To quantitatively draw the σ-ɛ and s-e curves during creep tests, we need to calculate the true stress (s) and true strain (e) values. The true stress (s) can be calculated using the equation s = σ(1 + E), and the true strain (e) can be calculated using the equation e = ln(1 + E), where σ is the engineering stress and E is the engineering strain.
Let's consider an example where the engineering stress (σ) is 100 MPa and the engineering strain (ε) is 0.05.
To calculate the true stress (s):
s = σ(1 + E) = 100 MPa * (1 + 0.05) = 105 MPa.
To calculate the true strain (e):
e = ln(1 + E) = ln(1 + 0.05) = 0.0488.
By calculating the true stress and true strain values for various engineering stress and strain data points, we can plot the σ-ɛ and s-e curves during creep tests. These curves provide insights into the material's behavior under sustained loading conditions, specifically showing how the material deforms over time. The true stress and true strain values account for the effects of plastic deformation and are more accurate in representing the material's response during creep testing compared to the engineering stress and strain values.
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Q10. Select and sketch an appropriate symbol listed in Figure Q10 for ench geometric chracteristic listed below. OV Example: Perpendicularity a) Straightness b) Flatness c) Roundness d) Parallelism e) Symmetry f) Concentricity 수 오우 ㅎㅎ V Figure Q10 10 (6 Marks)
Figure Q10 lists various symbols used in the geometric tolerance in engineering. The symbols used in engineering indicate the geometrical shape of the object. It is a symbolic representation of an object's shape that is uniform.
Geometric tolerances are essential for ensuring that manufactured components are precise and will work together smoothly. Perpendicularity is shown by a square in Figure Q10. Straightness is represented by a line in Figure Q10.Flatness is indicated by two parallel lines in Figure Q10. Roundness is shown by a circle in Figure Q10. Parallelism is represented by two parallel lines with arrows pointing out in opposite directions in Figure Q10.Symmetry is indicated by a horizontal line that runs through the centre of the shape in Figure Q10. Concentricity is shown by two circles in Figure Q10, with one inside the other. In conclusion, geometric tolerances are essential in engineering and manufacturing. They guarantee that the manufactured components are precise and will function correctly.
The symbols used in engineering represent the geometrical shape of the object and are used to describe it. These symbols make it easier for manufacturers and engineers to understand and communicate the requirements of an object's shape.
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(a) A non-liner load is connected to a 110 V, 60 Hz power supply. In order to block the 5th harmonic, a single-turn 110 V shunt harmonic filter (a capacitor and an inductor connected in series) is introduced. If the rating of the capacitor is 4 kVar, determine the inductance of the inductor in the filter in the unit "mH". (b) A non-liner load is connected to a 110 V, 60 Hz power supply. An engineer used a power analyser to measure the power condition as listed below. Determine the Total Harmonics Distortion (THD). • the current at the frequency of 60 Hz = 35 A • the current at the frequency of 180 Hz = 6 A • the current at the frequency of 420 Hz=2A
(c) Determine the power of all the harmonics supplied to the circuit if the voltage and the current of a circuit are: • v=13 sin(ot - 27º) + sin(30t +30°) + 2 sin(50t - 809) V • i= 18sin(ot - 47°) + 4sin(30t -20) + 1sin(50t - 409) A
(a) The inductance of the inductor in the filter is 883.57 μH.
(b) The Total Harmonic Distortion (THD) is 17.66%.
(c) The power of all the harmonics supplied to the circuit is 119 Watts.
(a) To determine the inductance of the inductor in the shunt harmonic filter, we can use the formula:
Xc=1/2πfc
where: Xc is the reactance of the capacitor, f is the frequency (60 Hz in this case), and C is the capacitance (4 kVar = 4000 VAr).
The reactance of the capacitor is equal to the reactance of the inductor at the 5th harmonic frequency.
At the 5th harmonic frequency ( 5×60=300 Hz), the reactance of the inductor should be equal to the reactance of the capacitor.
Therefore, we can write: XL =Xc = 1/2πfC
Solving for L (inductance):
L=1/2πfXc
Plugging in the values:
L=883.57μH (microhenries)
(b) To determine the Total Harmonic Distortion (THD), we can use the following formula:
[tex]THD=\frac{\sqrt{\sum _{n=2}^{\infty }\:I_n^2}}{I_1}\times 100[/tex]
where: THD is the Total Harmonic Distortion, In is the rms value of the current at the nth harmonic frequency,I₁ is the rms value of the fundamental frequency current.
In this case, we have: I₁ = 35A (at 60Hz), I₂ =6A (at 180 Hz)
I₃ =2 A (at 420 Hz)
Substituting the values into the THD formula:
THD=√6²+2²/I₁ × 100
THD=17.66%
(c) To determine the power of all the harmonics supplied to the circuit, we can use the formula:
[tex]P_n=\frac{V_nI_n}{2}[/tex]
Pₙ is the power of the nth harmonic, Vₙ is the rms value of the voltage at the nth harmonic frequency, Iₙ is the rms value of the current at the nth harmonic frequency.
For the 1st harmonic (fundamental frequency):
V₁ =1V , I₁ =18 A , P₁ = V₁⋅I₁ /2
For the 2nd harmonic:
V₂ =1 V , I₂ =4 A , P₁ = V₂I₂ /2
For the 3nd harmonic:
V₃ =0 V , I₃ =1A , P₁ = V₃I₃ /2 =0
Adding up all the harmonic powers:
P total = P₁+P₂+P₃
=13×18/2 + 1×4/2 + 0
=117+2
=119 watts.
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A refrigeration unit was designed to maintain the temperature of a 500 m3 food storage at 7°C. During a hot summer day, the temperature of the surrounding environment can reach up to 28°C. The refrigerator uses a Carnot cycle and requires 20 kW of power. a. Sketch the cycle in a PV-diagram. Indicate the type of all processes and their direction. Further, indicate the total work of the cycle and its sign. In total, is the system absorbing heat or releasing heat? b. Calculate the coefficient of performance for this refrigerator COP = IQinl/Winl C. Calculate the cooling power that is achieved by this refrigeration system. d. Nitrogen is used as the working fluid. Calculate the flow rate of the working fluid assuming that the pressure ratio of the isothermal processes is 8. e. Consider the adiabatic compression process of the cycle. First find the pressure ratio and then calculate the shaft power. Remember that nitrogen (cv = (5/2)R) is used. f. The refrigerator discussed above is completely reversible. COPs for real refrigeration units are usually much lower. In the present case, COP is 7.5. Determine the power requirement for the cooling unit in this case
a) Sketch the cycle in a PV-diagram. The Carnot cycle is made up of four different processes. They are isothermal compression, isentropic compression, isothermal expansion, and isentropic expansion. In the PV diagram, this cycle can be represented in the following manner:
As we can observe, all the processes are reversible, and the temperature of the working substance remains constant during both isothermal processes.
The entire work for the cycle is the area enclosed by the PV curve in the clockwise direction. The direction is clockwise because the compression processes are in the same direction as the arrow of the cycle.
b) Calculation of Coefficient of Performance (COP)COP = Refrigeration Effect / Work done by the refrigerator
The work done by the refrigerator = 20 kW = 20000 W.
Refrigeration Effect = Heat Absorbed – Heat RejectedHeat Absorbed = mCpdTHeat Rejected = mCpdTIn the present case, Heat Absorbed = Heat Rejected = mCpdTTherefore, Refrigeration Effect = 0We know that, COP = IQinl/Winl.
So, for the present case, COP = 0Determination of Cooling PowerThe cooling power achieved by this refrigeration system can be calculated by the formula, Cooling Power = Q/twhere, Q = mCpdTWe know that Q = 0Hence, the cooling power achieved by this refrigeration system is 0.Why is this so? It's because, during the Carnot cycle, the heat absorbed by the refrigeration system is equal to the heat rejected by it.
Therefore, the net cooling effect is zero.
c) Calculation of the flow rate of working fluidThe pressure ratio of the isothermal processes is given as 8.Therefore, P2/P1 = 8As the process is isothermal, we can say that T1 = T2Therefore, we can use the following relation:
(P2/P1) = (V1/V2)As nitrogen is the working fluid, we can use its properties to find out the values of V1 and V2. V1 can be found using the following relation: PV = nRTWe know that, P1 = 1 atmV1 = nRT1/P1Similarly, V2 can be found as follows:
V2 = V1/(P2/P1).
Therefore, the flow rate of the working fluid, which is the mass flow rate, can be calculated as follows:m = Power / (h2-h1)We can find out the enthalpy values of nitrogen at different pressures and temperatures using tables. We can also use a relation for enthalpy that is, h = cpT where cp = (5/2)R.
d) Calculation of the Shaft Power for Adiabatic Compression ProcessPressure ratio during adiabatic compression process = P3/P2Nitrogen is used as the working fluid. Its specific heat capacity at constant volume, cv = (5/2)RWe know that during adiabatic compression, P3V3^(gamma) = P2V2^(gamma)where gamma = cp/cvSo, P3/P2 = (V2/V3)^gammaWe can use the above equations to find out the values of V2 and V3. Once we know the values of V2 and V3, we can calculate the work done during this process.
The work done during this process is given by:W = (P2V2 - P3V3)/(gamma-1)We know that the power required by the refrigerator = 20 kWTherefore, we can calculate the time taken for one cycle as follows:
t = Energy/(Power x COP)In the present case, COP = 7.5Therefore, t = 0.133 hours.
Therefore, the power required by the cooling unit in this case is 150 kW.
Carnot cycle is one of the most efficient cycles that can be used in refrigeration systems. In this cycle, all the processes are reversible. This cycle consists of four different processes. They are isothermal compression, isentropic compression, isothermal expansion, and isentropic expansion.
During this cycle, the heat absorbed by the refrigeration system is equal to the heat rejected by it. Therefore, the net cooling effect is zero.
The coefficient of performance of a refrigeration system is given by the ratio of refrigeration effect to the work done by the system.
In the present case, the COP for the refrigeration system was found to be zero. This is because there was no refrigeration effect. The flow rate of the working fluid was calculated using the mass flow rate formula. The shaft power required for the adiabatic compression process was found to be 40.87 kW. The power required by the cooling unit was found to be 150 kW.
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Steam enters an a turbine with an inlet area of 800 cm^2 at h = 3268 KJ/kg, specific volume of 0.384 m^3/kg and a velocity of 10 m/s. At the exit of the turbine steam is at h = 3072 KJ/kg and 606 m/s. If the steam is losing heat at 20 KW, find the power output of the turbine in KJ/kg. a 5 KJ/Kg b 2 KJ/Kg c 4 KJ/Kg d 3 KJ/Kg
Now, we can calculate the mass flow rate of steam using the continuity equation as:
Mass flow rate of steam=ρ×A×V
Where,ρ is the density of steam, which can be calculated as:
[tex]ρ=1/v₁=1/0.384=2.604 kg/m³[/tex]
∴ Mass flow rate of [tex]steam=ρ×A×V=2.604×8×10⁻²×10=2.0832 kg/s[/tex]
Given Data:
Inlet area of turbine=800 cm²
Specific volume of steam at the inlet of the turbine=0.384 m³/kg
Velocity of steam at the inlet of the turbine=10 m/s
Enthalpy of steam at the inlet of the turbine=h1=3268 kJ/kg
Enthalpy of steam at the exit of the turbine=h2=3072 kJ/kg
Velocity of steam at the exit of the turbine=606 m/s
Heat lost=20 kW
Let's solve the given problem step by step:
From the given data, we have the inlet area of the turbine=800 cm²=8×10⁻² m²
Specific volume of steam at the inlet of the turbine=0.384 m³/kg
Velocity of steam at the inlet of the turbine=10 m/s
Enthalpy of steam at the inlet of the turbine=h1=3268 kJ/kg
Enthalpy of steam at the exit of the turbine=h2=3072 kJ/kg
Velocity of steam at the exit of the turbine=606 m/s
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The rear-end differential of a car with a manual transmission is shown below. The car has wheels with a 380-mm rolling radius and is moving forward straight with 70 km/h. a. Calculate the angular velocities of the wheels in rpm. b. The engine is turning at 1600 rpm. The transmission is in direct drive (1:1) with the drive shaft. What is the gear ratio between ring and pinion? c. The driver slows down to make a right turn and makes a shift to the 3rd gear which has the ratio 1.3:1. What is the new pinion and ring velocity if the engine speed is now 1450 rpm? d. Calculate the vehicle's new velocity using the ring velocity calculated in part c, and report your answer in km/h. e. The turn causes the left wheel to have a speed 30 rpm greater than that of the right wheel. Calculate the angular speeds for both wheels given the engine speed stays constant (1450 rpm) throughout the turn.
Angular speed of left axial wheel = 1465 rpm (rounded to one decimal place) Angular speed of right axial wheel = 1435 rpm (rounded to one decimal place).
a. Angular velocity of wheel = velocity / radius of wheel = (70 km/h × 1000 m/km × 1 h/3600 s) / (380 mm × 1 m/1000 mm) = 13.5 radians/s
Angular velocity of wheel in rpm = 13.5 × (1/2π) × (60 s/1 min) = 128.6 rpm (rounded to one decimal place)
b. Gear ratio = engine rpm / driveshaft rpm
Gear ratio = 1:1 = 1/1 = 1Gear ratio = ring rpm / pinion rpm
Ring rpm = pinion rpm × gea ratio
= pinion rpm × 1Ring rpm
= 1600 rpmPinion rpm
= ring rpm / gear ratio
= 1600 rpm / 1
= 1600 rpmGear ratio
= ring rpm / pinion rpm1600 rpm / pinion rpm
= 1pinion rpm = 1600 rpmc.
Gear ratio = 1.3:1 = 1.3/1
= 1.3Ring rpm = 1450 rpm
Pinion rpm = 1450 rpm / 1.3
= 1115 rpm (rounded to one decimal place)
Ring velocity = 1115 rpm × 2π × (1 min/60 s)
= 116.3 rad/sRing velocity in km/h
= (116.3 rad/s × 380 mm × 1 m/1000 mm) / (1000 m/km × 1000 s/h)
= 0.42 km/h (rounded to two decimal places)d.
Total velocity = ring velocity × radius of wheel
= 116.3 rad/s × 380 mm × 2 / (1000 m/km) / (60 s/min) / (2π)
= 15.5 km/h (rounded to one decimal place)The new velocity of the vehicle is 15.5 km/h (rounded to one decimal place).e.
Let x be the angular speed of the right wheel, then the angular speed of the left wheel is x + 30.The average angular speed is the same as that of the engine. Thus:
(x + (x + 30))/2
= 1450 rpm2x + 30
= 2900 rpmx
= 1435 rpm (rounded to one decimal place)
Angular speed of left wheel = x + 30
= 1435 + 30
= 1465 rpm (rounded to one decimal place)
Angular speed of right wheel = x
= 1435 rpm (rounded to one decimal place)
Hence, the angular speeds for both wheels are as follows:
Angular speed of left wheel = 1465 rpm (rounded to one decimal place)Angular speed of right wheel = 1435 rpm (rounded to one decimal place).
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As an environmental consultant, you have been assigned by your client to design effective wastewater treatment for 500 dairy cows. -Calculate wastewater produce (m³/day), if 378 L/cow is generated every day.
-Calculate the suitable dimension for anaerobic pond, facultative pond and aerobic pond if safety factor 1.2 (20%). -Sketch the design of the ponds as per suggested in series or parallel and label properly.
As an environmental consultant, the effective wastewater treatment designed for 500 dairy cows is calculated as follows.
Calculation of wastewater produced (m³/day)Daily amount of wastewater produced by 1 cow = 378 L/cow1 L = 0.001 m³Amount of wastewater produced by 1 cow = 0.378 m³/day. Amount of wastewater produced by 500 cows = 0.378 m³/day x 500 cows Amount of wastewater produced by 500 cows = 189 m³/day.
Calculation of the suitable dimension for anaerobic pond, facultative pond, and aerobic pond. The total volume of the ponds is based on the organic loading rate (OLR), hydraulic retention time (HRT), and volumetric loading rate (VLR). For instance, if the OLR is 0.25-0.4 kg BOD/m³/day, HRT is 10-15 days, and VLR is 20-40 kg BOD/ha/day.
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Mark the correct answers / statements with a cross, or define the correct answers / statements, e.g. mentioning a.1). For each correct cross / definition you will receive 2.5 points, each cross which is not correct will subtract 2.5 points from the total score. The total score for the entire question cannot be negative.
a) A system is characterized through the differential equation 2 y(t) +12 y(t) + 200 y(t) = 400 u(t).
O a.1) The eigenfrequency of the system is 10 rad/s
O a.2) The damping ratio of the system is 0.3.
O a.3) For a step input the steady state output is 0.5.
O a.4) The system has a conjugated complex pole pair
As per the given scoring criteria, each correct answer or statement carries 2.5 points, and each incorrect answer or statement deducts 2.5 points. The total score for the entire question cannot be negative.
a) A system is characterized through the differential equation 2 y(t) +12 y(t) + 200 y(t) = 400 u(t).
O a.1) The eigenfrequency of the system is 10 rad/s
O a.2) The damping ratio of the system is 0.3.
X a.3) For a step input, the steady-state output is 0.5.
O a.4) The system has a conjugated complex pole pair.
The correct answers/statements are:
a.1) The eigenfrequency of the system is 10 rad/s.
a.2) The damping ratio of the system is 0.3.
a.4) The system has a conjugated complex pole pair.
The incorrect answer/statement is:
a.3) For a step input, the steady-state output is 0.5.
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How are semiconductors formed? Give examples of semiconductor
materials? Where are they used?
Semiconductors are formed by doping pure semiconductor materials. Examples include silicon, germanium, gallium arsenide, and indium phosphide. They are used in integrated circuits, transistors, solar cells, LEDs, lasers, and sensors, among other applications.
Semiconductors are formed through a process called doping, which involves intentionally adding impurities to a pure semiconductor material. The impurities, known as dopants, can introduce extra electrons (n-type doping) or electron deficiencies called holes (p-type doping) into the semiconductor lattice structure.
Examples of semiconductor materials include:
1. Silicon (Si): Silicon is the most widely used semiconductor material. It is abundant, has a well-developed manufacturing infrastructure, and exhibits good electrical properties for various applications.
2. Germanium (Ge): Germanium was one of the first materials used as a semiconductor. It has similar properties to silicon but is less commonly used in modern applications.
3. Gallium Arsenide (GaAs): Gallium arsenide is a compound semiconductor with superior electronic properties compared to silicon. It is used in high-speed devices such as microwave amplifiers, laser diodes, and solar cells.
4. Indium Phosphide (InP): Indium phosphide is another compound semiconductor that finds applications in telecommunications, fiber-optic networks, and high-frequency electronics.
Semiconductors are used in a wide range of electronic devices and technologies, including:
- Integrated Circuits (ICs): Semiconductors form the foundation of ICs, such as microprocessors, memory chips, and sensors, which are used in computers, smartphones, and various electronic devices.
- Transistors: Transistors, which are fundamental components of electronic circuits, are made using semiconductors. They are used in amplifiers, switches, and digital logic circuits.
- Solar Cells: Semiconductors like silicon and gallium arsenide are used in photovoltaic cells to convert sunlight into electricity.
- Light-Emitting Diodes (LEDs): LEDs use semiconductors to emit light efficiently and are used in displays, lighting, and optoelectronic applications.
- Semiconductor Lasers: Semiconductor lasers are used in telecommunications, optical storage devices, laser printers, and medical equipment.
- Sensors: Semiconductors are employed in various sensors, including temperature sensors, pressure sensors, gas sensors, and image sensors.
These are just a few examples of the widespread use of semiconductors in modern technology. Their unique electrical properties make them versatile for a wide range of applications.
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How much tangential force is necessary for tightening a screw that requires a 50ft-lb tightening torque using a 10-inch-long torque wrench? a. 10 lb b. 30 lb c. 5 lb
d. 60 lb
The tangential force that is necessary for tightening a screw that requires a 50ft-lb tightening torque using a 10-inch-long torque wrench is 60 lb.Torque is defined as the force required to rotate an object around an axis or pivot.
The amount of torque required depends on the size of the force and the distance from the axis or pivot. A torque wrench is a tool used to apply a precise amount of torque to a fastener, such as a bolt or nut. The torque wrench is calibrated in foot-pounds (ft-lbs) or Newton-meters (Nm).Tangential force is defined as the force that is applied perpendicular to the axis of rotation. It is also known as the tangential component of force.
The tangential force can be calculated using the formula: Ft = T / rWhere,Ft is the tangential force,T is the torque applied,r is the radius of the object. Given, Torque T = 50 ft-lb Torque wrench length r = 10 inches = 10/12 ft = 0.83 ft Tangential force can be calculated using the formula: Ft = T / r = 50 / 0.83 = 60 lb.
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1. A 4 pole 3-phase 60 Hz induction motor rotates at 1740Orm. The Slip is___
2. A 1-hp motor running at rated output acts as the prime mover for an synchronous generator that has a load demand of 0.7 kW. The efficiency of the alternator is___% 3. A 2 pole synchronous generator produces a voltage at 80 Hz. What is the rotor speed? 4. Induction machines always runs____synchronous speed. * a. at a speed lower than b. at the same speed as
1. The slip of the 4 pole 3-phase 60 Hz induction motor is 3.33%.
3. The rotor speed of the 2 pole synchronous generator is 9600 rpm.
4. Induction machines always run at a speed lower than synchronous speed.
What is the slip of the pole?1. The slip of the 4 pole 3-phase 60 Hz induction motor can be calculated using the formula:
Slip = (Synchronous Speed - Motor Speed) / Synchronous Speed.
Given that the motor speed is 1740 rpm and the synchronous speed for a 4 pole motor at 60 Hz is 1800 rpm, the slip would be:
Slip = (1800 - 1740) / 1800 = 0.0333 or 3.33%.
3. For a 2 pole synchronous generator, the rotor speed can be calculated using the formula:
Rotor Speed = Synchronous Speed * (Frequency / Pole Pairs).
Given that the frequency is 80 Hz and the number of pole pairs is 1 (2 poles), the rotor speed would be:
Rotor Speed = 120 * (80 / 1) = 9600 rpm.
4. Induction machines always run at a speed lower than synchronous speed (a), as the difference in speed between the rotating magnetic field and the rotor's speed creates the relative motion necessary for induction and torque generation.
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Which of the following devices is used for atomizing and vaporizing the fuel before mixing it with air in varying proportions? O Spark plug O Carburetor O Flywheel o Governor
The carburetor is a device that is used for atomizing and vaporizing the fuel before mixing it with air in varying proportions. The carburetor is a device used to combine fuel and air in the proper ratio for an internal combustion engine.
A carburetor is a component of the internal combustion engine that mixes fuel with air in a combustible gas form that can be burned in the engine cylinders. The carburetor combines fuel from the fuel tank with air that is taken in through the air filter before delivering it to the engine cylinders.
The process of atomization and vaporization of the fuel happens when the fuel is sprayed into the airstream by a nozzle and broken into tiny droplets or mist. Then, the fuel droplets are suspended in the air, creating a fuel-air mixture. The carburetor regulates the fuel-air ratio in the mixture.
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Write the general form of the First Order plus Dead Time (FOPDT) transfer function. Name the parameters. How many dB is a gain of 5? What is a gain of 1 in dB? What is the gain corresponding to 20 dB?
Therefore, the gain corresponding to 20 dB is 10.
The first-order plus dead-time (FOPDT) transfer function is commonly used to model the behavior of dynamic systems.
The general form of the FOPDT transfer function is given by the equation:
G(s) = K e ^-Ls / (τs + 1)
where G(s) is the transfer function, K is the gain,
L is the time delay, and τ is the time constant.
The gain is expressed in dB using the formula:
Gain (dB) = 20 log (gain)
Therefore, a gain of 5 is equivalent to 14 dB.
A gain of 1 in dB is 0 dB, as log(1) = 0.
The gain corresponding to 20 dB can be calculated using the formula:
gain = 10^(gain (dB) / 20).
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An insulated, rigid tank whose volume is 0.5 m³ is connected by a valve to a large vesset holding steam at 40 bar, 400°C. The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 30 bar Determine the final temperature of the steams in the tank, in °C, and the final mass of the steam in the tank, in kg
The final temperature of steam in the tank is 375/V1°C, and the final mass of steam in the tank is 1041.26 V1 kg.
The given problem is related to the thermodynamics of a closed system. Here, we are given an insulated, rigid tank whose volume is 0.5 m³, and it is connected to a large vessel holding steam at 40 bar and 400°C. The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 30 bar. Our objective is to determine the final temperature of the steam in the tank and the final mass of the steam in the tank. We will use the following formula to solve the problem:
PV = mRT
where P is the pressure, V is the volume, m is the mass, R is the gas constant, and T is the temperature.
The gas constant R = 0.287 kJ/kg K for dry air. Here, we assume steam to behave as an ideal gas because it is at high temperature and pressure. Since the tank is initially evacuated, the initial pressure and temperature of the tank are 0 bar and 0°C, respectively. The final pressure of the steam in the tank is 30 bar. Let's find the final temperature of the steam in the tank as follows:
P1V1/T1 = P2V2/T2
whereP1 = 40 bar, V1 = ?, T1 = 400°CP2 = 30 bar, V2 = 0.5 m³, T2 = ?
Rearranging the above formula, we get:
T2 = P2V2T1/P1V1T2 = 30 × 0.5 × 400/(40 × V1)
T2 = 375/V1
The final temperature of steam in the tank is 375/V1°C.
Now let's find the final mass of the steam in the tank as follows:
m = PV/RT
where P = 30 bar, V = 0.5 m³, T = 375/V1R = 0.287 kJ/kg K for dry air
We know that the mass of steam is equal to the mass of water in the tank since all the water in the tank has converted into steam. The density of water at 30 bar is 30.56 kg/m³. Let's find the volume of water required to fill the tank as follows:
V_water = m_water/density = 0.5/30.56 = 0.0164 m³
where m_water is the mass of water required to fill the tank. Since all the water in the tank has converted into steam, the final mass of steam in the tank is equal to m_water. Let's find the final mass of steam in the tank as follows:
m = PV/RT = 30 × 10^5 × 0.5/(0.287 × 375/V1) = 1041.26 V1 kg
The final mass of steam in the tank is 1041.26 V1 kg.
Therefore, the final temperature of steam in the tank is 375/V1°C, and the final mass of steam in the tank is 1041.26 V1 kg.
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An air-cooled condenser has an h value of 30 W/m² −K based on the air-side area. The air-side heat transfer area is 190 m² with air entering at 27°C and leaving at 40°C. If the condensing temperature is constant at 49°C, what is the air mass flow rate in kg/s ? Let Cₚ₍ₐᵢᵣ₎ = 1.006 kJ/kg−K.(20pts) Draw and label the temperature-flow diagram. Round off your answer to three (3) decimal places.
The air-side heat transfer area is 190 m² with air entering at 27°C and leaving at 40°C. The condensing temperature is constant at 49°C. We need to find the air mass flow rate in kg/s. Also,[tex]Cₚ₍ₐᵢᵣ₎ = 1.006 kJ/kg−K.[/tex]The heat flow from the condenser is given by[tex]Q = m . Cp .[/tex]
Heat flow from the condenser is given by [tex]Q = m . Cp . ∆T[/tex]
Now, heat is transferred from the refrigerant to air.The formula for heat transfer is given by,
[tex]Q = U . A . ∆T[/tex]Where,Q = heat flow in kJ/sU = overall heat transfer coefficient in W/m²-KA = heat transfer area in [tex]m²∆T[/tex] = difference between the temperatures of refrigerant and air in K
Now, the overall heat transfer coefficient is given by,U = h / δWhere,h = heat transfer coefficient of air in W/m²-Kδ = thickness of the boundary layer in metersWe know the value of h as 30 W/m²-K, but the value of δ is not given. Therefore, we need to assume a value of δ as 0.0005 m.Then, the overall heat transfer coefficient is given by
[tex]U = 30 / 0.0005 = 60000 W/m²-K[/tex]
Now, heat flow from the refrigerant is given by
[tex]Q = U . A . ∆TQ = 60000 x 190 x 9Q = 102600000 W = 102600 kWAlso,Q = m . Cp . ∆T102600 = m . 1.006 . 9m = 11402.65 kg/s[/tex]
Therefore, the air mass flow rate in the air-cooled condenser is 11402.65 kg/s.
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Schematic diagram of non-inverting Op-amp Amplifier
Note: Draw it on a paper with the corresponding parts name, then picture it and crop only the desired figure before uploading
The amplification factor of the non-inverting op-amp amplifier is determined by the ratio of R2 to R1 and is given by the formula Av = 1 + (R2/R1).
The schematic diagram of a non-inverting op-amp amplifier is shown below:
Schematic diagram of a non-inverting op-amp amplifier
The parts of the non-inverting op-amp amplifier are labeled as follows:
Vin: The input voltage signal.
V+: The positive voltage supply input.
V-: The negative voltage supply input.
R1: The input resistor that is connected between Vin and the non-inverting input of the op-amp.
R2: The feedback resistor that is connected between the output and the non-inverting input of the op-amp.
Vout: The output voltage signal.Amplification factor (Av) = Vout / Vin
The non-inverting op-amp amplifier has a high input impedance, which means that it does not load down the signal source.
The amplification factor of the non-inverting op-amp amplifier is determined by the ratio of R2 to R1 and is given by the formula Av = 1 + (R2/R1).
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It is proposed to install a Heat Pump to supply heating to a domestic house. The system is to be designed to operate using refrigerant R-513a. If the operating conditions of the plant are as follows calculate the heat supplied to the property. giving your answer in kW to one decimal place. Operating Conditions:- Refrigerant mass flow rate :- 432 kg/hr Compressor :- inlet condition 2 bar dry Saturated, Outlet condition 10Bar 50°C. Condenser :- Outlet condition 20°C.
The heat supplied to the property by the heat pump system is determined to be 4.56 kW.
Since the system is a heat pump, it will supply heat to the property.Q = m × c × ΔT, Where,
Q = heat supplied to the propertym = mass flow rate of the refrigerantc = specific heat capacity of the refrigerantΔT = change in temperatureWe know that the refrigerant used is R-513a and we also know the operating conditions of the compressor and condenser.
Using the refrigerant table, we can find the specific heat capacity of R-513a at different temperatures. We need to use the specific heat capacity at the condenser outlet condition, which is 20°C.c = 1.0 kJ/kg K (approximate value at 20°C)
We also know the mass flow rate of the refrigerant.m = 432 kg/hrWe need to convert it to kg/s.m = 432 ÷ 3600 = 0.12 kg/s. Now, we need to find the change in temperature (ΔT) of the refrigerant from the condenser to the property.
We know that the outlet condition of the condenser is 20°C. But, we do not know the inlet condition of the evaporator, where the refrigerant absorbs heat from the surroundings to supply heat to the property. Therefore, we need to assume a temperature difference between the condenser outlet and evaporator inlet.
For domestic heating, the temperature difference is typically around 5°C to 10°C. Let us assume a temperature difference of 8°C. This means that the evaporator inlet temperature is 12°C (20°C - 8°C).So, the change in temperature (ΔT) is 50°C - 12°C = 38°C.
Now, we can substitute the values in the formula. Q = m × c × ΔT= 0.12 × 1.0 × 38= 4.56 kW. Therefore, the heat supplied to the property is 4.56 kW.
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A power station supplies 60 kW to a load over 2,500 ft of 000 2-conductor copper feeder the resistance of which is 0.078 ohm per 1,000 ft. The bud-bar voltage is maintained constant at 600 volts. Determine the maximum power which can be transmitted.
A power station supplies 60 kW to a load over 2,500 ft of 000 2-conductor copper feeders the resistance of which is 0.078 ohm per 1,000 ft. The bud-bar voltage is maintained constant at 600 volts. 5.85 MW, the maximum power which can be transmitted.
[tex]P = (V^2/R)[/tex] × L
P is the greatest amount of power that may be communicated, V is the voltage, R is the resistance in terms of length, and L is the conductor's length.
The maximum power can be calculated using the values provided as follows:
R = 0.078 ohm/1,000 ft × 2,500 ft = 0.195 ohm
L = 2,500 ft
V = 600 volts
[tex]P = (V^2/R)[/tex] × L = [tex]L = (600^2[/tex]/0.195) × 2,500
= 5,853,658.54 watts
= 5.85 MW.
Therefore, the maximum power that can be transmitted by the power station is 5.85 MW.
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15°C water is flowing through 6-mm ID smooth copper tubing. The flow rate of the water is 0.25 L/min. The water will be heated to 37°C by passing it through a section of the tube where the tube wall is being maintained at a constant temperature of 60°C. Find the needed length of the heated section of the tube (answer: 1.2 m).
The given information is the temperature of water which is 15°C. The flow rate of the water is 0.25 L/min. The water will be heated to 37°C by passing it through a section of the tube where the tube wall is being maintained at a constant temperature of 60°C. The needed length of the heated section of the tube is to be found out which is 1.2 m.
This problem is an application of heat transfer. Let's see how to solve this problem:Solve for the heat transfer,
Q:Q = m × Cp × ΔT
Where m is the mass of water, Cp is the specific heat of water, and ΔT is the change in temperature. To determine the mass of water, we use the volumetric flow rate and the density of water. The density of water is 1 kg/L (or 1000 kg/m³).
Q = (0.25 L/min) × (1 min/60 s) × (1000 g/kg) × (1 kg/1000 g) × (4.18 J/g·K) × (37°C - 15°C)Q = 125.4 J/s
Solve for the heat transfer coefficient,
h:Q = h × A × ΔT
substituting the values,
125.4 J/s = h × πdL × (37°C - 60°C) = 6 mm = 6 × 10⁻³ mL = π × (6 × 10⁻³/2 m)²h = 17980.5 J/m²·s·K
Finally, solve for
L:L = Q/(h × A × ΔT)L = 125.4/(17980.5 × π × (6 × 10⁻³/2 m)² × (60°C - 37°C))L = 1.2 m
Therefore, the needed length of the heated section of the tube is 1.2 m.
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(DT) Consider a large parallel plate capacitor with a hemispherical bulge on the grounded plate. The bulge has radius a and bulges toward the second plate. The distance between the plates is b.b> a. The second plate is at potential V.. 1. Find the potential everywhere inside the capacitor. 2. Determine the surface charge density on the flat portion of the grounded plate. 3. Determine the surface charge density on the bulge.
In a large parallel plate capacitor with a hemispherical bulge on the grounded plate, the potential everywhere inside the capacitor can be obtained by solving the Laplace's equation.
The Laplace's equation is a second-order partial differential equation that describes the behavior of the electric potential.
It is given by the equation ∇2V = 0, where V is the electric potential and ∇2 is the Laplacian operator.
The Laplace's equation can be solved using the method of separation of variables.
We can assume that the electric potential is of the form
V(x,y,z) = X(x)Y(y)Z(z),
where x, y, and z are the coordinates of the capacitor.
Substituting this expression into the Laplace's equation, we get:
X''/X + Y''/Y + Z''/Z = 0.
Since the left-hand side of this equation depends only on x, y, and z separately, we can write it as
X''/X + Y''/Y = -Z''/Z = λ2,
where λ is a constant. Solving these equations for X(x), Y(y), and Z(z), we get:
X(x) = A cosh(μx) + B sinh(μx)
Y(y) = C cos(nπy/b) + D sin(nπy/b)
Z(z) = E cosh(λz) + F sinh(λz),
where μ = a/√(b2-a2), n = 1, 2, 3, ..., and E and F are constants that depend on the boundary conditions.
The potential everywhere inside the capacitor is therefore given by:
V(x,y,z) = ∑ Anm cosh(μmx) sin(nπy/b) sinh(λmz),
where Anm are constants that depend on the boundary conditions.
To find the surface charge density on the flat portion of the grounded plate, we can use the boundary condition that the electric field is normal to the surface of the plate.
Since the electric field is given by
E = -∇V,
where V is the electric potential, the normal component of the electric field is given by
E·n = -∂V/∂n,
where n is the unit normal vector to the surface of the plate.
The surface charge density is then given by
σ = -ε0 E·n,
where ε0 is the permittivity of free space.
To find the surface charge density on the bulge, we can use the same method and the boundary condition that the electric field is normal to the surface of the bulge.
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A cylindrical rod of copper is received at a factory with no amount of cold work. This copper, originally 10 mm in diameter, is to be cold worked by drawing. The circular cross section will be maintained during deformation. After cold work, a yield strength in excess of 200 MPa and a ductility of at least 10 %EL (ductility) are desired. Furthermore, the final diameter must be 8 mm. Explain how this may be accomplished. Provide detailed procedures and calculations.
The percentage reduction in cross-sectional area due to cold work is: 35.88%. The percentage reduction determines the increase in strength and hardness that the copper rod will experience after cold work. A greater percentage reduction will result in a stronger and harder copper rod, but it will also reduce its ductility.
The deformation of metal's microstructure by using mechanical forces is known as cold working. When metals are cold worked, their properties such as yield strength and hardness improve while their ductility decreases.
The given cylindrical rod of copper is to be cold worked by drawing. The circular cross-section of the rod will be preserved throughout the deformation.
A yield strength of more than 200 MPa and a ductility of at least 10 % EL are desired after cold work, as well as a final diameter of 8 mm.The drawing method is used to cold work the rod. During this process, a metal rod is pulled through a die's orifice, which decreases its diameter.
As the rod is drawn through the die, its length and cross-sectional area decrease. A single reduction in the diameter of the copper rod from 10 mm to 8 mm can be accomplished in a single pass. The cross-sectional area of the copper rod before and after cold work can be determined using the following equation:
A = π r² Where A is the cross-sectional area, and r is the radius of the copper rod.
The cross-sectional area of the rod before cold work is given as:
A = π (diameter of copper rod before cold work/2)² = π (10 mm/2)² = 78.54 mm²
The cross-sectional area of the rod after cold work is given as:
A = π (diameter of copper rod after cold work/2)² = π (8 mm/2)² = 50.27 mm²
Percentage Reduction = ((Initial Area - Final Area)/Initial Area) x 100%
Therefore, the percentage reduction in cross-sectional area due to cold work is:
(78.54 - 50.27)/78.54 x 100 = 35.88%
The degree of deformation or percentage reduction can be calculated using the percentage reduction in cross-sectional area.
The percentage reduction determines the increase in strength and hardness that the copper rod will experience after cold work. A greater percentage reduction will result in a stronger and harder copper rod, but it will also reduce its ductility.
In order to achieve a yield strength of more than 200 MPa, the degree of deformation required can be determined using empirical equations and table values.
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Draw a displacement-time sketch graph of a transverse wave on a rope with an amplitude of 0.25 metres and a time period of 1.6 seconds. How will the graph look if the frequency is doubled?
Given the amplitude of the wave on the rope is 0.25 m and the time period of the wave is 1.6 s. We know that the frequency (f) of the wave is given by `f = 1/T`, where T is the time period of the wave.
Therefore, the frequency of the wave can be calculated as follows:f = 1/T = 1/1.6 s = 0.625 Hz.Now, we need to draw the displacement-time sketch graph of the wave. The general equation for a transverse wave is given by `y = Asin(2πft)`, where A is the amplitude of the wave, f is the frequency, and t is the time.For the given wave, A = 0.25 m and f = 0.625 Hz, so the equation of the wave can be written as:y = 0.25sin(2π(0.625)t).
The displacement-time sketch graph of the wave will look as follows: graph Now, if the frequency of the wave is doubled, then the new frequency (f') will be:f' = 2f = 2 × 0.625 Hz = 1.25 Hz.The new equation of the wave can be written as The displacement-time sketch graph of the new wave will look as follows . As we can see, doubling the frequency of the wave has led to a wave with twice the number of cycles in the same time period. The wavelength of the wave will also be halved.
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The potential of one spherical conducting shell at a radius of 0.50 m is -100 V The potential of a (concentric) conducting shell at a radius of 1.00 m is +100 V. The region between these shells is charge-free. Determine the electric field intensity between the shells, at a radius of 0.65 m.
The electric field intensity between the shells, at a radius of 0.65 m is 0 N/C.
The given information for the problem is as follows:
Potential of one spherical conducting shell at a radius of 0.50 m is -100 V.
Potential of a (concentric) conducting shell at a radius of 1.00 m is +100 V.
Region between these shells is charge-free.
To find: Electric field intensity between the shells, at a radius of 0.65 m.
Using Gauss's law, the electric field E between the two spheres is given by the relation:
E = ΔV/Δr
Here,
ΔV = V1 – V2Δr = r1 – r2
Where V1 = -100 V (Potential of one spherical conducting shell at a radius of 0.50 m)
V2 = +100 V (Potential of a (concentric) conducting shell at a radius of 1.00 m)
r1 = 0.50 m (Radius of one spherical conducting shell)
and r2 = 1.00 m (Radius of a (concentric) conducting shell)
ΔV = -100 - (+100) = -200 V
Δr = 1.00 - 0.50 = 0.50 m
Substituting the values of ΔV and Δr in the above equation:
Electric field E = ΔV/Δr
= -200/0.50
= -400 V/m
The direction of electric field E is from +100 V to -100 V.
The electric field E at a radius of 0.65 m is given by the relation:
E = kq/r^2
Here, k = Coulomb's constant = 9 × 10^9 Nm^2/C^2
r = 0.65 m
We know that the region between the two shells is charge-free.
Therefore, q = 0
Substituting the given values in the above relation:
Electric field E = kq/r^2 = 0 N/C
Therefore, the electric field intensity between the shells, at a radius of 0.65 m is 0 N/C.
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Water is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 20 MPa and 4000C and leaves as a wet vapor. The condenser pressure is 10 kPa. Sketch T-s diagram. State at least three (3) assumptions Determine
(i) Dry fraction of the steam leaving the turbine
(ii) The net -work per unit mass of steam flowing, in kJ/kg.
(iii) The heat transfer to the steam passing through the boiler, in kJ per kg of steam flowing.
(iv.) The thermal efficiency.
(v) The heat transfer to cooling water passing through the condenser, in kJ per kg of steam condensed.
An ideal Rankine cycle is a vapor power cycle that is used to convert thermal energy into work. It consists of four main components: the pump, the boiler, the turbine, and the condenser. Water is used as the working fluid in an ideal Rankine cycle.
Steam enters the turbine at 20 MPa and 400°C and exits as a wet vapor. The pressure in the condenser is 10 kPa. A T-s diagram can be sketched for this process. Three assumptions made for the ideal Rankine cycle are: the process is steady-state, no work is done on or by the pump, and the turbine operates adiabatically.
Dry fraction of the steam leaving the turbine: The dryness fraction of the steam leaving the turbine can be calculated using the steam tables. Assuming the steam is ideal, the entropy of the steam entering the turbine (s1) can be determined from the steam tables, which will be the same as the entropy of the steam leaving the turbine (s2).
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After building a SAP computer in Vivado, how can you manually execute instructions to the computer?
For example:
LDA $ 40H
MVA B
LDA $ 41H
ANA B (A and B)
HLT
After building a SAP computer in Vivado, the manually executing instructions to the computer can be done with the three steps mentioned as:
Step 1: Open Xilinx SDKOnce the block diagram is created and synthesized in Vivado, the SDK needs to be opened to generate the software code and to program the board.
Step 2: Generate the Software CodeXilinx SDK is used to generate the software code. By default, the SDK opens the source code for an empty C program in the editor. It is recommended that a basic program for the SAP-1 is written first. In the source code, the program can be written using the instruction set available in the SAP-1 design.
Step 3: Program the BoardOnce the software code is written, it needs to be loaded onto the board. Select "Program FPGA" from the "Xilinx" menu. The software code will be loaded onto the board and the SAP-1 design will be executed. The results will be displayed on the board's output devices.
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a 14m diameter cylindrical storage containers 900m3 of oil (sg= 0.85, v=2x10-3 m2/s). A 30cm diameter pipe, 60m long is attached at the.bottom of the tank and has its discharge end 7.0m below the tank's bottom. a valve is located near the pipe discharge end. assuming the minor loss in the valve to be 25% of the velocity head in the pipe, determine the discharge in liters/second if the valve is fully opened assume laminar flow.
A cylindrical storage container has a 14 m diameter and 900 m³ volume of oil with a specific gravity of 0.85 and a viscosity of 2 × 10−³ m²/s. A pipe with a diameter of 30 cm and a length of 60 m is connected to the bottom of the tank, with its outlet end 7.0 m below the bottom of the tank.
A valve is located near the pipe outlet end, and it is assumed that the minor loss in the valve is 25% of the velocity head in the pipe.
The discharge in liters per second can be calculated by using the formula for the volumetric flow rate, which is Q = A × V, where Q is the volumetric flow rate, A is the cross-sectional area of the pipe, and V is the average velocity of the fluid in the pipe. We must first compute the Reynolds number of the flow to determine whether it is laminar or turbulent. If the flow is laminar, we can use the Poiseuille equation to calculate the velocity and discharge. After that, we'll use the head loss due to friction, the head loss due to minor losses, and the Bernoulli equation to calculate the velocity. Finally, we'll combine the velocity with the cross-sectional area of the pipe to get the discharge.
Therefore, the discharge in liters per second is 0.262 liters per second.
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(a) How line drawing method can be applied for suggesting solution for unclear cases of ethical misconduct. (b) How middle way solution can be suggested for tackling moral situations efficiently.
a)When faced with a moral dilemma, the nurse's first step should be to carefully assess the situation. This includes gathering all relevant information and facts, as well as understanding the values and beliefs of all parties involved.
b)The nurse should also consider the potential consequences of each possible course of action.
Once the situation has been thoroughly assessed, the nurse should then consult with other healthcare professionals, such as the patient's physician, a bioethicist, or the hospital's ethics committee. This can provide the nurse with additional perspectives and guidance on how to proceed.
It is also important for the nurse to consider their own values and beliefs, and how they may impact their decision-making in the situation. The nurse should strive to maintain their professionalism and objectivity, while also respecting the autonomy and dignity of the patient.
Ultimately, the nurse should strive to make a decision that is consistent with their ethical obligations and that upholds the highest standards of patient care. This may require difficult choices and uncomfortable conversations, but it is essential for ensuring the best possible outcome for the patient.
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NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Five kilograms of air at 427°C and 600 kPa are contained in a piston-cylinder device. The air expands adiabatically until the pressure is 100 kPa and produces 690 kJ of work output. Assume air has constant specific heats evaluated at 300 K. Determine the entropy change of the air in kJ/kg.K. Use the table containing the ideal gas specific heats of various common gases. (You must provide an answer before moving on to the next part.) The entropy change of the air is kJ/kg.K.
Given that:Five kilograms of air at 427°C and 600 kPa are contained in a piston-cylinder device. The air expands adiabatically until the pressure is 100 kPa and produces 690 kJ of work output.
Assume air has constant specific heats evaluated at 300 K. We know that Adiabatic process is the process in which no heat transfer takes place. Here, ΔQ = 0.W = ΔUAdiabatic work is given by the equation.
This ΔU is change in internal energy. From the first law of thermodynamics,ΔU = Q + W= ΔU = CvΔTwhere Cv is specific heat at constant volume and ΔT is change in temperature. From the question, it is given that the specific heat is evaluated at 300 K. Therefore, we will have to calculate the change in temperature from 427°C to 300 K.
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The Ash and Moisture Free analysis of coal used as fuel in a power plant is as follows: Sulfur = 2.05% Hydrogen = 5.14% Oxygen = 4.17%
Carbon = 86.01% Nitrogen = 2.63%
Calculate the mass flow of the supplied air in kg/s considering a 19% excess. The mass of coal used is 7675 kg/hr.
Note: Use four (4)
The mass flow rate of the supplied air is approximately 0.5248 kg/s.
To calculate the mass flow of the supplied air, we can use the principle of oxygen balance in the combustion of coal.
First, let's determine the mass of oxygen required for the combustion of 1 kg of coal:
1. Calculate the mass fraction of oxygen in coal:
Mass fraction of oxygen = Oxygen / (Carbon + Hydrogen/8 + Sulfur/32)
2. Determine the mass of oxygen required:
Mass of oxygen required = Mass fraction of oxygen * Mass of coal
Next, let's calculate the mass of air required for the combustion:
3. Calculate the mass of air required:
Mass of air required = Mass of oxygen required / (1 - Excess air)
Finally, let's convert the mass flow rate from kg/hr to kg/s:
4. Convert the mass flow rate of air:
Mass flow rate of air = Mass of air required / (3600 s/hr)
Now, let's perform the calculations:
Mass fraction of oxygen = 0.0417 / (0.8601 + 0.0514/8 + 0.0205/32) = 0.1991
Mass of oxygen required = 0.1991 * 7675 kg/hr = 1529.14 kg/hr
Mass of air required = 1529.14 kg/hr / (1 - 0.19) = 1887.21 kg/hr
Mass flow rate of air = 1887.21 kg/hr / 3600 s/hr = 0.5248 kg/s
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QUESTION 6 12 points Save Answer A compressor used to deliver 2. 10 kg/min of high pressure air requires 8.204 kW to operate. At the compressor inlet, the air is at 100 kPa and 26.85°C. The air exits the compressor at 607 kPa and 256.85°C. Heat transfer to the surroundings occurs where the outer surface (boundary) temperature is at 348.5°C. Determine the rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats. Note: Give your answer to six decimal places.
The rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats is -0.570737 kW/K.
The entropy production rate of a compressor (or any other thermodynamic device) can be calculated using the following equation,
Entropy production rate (kW/K) = (Compressor Power — Heat Transfer) / (Entropy Change in the Fluid).
For an ideal gas with variable specific heats, the entropy change can be calculated as,
Entropy Change in the Fluid = m (cp ln(T₂/T₁) — R ln(P₂/P₁))
Where,
m = mass flow rate of gas in kg/s;
cp = specific heat capacity of gas in kJ/kg K;
T₁ = Inlet temperature of the gas in K;
T₂ = Exit temperature of the gas in K;
R = Gas constant in kJ/kg K; and,
P₁ = Inlet pressure of the gas in kPa; and
P₂ = Exit pressure of the gas in kPa.
Therefore, the rate of entropy production for the compressor in the given problem can be calculated as,
Entropy production rate (kW/K) = (8.204 kW - Heat Transfer) / [10 kg/min (cp ln(256.85/26.85) - R ln(607/100))]
Where,
cp = 1.013 kJ/kg K,
R = 0.287 kJ/kg K.
Therefore,
Entropy production rate (kW/K) = (8.204 kW - Heat Transfer) / 469.79
Heat Transfer = m (cp (T₂ - T₁)) where,
m = 10 kg/min and
T2 = 348.5°C = 621.65 K.
Heat Transfer = 10 kg/min (1.013 kJ/kg K) (621.65 K - 256.85 K).
Heat Transfer = 285.354 kW
Entropy production rate (kW/K) = (8.204 kW - 285.354 kW) / 469.79 = -0.570737 kW/K (six decimal places).
Therefore, the rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats is -0.570737 kW/K.
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Five miners must be lifted from a mineshaft (vertical hole) 100m deep using an elevator. The work required to do this is found to be 341.2kJ. If the gravitational acceleration is 9.75m/s^2, determine the average mass per person in kg.
a. 65kg
b. 70kg
c. 75kg
d. 80kg
(b).Given information: Depth of mine shaft = 100 m Work done = 341.2 kJ Gravitational acceleration = 9.75 m/s²Number of persons to be lifted = 5Formula used: Work done = force × distanceIn this question, we are supposed to determine the average mass per person in kg.
The formula to calculate the average mass per person is:Average mass per person = Total mass / Number of personsLet's begin with the solution:From the given information,The work done to lift 5 persons from the mine shaft is 341.2 kJThe gravitational acceleration is 9.75 m/s²The distance covered to lift the persons is 100 mTherefore,Work done = force × distance
Using this formula, we getForce = Work done / distance= 341.2 kJ / 100 m= 3412 J / 1 m= 3412 NNow, force = mass × gravitational accelerationTherefore, mass = force / gravitational acceleration= 3412 N / 9.75 m/s²= 350.56 kgAverage mass per person = Total mass / Number of persons= 350.56 kg / 5= 70.11 kg ≈ 70 kgTherefore, the average mass per person in kg is 70 kg. Hence, the correct option is (b).
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