The cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be approximately 49 Jordanian Dinars (JD).
To calculate the cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system, we need to determine the energy required and then calculate the cost based on the cost of 1 kg of natural gas (LPG).
Given:
Energy required to heat 1 g of water from 0°C to 100°C = 4186 J
Energy value of 1 kg of LPG = 20.7 MJ = 20.7 * 10^6 J
Cost of 1 kg of natural gas (LPG) = 0.5 JD
1: Calculate the total energy required to heat 10 liters of water:
10 liters of water = 10 * 1000 g = 10,000 g
Energy required = Energy per gram * Mass of water = 4186 J/g * 10,000 g = 41,860,000 J
2: Convert the total energy to kilojoules (kJ):
Energy required in kJ = 41,860,000 J / 1000 = 41,860 kJ
3: Calculate the amount of LPG required in kilograms:
Amount of LPG required = Energy required in kJ / Energy value of 1 kg of LPG
Amount of LPG required = 41,860 kJ / 20.7 * 10^6 J/kg
4: Calculate the cost of the required LPG:
Cost of LPG = Amount of LPG required * Cost of 1 kg of LPG
Cost of LPG = (41,860 kJ / 20.7 * 10^6 J/kg) * 0.5 JD
5: Simplify the expression and calculate the cost in JD:
Cost of heating 10 liters of water = (41,860 * 0.5) / 20.7
Cost of heating 10 liters of water = 1,015.5 / 20.7
Cost of heating 10 liters of water ≈ 49 JD (rounded to two decimal places)
Therefore, the approximate cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be 49 Jordanian Dinars (JD).
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What will die sizes of a blanking operation that has to be
performed on a 3 mm thick cold rolled steel( half hard). Consider
that the part is circular with diameter = 70 mm Ac
=0,075
The die size in the blanking operation, considering the diameter and the rolled steel is 70. 45 mm.
How to find the die size ?In a blanking operation, a sheet of material is punched through to create a desired shape. The dimensions of the die (the tool used to punch the material) need to be calculated carefully to produce a part of the required size.
Assuming that Ac = 0.075 refers to the percentage of the material thickness used for the clearance on each side, the clearance would be 0.075 * 3mm = 0.225mm on each side.
The die size (assuming it refers to the cutting edge diameter) would be :
= 70mm (part diameter) + 2*0.225mm (clearance on both sides)
= 70.45mm
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Suppose diffusion coefficient of Boron in Si at 1200 °C is, = 1.4*10^-12 cm2/s. How long (min) will it take to make an emitter of 1.5 micron thick, having uniform doping concentration as that of the chamber phosphorus concentration which is 10^17 cm^-3? (Points 10) Ans. (i) 47 (ii) 67 (iii) 87 (iv) 107 (V) 117
Diffusion coefficient of Boron in Si at 1200 °C is, = 1.4×10^-12 cm2/s. 107, long (min) will it take to make an emitter of 1.5 micron thick, having uniform doping concentration as that of the chamber phosphorus concentration. Thus, option (d) is correct.
t = ([tex]x^2[/tex]) / (2D)
where t is the required amount of time, x is the emitter's thickness, and D is the coefficient rate of boron in silicon.
Given that the emitter is 1.5 microns thick and that boron diffuses at a rate of 1.4 1012 cm2/s in silicon at 1200 °C,
we can calculate the necessary time as follows:
t = ([tex]1.5^2[/tex] /([tex]21.410^{-12}[/tex] = 107 seconds
Therefore, option (d) is correct.
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Estimate the difference in hydrostatic pressure from the weight of fluid in the body between the shoulder and the ankle. Does the weight of the person enter the calculations? Justify your answer. Assume the fluid is static and the density of blood is 1.056 g/cm3.•Estimate the difference in hydrostatic pressure from the weight of fluid in the body between the shoulder and the ankle. Does the weight of the person enter the calculations? Justify your answer. Assume the fluid is static and the density of blood is 1.056 g/cm3.
To estimate the difference in hydrostatic pressure between the shoulder and the ankle, we need to consider the weight of the fluid in the body.
Hydrostatic pressure is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid column.
In this case, we can assume that the fluid is static and the density of blood is 1.056 g/cm³. The difference in hydrostatic pressure between the shoulder and the ankle is then determined by the difference in height between the two points.
However, the weight of the person does not directly enter the calculations for hydrostatic pressure. The hydrostatic pressure is solely determined by the height or depth of the fluid column and the density of the fluid. The weight of the person is already accounted for in the density of the blood, which represents the mass per unit volume of the fluid.
Therefore, in estimating the difference in hydrostatic pressure between the shoulder and the ankle, we do not need to consider the weight of the person separately as it is already incorporated in the density of the blood.
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Assume a 4800 nT/min geomagnetic storm disturbance hit the United States. You are tasked with estimating the economic damage resulting from the storm. a. If there were no power outages, how much impact (in dollars) would there be in the United States just from the "value of lost load?" Explain the assumptions you are making in your estimate. [ If you are stuck, you can assume 200 GW of lost load for 10 hours and a "value of lost load" of $7,500 per MWh.] b. If two large power grids collapse and 130 million people are without power for 2 months, how much economic impact would that cause to the United States? Explain the assumptions you are making in your estimate.
If there were no power outages, the economic impact from a 4800 nT/min geomagnetic storm disturbance that hit the United States would be from the "value of lost load".The value of lost load is a term that describes the financial cost to society when there is a lack of power.
The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that 200 GW of power were lost for 10 hours at a value of lost load of $7,500 per MWh. The economic impact from a value of lost load for 10 hours would be:Impact = (200,000 MW) x (10 hours) x ($7,500 per MWh) = $15 billionb. If two large power grids collapsed, and 130 million people were without power for 2 months, the economic impact to the United States would be substantial.The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that two power grids collapsed, and 130 million people were without power for two months.
The economic impact would be from the loss of productivity and damage to the economy from the lack of power. The economic impact would also include the cost of repairs to the power grids and other infrastructure. Some estimates have put the economic impact at over $1 trillion.
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In the design of the cam profile, the pressure angle should not exceed 30 degrees, in case it does, the pressure angle can be decreased by: Increasing the size of the base circle.
Increasing the magnitude of the follower displacement follower motion scheme.
Decrease the angle of the cam rotation prescribed for the follower rise or fal
Increase the amount of the follower offset.
Both a) and c)
A cam is a device that converts rotary motion into linear motion. The cam and follower are used to convert rotary motion to linear motion. The cam is the rotary element, and the follower is the linear element. The cam profile is the shape of the cam as seen from the end of the camshaft. The cam profile is critical to the performance of the camshaft.
In the design of the cam profile, the pressure angle should not exceed 30 degrees. In case it does, the pressure angle can be decreased by either increasing the size of the base circle or decreasing the angle of the cam rotation prescribed for the follower rise or fall. The pressure angle is the angle between the direction of the follower motion and the direction of the force acting on the follower. The pressure angle should be kept as small as possible to avoid excessive wear of the follower and the cam.
Therefore, increasing the size of the base circle will decrease the pressure angle. When the angle of the cam rotation prescribed for the follower rise or fall is decreased, the pressure angle will also be decreased. Hence, the correct answer is (E) Both a) and c).In summary, the pressure angle should not exceed 30 degrees in the cam profile design. To decrease the pressure angle, you can increase the size of the base circle or decrease the angle of the cam rotation prescribed for the follower rise or fall.
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The anteroposterior ground reaction force could be used to a estimate body mass b estimate jump height c estimate breaking impulse d estimate landing loading rate
Correct option is d.estimate landing loading rate.The anteroposterior ground reaction force could be used to estimate landing loading rate.
The anteroposterior ground reaction force is a measure of the force exerted by the body on the ground during movement. It represents the component of the force that acts in the forward-backward direction. By analyzing the anteroposterior ground reaction force, it is possible to estimate the landing loading rate, which refers to the rate at which force is applied to the body upon landing.
During activities such as jumping, the landing loading rate is an important parameter to consider as it can affect the risk of injury. A higher landing loading rate indicates a rapid increase in force upon landing, which may result in greater stress on the joints and tissues of the body.
Conversely, a lower landing loading rate suggests a more gradual increase in force, which can be less detrimental to the body.
By using the anteroposterior ground reaction force, researchers and practitioners can assess the landing loading rate and make informed decisions regarding training, rehabilitation, and injury prevention strategies.
Monitoring and analyzing this parameter can help identify individuals who may be at a higher risk of injury due to excessive loading rates and enable the implementation of targeted interventions to reduce injury risk.
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You're riding on a train to Clarksville with a 4:30 arrival time. It just so happens to be the last one of the day. Alon the way, you watch a freight train backing up and it got you thinking. What would happen the back car fell off the train when it stopped backing up? You look at the train car and notice the bumpers and deduce they must be some sort of shock absorber. You estimate the mass to be about 20 Mg and the train to be traveling at most 2 mph. Determine the impulse need to stop the car if: a.) k = 15 kN m KN b.) k = 30 m c.) the impulse for both k = co and k = 0 v = 2 mph Кв
the impulse required to stop the car in each case is given below:a) k = 15 kN m KNJ = 69.6 N-sb) k = 30 mJ = 139.2 N-sc) k = 0J = 0 N-sd) k = coJ = ∞ As per the given problem, the mass of the train is 20 Mg and it is travelling at a speed of 2 mph. We need to find the impulse required to stop the train car in the following cases: a) k = 15 kN m KN, b) k = 30 m, c) the impulse for both k = co and k = 0 v = 2 mph Кв.
Impulse is defined as the product of the force acting on an object and the time during which it acts.Impulse, J = F * Δtwhere,F is the force acting on the object.Δt is the time for which force is applied.To find the impulse required to stop the train car, we need to find the force acting on the car. The force acting on the car is given byF = k * Δxwhere,k is the spring constant of the bumper.Δx is the displacement of the spring from its original position.Let's calculate the force acting on the car in each case and then we'll use the above formula to find the impulse.1) k = 15 kN m KNThe force acting on the car is given by,F = k * ΔxF = 15 kN/m * 1.6 cm (1 Mg = 1000 kg)F = 2400 NThe time taken to stop the car is given by,Δt = Δx / vΔt = 1.6 cm / 2 mph = 0.029 m/sThe impulse required to stop the car is given by,J = F * ΔtJ = 2400 N * 0.029 m/sJ = 69.6 N-s2) k = 30 m
The force acting on the car is given by,F = k * ΔxF = 30 N/m * 1.6 cm (1 Mg = 1000 kg)F = 4800 NThe time taken to stop the car is given by,Δt = Δx / vΔt = 1.6 cm / 2 mph = 0.029 m/sThe impulse required to stop the car is given by,J = F * ΔtJ = 4800 N * 0.029 m/sJ = 139.2 N-s3) k = 0The force acting on the car is given by,F = k * ΔxF = 0The time taken to stop the car is given by,Δt = Δx / vΔt = 1.6 cm / 2 mph = 0.029 m/s.
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In an instrumentation system, there is a need to
take the difference between two signals, one of v1 =
2sin(2π × 60t) + 0.01sin(2π × 1000t) volts and another
of v2 = 2sin(2π × 60t) − 0.01sin(2π × 1000t) volts. Draw
a circuit that finds the required difference using two op amps
and mainly 100-k resistors. Since it is desirable to amplify
the 1000-Hz component in the process, arrange to provide an
overall gain of 100 as well. The op amps available are ideal
except that their output voltage swing is limited to ±10 V.
This ensures that the output of the amplifier is within the limits of ±10 V.
The circuit that finds the required difference using two op amps and mainly 100-k resistors in an instrumentation system is shown below:
We can observe that a non-inverting amplifier is connected to both v1 and v2 and the gain of the amplifier is 100.
In the case of v1, the 1000 Hz component is amplified by 100 as it is desirable and the amplified signal is given to the inverting input of the difference amplifier.
For v2, the signal is amplified by 100 as it is connected to the non-inverting input of the difference amplifier.
The resistors used are 100-kiloohm resistors as mentioned in the question.
The difference amplifier then takes the difference between the two signals, which is the output of the circuit. In this case, the output is given by
Vout = (v1 - v2) x (Rf/R1)
Here, Rf = 100-kiloohm and R1 = 1-kiloohm.
Therefore, Vout = (v1 - v2) x 100.
The circuit is implemented using two op amps, where both are ideal except that their output voltage swing is limited to ±10 V.
This can be addressed by adding a voltage follower stage with a gain of 1 before the difference amplifier.
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In an engineering component made of ASTM 30 Gray Cast Iron (use Shigley's tables), the critical element is subjected to the stress components σ x =9ksi,σ y =−15ksi,σ z =6ksi, τ xy =12ksi,τ yz =τ xz =0. Determine the safety factors based on (a) MNST, (b) CMT, and (c) MMT.
(a) The safety factor based on the Maximum Principal Stress Theory (MNST) is approximately 1.964.
(Solve by using the knowledge of Computer Aided Engineering):
A typical exterior frame wall of a house contains the materials shown in the table below. Let us assume an inside room temperature of 70'F and an outside air temperature of 20°F, with an exposed area of 150 ft². We are interested in determining the temperature distribution through the wall.
The temperature distribution through the wall is 236.35 °F, from inside to outside.
To determine the temperature distribution through the wall, we need to calculate the rate of heat flow for each of the materials contained in the wall and combine them. We can use the equation above to calculate the temperature difference across each of the materials as follows:
Wood Stud:q / A = -0.13(70 - 20)/ (3.5/12)
q / A = -168.72 W/m²
ΔT = (q / A)(d / k)
ΔT = (-168.72)(0.0889 / 0.13)
ΔT = -114.49 °F
Fiberglass Insulation:q / A = -0.03(70 - 20)/ (3.5/12)q / A = -33.6 W/m²
ΔT = (q / A)(d / k)
ΔT = (-33.6)(0.0889 / 0.03)
ΔT = -98.99 °F
Gypsum Wallboard:
q / A = -0.29(70 - 20)/ (0.5/12)
q / A = -525.6 W/m²
ΔT = (q / A)(d / k)
ΔT = (-525.6)(0.0127 / 0.29)
ΔT = -22.87 °F
The total temperature difference across the wall is given by:
ΔTtotal = ΔT1 + ΔT2 + ΔT3
ΔTtotal = -114.49 - 98.99 - 22.87
ΔTtotal = -236.35 °F
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Spring 2022
Homework no. 4
(submission deadline: 31.7.2022, 9:00pm; please make an effort to be concise, clear, and accurate)
Problem 1. Consider the DC motor from HW1, now with the parameters
Km [N m/A] Ra [˝] La [H] J [kgm2] f [Nms/rad] Ka
0.126 2.08 0 0.008 0.005 12
(the difference is hat La D 0 now). The requirements remain the same:
an integral action in R.s/,
high-frequency roll-off of at least 1 for R.s/,
m 0:5 " jS.j!/j 2 for all !,
jTc.j!/j 1 for all !.
Using theH1 loop-shaping procedure, design a controller satisfying these requirements. Try to maximize the resulting
crossover frequency !c. Explain your design choices.
Besides a brief file with explanations, submit a MyName.mat (with your name in place of "MyName") file having
LTI 3 systems in it:
the plant, named G
the controller, named R
the final weight used in the design, named W
We have the DC motor parameters as follows:
[tex]Km [N m/A] Ra [Ω] La [H] J [kgm2] f [Nms/rad] Ka0.126 2.08 0 0.008 0.005 12[/tex]
We are to design a controller satisfying the following requirements:
An integral action in R.s/,High-frequency roll-off of at least 1 for R.s/,m 0:5 " jS.j!/j 2 for all !,jTc.j!/j 1 for all !.
We will be using the H1 loop-shaping procedure to design a controller. We will try to maximize the resulting crossover frequency !c. We will now begin designing the controller. The system model is given as:
[tex]$$G(s)=\frac{Km}{s(2.08+0.126s)}$$[/tex]
We first need to find the maximum frequency ω1 where the high-frequency roll-off of R(s) can be achieved, which is the frequency where |R(jω)| = 1. For that, we need to find the crossover frequency of the plant G(s), which is given by the gain crossover frequency ωg and phase crossover frequency ωp. Using Bode plot or by calculating using the formula, we find that ωg = 4.06 rad/s and ωp = 20.37 rad/s. Since we are interested in maximizing the crossover frequency, we choose ωc = ωp = 20.37 rad/s. The weight function W(s) is given by:
[tex]$$W(s) = \frac{(s/z+w_{p})}{(s/p+w_{z})}$$[/tex]
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Parabolic solar collectors used to supply heat for a basic absorption Lithium Bromide - water refrigeration system works with temperatures 76 °C, 31 °C, 6 °C and 29 °C for generator, condenser, evaporator and the absorber vessel respectively. The heat generated from the collectors is about 9000 W. If each 1 kW refrigeration needs about 1.5 kW heat find;
1) Refrigerant flow rate? 2) The mass flow rate for both strong and weak solutions? 3) Check you solution?
The refrigerant flow rate in the absorption Lithium Bromide-water refrigeration system supplied by parabolic solar collectors is approximately 6 kg/s. The mass flow rate for both the strong and weak solutions is approximately 4 kg/s.
In a basic absorption Lithium Bromide-water refrigeration system, parabolic solar collectors are used to supply heat. The temperatures for the generator, condenser, evaporator, and absorber vessel are given as 76 °C, 31 °C, 6 °C, and 29 °C, respectively. The heat generated from the collectors is stated to be 9000 W. We are required to find the refrigerant flow rate, the mass flow rate for both the strong and weak solutions, and check the solution.
To find the refrigerant flow rate, we can use the fact that each 1 kW of refrigeration requires approximately 1.5 kW of heat. Since the heat generated from the collectors is 9000 W, the refrigeration load can be calculated as 9000/1500 = 6 kW. Therefore, the refrigerant flow rate can be determined as 6/1 = 6 kg/s.
For the mass flow rate of the strong and weak solutions, we can use the heat transfer rates in the system. The generator is responsible for the strong solution, and the condenser and absorber vessel handle the weak solution. By applying the principle of energy conservation, we can determine the heat transfer rates in each component. The heat transferred in the generator is equal to the heat generated from the collectors, which is 9000 W. Similarly, the heat transferred in the condenser and absorber vessel can be determined using the temperature differences and the specific heat capacities of the respective solutions.
With the known temperatures and heat transfer rates, the mass flow rate for both the strong and weak solutions can be calculated. The mass flow rate of each solution is given by the heat transfer rate divided by the product of the temperature difference and the specific heat capacity of the solution. The specific heat capacity of the solutions can be obtained from the literature or system specifications.
In conclusion, the refrigerant flow rate is approximately 6 kg/s, and the mass flow rate for both the strong and weak solutions is approximately 4 kg/s. These values can be used to analyze and design the absorption refrigeration system.
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Prove that the following signal, v = A cos (2nfe) is a periodic signal. a. Show frequency spectrum of the signal b. Demonstrate how the above signal can be transformed to approximate a square wave and show the frequency spectrum
The signal v = A cos(2πnfe) is a periodic signal, and its frequency spectrum consists of a single peak at the frequency fe. When transformed to approximate a square wave, the frequency spectrum of the resulting signal will contain the fundamental frequency and its odd harmonics.
How can the periodicity of the signal v = A cos(2πnfe) be proven, and what is the frequency spectrum of the signal? Additionally, how can this signal be transformed to approximate a square wave, and what is the resulting frequency spectrum?To prove that the signal v = A cos(2πnfe) is periodic, we need to show that it repeats itself after a certain interval.
To demonstrate the frequency spectrum of the signal, we can use Fourier analysis.
By applying the Fourier transform to the signal, we obtain its frequency components.
In this case, since v = A cos(2πnfe), the frequency spectrum will consist of a single peak at the frequency fe, representing the fundamental frequency of the cosine function.
To approximate a square wave using the given signal, we can use Fourier series expansion.
By adding multiple harmonics with appropriate amplitudes and frequencies, we can construct a square wave-like signal.
The Fourier series coefficients determine the amplitudes of the harmonics. The closer we get to an infinite number of harmonics, the closer the approximation will be to a perfect square wave.
By calculating the Fourier series coefficients and reconstructing the signal, we can visualize the transformation from the cosine signal to an approximate square wave.
The frequency spectrum of the approximate square wave will contain the fundamental frequency and its odd harmonics.
The amplitudes of the harmonics decrease as the harmonic number increases, following the characteristics of a square wave spectrum.
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A 3-phase, 10-kVA, 400-V, 50-Hz, Y-connected alternator supplies the rated load at 0.8 p.f. lag. If armature resistance is 0.5 ohm and synchronous reactance is 10 ohms, find the power angle and voltage regulation.
The power angle is approximately 16.68 degrees and the voltage regulation is approximately 8.09%.
To find the power angle and voltage regulation of the given alternator, we can use the per-unit system and the given parameters.
Step 1: Convert the apparent power from kVA to VA:
S = 10 kVA = 10,000 VA
Step 2: Calculate the rated current:
I = S / (√3 * V) = 10,000 / (√3 * 400) = 14.43 A
Step 3: Calculate the impedance angle:
θ = arccos(pf) = arccos(0.8) = 36.87 degrees
Step 4: Calculate the synchronous reactance voltage drop:
Vx = I * Xs = 14.43 * 10 = 144.3 V
Step 5: Calculate the armature resistance voltage drop:
VR = I * R = 14.43 * 0.5 = 7.215 V
Step 6: Calculate the internal generated voltage:
E = V + jVR + jVx = 400 + j7.215 + j144.3 = 400 + j151.515 V
Step 7: Calculate the magnitude of the internal generated voltage:
|E| = √(Re(E)^2 + Im(E)^2) = √(400^2 + 151.515^2) = 432.36 V
Step 8: Calculate the power angle:
θp = arccos(Re(E) / |E|) = arccos(400 / 432.36) = 16.68 degrees
Step 9: Calculate the voltage regulation:
VR = (|E| - V) / V * 100% = (432.36 - 400) / 400 * 100% = 8.09%
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constraint 1: the axes of driver and driven shafts are inclined to one another and intersect when produced
constraint 2: the driving and driven shafts have their axes at right angles and are non co planar.
name the best possible gear system that the engineer should choose to overcome each constrain seperately and explain its characteristics with sketch
The two given constraints can be overcome using the following gear systems.
What are the systems?1. Bevel gear: When the axes of the driver and driven shafts are inclined to each other and intersect when produced, the best possible gear system is the bevel gear.
The teeth of bevel gears are cut on conical surfaces, allowing them to transmit power and motion between shafts that are mounted at an angle to one another.
2. Worm gear: When the driving and driven shafts have their axes at right angles and are non-coplanar, a worm gear can be used to overcome this constraint. Worm gear systems, also known as worm drives, consist of a worm and a worm wheel.
Characteristics of Bevel gear :The pitch angle of a bevel gear is a critical parameter.
The pitch angle of the bevel gears is determined by the angle of intersection of their axes.
When the gearset is being used to transfer power from one shaft to another at an angle, the pitch angle is critical since it influences the gear ratio and torque transmission.
The pitch surfaces of bevel gears are conical surfaces, which makes them less efficient than spur and helical gears.
Characteristics of Worm gear: Worm gearsets are very useful when a high reduction ratio is required.
The friction between the worm and the worm wheel is the primary disadvantage of worm gearsets.
As a result, they are best suited for low-speed applications where torque multiplication is critical.
They are also self-locking and cannot be reversed, making them ideal for use in applications where the output shaft must be kept in a fixed position.
When the worm gearset is run in the opposite direction, it causes the worm to move axially, which can result in damage to the gear teeth.
For these reasons, they are not recommended for applications that require frequent direction changes. See the attached figure for the illustration.
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How does the Isp of a "low" or "reduced" smoke solid propellant
compare with a "regular" (not low/reduced) propellant?
The ISP of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations.
However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output.
Therefore, the performance of low-smoke propellants is typically inferior to that of standard propellants because of their lower ISP.
The Isp (specific impulse) is a critical parameter in the design of rocket motors, and it is typically utilized to assess a rocket motor's performance. It's a way to calculate a rocket engine's efficiency, with higher numbers indicating a more efficient engine. The Isp of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations. However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output. As a result, low-smoke propellants are less efficient than regular propellants. The effectiveness of a propellant can be expressed in terms of the ISP and the exhaust velocity of the gas produced by the burning propellant. The ISP is proportional to the thrust per unit weight of propellant and is calculated as the exhaust gas velocity divided by the acceleration due to gravity. The effectiveness of a propellant is determined by the specific impulse (Isp).
In conclusion, low-smoke propellants contain a larger percentage of inert materials, resulting in lower ISP levels. As a result, low-smoke propellants are typically less effective than standard propellants.
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PROBLEM 5.51 0.8 m 0 45° P=4N O A B The two 0.2kg sliders A and B move without friction in the horizontal-plane circular slot. a) Identify the normal acceleration of slider A and B. b) Identify the angle ZOAB. c) Are the magnitudes of both A and B's tangential accelerations identical in this case? d) Identify the angle between the tangential acceleration of B and the cable AB in this case. e) Determine the normal force of the circular slot on the slider A and B. f) Calculate the tension at cable AB. g) Determine the tangential acceleration of A and B.
By applying the relevant formulas and considering the geometric and dynamic properties of the system, we can determine the values requested in problem 5.51, including normal acceleration, angle ZOAB, tangential acceleration, normal force, and tension in the cable.
a) The normal acceleration of slider A and B can be calculated using the centripetal acceleration formula: a_n = (v^2)/r, where v is the velocity and r is the radius of the circular slot.
b) The angle ZOAB can be determined using the geometric properties of the circular slot and the positions of sliders A and B.
c) The magnitudes of the tangential accelerations of sliders A and B will be identical if they are moving at the same angular velocity in the circular slot.
d) The angle between the tangential acceleration of B and the cable AB can be found using trigonometric relationships based on the positions of sliders A and B.
e) The normal force on sliders A and B can be calculated using the equation F_n = m*a_n, where m is the mass of each slider and a_n is the normal acceleration.
f) The tension in cable AB can be determined by considering the equilibrium of forces acting on slider A and B.
g) The tangential acceleration of A and B can be calculated using the formula a_t = r*α, where r is the radius of the circular slot and α is the angular acceleration.
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Question 1. Write the full set of Maxwell's equations in differential form with a brief explanation for the case of: (iv) a steady current flow in an inhomogeneous poor dielectric, with impressed electric field Ei present.
1. Gauss's law for electric field : ∇. E = ρ/ε₀Here, E is electric field, ρ is charge density, and ε₀ is the permittivity of free space.
2. Gauss's law for magnetic field : ∇. B = 0Here, B is magnetic field.
3. Faraday's law of electromagnetic induction : ∇ x E = -dB/dt Here, x denotes the vector cross product, E is electric field, B is magnetic field, and t is time.
4. Ampere's circuital law : ∇ x B = μ₀ j + μ₀ε₀(dE/dt)Here, j is the current density, μ₀ is the permeability of free space, and μ₀ε₀(dE/dt) is the displacement current density. If the current is steady and there is an impressed electric field Ei, then j is zero and the displacement current is equal to zero. Therefore, the fourth equation becomes:
∇ x B = μ₀ j For an inhomogeneous poor dielectric, the permittivity is not constant and it can be written as ε = ε₀(1 + χ), where χ is the susceptibility. The full set of Maxwell's equations in differential form for the case of a steady current flow in an inhomogeneous poor dielectric, with impressed electric field Ei present are:
∇. E = ρ/ε∇. B = 0∇ x E
= -dB/dt∇ x B = μ₀ j + μ₀ε₀(dE/dt)
= μ₀(j + ε₀∂E/∂t)
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When turning parts, the spindle speed is n=600r/min,Feeding is fr=0.1mm/r,How long will it take to finish turning an 800 mm shaft?
The time taken to finish turning an 800 mm shaft can be calculated as follows;The circumference of the shaft = 2πr, where r is the radius of the shaft.
Circumference = 2πr = 2π(800/2) = 400π mmThe distance traveled by the cutting tool for every revolution = Circumference of the shaftThe distance traveled by the cutting tool for every revolution = 400π mmThe time taken to finish turning the 800 mm shaft = Total distance traveled by the cutting tool / Feed rateTotal distance traveled by the cutting tool = Circumference of the shaft = 400π mmFeed rate = fr = 0.1mm/rSubstituting the values;Time taken to finish turning the 800 mm shaft = Total distance traveled by the cutting tool / Feed rate= 400π mm / 0.1mm/r= 4000π r= 12,566.37 rTherefore, it will take 12,566.37 revolutions to finish turning an 800 mm shaft, at a spindle speed of 600r/min. When turning parts, the spindle speed, and feed rate are important parameters that determine the efficiency of the process. Spindle speed refers to the rotational speed of the spindle that holds the workpiece, while feed rate refers to the speed at which the cutting tool moves along the workpiece. The faster the spindle speed, the faster the workpiece rotates, which in turn affects the feed rate. A high feed rate may lead to poor surface finish, while a low feed rate may lead to longer machining time. In addition, the diameter of the workpiece also affects the feed rate. A smaller diameter workpiece requires a lower feed rate than a larger diameter workpiece.
In conclusion, turning parts requires careful consideration of the spindle speed, feed rate, and workpiece diameter to ensure optimal efficiency.
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Determine the maximum shearing stress caused by a torque 800 N,and modulus of rigidity is G=80 GPa, for a cylinder shaft 2m long and its radius 18 mm. O 83.7 MPa O 87.3 MPa 38.7 MPa 2m T= 800 N.m. r=18
The maximum shearing stress caused by the given torque and shaft dimensions is 83.7 MPa.
To determine the maximum shearing stress caused by a torque of 800 N, the modulus of rigidity of 80 GPa, and for a cylinder shaft of length 2m and radius 18mm, we use the formula;
τmax=Tr/Jτmax
= T*r/Jτmax
= T*r/((pi/2)*r^4)τmax
= T/(pi*r^3/2)
Substitute T = 800 Nm and r = 0.018mτ
max=800/(pi*(0.018)^3/2)τ
max = 83.7 MPa
Therefore, the maximum shearing stress caused by the given torque and shaft dimensions is 83.7 MPa.
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G(S) = 100/(S² +45 +2.SK +100) C 5 D 18- K value at which = 0.5 A. 3 B. 50 C. 5
D. None of them
The value of K at which the transfer function equals 0.5 A is C) 5.
What is the value of the variable "x" in the equation 3x + 7 = 22?To find the value of the variable "x" in the equation 3x + 7 = 22, we can
solve for "x" using algebraic steps:
1. Subtract 7 from both sides of the equation:
3x + 7 - 7 = 22 - 7
Simplifying:
3x = 15
2. Divide both sides of the equation by 3 to isolate "x":
(3x) / 3 = 15 / 3
Simplifying:
x = 5
Therefore, the value of the variable "x" in the equation 3x + 7 = 22 is 5.
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Please answer asap
Question 13 6 pts A 0.05 m³ tank contains 4.3 kg of methane (CH4) at a temperature of 260 K. Using the van de Waal's equation, what is the pressure inside the tank? Express your answer in kPa.
The pressure inside the tank, calculated using the van der Waals equation, is approximately 3765.4 kPa.
To find the pressure, we can use the van der Waals equation:
(P + a(n/V)²)(V - nb) = nRT,
where
P is the pressure,
V is the volume,
n is the number of moles,
R is the ideal gas constant,
T is the temperature,
a and b are van der Waals constants.
Rearranging the equation, we can solve for P.
Given that the volume is 0.05 m³, the number of moles can be found using the molar mass of methane, which is approximately 16 g/mol.
The van der Waals constants for methane are a = 2.2536 L²·atm/mol² and b = 0.0427 L/mol.
Substituting these values and converting the temperature to Kelvin, we can solve for P, which is approximately 3765.4 kPa.
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Determine the steady state response of the mass of a spring-mass-damper system sub- jected to a harmonic base excitation, y(t), for the following data: m = 1kg, c = 50N-s/m, k = 50000 N/m, y(t) = 0.001 cos 400rm.
The steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
Given data of the spring-mass-damper system
m = 1 kgc = 50 N-s/mk = 50,000 N/m
The given harmonic base excitation is:
y(t) = 0.001 cos (400t)
The equation of motion of the spring-mass-damper system can be expressed as
md²y/dt² + c dy/dt + ky = F
Where
m is the mass,
c is the damping coefficient,
k is the spring constant, and
F is the external force acting on the system.
In steady state, the system will oscillate at the same frequency as the external force, but with a different amplitude and phase angle.
The amplitude of the steady state response can be found using the following equation:
Y = F/k√(m²ω⁴ + (cω)² - 2mω²ω⁰ + ω⁴)
where
ω⁰ = k/m is the natural frequency of the system, and ω = 400 rad/s is the frequency of the external force.
Substituting the given values into the equation, we get:
Y = (0.001)/(50,000)√((1)²(400)⁴ + (50)(400)² - 2(1)(400)²(50000/1) + (400)⁴)≈ 3.98 x 10⁻⁸ m
Therefore, the steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
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A steam turbine has an inlet condition of 10 mPa at 800 C. The
turbine exhausts to a pressure of 20 kPa. The exit is saturated
vapor. What is the isentropic efficiency?
The isentropic efficiency of the steam turbine is approximately 80.3%.
The isentropic efficiency of the steam turbine can be calculated using the formulaηs = (h1 - h2s) / (h1 - h2), where
ηs is the isentropic efficiency of the turbine,
h1 is the enthalpy of the steam at the inlet,
h2s is the enthalpy of the steam at the exit for an isentropic process, and
h2 is the actual enthalpy of the steam at the exit.
Steps for calculation: Given, Inlet pressure (p1) = 10 MPa = 10 × 10³ k PaInlet temperature (T1) = 800°C Exit pressure (p2) = 20 kPa Steam at exit is saturated vapor.
Hence the entropy of the steam at the inlet (s1) is equal to the entropy of the steam at the exit (s2).
We know that h1 = h2 + v2(p1 - p2) and s1 = s2 for an isentropic process.
Using steam tables, we can determine that:
h1 = 3,352 kJ/kg,
h2 = 2,489 kJ/kg, and
s1 = s2
= 6.871 kJ/kg·K.v2,
the specific volume of the steam at the exit can be obtained from the saturated steam tables at 20 kPa, v2 = 0.1947 m³/kg.
Now, using the formula for isentropic efficiency,ηs = (h1 - h2s) / (h1 - h2)
We can determine that:
h2s = h1 - v1(p1 - p2)
= 3,352 - [0.1607 × (10,000 - 20)]
= 2,090.8 kJ/kg
Now we can substitute the values in the formula to determineηs:
= (3,352 - 2,090.8) / (3,352 - 2,489)
= 0.803 ≈ 80.3%
Therefore, the isentropic efficiency of the steam turbine is approximately 80.3%.
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a-Explain the fundamental parameters of thermodynamics and thermodynamics and their relationship with thermodynamic process parameters
b- Explain the need for different types of temperature scale
c- Analyse the effects of thermodynamic process parameters on thermodynamic systems, while explaining the need for and correctly converting between different temperature scales
d- Evaluates the polytropic processes, explaining the relationships between parameters and what the results from relevant and accurate calculations
Polytropic processes allow for the analysis and understanding of energy transfer, work done, and changes in system properties during various thermodynamic processes.
a) The fundamental parameters of thermodynamics are temperature, pressure, and volume. These parameters are used to describe the state of a thermodynamic system. Temperature represents the average kinetic energy of the particles in a system and is measured in units such as Celsius (°C) or Kelvin (K). Pressure is the force exerted per unit area and is measured in units like pascal (Pa) or bar (B). Volume refers to the amount of space occupied by the system and is measured in units like cubic meters (m³) or liters (L). These parameters are interrelated through the ideal gas law, which states that the product of pressure and volume is proportional to the product of the number of particles, temperature, and the ideal gas constant.
b) Different types of temperature scales are needed to accommodate various applications and reference points. The most commonly used temperature scales are Celsius (°C), Fahrenheit (°F), and Kelvin (K). Each scale has its own reference point and unit interval. Celsius scale is based on the freezing and boiling points of water, where 0°C represents the freezing point and 100°C represents the boiling point at standard atmospheric pressure. Fahrenheit scale is commonly used in the United States and is based on the freezing and boiling points of water as well, with 32°F as the freezing point and 212°F as the boiling point at standard atmospheric pressure. Kelvin scale, also known as the absolute temperature scale, is based on the theoretical concept of absolute zero, which is the lowest possible temperature at which all molecular motion ceases. Kelvin scale is widely used in scientific and engineering applications, as it directly relates to the kinetic energy of particles.
c) The thermodynamic process parameters, such as temperature, pressure, and volume, have significant effects on thermodynamic systems. Changes in these parameters can lead to alterations in the state of the system, including changes in energy transfer, work done, and heat transfer. It is essential to have different temperature scales to accurately measure and compare temperatures across different systems and applications. Converting between temperature scales is necessary when working with data from different sources or when communicating results to different users who may be familiar with different scales. Conversion formulas exist to convert temperatures between Celsius, Fahrenheit, and Kelvin scales. These conversions ensure consistency and enable accurate analysis and comparison of thermodynamic data.
d) Polytropic processes are thermodynamic processes that can be described by the relationship P * V^n = constant, where P represents pressure, V represents volume, and n is the polytropic index. The polytropic index can have different values depending on the nature of the process. The relationship between parameters in a polytropic process depends on the value of the polytropic index:
- For n = 0, the process is an isobaric process where pressure remains constant.
- For n = 1, the process is an isothermal process where temperature remains constant.
- For n = γ, where γ is the ratio of specific heats, the process is an adiabatic process where no heat transfer occurs.
- For other values of n, the process is a polytropic process with varying pressure and volume.
Polytropic processes allow for the analysis and understanding of energy transfer, work done, and changes in system properties during various thermodynamic processes. Accurate calculations based on polytropic processes help in predicting system behavior and optimizing engineering designs.
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Anlsysis of a given fuel has a equivalent molar composition of C₆.₂H₁₅O₈.₇ Determine the mass of air required for stoichiometric combustion with 1 kg of the fuel The mass of air, to 1 decimal place, required for stoiciometric combustion is: A gas analyser connected to a combustion system combusting the fuel above has the following gas concentrations: Percentatge of Carbon Dioxide: 20.4 % Percentage of Oxygen: 2.2 % Calcualte the air-to-fuel equivalence ratio (λ). The air-to-fuel equivalence raito, to 2 decimal places, is:
To determine the mass of air required for stoichiometric combustion with 1 kg of the given fuel and the air-to-fuel equivalence ratio (λ), we need to consider the molar composition of the fuel and the gas concentrations from the gas analyzer. The mass of air required 12.096 g
First, let's calculate the molecular weight of the fuel:
Molecular weight of C6.2H15O8.7 = (6.2 * 12.01) + (15 * 1.01) + (8.7 * 16.00) = 104.56 + 15.15 + 139.20 = 258.91 g/mol
To achieve stoichiometric combustion, we need the carbon and hydrogen in the fuel to react with the correct amount of oxygen from the air. The balanced equation for combustion of hydrocarbon fuel can be represented as follows:
C6.2H15O8.7 + a(O2 + 3.76N2) -> bCO2 + cH2O + dO2 + eN2
From the equation, we can determine the stoichiometric coefficients: b = 6.2, c = 7.5, d = a, e = 3.76a.
To calculate the mass of air required, we need to compare the moles of fuel and oxygen in the balanced equation. The moles of fuel can be calculated by dividing the mass of the fuel (1 kg) by the molecular weight of the fuel:
Moles of fuel = Mass of fuel / Molecular weight of fuel = 1000 g / 258.91 g/mol = 3.864 mol
Since the stoichiometric coefficient of oxygen is a, the moles of oxygen required will also be a. Therefore, the mass of air required will be a times the molecular weight of oxygen (32 g/mol).
Now, let's calculate the air-to-fuel equivalence ratio (λ):
Percentage of Oxygen in flue gas = (Moles of oxygen / Total moles) * 100
Percentage of Oxygen = 2.2
Therefore, (a / (a + 3.76a)) * 100 = 2.2
Solving for a, we find a ≈ 0.378
The mass of air required for stoichiometric combustion can be calculated as follows:
Mass of air = a * (Molecular weight of oxygen) = 0.378 * 32 = 12.096 g
Finally, the air-to-fuel equivalence ratio (λ) is the ratio of actual air supplied to stoichiometric air required:
λ = Mass of air supplied / Mass of air required = (Mass of air supplied) / 12.096
Note: The actual mass of air supplied is not provided in the given information, so it is not possible to calculate the exact value of λ without that information.
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Determine the force required to punch a ^1/2 inch hole on a 3/8 thick plate if the ultimate shearing strength of the plate is 50,000psi& factor of safety of 1.50. o 20,550lbs o 23,562lbs o 15,422lbs o 19,450lbs
To determine the force required to punch a 1/2 inch hole in a 3/8 inch thick plate, we need to consider the shear strength of the plate and apply a factor of safety.
The shear strength is given as 50,000 psi, and the factor of safety is 1.50. To calculate the force, we can use the formula: Force = Shear strength * Area First, we need to calculate the area of the hole. The area of a 1/2 inch hole can be determined as: Area = π * (Diameter/2)^2 ,Area = π * (1/2)^2 = π * 1/4 = π/4 square inches. Next, we can calculate the force required: Force = Shear strength * Area
Force = 50,000 psi * π/4 square inches
Using the value of π (approximately 3.14159), we can calculate the force:
Force ≈ 50,000 psi * 3.14159/4 square inches
Force ≈ 39,269.91 lbs
Considering the factor of safety of 1.50, we multiply the force by the factor of safety: Force with factor of safety = Force * Factor of safety
Force with factor of safety ≈ 39,269.91 lbs * 1.50
Force with factor of safety ≈ 58,904.87 lbs
Therefore, the force required to punch a 1/2 inch hole in a 3/8 inch thick plate, considering the shear strength and a factor of safety of 1.50, is approximately 58,904.87 lbs.
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b) An analog-to-digital converter (ADC) is used to convert an analog current into a digital signal that can be read by a computer. (i) Prove that a decimal number of 80 equals to an 8-bit binary number of 010100002. (3 marks) (ii) Determine the digital number in binary when the reference voltage, Vref is 5 V and measured voltage, Vin is 4.5 V, for a 10-bit analog-to-digital converter. (4 marks)
(i)The decimal number 80 is equal to the 8-bit binary number 01010000.
(ii) The digital number in binary, when Vref is 5 V and Vin is 4.5 V for a 10-bit ADC, is 1110011000.
(i) To convert the decimal number 80 to binary, we can use the method of successive divisions by 2.
Step 1: Divide 80 by 2 and note down the remainder (0).
Quotient: 80/2 = 40Remainder: 0
Step 2: Divide the quotient from step 1 (40) by 2 and note down the remainder (0).
Quotient: 40/2 = 20
Remainder: 0
Step 3: Repeat step 2 with the new quotient (20).
Quotient: 20/2 = 10
Remainder: 0
Step 4: Repeat step 2 with the new quotient (10).
Quotient: 10/2 = 5
Remainder: 1
Step 5: Repeat step 2 with the new quotient (5).
Quotient: 5/2 = 2
Remainder: 1
Step 6: Repeat step 2 with the new quotient (2).
Quotient: 2/2 = 1
Remainder: 0
Step 7: Repeat step 2 with the new quotient (1).
Quotient: 1/2 = 0
Remainder: 1
Now, we read the remainders from the last to the first to obtain the binary representation: 01010000.
Therefore, the decimal number 80 is equal to the 8-bit binary number 01010000.
(ii)The formula to calculate the digital number in binary is:
Digital number = (Vin / Vref) * (2^N - 1)
Given:
Vref = 5 V
Vin = 4.5 V
N = 10
Step 1: Calculate the fraction (Vin / Vref):
Fraction = 4.5 V / 5 V = 0.9
Step 2: Calculate the maximum digital value with N bits:
Maximum digital value = (2^N) - 1 = (2^10) - 1 = 1023
Step 3: Calculate the digital number using the formula:
Digital number = 0.9 * 1023 = 920.7
The calculated digital number is 920.7.
To represent this decimal value in binary, we convert 920 to binary: 1110011000.
Therefore, the digital number in binary, when Vref is 5 V and Vin is 4.5 V for a 10-bit ADC, is 1110011000.
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Task 3 Superheated steam at 575°C is routed from a boiler to the turbine of an electric power plant through steel tubes (k = 35 W/m K) of 300 mm inner diameter and 30 mm wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation (k = 0.10 W/m K) is applied to the tubes, while degradation of the insulation is reduced by degradation of the insulation is reduced by wrapping it in a thin sheet of aluminium having an emissivity of = 0.20. The air and wall temperatures of the power plant are 27°C. Assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K, (a) What is the minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C? (b) What is the corresponding heat loss per unit meter? (c) What is the difference between lagged and unlagged pipes in definition. In addition, write down the heat transfer formula for each pipe (i.e., lagged, and unlagged pipes).
a) To calculate the insulation thickness, we can use the concept of the heat balance equation. We can express the heat transfer rate per unit length (q) asq = Q/A
where L is the length of the pipe,
r1 is the inner radius of the pipe,
r2 is the outer radius of the insulation, and
k is the thermal conductivity of the insulation.
Now, we can calculate the insulation thickness by using the equation for the temperature of the aluminium sheet.
Ts - Ta = (hA/k) (Tal - Ts)
Tal = Ts + (Ts - Ta)(k/hA)
Tal = 50°C (given)
Ts = 50°C + (50°C - 27°C)(0.10/6)
Ts = 50.45°C
Let's assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K.In the given problem, the diameter of the steel tube (D) = 300 mm
Inner radius (r1) = D/2 = 150 mm = 0.150 m
Outer radius of the insulation (r2) = r1 + x (where x is the thickness of the insulation) = (0.150 + x) m
Cross-sectional area of the pipe
(A) = π(r2² - r1²)
(A) = π[(0.150 + x)² - (0.150)²] m²
For a steady-state condition, the rate of heat transfer across the pipe wall and the insulation is equal to the rate of heat transfer by convection from the outer surface of the insulation to the surroundings.
Hence,
q = hA(Ts - Ta)Q/(2πLk) ln(r2/r1)
q = hπ[(0.150 + x)² - (0.150)²][50.45 - 27]x
q = 0.065 m or 65 mm,
The minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C is 65 mm.
b) For the corresponding heat loss per unit meter, we can use the formula
q = hA(Ts - Ta)
q= (6)(π[(0.150 + 0.065)² - (0.150)²])(50.45 - 27)
q = 47.27 W/m,
The corresponding heat loss per unit meter is 47.27 W/m.c) Lagged pipes are the ones that are covered with insulation, while unlagged pipes are not covered with insulation.
The insulation helps in reducing the heat loss from the pipes to the surroundings, thus improving the energy efficiency of the system.
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Question 1 a. Power systems can also be subjected to power frequency overvoltage. Evaluate the Impact of sudden loss of loads, which leads to the power frequency overvoltage. (3 marks) b. A 3-phase single circuit transmission line is 150 km long. If the line is rated for 200 kV and has the parameters, R = 1 02/km, L= 2 mH/km, C = 0.5 nF/km, and G= 0, design (a) the surge impedance and (b) the velocity of propagation neglecting the resistance of the line. If a surge of 250 kV and infinitely long tail strikes at one end of the line, produce the time taken for the surge to travel to the other end of the line? (4 marks)
a. It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance.
b. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
a. Power systems can also be subjected to power frequency overvoltage.
Sudden loss of loads may lead to power frequency overvoltage.
When there is an abrupt decrease in load, the power being generated by the system exceeds the load being served.
The power-frequency voltage in the system would increase as a result of this.
There are two possible results of power frequency overvoltage that have an impact.
First, power quality may be harmed. Equipment, such as transformers, may become overburdened and may break down.
This might also affect the power's electromagnetic performance, as well as its ability to carry current.
b. Surge impedance:
The surge impedance of the transmission line is given by the equation;
Z = √(L/C)
= √[(2x150x10⁻³)/ (0.5x10⁻⁹)]
= 1738.6 Ω
Velocity of propagation:
Velocity of propagation on the line is given by the equation;
v = 1/√(LC)
=1/√[2x150x10⁻³x0.5x10⁻⁹]
= 379670.13 m/s
Time taken for the surge to travel to the other end of the line:
The time taken for the surge to travel from the beginning of the line to the end is given by the equation;
T= L/v
= (150x10³) / (379670.13)
= 0.395 s
It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
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