The rate of evaporation of water per unit width of the container is 5.45 × 10^-6 mol/(m.s).
Given data:
Temperature of air, T_1 = 32.0 ℃
Length of the container, L = 1.2 m
Temperature at the air-water interface, T2 = 20.0 ℃
Initial humidity of air, H_1 = 25.0%
Speed of air, V = 0.15 m/s
Water vapour pressure at T2,
Psat = 0.02308 atm
Water vapour pressure at T1,
P = 0.04696 atm
Gas constant, R = 0.082 atm l/Kmol
Diffusion coefficient of water in air, Dwater = 2.77 × 10^-5 m^2⁄s
Using the Sherwood Number equation:
Sℎ = 0.664Re^0.5Sc^0.333
Where Re is Reynolds's Number and Sc is Schmidt's Number.
Mass transfer coefficient = Dwater / L ShSc= 0.7
for air-water interface at 25°CSc = 2.14 × 10^-5 / 0.0343 = 6.23 × 10^-4 (calculated from Sc = v/D)
Re = ρvd/μ = 1092.8 (calculated from Re = VDwater/ν, where ν = viscosity of air = 1.81 × 10^-5 kg/m.s)
Therefore, Sh = 2.0 (calculated from Sherwood Number equation)
Mass transfer coefficient = Dwater / L Sh
= 2.77 × 10^-5 / (1.2 × 2) = 1.15 × 10^-5 m/s
Calculating the rate of evaporation of water per unit width of the container:
RH1 = H1 Psat / P - Psat
= 6.85% (Relative humidity)
Mass transfer rate = KH2O A RH = KH2O L RH1
W= 1.15 × 10^-5 × 1.2 × 6.85 / 18
= 5.45 × 10^-6 mol/(m.s)
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Write any five Verilog and VHDL code Simulate and realize the following applications using Xilinx Spartan 6 FPGA PROCESSOR. (using structural/dataflow /behavioural modelling)
1. BCD counter
2. 7 segment display
Verilog and VHDL are two of the most popular hardware description languages used in the electronic industry. They are used to design digital systems. Spartan 6 FPGA PROCESSOR is an integrated circuit that is programmable, hence can be used in a wide range of applications.
Some of the applications that can be realized using Spartan 6 FPGA PROCESSOR include BCD counter and 7 segment display. The applications can be realized using structural, dataflow, or behavioural modelling. Here are five Verilog and VHDL code simulate for the applications using Xilinx Spartan 6 FPGA PROCESSOR.
These are some of the Verilog and VHDL codes that can be used to simulate and realize BCD counter and 7 segment display using Xilinx Spartan 6 FPGA PROCESSOR. Note that the code can be modified to meet specific design requirements.
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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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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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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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A one kilogram of moist air has a dry bulb temperature and relative humidity of 35 °C and 70%, respectively. The air is cooled until its temperature reaches 5 °C. The air pressure is 1 bar and remains constant during the cooling process. Calculate the amount of the condensed water. Also find the amount of heat transferred per kg of dry air (sensible, latent, and total). (0.02 kg, -30.105 kJ, -48.372 kJ, -78.477 kJ)
Given parameters, Dry bulb temperature, T1 = 35 °C Relative Humidity, φ1 = 70%Mass of air, m = 1kgPressure, P = 1 bar, Final temperature, T2 = 5 °C Solution :First, we will find out the dew point temperature (Tdp) at T1Step 1: Calculation of Dew Point Temperature (Tdp).
We can use the formula:T[tex]dp=243.04×[lnφ1/100 + (17.625T1)/(243.04+T1)]\\[/tex]
We will substitute the values in the above equation:T
[tex]dp=243.04×[ln(70/100) + (17.625 × 35)/(243.04+35)] = 25.34 °C[/tex]
Now, we have Tdp and T1, so we can calculate the moisture content (ω1) in the air.Step 2: Calculation of moisture content (ω1)The formula to calculate ω is given by:
[tex]ω1=0.622×[e/(P−e)]Here,e= (0.611×exp((17.502×Tdp)/(Tdp+240.97)))…[/tex]
(1)We will put Tdp value in the equation (1):
[tex]e= (0.611×exp((17.502×25.34)/(25.34+240.97))) = 3.283 k PaPut the value of e in the equation (2):ω1=0.622×[3.283/(100−3.283)] = 0.0215 kg/kg[/tex]
Total heat transferred, Q = Q sensible + Qlatent. Sensible heat is responsible for temperature change, while latent heat is responsible for the phase change of the moisture present. We can find Qlatent by using the formula:Qlatent=mc×hfg(T1)Here hfg(T1) is the latent heat of vaporization of water at T1. It can be calculated using the formula:hfg(T1)=2501−2.361T1Now, we can calculate the latent heat of vaporization,
[tex]hfg(T1)hfg(T1)=2501−2.361×35 = 2471.89 J/gSo, Qlatent=0.0168×2471.89 = -41.561 kJ/kg[/tex]
We can find the sensible heat by using the formula:Qsensible = mCpd (T1 - T2)Here Cp is the specific heat capacity of dry air at constant pressure. We can find the value of Cp by using the following formula
[tex]Cp=1.005+1.82ω1Here, ω1 = 0.0215, so,Cp = 1.005+1.82×0.0215 = 1.046 J/g/[/tex]
K Now, we can find Q sensible by using the formula:
[tex]Q sensible = m Cpd(T1 - T2)Q sensible = 1×1.046×(35-5) = 31.38 kJ/kg[/tex]
Total Heat transfer is [tex]Qsensible + Qlatent = -41.561 + 31.38 = -10.181 kJ[/tex]/kg.
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A carbon steel shaft has a length of 700 mm and a diameter of 50 mm determine the first shaft critical of the shaft due to its weight ?
When a slender structure such as a shaft is subjected to torsional loading, it will exhibit a critical speed known as the shaft's critical speed. The critical speed of a shaft is the speed at which it vibrates the most when subjected to an external force or torque.
The shaft's natural frequency is related to its stiffness and mass, and it is critical because if the shaft is allowed to spin at or near its critical speed, it may undergo significant torsional vibration, which can lead to failure. The critical speed of a shaft can be calculated by the following formula:ncr = (c/2*pi)*sqrt((D/d)^4/(1-(D/d)^4))
Where:ncr is the critical speed of the shaft in RPMsD is the diameter of the shaft in metersd is the length of the shaft in metersc is the speed of sound in meters per secondWe have the following data from the given problem:A carbon steel shaft has a length of 700 mm and a diameter of 50 mm. We will convert these units to meters so that the calculations can be done consistently in SI units.Length of the shaft, l = 700 mm = 0.7 mDiameter of the shaft, D = 50 mm = 0.05 m.
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(a) If the current flowing in a circuit is related to time by the formula i(t)=lde−5t cos5t and is applied to a capacitor with capacitance C=0.2 F. The voltage drops across the capacitor is given by VC=
1/c
∫i(t)dt, (i) Approximate VC,0≤t≤0.8 with h=0.1 by using trapezoidal rule and suitable Simpson's rule. (13 marks) (ii) Find the absolute error for each method from Q2(a)(i) if the actual value of VC is 0.498 V. (2 marks) (iii) Determine the best approximation method. (1 mark)
Thus, the best approximation method for approximating VC is Simpson's rule.
(i) Approximate VC, 0≤t≤0.8 with h=0.1 by using trapezoidal rule and suitable Simpson's rule.
Let us apply the trapezoidal rule to obtain the approximate value of VC at t = 0.8.Taking h = 0.1,i.e., n = (0.8 - 0) / 0.1 = 8, we have
VC = 1/C ∫ i(t) dt
VC = 1/0.2 ∫lde−5t cos5t dt
VC = 5 [0.0032 + 0.0198 + 0.0319 + 0.0362 + 0.0343 + 0.0281 + 0.0186 + 0.0077]
VC = 0.491 V
Now, let us apply the Simpson’s rule to obtain the approximate value of VC at t = 0.8. Taking h = 0.1,i.e., n = (0.8 - 0) / 0.1 = 8, we have
VC = 1/C ∫ i(t) dt
VC = 1/0.2 ∫lde−5t cos5t dt
VC = (h/3C) [i(0) + 4i(0.1) + 2i(0.2) + 4i(0.3) + 2i(0.4) + 4i(0.5) + 2i(0.6) + 4i(0.7) + i(0.8)]
VC = 0.497 V
(ii) Find the absolute error for each method from Q2(a)(i) if the actual value of VC is 0.498 V.
For trapezoidal rule
Absolute error = actual value - approximate value
Absolute error = 0.498 - 0.491 = 0.007 V
For Simpson’s rule
Absolute error = actual value - approximate value
Absolute error = 0.498 - 0.497 = 0.001 V
(iii) Determine the best approximation method.
The smaller the error, the better the method. From (ii) the error in Simpson’s rule is smaller. Hence, Simpson’s rule is the better approximation method.
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Chopped hemp fibre reinforced polyester with 55% volume fraction of fibres: • hemp fiber radius is 7.2 x 10-2 mm • an average fiber length of 8.3 mm fiber fracture strength of 2.8 GPa • matrix stress at the composite failure of 5.9 MPa • matrix tensile strength of 72 MPa • shear yielding strength of matrix 35 MPa (a) Calculate the critical fibre length. (6 marks) (b) With the aid of graph for stress vs. length, state whether the existing fibre length is enough for effective strengthening and stiffening of the composite material or not. (5 marks) (c) Glass fibre lamina with a 75% fibre volume fraction with Pglass = pr=2.5 gem?, ve=0.2, Vm = 0.3, Pepoxy = Pm= 1.35 gem?, Er= 70 GPa and Em = 3.6 GPa. Calculate the density of the composite and the mass fractions (in %) of the fibre and matrix. (14 marks)
The mass fractions of fiber and matrix are 74.53% and 25.47%, respectively.
(a) Calculation of critical fiber length:
Critical fiber length can be given by the following equation-:
lf = (tau_m / tau_f)^2 (Em / Ef)
Where,
tau_m = Matrix stress at composite failure
5.9 MPa;
tau_f = Fiber fracture strength
= 2.8 GPa;
Em = Matrix modulus
= 3.6 GPa;
Ef = Fiber modulus
= 70 GPa;
lf = critical fiber length.
So, putting the values in the formula, we get-:
lf = (5.9*10^6 / 2.8*10^9)^2 * (3.6*10^9 / 70*10^9)
= 0.0153 mm
Thus, the critical fiber length is 0.0153 mm.
(b) It is required to draw the stress-length graph first. Stress and length of fibers in the composite material are inversely proportional, thus as the length increases, the stress decreases.
The graph thus obtained is a straight line and the point where it intersects the horizontal line at 5.9 MPa gives the required length. So, the existing fiber length is not enough for effective strengthening and stiffening of the composite material.(c) Calculation of composite density: Composite density can be calculated using the following formula-:
Pcomposite = Vf * Pglass + Vm * Pm
Where,
Pcomposite = composite density;
Vf = fiber volume fraction = 0.75;
Pglass = density of glass fiber
= 2500 kg/m³;
Vm = matrix volume fraction
= 0.25;
Pm = density of matrix
= 1350 kg/m³.
So, putting the values in the formula, we get-:
Pcomposite = 0.75*2500 + 0.25*1350
= 2137.5 kg/m³
Calculation of mass fractions of fiber and matrix:
Mass fraction of fiber can be given by-:
mf = (Vf * Pglass) / (Vf * Pglass + Vm * Pm) * 100%
And, mass fraction of matrix can be given by-:
mm = (Vm * Pm) / (Vf * Pglass + Vm * Pm) * 100%
So, putting the values in the formulae, we get-:
mf = (0.75*2500) / (0.75*2500 + 0.25*1350) * 100%
= 74.53%
And,
mm = (0.25*1350) / (0.75*2500 + 0.25*1350) * 100%
= 25.47%
Therefore, the mass fractions of fiber and matrix are 74.53% and 25.47%, respectively.
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A drying chamber is maintained at 40.5 to 50.5 Centigrade having air outlet humidity ratio of 75 to 92 centigrade. If 800 to 1300 kg/hr of material at 10 to 15 percent moisture content determine the amount of feed material in kg/hr. If ambient air is recorded at 30 to 34 centigrade and 23 to 25 centigrade wet bulb temperature and if 1.6 to 2.3 MPa pressure steam is used, determine the volumetric flowrate of air supplied to the dryer in m3/hr, heat supplied to the heater, amount of steam used in kg/hr, Effeciency of the dryer, and the temperature of the hot air from the dryer in degrees centigrade. Draw the necessary schematic diagram of the system and the psychrometric diagrams of air.
The amount of feed material in kg/hr can be determined based on the given range of material flow rates (800 to 1300 kg/hr) at 10 to 15 percent moisture content.
To determine the volumetric flowrate of air supplied to the dryer in m3/hr, the specific volume of air at the given ambient conditions needs to be calculated using psychrometric properties.The heat supplied to the heater can be determined by considering the amount of moisture to be evaporated from the feed material and the specific heat capacity of water.The amount of steam used in kg/hr can be determined by considering the energy required to heat the air and evaporate moisture from the feed material.The efficiency of the dryer can be calculated by comparing the heat input (energy supplied) to the heat output (energy used for drying). The temperature of the hot air from the dryer in degrees centigrade can be determined by analyzing the energy balance and considering the specific heat capacities of air and moisture.
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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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steel shelf is used to support a motor at the middle. The shelf is 1 m long, 0.3 m wide and 2 mm thick and the boundary conditions can be considered as fixed-fixed. Find the equivalent stiffness and the natural frequency of the shelf considering it as a SDOF system. Assume that the mass of the motor is 10 kg and operating speed is 1800 rpm. Given, Mass, m= 10 kg Length, L = 1 m Rotating speed, N = 1800 rpm Modulus's Young, E = 200 GPa
A steel shelf is used to support a motor, and it is treated as a (SDOF) Single Degree of Freedom system. The objective is to find the equivalent stiffness and natural frequency of the shelf.
To determine the equivalent stiffness of the steel shelf, we need to consider its geometry and material properties. The formula for the equivalent stiffness of a rectangular beam with fixed-fixed boundary conditions is:
k = (3 * E * w * h^3) / (4 * L^3)
Where:
k is the equivalent stiffness,
E is the modulus of elasticity (Young's modulus) of the steel material,
w is the width of the shelf,
h is the thickness of the shelf,
L is the length of the shelf.
Once we have the equivalent stiffness, we can calculate the natural frequency of the shelf using the formula:
f_n = (1 / (2 * π)) * √(k / m)
Where:
f_n is the natural frequency,
k is the equivalent stiffness,
m is the mass of the motor.
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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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25. Uncertainty: (10 points) Calculate the minimum uncertainty of position for a particle whose momentum is known to within 2x10-25 kg. m/s.
The minimum uncertainty of position for a particle whose momentum is known to within 2 x 10^-25 kg.m/s is calculated using the Uncertainty Principle of Heisenberg.Uncertainty Principle states that it is impossible to measure the exact position and momentum of an object simultaneously.
Mathematically, the principle is expressed as follows: Δx.Δp >= h/4π, where Δx is the uncertainty of position, Δp is the uncertainty of momentum, and h is Planck's constant, which has a value of 6.626 x 10^-34 J.s.Solving for Δx, the formula becomes:Δx >= h/4πΔp
Substituting the given values, we get:Δx >= (6.626 x 10^-34 J.s)/(4π x 2 x 10^-25 kg.m/s)≈ 2.65 x 10^-9 mTherefore, the minimum uncertainty of position for a particle whose momentum is known to within 2 x 10^-25 kg.m/s is approximately 2.65 x 10^-9 m.
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A) Draw and explain different type of material dislocation.
B) Explain the stages of Creep Test with aid of diagram.
C) Sketch and discuss creep strain and stress relaxation.
A- Material dislocation refers to the defects in the crystal lattice structure of a material. B- stages of a creep test include primary, secondary, and tertiary creep
A) Material Dislocation:
Dislocations are line defects in the crystal lattice of a material that affect its mechanical properties. There are three main types of dislocations:
Edge Dislocation: This type of dislocation occurs when an extra half-plane of atoms is introduced into the crystal lattice. It creates a step or edge along the lattice planes.
Screw Dislocation: A screw dislocation forms when the atomic planes of a crystal are displaced along a helical path, resulting in a spiral-like defect in the lattice structure.
Mixed Dislocation: Mixed dislocations possess characteristics of both edge and screw dislocations. They have components of edge motion along one direction and screw motion along another.
B) Stages of Creep Test:
Creep testing is performed to assess the time-dependent deformation behavior of a material under a constant load at elevated temperatures. The test typically consists of three stages:
Primary Creep: In this stage, the strain increases rapidly initially, but the rate of strain gradually decreases over time. It is associated with the adjustment and rearrangement of dislocations in the material.
Secondary Creep: The secondary stage is characterized by a relatively constant strain rate. During this stage, the rate of strain is balanced by the recovery processes occurring within the material, such as dislocation annihilation and grain boundary sliding.
Tertiary Creep: In the tertiary stage, the strain rate accelerates, leading to accelerated deformation and eventual failure. This stage is characterized by the development of localized necking, microstructural changes, and the occurrence of cracks or voids.
C) Creep Strain and Stress Relaxation:
Creep strain refers to the time-dependent and permanent deformation that occurs under constant stress and elevated temperatures. It is commonly represented by a logarithmic strain-time curve, exhibiting the different stages of creep.
Stress relaxation, on the other hand, refers to the decrease in stress over time under a constant strain. It is observed when a material is subjected to a constant strain and the stress required to maintain that strain gradually reduces.
Both creep strain and stress relaxation are important phenomena in materials science and engineering, especially for materials exposed to long-term loads at elevated temperatures. These processes can lead to significant deformation and structural changes in materials, which must be considered for design and reliability purposes.
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A- Material dislocation refers to the defects in the crystal lattice structure of a material. B- stages of a creep test include primary, secondary, and tertiary creep
A) Material Dislocation:
Dislocations are line defects in the crystal lattice of a material that affect its mechanical properties. There are three main types of dislocations:
Edge Dislocation: This type of dislocation occurs when an extra half-plane of atoms is introduced into the crystal lattice. It creates a step or edge along the lattice planes.
Screw Dislocation: A screw dislocation forms when the atomic planes of a crystal are displaced along a helical path, resulting in a spiral-like defect in the lattice structure.
Mixed Dislocation: Mixed dislocations possess characteristics of both edge and screw dislocations. They have components of edge motion along one direction and screw motion along another.
B) Stages of Creep Test:
Creep testing is performed to assess the time-dependent deformation behavior of a material under a constant load at elevated temperatures. The test typically consists of three stages:
Primary Creep: In this stage, the strain increases rapidly initially, but the rate of strain gradually decreases over time. It is associated with the adjustment and rearrangement of dislocations in the material.
Secondary Creep: The secondary stage is characterized by a relatively constant strain rate. During this stage, the rate of strain is balanced by the recovery processes occurring within the material, such as dislocation annihilation and grain boundary sliding.
Tertiary Creep: In the tertiary stage, the strain rate accelerates, leading to accelerated deformation and eventual failure. This stage is characterized by the development of localized necking, microstructural changes, and the occurrence of cracks or voids.
C) Creep Strain and Stress Relaxation:
Creep strain refers to the time-dependent and permanent deformation that occurs under constant stress and elevated temperatures. It is commonly represented by a logarithmic strain-time curve, exhibiting the different stages of creep.
Stress relaxation, on the other hand, refers to the decrease in stress over time under a constant strain. It is observed when a material is subjected to a constant strain and the stress required to maintain that strain gradually reduces.
Both creep strain and stress relaxation are important phenomena in materials science and engineering, especially for materials exposed to long-term loads at elevated temperatures.
These processes can lead to significant deformation and structural changes in materials, which must be considered for design and reliability purposes.
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Which statement is NOT true about fatigue crack?
(a) In low-cycle fatigue, crack generally propagates parallel to the tensile axis.
(b) The shape of fatigue crack at any given time can be indicated by the appearance of "beach marks’. (c) Sudden changes of section or scratches are very dangerous in high-cycle fatigue as it can ultimately initiate the crack there.
(d) Crack propagate slowly at first and then very rapidly once it reaches the critical size.
The statement that is NOT true about fatigue crack is (c) Sudden changes of section or scratches are very dangerous in high-cycle fatigue as it can ultimately initiate the crack there.
In high-cycle fatigue, sudden changes of section or scratches are generally not considered as significant factors in initiating fatigue cracks. High-cycle fatigue is characterized by a large number of stress cycles, typically in the order of thousands or millions, where the stress amplitude is relatively low. Cracks in high-cycle fatigue often initiate at stress concentration points or material defects rather than sudden changes of section or scratches.
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7. Given definitions of gm and ra as partial derivatives.
Partial derivatives allow us to see how the rate of change of a function changes with respect to a particular variable.
gm and ra are partial derivatives. The definitions of these terms are given below:gm: This is the transconductance of a device, and it measures the gain of the device with regards to the current. It can be expressed in units of amperes per volt or siemens. Transconductance (gm) = ∂iout/∂vgsra: This is the output resistance of the device, and it measures the change in output voltage with regards to the change in output current. It can be expressed in ohms.
Output resistance (ra) = ∂vout/∂ioutIf we look at the above definitions of gm and ra, we can see that both are partial derivatives. Partial derivatives are a type of derivative used in calculus. They are used to calculate how a function changes as a result of changes in one or more of its variables. In other words, partial derivatives allow us to see how the rate of change of a function changes with respect to a particular variable.
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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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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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Indicate the incorrect:
a. The change in length of a stressed material has units
b. Stress and Young’s modulus have the same units
c. Tensile and shear stress have different units
d. Tension and compression have the same units
e. NoA
The incorrect statement is (b) Stress and Young’s modulus have the same units. Stress and Young’s modulus are mechanical properties that are used to describe the behavior of materials under stress.
Stress is defined as the amount of force per unit area, while Young's modulus is defined as the ratio of stress to strain for a particular material.
Stress is measured in pascals (Pa), whereas Young’s modulus is measured in pascals (Pa) as well.The change in length of a stressed material has units
The unit of strain is the same as that of stress. Because strain is the change in length per unit length, there are no units for strain.
When a material is stretched, the stress is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa), while the change in length is measured in units of length, such as inches or meters.
Tensile and shear stress have different unitsTensile stress and shear stress, for example, have different units. Tensile stress is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa), while shear stress is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa).
Tension and compression have the same units
Both tension and compression are types of stress that are commonly used to describe the behavior of materials under different types of stress.
Tension is defined as the force that is applied to a material that causes it to stretch, while compression is defined as the force that is applied to a material that causes it to compress.
Both of these types of stress are measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa).NoAThere is no context given to define NoA.
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(c) A typical plastic shopping bag made by blown film has a lateral dimension (width) of 550 mm. Assume the tube is expended from 1.5 to 2.5 times the extrusion die diameter. Calculate the extrusion die diameter size. (5 marks; C3) (d) Elaborate how these plastic shopping bag achieved the strength and toughness from blow molding process. (4 marks; C2)
The formula to find the diameter of the extrusion die size of the plastic bag is Diameter of Extrusion Die = Lateral Dimension / (Expansion Ratio x 2). So, the extrusion die diameter size is (550/4) = 137.5 mm.
Given that the lateral dimension (width) of the typical plastic shopping bag is 550 mm.Assume the tube is expended from 1.5 to 2.5 times the extrusion die diameter.
Expansion ratio is given as (1.5 to 2.5).To find the extrusion die diameter size,
use the formula Diameter of Extrusion Die = Lateral Dimension / (Expansion Ratio x 2).
The extrusion die diameter size is (550/4) = 137.5 mm.
The plastic shopping bags achieve strength and toughness from blow molding process by the introduction of the right amount of chemicals that make the plastic bags resistant to wear and tear.
Moreover, the method of blow molding also allows the bags to be created with unique features such as handles and shapes
Blow molding is an innovative manufacturing process used in the production of plastic products such as shopping bags. The process involves inflating a hollow plastic tube with compressed air, which makes it assume the shape of a mold.
Blown film extrusion process involves the use of high-density polyethylene (HDPE), low-density polyethylene (LDPE), and linear low-density polyethylene (LLDPE) materials, which are suitable for making plastic bags.
During the blow molding process, the right amount of chemicals is introduced to make the plastic bags resistant to wear and tear. Additionally, the process allows the bags to be created with unique features such as handles and shapes.
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What are Microwaves? Bring out the basic advantage of Microwaves
over Co-axial cables and the Fiber optics.
Microwaves are a type of electromagnetic radiation characterized by wavelengths ranging from one millimeter to one meter. They are widely utilized in communication systems due to their high frequency and short wavelength, which enable efficient transmission of data and information over long distances with minimal signal degradation.
Microwaves offer several advantages over coaxial cables and fiber optics. Firstly, they can transmit signals over extensive distances without the need for repeaters. This is made possible by their high frequency and short wavelength, enabling them to maintain signal strength over long stretches. Secondly, microwaves are unaffected by adverse weather conditions such as rain, fog, or snow. This resilience allows their use in outdoor environments without experiencing signal loss or degradation. Thirdly, microwaves possess high-speed transmission capabilities, enabling rapid data and information transfer. These characteristics make microwaves well-suited for applications like internet connectivity, mobile communication, and satellite communication.
To summarize, microwaves represent a form of electromagnetic radiation that offers numerous advantages over coaxial cables and fiber optics. These advantages include long-distance transmission capabilities, resilience to weather conditions, and high-speed data transfer.
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The maximum dry unit weight obtained from a standard proctor test for a soil used in the field is 102.1 pcf, while the maximum dry unit weight obtained from the modified proctor test for the same soil is 107.5 pcf. What is the relative compaction with respect to the modified test if the sand cone test yielded a dry unit weight of 99 pcf? O 103.1% 97.0% 92.1% O 95.0%
The relative compaction with respect to the modified proctor test is approximately 92.1%.
To calculate the relative compaction with respect to the modified proctor test, we can use the formula:
Relative Compaction (%) = (Dry Unit Weight from Field Test / Maximum Dry Unit Weight from Modified Proctor Test) * 100
Given:
Maximum Dry Unit Weight from Modified Proctor Test = 107.5 pcf
Dry Unit Weight from Field Test = 99 pcf
Relative Compaction (%) = (99 / 107.5) * 100
Relative Compaction (%) ≈ 92.1%
Therefore, the relative compaction with respect to the modified proctor test is approximately 92.1%.
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The solar collector having the highest efficiency for high temperatures is:
Select one or more:
a. Unglazed type
b. Glazed type
C. Evacuated Thoes type
d. The 3 types have the same efficiency
Option C, the evacuated tube type, is the solar collector with the highest efficiency for high temperatures.
The evacuated tube type solar collector generally has the highest efficiency for high temperatures compared to unglazed and glazed types. The evacuated tube collector consists of multiple glass tubes, each containing a metal absorber tube surrounded by a vacuum. This design minimizes heat loss and provides better insulation, allowing the collector to achieve higher temperatures and maintain higher thermal efficiency.
On the other hand, unglazed collectors are typically used for lower temperature applications and do not have a glass covering, resulting in lower efficiency for high temperatures. Glazed collectors have a glass cover that helps to trap and retain heat, but they may not match the efficiency of evacuated tube collectors in high-temperature applications.
Therefore, option C, the evacuated tube type, is the solar collector with the highest efficiency for high temperatures.
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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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(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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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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Differentiate resilience from proof resilience.
A mild steel shaft 120mm diameter is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at particular section. Calculate the factor of safety according to the maximum shear stress theory if the elastic limit in simple tension is 220 MN/m²
A uniform metal bar has a cross-sectional area of 7 cm² and a length of 1.5m. With an elastic limit of 160 MN/m², what will be its proof resilience? Determine also the maximum value of an applied load which may be suddenly applied without exceeding the elastic limit. Calculate the value of gradually applied load which will produce the same extension as that produced by the suddenly applied load above. Take: E-200 GN/m².
Resilience refers to the ability of a material to absorb energy without undergoing permanent deformation, while proof resilience specifically measures the energy absorbed per unit volume up to the elastic limit. In the given scenario, the factor of safety based on the maximum shear stress theory can be calculated using the provided data. For a mild steel shaft with a diameter of 120mm, subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm, the factor of safety can be determined based on the elastic limit in simple tension.
Resilience is a material's ability to absorb energy when subjected to stress without experiencing permanent deformation. It is typically measured as the area under the stress-strain curve up to the elastic limit. On the other hand, proof resilience specifically quantifies the amount of energy absorbed per unit volume up to the elastic limit.
In the given case, a mild steel shaft with a diameter of 120mm is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at a particular section. To calculate the factor of safety based on the maximum shear stress theory, we need to compare the maximum shear stress experienced by the shaft with the elastic limit in simple tension.
The maximum shear stress (τ) can be calculated using the formula:
τ = (16 * T) / (π * d^3)
Where T is the maximum torque and d is the diameter of the shaft.
Substituting the values, we have:
τ = (16 * 20 kNm) / (π * (120mm)^3)
Next, we can compare this shear stress with the elastic limit in simple tension, which is given as 220 MN/m².
To find the factor of safety, we divide the elastic limit by the calculated maximum shear stress:
Factor of Safety = Elastic Limit / Maximum Shear Stress
Now, let's proceed to the second scenario:
We have a uniform metal bar with a cross-sectional area of 7 cm² and a length of 1.5m. The elastic limit of the material is 160 MN/m². We need to determine the proof resilience of the bar, which is the energy absorbed per unit volume up to the elastic limit.
Proof resilience (U) can be calculated using the formula:
U = (σ²) / (2E)
Where σ is the elastic limit and E is the Young's modulus of the material.
Substituting the values, we have:
U = (160 MN/m²)² / (2 * 200 GN/m²)
To calculate the maximum value of an applied load that can be suddenly applied without exceeding the elastic limit, we need to consider the stress caused by this load. Assuming the load is uniformly distributed over the cross-sectional area, the stress (σ) can be calculated as:
σ = F / A
Where F is the applied load and A is the cross-sectional area of the bar.
To find the maximum load without exceeding the elastic limit, we set the stress equal to the elastic limit and solve for F.
Finally, to determine the gradually applied load that produces the same extension as the suddenly applied load, we consider Hooke's Law, which states that stress is directly proportional to strain within the elastic limit. We can set up an equation equating the strain caused by the suddenly applied load to the strain caused by the gradually applied load and solve for the gradually applied load value.
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Resilience refers to the ability of a material to absorb energy without undergoing permanent deformation, while proof resilience specifically measures the energy absorbed per unit volume up to the elastic limit.
In the given scenario, the factor of safety based on the maximum shear stress theory can be calculated using the provided data. For a mild steel shaft with a diameter of 120mm, subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm, the factor of safety can be determined based on the elastic limit in simple tension.
Resilience is a material's ability to absorb energy when subjected to stress without experiencing permanent deformation. It is typically measured as the area under the stress-strain curve up to the elastic limit. On the other hand, proof resilience specifically quantifies the amount of energy absorbed per unit volume up to the elastic limit.
In the given case, a mild steel shaft with a diameter of 120mm is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at a particular section.
To calculate the factor of safety based on the maximum shear stress theory, we need to compare the maximum shear stress experienced by the shaft with the elastic limit in simple tension.
The maximum shear stress (τ) can be calculated using the formula:
τ = (16 * T) / (π * d^3)
Where T is the maximum torque and d is the diameter of the shaft.
Substituting the values, we have:
τ = (16 * 20 kNm) / (π * (120mm)^3)
Next, we can compare this shear stress with the elastic limit in simple tension, which is given as 220 MN/m².
To find the factor of safety, we divide the elastic limit by the calculated maximum shear stress: Factor of Safety = Elastic Limit / Maximum Shear Stress
Now, let's proceed to the second scenario:
We have a uniform metal bar with a cross-sectional area of 7 cm² and a length of 1.5m. The elastic limit of the material is 160 MN/m². We need to determine the proof resilience of the bar, which is the energy absorbed per unit volume up to the elastic limit.
Proof resilience (U) can be calculated using the formula:
U = (σ²) / (2E)
Where σ is the elastic limit and E is the Young's modulus of the material.
Substituting the values, we have:
U = (160 MN/m²)² / (2 * 200 GN/m²)
To calculate the maximum value of an applied load that can be suddenly applied without exceeding the elastic limit, we need to consider the stress caused by this load.
Assuming the load is uniformly distributed over the cross-sectional area, the stress (σ) can be calculated as: σ = F / A Where F is the applied load and A is the cross-sectional area of the bar.
To find the maximum load without exceeding the elastic limit, we set the stress equal to the elastic limit and solve for F.
Finally, to determine the gradually applied load that produces the same extension as the suddenly applied load, we consider Hooke's Law, which states that stress is directly proportional to strain within the elastic limit.
We can set up an equation equating the strain caused by the suddenly applied load to the strain caused by the gradually applied load and solve for the gradually applied load value.
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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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4) Disc brakes are used on vehicles of various types (cars, trucks, motorcycles). The discs are mounted on wheel hubs and rotate with the wheels. When the brakes are applied, pads are pushed against the faces of the disc causing frictional heating. The energy is transferred to the disc and wheel hub through heat conduction raising its temperature. It is then heat transfer through conduction and radiation to the surroundings which prevents the disc (and pads) from overheating. If the combined rate of heat transfer is too low, the temperature of the disc and working pads will exceed working limits and brake fade or failure can occur. A car weighing 1200 kg has four disc brakes. The car travels at 100 km/h and is braked to rest in a period of 10 seconds. The dissipation of the kinetic energy can be assumed constant during the braking period. Approximately 80% of the heat transfer from the disc occurs by convection and radiation. If the surface area of each disc is 0.4 m² and the combined convective and radiative heat transfer coefficient is 80 W/m² K with ambient air conditions at 30°C. Estimate the maximum disc temperature.
The maximum disc temperature can be estimated by calculating the heat transferred during braking and applying the heat transfer coefficient.
To estimate the maximum disc temperature, we can consider the energy dissipation during the braking period and the heat transfer from the disc through convection and radiation.
Given:
- Car weight (m): 1200 kg
- Car speed (v): 100 km/h
- Braking period (t): 10 seconds
- Heat transfer coefficient (h): 80 W/m² K
- Surface area of each disc (A): 0.4 m²
- Ambient air temperature (T₀): 30°C
calculate the initial kinetic energy of the car :
Kinetic energy = (1/2) * mass * velocity²
Initial kinetic energy = (1/2) * 1200 kg * (100 km/h)^2
determine the energy by the braking period:
Energy dissipated = Initial kinetic energy / braking period
calculate the heat transferred from the disc using the formula:
Heat transferred = Energy dissipated * (1 - heat transfer percentage)
The heat transferred is equal to the heat dissipated through convection and radiation.
Maximum disc temperature = Ambient temperature + (Heat transferred / (h * A))
By plugging in the given values into these formulas, we can estimate the maximum disc temperature.
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