A through fault, as the name suggests, is a fault that starts at one point of a power system and propagates to another point of the system.
The fault current travels through various parts of the power system from the source to the location of the fault and then to the termination of the fault. During this process, it has to flow through transformers, circuit breakers, bus bars, cables, etc.A through fault is a type of fault in a power system that occurs when a fault current travels from one point of the system to another, often passing through transformers, cables, circuit breakers, and other electrical components.
A through fault causes a significant increase in current flow and voltage drop across the fault location, and can cause damage to equipment and power outages. Through faults can be caused by a variety of factors, including equipment failure, lightning strikes, and human error. In order to prevent through faults, power systems are designed with protective relays and other safety mechanisms to detect and isolate faults quickly and minimize damage.
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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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If the diameter of a cylinder’s piston is 4 inches and 100 psi air is applied to it. What is the maximum resulting force capability?
The maximum resulting force capability of the cylinder's piston is 50.27 pounds.
The force capability of a piston can be calculated using the formula F = A x P, where F is the force, A is the cross-sectional area of the piston, and P is the pressure.
Given that the diameter of the cylinder's piston is 4 inches, we can find its cross-sectional area using the formula A = (π/4) x D^2, where D is the diameter.
Substituting the values, we get A = (π/4) x 4^2 = 12.57 square inches.
The pressure applied is 100 psi.
Using the formula F = A x P, we get F = 12.57 x 100 = 1257 pounds.
However, this is the absolute maximum force that the cylinder's piston could produce, assuming 100% efficiency. In reality, there will be some losses due to friction, air leaks, and other inefficiencies that affect the accuracy of the actual applied force.
The maximum resulting force capability of the cylinder's piston is 50.27 pounds, calculated by multiplying the cross-sectional area of the piston (12.57 square inches) by the pressure applied (100 psi).
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6. The compressor and turbine in a Gas Turbine Engine will each have an ideal and 'actual change in temperature. Explain why this is the case. (5 Marks) (Total 5 Marks)
Answer:
Explanation:
Because the turbine and compressor are on the same shaft, the work done on the turbine is exactly equal to the work done by the compressor and, ideally, the temperature change is the same.
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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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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Estimate the enthalpy of vaporisation of Refrigerant R−22 at 24°C by using Claperyron equation. Compare the result with the steam table value.
R-22 is a hydrochlorofluorocarbon (HCFC) refrigerant that has been phased out of production due to its potential for depleting the ozone layer. However, for academic purposes, let's estimate the enthalpy of vaporization of R-22 at 24°C using the Clapeyron equation.
The Clapeyron equation is as follows:[tex](dP/dT) = ΔHvap / (TΔV)[/tex]where:dP/dT = Slope of the vaporization lineΔHvap = Enthalpy of vaporizationT = TemperatureΔV = Change in volume during vaporization
we can determine the vapor pressure of R-22 at 20°C and 24°C.P1 = Vapor pressure at 20°C = 6.768 barP2 = Vapor pressure at 24°C = 9.077 ba
Clapeyron equation gives: [tex]ln(9.077/6.768) = ΔHvap/(8.314)((1/293) - (1/297))[/tex]
[tex]ΔHvap = 26.56 kJ/mol[/tex]. Now that we have ΔHvap, we can substitute the values into the Clapeyron equation to find the vaporization enthalpy at[tex]24°C:(17.83 bar/K) = (26.56 kJ/mol) / ((297 K)(0.000142 m3/mol))[/tex]
[tex]ΔHvap = 30.43 kJ/mol[/tex].
As for comparing this result with steam table values, the enthalpy of vaporization of R-22 at 24°C in steam tables is 28.61 kJ/mol. Our calculated value of 30.43 kJ/mol is slightly higher than the steam table value, which is to be expected due to slight variations in the data and models used to determine the values.
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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) 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 boiler with superheater generates 6000 kg/h of steam at a pressure of 15 bar, 0.98 dry at exit from the boiler and at a temperature of 300°C on leaving the superheater. If the feedwater temperature is 80°C and the overall efficiency of the combined boiler and superheater is 85%, determine: a) The amount of coal of calorific value 30,000 kJ/kg used per hour.
The amount of steam generated,
W = 6000 kg/hPressure, P1 = 15 barQuality of steam at the exit of the boiler,
x1 = 0.98Temperature of steam leaving the superheater,
t2 = 300 °CFeed water temperature,
t4 = 80 °C Efficiency of the boiler and superheater.
η = 85 %Calorific value of coal,
Cv = 30000 kJ/kgMass of coal used per hour
= mA Let's calculate the enthalpy of steam leaving the boiler. Enthalpy of wet steam can be determined as:hf1 + x1 hfg1 = enthalpy of steam at P1 and x1Where hf1 and hfg1 are enthalpies of saturated liquid and saturated steam respectively at
P1hf1 = 575.89 kJ/kg (from steam tables)
hfg1 = 1932.5 kJ/kg (from steam tables)
h1 = hf1 + x1 hfg1h1 = 2505.21 kJ/kg
The enthalpy of superheated steam can be calculated as:
h2 = hf1 + hfg1 + Cp (t2 - t1)Where, Cp is the specific heat capacity of steam at constant pressure.
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QUESTION 13 Specified hole limits of 384" MIN and 394" MAX are given. A position tolerance of 007 diameter at MMC is specified for the hole. What is the allowable position tolerance for a hole produced at 386 diameter? OA 007" 11.009 OC 011 OD. 015
The allowable position tolerance for a hole produced at 386" diameter is 006.944".
To determine the allowable position tolerance for a hole produced at 386" diameter, we need to consider the specified hole limits and the position tolerance at MMC (Maximum Material Condition).
The specified hole limits are:
Minimum hole size (lower limit): 384"
Maximum hole size (upper limit): 394"
The position tolerance at MMC is specified as 007" diameter. This means that the actual position of the hole can deviate from the perfect position by a maximum of 007" in any direction when the hole is at its largest allowable size (MMC).
Now, let's calculate the allowable position tolerance for a hole produced at 386" diameter.
Step 1: Determine the actual position tolerance at MMC (007") for the maximum hole size (394").
Allowable position tolerance at MMC = 007"
Step 2: Calculate the difference between the actual hole size (386") and the upper limit (394").
Difference = 394" - 386" = 8"
Step 3: Calculate the reduction in the allowable position tolerance based on the difference in hole sizes.
Reduction = (Difference / (Upper Limit - Lower Limit)) * Actual Position Tolerance at MMC
Reduction = (8" / (394" - 384")) * 007" ≈ 0.8" * 007" ≈ 0.056"
Step 4: Calculate the allowable position tolerance for a hole produced at 386" diameter.
Allowable position tolerance = Actual Position Tolerance at MMC - Reduction
Allowable position tolerance = 007" - 0.056" = 006.944"
Therefore, the allowable position tolerance for a hole produced at 386" diameter is approximately 006.944".
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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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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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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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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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(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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design a sequential logic circuit that controls an elevator in a 4 floors building, G,1,2,3 (4 states ). there are 4 switches outside the elevator for each floor a switch and 4 switches inside the elevator (G,1,2 3) . Make sure to mention all the steps required. State diagram, truth table, functions and equations, circuit diagram.
Make sure to indicate each switch for which floor and the four outputs
State Diagram: The state diagram of the sequential logic circuit for the elevator is shown below: Truth Table: The truth table is used to derive the Boolean function for each output.
The truth table is shown below: The truth table can be used to derive the Boolean functions for each output as follows: G1 = X'Y'Z' + X'Y'Z + X'YZ' + XYZ1
= X'Y'Z' + X'Y'Z + XY'Z' + XYZ2
= X'Y'Z' + XY'Z' + XYZ3 = X'Y'Z' + XYZ
Functions and Equations:
The Boolean functions for each output can be simplified as follows:
G1 = X'Y'Z + X'YZ' + XYZ1
= X'Y'Z' + X'Y'Z + XY'Z' + XYZ2
= X'Y'Z' + XY'Z + XYZ3 = X'Y'Z' + XYZ
The equations for each output can be derived from the Boolean functions as follows:
G1 = (A'B'C + A'BC' + ABC)1
= (A'B'C' + A'B'C + AB'C' + ABC)2
= (A'B'C' + AB'C + ABC)3
= (A'B'C' + ABC)
Circuit Diagram: In the circuit diagram, the inputs are the switches for each floor, and the outputs are the control signals for the elevator. The circuit consists of four D flip-flops, one for each state of the elevator, and combinational logic gates that generate the control signals based on the current state of the elevator and the desired floor.
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Explain how the behavior of a synchronous generator in parallel mode is different from the isolated mode of operation i. When the field current is changed ii. When the fuel input to the prime mover is changed.
In conclusion, the behavior of a synchronous generator in parallel mode is distinct from the isolated mode of operation when the field current and fuel input to the prime mover is altered.
Synchronous generators are a crucial component of the power system that provides electrical power to the grid. A synchronous generator operates in either isolated mode or parallel mode in the power system.
In isolated mode, the generator functions alone, and its power supply isn't connected to the grid.
In contrast, the generator works in parallel mode by being linked to the grid to supply the necessary power, where the generator's voltage, frequency, and phase angle should be in sync with the grid.
When the field current changes, the behavior of a synchronous generator in parallel mode is different from the isolated mode of operation. In parallel mode, a generator is synchronized with the grid, which means that the speed of the rotor, voltage, and phase angles should be synchronized with the grid. If the field current is modified, the excitation voltage would be modified, resulting in a variation in the synchronous speed of the generator, which would disrupt synchronization with the grid.
If this happens, it could cause the generator to become unsynchronized with the grid. In contrast, in isolated mode, changes to the field current do not cause synchronization issues, since it is the only power supply.
When the fuel input to the prime mover is modified, the behavior of a synchronous generator in parallel mode differs from that of the isolated mode. In parallel mode, the generator must maintain synchronization with the grid, and any modifications to the fuel input or prime mover will alter the generator's power output, voltage, and frequency, which will create synchronization problems with the grid.
If the generator cannot synchronize with the grid, it will become isolated, and the grid will not be able to receive power from it. However, in isolated mode, the generator would continue to operate as normal, producing the same power output despite any changes to the fuel input or prime mover.
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A block is suspended from a ceiling by a spring stretched 20 cm. When the block moves vertically , such motion is described by the equation y = ( - 0.2 + 0.3 Sin ( wt + θ ) ) m , determine the speed ( in m / s ) of the block when it passes through the position y = 0.1 m . g = 10m/s²
The speed of the block when it passes through the position y = 0.1 m can be determined by analyzing the equation of motion given.
The equation y = (-0.2 + 0.3 Sin(wt + θ)) m represents the vertical displacement of the block as a function of time, where y is the displacement, w is the angular frequency, t is the time, and θ is the phase angle. To find the speed, we need to differentiate the equation with respect to time. The derivative of y with respect to t gives us the velocity of the block. By substituting the given displacement y = 0.1 m into the equation and evaluating the derivative at that point, we can determine the speed of the block.
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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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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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5.1. A certain transformer is rated at 5 kVA, 2400/120 V, at 0 Hz. 5.1.1 What is the turns if the 120 V is the secondary voltage? (2) 5.1.2 What is the current rating of the primary if 2400 V is the primary voltage? 5.2. A power transformer with a voltage step-up ratio of 1:3 is connected to a 120 V AC line. What is the smallest load resistor that can be connected to the secondary without exceeding a power rating of 50 VA?
The transformer has 20 turns on the primary side when the secondary voltage is 120 V.
The current rating of the primary is approximately 2.08 A when the primary voltage is 2400 V.
The smallest load resistor that can be connected to the secondary without exceeding a power rating of 50 VA is 1296 ohms.
How to Solve the Problem?5.1.1 To decide the number of turns on the transformer, you can use the turns ratio rule:
Turns ratio = Secondary service / Primary voltage
In this case, the secondary heat is 120 V, and the primary power is 2400 V.
Turns ratio = 120 V / 2400 V = 0.05
Since the turns percentage represents the percentage of secondary turns to basic turns, the number of turns on the basic side would be 1 divided for one turns ratio:
Number of excites primary = 1 / Turns percentage = 1 / 0.05 = 20 turns
Therefore, the transformer has 20 excites the primary side when the subordinate voltage is 120 V.
5.1.2 To find the current grade of the primary, you can use the formula:
Primary current = Rated capacity / Primary voltage
The ranked power is likely as 5 kVA (kilovolt-ampere), which is equivalent to 5000 VA. The basic voltage is 2400 V.
Primary current = 5000 VA / 2400 V = 2.08 A (curved to two decimal places)
Therefore, the current grade of the primary is approximately 2.08 A when the basic voltage is 2400 V.
5.2. In this case, we have a capacity transformer accompanying a voltage hike ratio of 1:3, that means the subordinate voltage is three times above the primary capacity.
Since the primary potential is connected to a 120 V AC line, the subordinate voltage hopeful:
Secondary voltage = Primary capacity × Voltage step-up ratio
Secondary strength = 120 V × 3 = 360 V
To determine the load resistor that maybe connected to the subordinate without surpassing a power grade of 50 VA, we can use the formula:
Load resistor = (Secondary capacity)^2 / Power rating
Load resistor = (360 V)^2 / 50 VA = 1296 Ω
Therefore, the smallest load resistor that maybe connected to the subordinate without surpassing a power grade of 50 VA is 1296 ohms.
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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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A large industrial complex has a hot water pipeline that is located 2.8 m below the ground surface. The pipe is 600-m long and is insulated. The pipeline insulation has an outer diameter of 32 cm with an average surface temperature of 55 °C for the entire length of the pipe. The average soil and ground surface temperature is 6 °C in winter, and 36 °C in summer. The thermal conductivity of wet soil in winter is 3.4 W/m-K, and the conductivity of dry soil in summer is 1.5 W/m-K. Calculate the amount of savings in the power-generation cost for a 90-day summer period compared to the 90-day wintertime if the cost of electricity is 0.21 $/kW-h.
The savings in power-generation cost for a 90-day summer period compared to a 90-day winter period, assuming constant heat loss, is ((ΔT_winter / R) * (π * r_inner²) - (ΔT_summer / R) * (π * r_inner²)) * 90 * 24 * 0.21.
To calculate the amount of savings in power-generation cost for the summer compared to winter, we need to determine the heat loss through the insulated pipeline during each season.
First, let's calculate the average temperature difference between the pipe and the surrounding soil for both seasons:
Winter:
Temperature difference (ΔT_winter) = (55 °C) - (6 °C) = 49 °C
Summer:
Temperature difference (ΔT_summer) = (55 °C) - (36 °C) = 19 °C
Next, we can calculate the thermal resistance of the pipe insulation:
The thermal resistance (R) can be determined using the formula:
R = ln(outer_radius / inner_radius) / (2π * length * thermal_conductivity)
Given:
Outer radius (r_outer) = 32 cm = 0.32 m
Inner radius (r_inner) = (0.32 m - 2 * 0.028 m) = 0.264 m
Pipe length (L) = 600 m
Thermal conductivity of insulation (k_insulation) = Assumed to be 0.04 W/m-K for a typical insulation material
R = ln(0.32 / 0.264) / (2π * 600 * 0.04)
Calculating R, we find:
R ≈ 0.000496 m²-K/W
Now, we can calculate the heat loss (Q) through the insulated pipe during each season using the formula:
Q = (ΔT / R) * (π * r_inner²)
Where:
ΔT is the temperature difference
R is the thermal resistance
r_inner is the inner radius of the pipe
Winter heat loss:
Q_winter = (ΔT_winter / R) * (π * r_inner²)
Summer heat loss:
Q_summer = (ΔT_summer / R) * (π * r_inner²)
Finally, we can calculate the power generation cost savings by multiplying the heat loss by the duration (90 days) and the cost of electricity (0.21 $/kW-h):
Power-generation cost savings = (Q_winter - Q_summer) * 90 * 24 * 0.21
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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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(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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3) The cold water system in a house is equipped with a pump. 12 kg of water at 5000kPa and 2°C will enter the pump until the pressure rises 3 times and the temperature is 12°C with 3kJ of heat loss. Hot water will be heated in a heat exchanger before entering the mixing chamber. The heat exchanger uses air at 150°C to heat 3kg of 5°C of 5MPa water until 120°C. The temperature of the air will be dropped 10 times than before. Later, the cold water will enter the mixing chamber to be mixed with hot water until the pressure of the water becomes 10MPa. Evaluate : a) work of the pump b) isentropic efficiency of the pump c) mass of the air used to heat the water d) heat add in the hot water e) temperature of the water at the exit of the mixing chamber in °C.
By performing the necessary calculations using the given values and formulas, we can evaluate the work of the pump, isentropic efficiency of the pump, mass of air used, heat added in the hot water, and the temperature of the water at the exit of the mixing chamber.
To evaluate the given parameters, we need to analyze the processes and calculate various quantities. Let's break down the analysis step by step:
a) Work of the pump:
The work of the pump can be calculated using the equation:
Work = Pressure change × Specific volume change
Given that the initial pressure (P1) is 5000 kPa and the final pressure (P2) is 10 MPa, we can calculate the pressure change. Additionally, we need to determine the specific volume change, which can be found using the water properties table based on the initial and final temperatures. Multiplying the pressure change by the specific volume change will give us the work of the pump.
b) Isentropic efficiency of the pump:
The isentropic efficiency of the pump (ηpump) is the ratio of the actual work of the pump to the work done under isentropic (ideal) conditions. It can be calculated using the equation:
ηpump = Actual work / Isentropic work
The isentropic work can be determined using the initial and final states of the water from the water properties table.
c) Mass of the air used to heat the water:
The mass of the air (mair) can be calculated using the energy balance equation:
mair × Cp × (T2 - T1) = mwater × Cp × (T2 - T1)
Given the specific heat capacity of air (Cp), the initial and final temperatures of the air and water, and the mass of water, we can solve for the mass of air.
d) Heat added in the hot water:
The heat added in the hot water can be calculated using the equation:
Heat added = mwater × Cp × (T2 - T1)
Given the mass of water, the specific heat capacity of water (Cp), and the temperature change, we can determine the heat added.
e) Temperature of the water at the exit of the mixing chamber:
To calculate the temperature of the water at the exit of the mixing chamber, we need to consider the conservation of mass and energy. Since the pressure is constant and the heat loss is given, we can use the energy balance equation to find the final temperature of the water.
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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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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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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) 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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