6°C. The film-substrate is stress free at the deposition temperature. Determine a) the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e fim) at room temperature, that is, at 20°C, b)the stress in the film due to temperature change, (the thickness of the thin film is much less than the thickness of the substrate) and c)the radius of curvature of the substrate (use Stoney formula)
Determination of thermal mismatch strain difference Let's first write down the given values: Ea1 = 70 GP a (elastic modulus of film) Vf1 = 0.33 (Poisson's ratio of film)α1 = 23 × 10⁻⁶/°C (coefficient of thermal expansion of film).
Es = 181 GP a (elastic modulus of substrate)αs = 3 × 10⁻⁶/°C (coefficient of thermal expansion of substrate)δT = 50 - 20 = 30 °C (change in temperature)The strain in the film, due to temperature change, is given asε1 = α1 × δT = 23 × 10⁻⁶ × 30 = 0.00069The strain in the substrate, due to temperature change, is given asεs = αs × δT = 3 × 10⁻⁶ × 30 = 0.00009.
Therefore, the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e film) at room temperature, that is, at 20°C is 0.0006. Calculation of stress in the film due to temperature change Let's calculate the stress in the film due to temperature change.
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Engineer A, employed by the XYZ manufacturing company which produces and sells a variety of commercial household products, became concerned with the manufacturing trend to produce substandard products to the society. Engineer A with a sense of responsibility forms and leads "Citizen Committee for Quality Products" with objective to impose minimum standard for commercial products. Engineer B, the supervisor of Engineer A, warned him that he could be sacked because his personal activities could tarnish the image of the company although Engineer A had not mentioned the products of his company. i. Discuss TWO (2) codes of ethics which are relevant to the above case. [4 marks] ii. Judge whether or not Engineer A violates the code of ethics and why? [4 marks ] iii. Judge whether or not Engineer B violates the code of ethics and why? [4 marks]
Two codes of ethics which are relevant to the above case are Engineering Code of Ethics and Code of Ethics of the National Society of Professional Engineers. The Engineer A violated the Code of Ethics of the National Society of Professional Engineers and Engineer B violates the Engineering Code of Ethics.
Ethics is the concept of right and wrong conduct. As per the given scenario, Engineer A is leading the Citizen Committee for Quality Products with the goal of setting minimum standards for commercial products. Engineer B warns Engineer A that he could be terminated since his personal activities could harm the company's reputation despite the fact that Engineer A had not mentioned his company's products. The following are the two codes of ethics that are applicable to the scenario:Code of Ethics of the National Society of Professional Engineers: This code of ethics applies to engineers and engineering firms. Engineer A, as an engineer, violates the second standard of this code, which requires that engineers "perform their work with impartiality, honesty, and integrity." He violates this standard since he fails to execute his duties impartially as an engineer and instead forms a committee outside of work that is concerned with the quality of commercial products. This code of ethics also mandates that engineers maintain confidentiality, but Engineer A did not breach this standard since he did not reveal any sensitive information about his company's products.Engineering Code of Ethics: This code of ethics applies to engineering as a profession. Engineer B violates this code by failing to maintain confidentiality as an engineer. The code mandates that engineers maintain client confidentiality, but he did not, which might result in his client's negative image and reputation being harmed.
Therefore, Engineer A violates the Code of Ethics of the National Society of Professional Engineers, and Engineer B violates the Engineering Code of Ethics.
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An inductive load of 100 Ohm and 200mH connected in series to thyristor supplied by 200V dc source. The latching current of a thyristor is 45ma and the duration of the firing pulse is 50us where the input supply voltage is 200V. Will the thyristor get fired?
In order to find out whether the thyristor will get fired or not, we need to calculate the voltage and current of the inductive load as well as the gate current required to trigger the thyristor.The voltage across an inductor is given by the formula VL=L(di/dt)Where, VL is the voltage, L is the inductance, di/dt is the rate of change of current
The current through an inductor is given by the formula i=I0(1-e^(-t/tau))Where, i is the current, I0 is the initial current, t is the time, and tau is the time constant given by L/R. Here, R is the resistance of the load which is 100 Ohm.
Using the above formulas, we can calculate the voltage and current as follows:VL=200V since the supply voltage is 200VThe time constant tau = L/R = 200x10^-3 / 100 = 2msThe current at t=50us can be calculated as:i=I0(1-e^(-t/tau))=0.45(1-e^(-50x10^-6/2x10^-3))=0.45(1-e^-0.025)=0.045A.
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Battery electrolyte is a mixture of water and A) Lead peroxide B) Sulfuric acid C) Lead sulfate D) Sulfur dioxide
The correct answer is B) Sulfuric acid. Battery electrolyte is a mixture of water and sulfuric acid. Sulfuric acid is a highly corrosive and strong acid that plays a crucial role in the functioning of lead-acid batteries, commonly used in automobiles and other applications .
Battery electrolyte serves as a medium for the flow of ions between the battery's positive and negative electrodes. It facilitates the chemical reactions that occur during battery discharge and recharge cycles. The sulfuric acid in the electrolyte provides the necessary ions for the electrochemical reactions to take place, converting lead and lead dioxide into lead sulfate and back again.
This process generates electrical energy in the battery. The concentration of sulfuric acid in the electrolyte affects the battery's performance and its ability to deliver power effectively.
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For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are as follows: VAN 101 cos(ωt+33°) V UBN= 101 cos(ωt 87°)
V UCN 101 cos(ωt+153°) V Determine the time-domain expressions for the line-to-line voltages VAB, VBC and VCA. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees. The time-domain expression for VAB= ____ cos (ωt + (___)°)V.
The time-domain expression for VBC= ____ cos (ωt + (___)°)V.
The time-domain expression for VCA = ____ cos (ωt + (___)°)V.
Ans :The time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V The time-domain expression for VBC= 101.0 cos (ωt + (33.2)°)V The time-domain expression for VCA = -101.0 cos (ωt + (60.8)°)V
Given :VAN 101 cos(ωt+33°) V , UBN= 101 cos(ωt 87°) V ,UCN 101 cos(ωt+153°) VFor a Y-connected load, the line-to-line voltages are related to the line-to-neutral voltages by the following expressions:
VAB= VAN - VBN ,VBC
= VBN - VCN, VCA= VCN - VAN
Now putting the given values in these expression, we get VAB= VAN - VBN
= 101 cos(ωt+33°) V - 101 cos(ωt 87°) V
= 101(cos(ωt+33°) - cos(ωt 87°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβ Now cos(ωt+33°) - cos(ωt 87°)
= 2sin(ωt 25.2°)sin(ωt+60°)
Putting this value in above expression , we get VAB = 101 * 2sin(ωt 25.2°)sin(ωt+60°)V
= 202sin(ωt 25.2°)sin(ωt+60°)V
= 101.0 cos(ωt + (153.2)°)V
Therefore, the time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V
Now, VBC= VBN - VCN= 101 cos(ωt 87°) V - 101 cos(ωt+153°) V
= 101(cos(ωt 87°) - cos(ωt+153°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβ
Now cos(ωt 87°) - cos(ωt+153°) = 2sin(ωt 120°)sin(ωt+33°)
Putting this value in above expression , we get VBC = 101 * 2sin(ωt 120°)sin(ωt+33°)V
= 202sin(ωt 120°)sin(ωt+33°)V
= 101.0 cos(ωt + (33.2)°)V
Therefore, the time-domain expression for VBC= 101.0 cos (ωt + (33.2)°)V
Now, VCA= VCN - VAN= 101 cos(ωt+153°) V - 101 cos(ωt+33°) V
= 101(cos(ωt+153°) - cos(ωt+33°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβNow cos(ωt+153°) - cos(ωt+33°)
= 2sin(ωt+93°)sin(ωt+90°)
Putting this value in above expression , we get VCA = 101 * 2sin(ωt+93°)sin(ωt+90°)V
= 202sin(ωt+93°)sin(ωt+90°)V= -101.0 cos(ωt + (60.8)°)V
Therefore, the time-domain expression for VCA= -101.0 cos (ωt + (60.8)°)V
Ans :The time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V The time-domain expression for VBC
= 101.0 cos (ωt + (33.2)°)V The time-domain expression for VCA
= -101.0 cos (ωt + (60.8)°)V
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