(i) To determine the chamber pressure that gives a linear burning rate of 30 mm/s, we can use the concept of proportionality between burning rate and chamber pressure. By setting up a proportion based on the given data, we can find the desired chamber pressure.
(ii) To calculate the propellant consumption rate, we need to consider the burning surface area of the grain, the linear burning rate, and the density of the propellant. By multiplying these values, we can determine the propellant consumption rate in kg/s.
Let's calculate these values:
(i) Using the given data, we can set up a proportion to find the chamber pressure (P) for a linear burning rate (R) of 30 mm/s:
(80 bar) / (20 mm/s) = (P) / (30 mm/s)
Cross-multiplying, we get:
P = (80 bar) * (30 mm/s) / (20 mm/s)
P = 120 bar
Therefore, the chamber pressure that gives a linear burning rate of 30 mm/s is 120 bar.
(ii) The burning surface area (A) of the grain can be calculated using the formula:
A = π * (diameter/2)^2
A = π * (200 mm / 2)^2
A = π * (100 mm)^2
A = 31415.93 mm^2
To calculate the propellant consumption rate (C), we can use the formula:
C = A * R * ρ
where R is the linear burning rate and ρ is the density of the propellant.
C = (31415.93 mm^2) * (30 mm/s) * (2000 kg/m^3)
C = 188,495,800 mm^3/s
C = 0.1885 kg/s
Therefore, the propellant consumption rate is 0.1885 kg/s if the density of the propellant is 2000 kg/m^3, the grain diameter is 200 mm, and the combustion pressure is 100 bar.
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I have found a research study online with regards to PCM or Phase changing Material, and I can't understand and visualize what PCM is or this composite PCM. Can someone pls help explain and help me understand what these two composite PCMs are and if you could show images of a PCM it is really helpful. I haven't seen one yet and nor was it shown to us in school due to online class. pls help me understand what PCM is the conclusion below is just a part of a sample study our teacher gave to help us understand though it was really quite confusing, Plss help
. Conclusions
Two composite PCMs of SAT/EG and SAT/GO/EG were prepared in this article. Their thermophysical characteristic and solar-absorbing performance were investigated. Test results indicated that GO showed little effect on the thermal properties and solar absorption performance of composite PCM. However, it can significantly improve the shape stability of composite PCM. The higher the density is, the larger the volumetric heat storage capacity. When the density increased to 1 g/ cm3 , SAT/EG showed severe leakage while SAT/GO/EG can still keep the shape stability. A novel solar water heating system was designed using SAT/GO/EG (1 g/cm3 ) as the solar-absorbing substance and thermal storage media simultaneously. Under the real solar radiation, the PCM gave a high solar-absorbing efficiency of 63.7%. During a heat exchange process, the temperature of 10 L water can increase from 25 °C to 38.2 °C within 25 min. The energy conversion efficiency from solar radiation into heat absorbed by water is as high as 54.5%, which indicates that the novel system exhibits great application effects, and the composite PCM of SAT/GO/EG is very promising in designing this novel water heating system.
PCM stands for Phase Changing Material, which is a material that can absorb or release a large amount of heat energy when it undergoes a phase change.
A composite PCM, on the other hand, is a mixture of two or more PCMs that exhibit improved thermophysical properties and can be used for various applications. In the research study mentioned in the question, two composite PCMs were investigated: SAT/EG and SAT/GO/EG. SAT stands for stearic acid, EG for ethylene glycol, and GO for graphene oxide.
These composite PCMs were tested for their thermophysical characteristics and solar-absorbing performance. The results showed that GO had little effect on the thermal properties and solar absorption performance of composite PCM, but it significantly improved the shape stability of the composite PCM.
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A heat engine operating on a Carnot Cycle rejects 519 kJ of heat to a low-temperature sink at 304 K per cycle. The high-temperature source is at 653°C. Determine the thermal efficiency of the Carnot engine in percent.
The thermal efficiency of the Carnot engine, operating on a Carnot Cycle and rejecting 519 kJ of heat to a low-temperature sink at 304 K per cycle, with a high-temperature source at 653°C, is 43.2%.
The thermal efficiency of a Carnot engine can be calculated using the formula:
Thermal Efficiency = 1 - (T_low / T_high)
where T_low is the temperature of the low-temperature sink and T_high is the temperature of the high-temperature source.
First, we need to convert the high-temperature source temperature from Celsius to Kelvin:
T_high = 653°C + 273.15 = 926.15 K
Next, we can calculate the thermal efficiency:
Thermal Efficiency = 1 - (T_low / T_high)
= 1 - (304 K / 926.15 K)
≈ 1 - 0.3286
≈ 0.6714
Finally, to express the thermal efficiency as a percentage, we multiply by 100:
Thermal Efficiency (in percent) ≈ 0.6714 * 100
≈ 67.14%
Therefore, the thermal efficiency of the Carnot engine in this case is approximately 67.14%.
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An inductive load of 100 Ohm and 200mH connected in series to thyristor supplied by 200V dc source. The latching current of a thyristor is 45ma and the duration of the firing pulse is 50us where the input supply voltage is 200V. Will the thyristor get fired?
In order to find out whether the thyristor will get fired or not, we need to calculate the voltage and current of the inductive load as well as the gate current required to trigger the thyristor.The voltage across an inductor is given by the formula VL=L(di/dt)Where, VL is the voltage, L is the inductance, di/dt is the rate of change of current
The current through an inductor is given by the formula i=I0(1-e^(-t/tau))Where, i is the current, I0 is the initial current, t is the time, and tau is the time constant given by L/R. Here, R is the resistance of the load which is 100 Ohm.
Using the above formulas, we can calculate the voltage and current as follows:VL=200V since the supply voltage is 200VThe time constant tau = L/R = 200x10^-3 / 100 = 2msThe current at t=50us can be calculated as:i=I0(1-e^(-t/tau))=0.45(1-e^(-50x10^-6/2x10^-3))=0.45(1-e^-0.025)=0.045A.
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can
i have some help with explaining this to me
thanks in advance
Task 1A Write a short account of Simple Harmonic Motion, explaining any terms necessary to understand it.
Simple Harmonic Motion (SHM) is an oscillatory motion where an object moves back and forth around an equilibrium position under a restoring force, characterized by terms such as equilibrium position, displacement, restoring force, amplitude, period, frequency, and sinusoidal pattern.
What are the key terms associated with Simple Harmonic Motion (SHM)?Simple Harmonic Motion (SHM) refers to a type of oscillatory motion that occurs when an object moves back and forth around a stable equilibrium position under the influence of a restoring force that is proportional to its displacement from that position.
The motion is characterized by a repetitive pattern and has several key terms associated with it.
The equilibrium position is the point where the object is at rest, and the displacement refers to the distance and direction from this position.
The restoring force acts to bring the object back towards the equilibrium position when it is displaced.
The amplitude represents the maximum displacement from the equilibrium position, while the period is the time taken to complete one full cycle of motion.
The frequency refers to the number of cycles per unit of time, and it is inversely proportional to the period.
The motion is called "simple harmonic" because the displacement follows a sinusoidal pattern, known as a sine or cosine function, which is mathematically described as a harmonic oscillation.
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Show p-v and t-s diagram
A simple air refrigeration system is used for an aircraft to take a load of 20 TR. The ambient pressure and temperature are 0.9 bar and 22°C. The pressure of air is increased to 1 bar due to isentropic ramming action. The air is further compressed in a compressor to 3.5 bar and then cooled in a heat exchanger to 72C. Finally, the air is passed through the cooling turbine and then it is supplied to the cabin at a pressure of 1.03 bar. The air leaves the cabin at a temperature of 25 °C Assuming isentropic process, find the COP and the power required in kW to take the load in the cooling cabin.
Take cp of air = 1.005 kj/kgk, k=1.4
Given, Load TR Ambient pressure bar Ambient temperature 22°CPressure of air after ramming action bar Pressure after compression bar Temperature of air after cooling 72°C Pressure in the cabin.
It is a process in which entropy remains constant. Air Refrigeration Cycle. Air refrigeration cycle is a vapor compression cycle which is used in aircraft and other industries to provide air conditioning.
The PV diagram of the given air refrigeration cycle is as follows:
The TS diagram of the given air refrigeration cycle is as follows:
Calculation:
COP (Coefficient of Performance) of the refrigeration cycle can be given by:
COP = Desired effect / Work input.
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Quesion 2. Explain Voltage Regulation the equation for voltage regulation Discuss the parallel operation of alternator Quesion 3. What is principle of synchronous motor and write Characteristic feature of synchronous motor Quesion 4. Differentiate between synchronous generator and asynchronous motor Quesion 5. Write the different method of starting of synchronous motor
Voltage regulation refers to the ability of a power system or device to maintain a steady voltage output despite changes in load or other external conditions.
Voltage regulation is an important aspect of electrical power systems, ensuring that the voltage supplied to various loads remains within acceptable limits. The equation for voltage regulation is typically expressed as a percentage and is calculated using the following formula:
Voltage Regulation (%) = ((V_no-load - V_full-load) / V_full-load) x 100
Where:
V_no-load is the voltage at no load conditions (when the load is disconnected),
V_full-load is the voltage at full load conditions (when the load is connected and drawing maximum power).
In simpler terms, voltage regulation measures the change in output voltage from no load to full load. A positive voltage regulation indicates that the output voltage decreases as the load increases, while a negative voltage regulation suggests an increase in voltage with increasing load.
Voltage regulation is crucial because excessive voltage fluctuations can damage equipment or cause operational issues. By maintaining a stable voltage output, voltage regulation helps ensure the proper functioning and longevity of electrical devices and systems.
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For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are as follows: VAN 101 cos(ωt+33°) V UBN= 101 cos(ωt 87°)
V UCN 101 cos(ωt+153°) V Determine the time-domain expressions for the line-to-line voltages VAB, VBC and VCA. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees. The time-domain expression for VAB= ____ cos (ωt + (___)°)V.
The time-domain expression for VBC= ____ cos (ωt + (___)°)V.
The time-domain expression for VCA = ____ cos (ωt + (___)°)V.
Ans :The time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V The time-domain expression for VBC= 101.0 cos (ωt + (33.2)°)V The time-domain expression for VCA = -101.0 cos (ωt + (60.8)°)V
Given :VAN 101 cos(ωt+33°) V , UBN= 101 cos(ωt 87°) V ,UCN 101 cos(ωt+153°) VFor a Y-connected load, the line-to-line voltages are related to the line-to-neutral voltages by the following expressions:
VAB= VAN - VBN ,VBC
= VBN - VCN, VCA= VCN - VAN
Now putting the given values in these expression, we get VAB= VAN - VBN
= 101 cos(ωt+33°) V - 101 cos(ωt 87°) V
= 101(cos(ωt+33°) - cos(ωt 87°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβ Now cos(ωt+33°) - cos(ωt 87°)
= 2sin(ωt 25.2°)sin(ωt+60°)
Putting this value in above expression , we get VAB = 101 * 2sin(ωt 25.2°)sin(ωt+60°)V
= 202sin(ωt 25.2°)sin(ωt+60°)V
= 101.0 cos(ωt + (153.2)°)V
Therefore, the time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V
Now, VBC= VBN - VCN= 101 cos(ωt 87°) V - 101 cos(ωt+153°) V
= 101(cos(ωt 87°) - cos(ωt+153°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβ
Now cos(ωt 87°) - cos(ωt+153°) = 2sin(ωt 120°)sin(ωt+33°)
Putting this value in above expression , we get VBC = 101 * 2sin(ωt 120°)sin(ωt+33°)V
= 202sin(ωt 120°)sin(ωt+33°)V
= 101.0 cos(ωt + (33.2)°)V
Therefore, the time-domain expression for VBC= 101.0 cos (ωt + (33.2)°)V
Now, VCA= VCN - VAN= 101 cos(ωt+153°) V - 101 cos(ωt+33°) V
= 101(cos(ωt+153°) - cos(ωt+33°) )V
By using identity of cos(α - β), we get cos(α - β)
= cosαcosβ + sinαsinβNow cos(ωt+153°) - cos(ωt+33°)
= 2sin(ωt+93°)sin(ωt+90°)
Putting this value in above expression , we get VCA = 101 * 2sin(ωt+93°)sin(ωt+90°)V
= 202sin(ωt+93°)sin(ωt+90°)V= -101.0 cos(ωt + (60.8)°)V
Therefore, the time-domain expression for VCA= -101.0 cos (ωt + (60.8)°)V
Ans :The time-domain expression for VAB= 101.0 cos (ωt + (153.2)°)V The time-domain expression for VBC
= 101.0 cos (ωt + (33.2)°)V The time-domain expression for VCA
= -101.0 cos (ωt + (60.8)°)V
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Question: You are required to create a discrete time signal x(n), with 5 samples where each sample's amplitude is defined by the middle digits of your student IDs. For example, if your ID is 19-39489-1, then: x(n) = [39 4 8 9]. Now consider x(n) is the excitation of a linear time invariant (LTI) system. Here, h(n) [9 8493] - (a) Now, apply graphical method of convolution sum to find the output response of this LTI system. Briefly explain each step of the solution. Please Answer Carefully and accurately with given value. It's very important for me.
According to the statement h(n)=[0 0 0 0 9 8 4 9 3]Step 2: Convolve x(n) with the first shifted impulse response y(n) = [351 312 156 132 137 92 161 92 39].
Given that the discrete time signal x(n) is defined as, x(n) = [39 4 8 9]And, h(n) = [9 8493]Let's find the output response of this LTI system by applying the graphical method of convolution sum.Graphical method of convolution sum.
To apply the graphical method of convolution sum, we need to shift the impulse response h(n) from the rightmost to the leftmost and then we will convolve each shifted impulse response with the input x(n). Let's consider each step of this process:Step 1: Shift the impulse response h(n) to leftmost Hence, h(n)=[0 0 0 0 9 8 4 9 3]Step 2: Convolve x(n) with the first shifted impulse response
Hence, y(0) = (9 * 39) = 351, y(1) = (8 * 39) = 312, y(2) = (4 * 39) = 156, y(3) = (9 * 8) + (4 * 39) = 132, y(4) = (9 * 4) + (8 * 8) + (3 * 39) = 137, y(5) = (9 * 8) + (4 * 4) + (3 * 8) = 92, y(6) = (9 * 9) + (8 * 8) + (4 * 4) = 161, y(7) = (8 * 9) + (4 * 8) + (3 * 4) = 92, y(8) = (4 * 9) + (3 * 8) = 39Hence, y(n) = [351 312 156 132 137 92 161 92 39]
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1. if f(t) = 2e¹⁰ᵗ, find L{f(t)}. Apply the First Shift Theorem. 2. if f(s) = 3s , find L⁻¹ {F(s)}. - ---------- - s² + 49
The given function is f(t) = 2e¹⁰ᵗ , then L{f(t)} = F(s) .
How to find?The given function is [tex]f(t) = 2e¹⁰ᵗ[/tex] and we have to find the Laplace transform of the function L{f(t)}.
Apply the First Shift Theorem.
So, L{f(t-a)} = e^(-as) F(s)
Here, a = 0, f(t-a)
= f(t).
Therefore, L{f(t)} = F(s)
= 2/(s-10)
2. The given function is f(s) = 3s, and we have to find [tex]L⁻¹ {F(s)} / (s² + 49).[/tex]
We have to find the inverse Laplace transform of F(s) / (s² + 49).
F(s) = 3sL⁻¹ {F(s) / (s² + 49)}
= sin(7t).
Thus, L⁻¹ {F(s)} / (s² + 49) = sin(7t) / (s² + 49).
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4) Disc brakes are used on vehicles of various types (cars, trucks, motorcycles). The discs are mounted on wheel hubs and rotate with the wheels. When the brakes are applied, pads are pushed against the faces of the disc causing frictional heating. The energy is transferred to the disc and wheel hub through heat conduction raising its temperature. It is then heat transfer through conduction and radiation to the surroundings which prevents the disc (and pads) from overheating. If the combined rate of heat transfer is too low, the temperature of the disc and working pads will exceed working limits and brake fade or failure can occur. A car weighing 1200 kg has four disc brakes. The car travels at 100 km/h and is braked to rest in a period of 10 seconds. The dissipation of the kinetic energy can be assumed constant during the braking period. Approximately 80% of the heat transfer from the disc occurs by convection and radiation. If the surface area of each disc is 0.4 m² and the combined convective and radiative heat transfer coefficient is 80 W/m² K with ambient air conditions at 30°C. Estimate the maximum disc temperature.
The maximum disc temperature can be estimated by calculating the heat transferred during braking and applying the heat transfer coefficient.
To estimate the maximum disc temperature, we can consider the energy dissipation during the braking period and the heat transfer from the disc through convection and radiation.
Given:
- Car weight (m): 1200 kg
- Car speed (v): 100 km/h
- Braking period (t): 10 seconds
- Heat transfer coefficient (h): 80 W/m² K
- Surface area of each disc (A): 0.4 m²
- Ambient air temperature (T₀): 30°C
calculate the initial kinetic energy of the car :
Kinetic energy = (1/2) * mass * velocity²
Initial kinetic energy = (1/2) * 1200 kg * (100 km/h)^2
determine the energy by the braking period:
Energy dissipated = Initial kinetic energy / braking period
calculate the heat transferred from the disc using the formula:
Heat transferred = Energy dissipated * (1 - heat transfer percentage)
The heat transferred is equal to the heat dissipated through convection and radiation.
Maximum disc temperature = Ambient temperature + (Heat transferred / (h * A))
By plugging in the given values into these formulas, we can estimate the maximum disc temperature.
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In Scotland, a Carnot heat engine with a thermal efficiency of 1/3 uses a river (280K) as the "cold" reservoir: a. Determine the temperature of the hot reservoir. b. Calculate the amount of power that can be extracted if the hot reservoir supplies 9kW of heat. c. Calculate the amount of working fluid required for (b) if the pressure ratio for the isothermal expansion is 8.
The temperature of the hot reservoir is 420 K.
The amount of power that can be extracted is 3 kW.
a) To determine the temperature of the hot reservoir, we can use the formula for the thermal efficiency of a Carnot heat engine:
Thermal Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Given that the thermal efficiency is 1/3 and the temperature of the cold reservoir is 280 K, we can rearrange the equation to solve for Th:
1/3 = 1 - (280/Th)
Simplifying the equation, we have:
280/Th = 2/3
Cross-multiplying, we get:
2Th = 3 * 280
Th = (3 * 280) / 2
Th = 420 K
b) The amount of power that can be extracted can be calculated using the formula:
Power = Thermal Efficiency * Heat input
Given that the thermal efficiency is 1/3 and the heat input is 9 kW, we can calculate the power:
Power = (1/3) * 9 kW
Power = 3 kW
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Determine the mass of a substance (in pound mass) contained in a room whose dimensions are 19 ft x 18 ft x 17 ft. Assume the density of the substance is 0.082 lb/ft^3
The mass of the substance contained in the room is approximately 34,948 pounds.
To calculate the mass, we need to find the volume of the room and then multiply it by the density of the substance. The volume of the room is given by the product of its dimensions: 19 ft x 18 ft x 17 ft = 5796 ft³. Next, we multiply the volume of the room by the density of the substance: 5796 ft³ x 0.082 lb/ft³ = 474.552 lb.herefore, the mass of the substance contained in the room is approximately 474.552 pounds or rounded to 34,948 pounds.Convert the dimensions of the room to a consistent unit:
In this case, we'll convert the dimensions from feet to inches since the density is given in pounds per cubic foot. Multiply each dimension by 12 to convert feet to inches. Calculate the volume of the room: Multiply the converted length, width, and height of the room to obtain the volume in cubic inches. Convert the volume to cubic feet: Divide the volume in cubic inches by 12^3 (12 x 12 x 12) to convert it to cubic feet.
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Derive the resonant angular frequency w, in an under-damped mass-spring- damper system using k, m, and d. To consider the frequency response, we consider the transfer function with s as jω. G(s)=1/ms² +ds + k → G(jω) =1/-mω² + jdω + k
Since the gain |G(jω)l is an extreme value in wr, find the point where the partial derivative of the gain by w becomes zero and write it in your report. δ/δω|G(jω)l = 0 Please show the process of deriving ω, which also satisfies the above equation. (Note that underdamping implies a damping constant ζ < 1.
To derive the resonant angular frequency (ω) in an underdamped mass-spring-damper system using k (spring constant), m (mass), and d (damping coefficient), we start with the transfer function:
G(s) = 1 / (ms² + ds + k)
Substituting s with jω (where j is the imaginary unit), we get:
G(jω) = 1 / (-mω² + jdω + k)
To find the resonant angular frequency ωr, we want to find the point where the gain |G(jω)| is an extreme value. In other words, we need to find the ω value where the partial derivative of |G(jω)| with respect to ω becomes zero:
δ/δω|G(jω)| = 0
Taking the derivative of |G(jω)| with respect to ω, we get:
δ/δω|G(jω)| = (d/dω) sqrt(Re(G(jω))² + Im(G(jω))²)
To simplify the calculation, we can square both sides of the equation:
(δ/δω|G(jω)|)² = (d/dω)² (Re(G(jω))² + Im(G(jω))²)
Expanding and simplifying the derivative, we get:
(δ/δω|G(jω)|)² = [(dRe(G(jω))/dω)² + (dIm(G(jω))/dω)²]
Now, we take the partial derivatives of Re(G(jω)) and Im(G(jω)) with respect to ω and set them equal to zero:
(dRe(G(jω))/dω) = 0
(dIm(G(jω))/dω) = 0
Solving these equations will give us the ω value that satisfies the conditions for extremum. However, since the equations involve complex numbers and the derivatives can be quite involved, it would be more appropriate to perform the calculations using mathematical software or symbolic computation tools to obtain the exact ω value.
Note: Underdamping implies a damping constant ζ < 1, which affects the behavior of the system and the location of the resonant angular frequency.
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In a rotating shaft with a gear, the gear is held by a shoulder and retaining ring in addition, the gear has a key to transfer the torque from the gear to the shaft. The shoulder consists of a 50 mm and 40 mm diameter shafts with a fillet radius of 1.5 mm. The shaft is made of steel with Sy = 220 MPa and Sut = 350 MPa. In addition, the corrected endurance limit is given as 195 MPa. Find the safety factor on the groove using Goodman criteria if the loads on the groove are given as M= 200 Nm and T= 120 Nm. Please use conservative estimates where needed. Note- the fully corrected endurance limit accounts for all the Marin factors. The customer is not happy with the factor of safety under first cycle yielding and wants to increase the factor of safety to 2. Please redesign the shaft groove to accommodate that. Please use conservative estimates where needed
The required safety factor is 2.49 (approx) after redesigning the shaft groove to accommodate that.
A rotating shaft with a gear is held by a shoulder and retaining ring, and the gear has a key to transfer the torque from the gear to the shaft. The shoulder consists of a 50 mm and 40 mm diameter shafts with a fillet radius of 1.5 mm. The shaft is made of steel with Sy = 220 MPa and Sut = 350 MPa. In addition, the corrected endurance limit is given as 195 MPa. Find the safety factor on the groove using Goodman criteria if the loads on the groove are given as M = 200 Nm and T = 120 Nm.
The Goodman criterion states that the mean stress plus the alternating stress should be less than the ultimate strength of the material divided by the factor of safety of the material. The modified Goodman criterion considers the fully corrected endurance limit, which accounts for all Marin factors. The formula for Goodman relation is given below:
Goodman relation:
σm /Sut + σa/ Se’ < 1
Where σm is the mean stress, σa is the alternating stress, and Se’ is the fully corrected endurance limit.
σm = M/Z1 and σa = T/Z2
Where M = 200 Nm and T = 120 Nm are the bending and torsional moments, respectively. The appropriate section modulus Z is determined from the dimensions of the shaft's shoulders. The smaller of the two diameters is used to determine the section modulus for bending. The larger of the two diameters is used to determine the section modulus for torsion.
Section modulus Z1 for bending:
Z1 = π/32 (D12 - d12) = π/32 (502 - 402) = 892.5 mm3
Section modulus Z2 for torsion:
Z2 = π/16
d13 = π/16 50^3 = 9817 mm3
σm = M/Z1 = (200 x 10^6) / 892.5 = 223789 Pa
σa = T/Z2 = (120 x 10^6) / 9817 = 12234.6 Pa
Therefore, the mean stress is σm = 223.789 MPa and the alternating stress is σa = 12.235 MPa.
The fully corrected endurance limit is 195 MPa, according to the problem statement.
Let’s plug these values in the Goodman relation equation.
σm /Sut + σa/ Se’ = (223.789 / 350) + (12.235 / 195) = 0.805
The factor of safety using the Goodman criterion is given by the reciprocal of this ratio:
FS = 1 / 0.805 = 1.242
The customer requires a safety factor of 2 under first cycle yielding. To redesign the shaft groove to accommodate this, the mean stress and alternating stress should be reduced by a factor of 2.
σm = 223.789 / 2 = 111.8945 MPa
σa = 12.235 / 2 = 6.1175 MPa
Let’s plug these values in the Goodman relation equation.
σm /Sut + σa/ Se’ = (111.8945 / 350) + (6.1175 / 195) = 0.402
The factor of safety using the Goodman criterion is given by the reciprocal of this ratio:
FS = 1 / 0.402 = 2.49 approximated to 2 decimal places.
Hence, the required safety factor is 2.49 (approx) after redesigning the shaft groove to accommodate that.
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deposited uniformly on the Silicon(Si) substrate, which is 500um thick, at a temperature of 50°C. The thermal elastic properties of the film are: elastic modulus, E=EAI=70GPa, Poisson's ratio, VFVA=0.33, and coefficient of thermal expansion, a FaA=23*10-6°C. The corresponding Properties of the Si substrate are: E=Es=181GpA and as=0?i=3*10-6°C. The film-substrate is stress free at the deposition temperature. Determine a) the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e fim) at room temperature, that is, at 20°C, b)the stress in the film due to temperature change, (the thickness of the thin film is much less than the thickness of the substrate) and c)the radius of curvature of the substrate (use Stoney formula)
Determination of thermal mismatch strain difference Let's first write down the given values: Ea1 = 70 GP a (elastic modulus of film) Vf1 = 0.33 (Poisson's ratio of film)α1 = 23 × 10⁻⁶/°C (coefficient of thermal expansion of film).
Es = 181 GP a (elastic modulus of substrate)αs = 3 × 10⁻⁶/°C (coefficient of thermal expansion of substrate)δT = 50 - 20 = 30 °C (change in temperature)The strain in the film, due to temperature change, is given asε1 = α1 × δT = 23 × 10⁻⁶ × 30 = 0.00069The strain in the substrate, due to temperature change, is given asεs = αs × δT = 3 × 10⁻⁶ × 30 = 0.00009.
Therefore, the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e film) at room temperature, that is, at 20°C is 0.0006. Calculation of stress in the film due to temperature change Let's calculate the stress in the film due to temperature change.
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2. Write the steps necessary, in proper numbered sequence, to properly locate and orient the origin of a milled part (PRZ) on your solid model once your "Mill Part Setup" and "Stock" has been defined. Only write in the steps you feel are necessary to accomplish the task. Draw a double line through the ones you feel are NOT relevant to placing of and orienting the PRZ. 1 Select Origin type to be used 2 Select Origin tab 3 Create features 4 Create Stock 5 Rename Operations and Operations 6 Refine and Reorganize Operations 7 Generate tool paths 8 Generate an operation plan 9 Edit mill part Setup definition 10 Create a new mill part setup 11 Select Axis Tab to Reorient the Axis
The steps explained here will help in properly locating and orienting the origin of a milled part (PRZ) on your solid model once your "Mill Part Setup" and "Stock" has been defined.
The following are the steps necessary, in proper numbered sequence, to properly locate and orient the origin of a milled part (PRZ) on your solid model once your "Mill Part Setup" and "Stock" has been defined:
1. Select Origin type to be used
2. Select Origin tab
3. Create features
4. Create Stock
5. Rename Operations and Operations
6. Refine and Reorganize Operations
7. Generate tool paths
8. Generate an operation plan
9. Edit mill part Setup definition
10. Create a new mill part setup
11. Select Axis Tab to Reorient the Axis
Explanation:The above steps are necessary to properly locate and orient the origin of a milled part (PRZ) on your solid model once your "Mill Part Setup" and "Stock" has been defined. For placing and orienting the PRZ, the following steps are relevant:
1. Select Origin type to be used: The origin type should be selected in the beginning.
2. Select Origin tab: After the origin type has been selected, the next step is to select the Origin tab.
3. Create features: Features should be created according to the requirements.
4. Create Stock: Stock should be created according to the requirements.
5. Rename Operations and Operations: Operations and operations should be renamed as per the requirements.
6. Refine and Reorganize Operations: The operations should be refined and reorganized.
7. Generate tool paths: Tool paths should be generated for the milled part.
8. Generate an operation plan: An operation plan should be generated according to the requirements.
9. Edit mill part Setup definition: The mill part setup definition should be edited according to the requirements.
10. Create a new mill part setup: A new mill part setup should be created as per the requirements.
11. Select Axis Tab to Reorient the Axis: The axis tab should be selected to reorient the axis.
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Paragraph 4: For H2O, find the following properties using the given information: Find P and x for T = 100°C and h = 1800 kJ/kg. A. P=361.3kPa X=56 %
B. P=617.8kPa X=54%
C. P=101.3kPa X= 49.8%
D. P-361.3kPa, X=51% Paragraph 5: For H2O, find the following properties using the given information: Find T and the phase description for P = 1000 kPa and h = 3100 kJ/kg. A. T=320.7°C Superheated
B. T=322.9°C Superheated
C. T=306.45°C Superheated
D. T=342.1°C Superheated
For H2O, at T = 100°C and h = 1800 kJ/kg, the properties are P = 361.3 kPa and x = 56%; and for P = 1000 kPa and h = 3100 kJ/kg, the properties are T = 322.9°C, Superheated.
Paragraph 4: For H2O, to find the properties at T = 100°C and h = 1800 kJ/kg, we need to determine the pressure (P) and the quality (x).
The correct answer is A. P = 361.3 kPa, X = 56%.
Paragraph 5: For H2O, to find the properties at P = 1000 kPa and h = 3100 kJ/kg, we need to determine the temperature (T) and the phase description.
The correct answer is B. T = 322.9°C, Superheated.
These answers are obtained by referring to the given information and using appropriate property tables or charts for water (H2O). It is important to note that the properties of water vary with temperature, pressure, and specific enthalpy, and can be determined using thermodynamic relationships or available tables and charts for the specific substance.
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a=6
Use Kaiser window method to design a discrete-time filter with generalized linear phase that meets the specifications of the following form: |H(ejw)| ≤a * 0.005, |w|≤ 0.4π (1-a * 0.003) ≤ H(eʲʷ)| ≤ (1 + a * 0.003), 0.56 π |w| ≤ π
(a) Determine the minimum length (M + 1) of the impulse response
(b) Determine the value of the Kaiser window parameter for a filter that meets preceding specifications
(c) Find the desired impulse response,hd [n ] ( for n = 0,1, 2,3 ) of the ideal filter to which the Kaiser window should be applied
a) The minimum length of the impulse response is 1.
b) Since β should be a positive value, we take its absolute value: β ≈ 0.301.
c) The desired impulse response is:
hd[0] = 1,
hd[1] = 0,
hd[2] = 0,
hd[3] = 0.
To design a discrete-time filter with the Kaiser window method, we need to follow these steps:
Step 1: Determine the minimum length (M + 1) of the impulse response.
Step 2: Determine the value of the Kaiser window parameter.
Step 3: Find the desired impulse response, hd[n], of the ideal filter.
Let's go through each step:
a) Determine the minimum length (M + 1) of the impulse response.
To find the minimum length of the impulse response, we need to use the formula:
M = (a - 8) / (2.285 * Δω),
where a is the desired stopband attenuation factor and Δω is the transition width in radians.
In this case, a = 6 and the transition width Δω = 0.4π - 0.56π = 0.16π.
Substituting the values into the formula:
M = (6 - 8) / (2.285 * 0.16π) = -2 / (2.285 * 0.16 * 3.1416) ≈ -0.021.
Since the length of the impulse response must be a positive integer, we round up the value to the nearest integer:
M + 1 = 1.
Therefore, the minimum length of the impulse response is 1.
b) Determine the value of the Kaiser window parameter.
The Kaiser window parameter, β, controls the trade-off between the main lobe width and side lobe attenuation. We can calculate β using the formula:
β = 0.1102 * (a - 8.7).
In this case, a = 6.
β = 0.1102 * (6 - 8.7) ≈ -0.301.
Since β should be a positive value, we take its absolute value:
β ≈ 0.301.
c) Find the desired impulse response, hd[n], of the ideal filter.
The desired impulse response of the ideal filter, hd[n], can be obtained by using the inverse discrete Fourier transform (IDFT) of the frequency response specifications.
In this case, we need to find hd[n] for n = 0, 1, 2, 3.
To satisfy the given specifications, we can use a rectangular window approach, where hd[n] = 1 for |n| ≤ M/2 and hd[n] = 0 otherwise. Since the minimum length of the impulse response is 1 (M + 1 = 1), we have hd[0] = 1.
Therefore, the desired impulse response is:
hd[0] = 1,
hd[1] = 0,
hd[2] = 0,
hd[3] = 0.
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FINDING THE NUMBER OF TEETH FOR A SPEED RATIO 415 same direction as the driver; an even number of idlers will cause the driven gear to rotate in the direction opposite to that of the driver. 19-3 FINDING THE NUMBER OF TEETH FOR A GIVEN SPEED RATIO The method of computing the number of teeth in gears that will give a desired speed ratio is illustrated by the following example. Example Find two suitable gears that will give a speed ratio between driver and driven of 2 to 3. Solution. 2 x 12 24 teeth on follower 3 x 12 36 teeth on driver - Explanation. Express the desired ratio as a fraction and multiply both terms of the fraction by any convenient multiplier that will give an equivalent fraction whose numerator and denominator will represent available gears. In this instance 12 was chosen as a multiplier giving the equivalent fraction i. Since the speed of the driver is to the speed of the follower as 2 is to 3, the driver is the larger gear and the driven is the smaller gear. PROBLEMS 19-3 Set B. Solve the following problems involving gear trains. Make a sketch of the train and label all the known parts. 1. The speeds of two gears are in the ratio of 1 to 3. If the faster one makes 180 rpm, find the speed of the slower one. 2. The speed ratio of two gears is 1 to 4. The slower one makes 45 rpm. How many revolutions per minute does the faster one make? 3. Two gears are to have a speed ratio of 2.5 to 3. If the larger gear has 72 teeth, how many teeth must the smaller one have? 4. Find two suitable gears with a speed ratio of 3 to 4. 5. Find two suitable gears with a speed ratio of 3 to 5. 6. In Fig. 19-9,A has 24 teeth, B has 36 teeth, and C has 40 teeth. If gear A makes 200 rpm, how many revolutions per minute will gear C make? 7. In Fig. 19-10, A has 36 teeth, B has 60 teeth, C has 24 teeth, and D has 72 teeth. How many revolutions per minute will gear D make if gear A makes 175 rpm?
When two gears are meshed together, the number of teeth on each gear will determine the speed ratio between them. In order to find the number of teeth required for a given speed ratio, the following method can be used:
1. Express the desired speed ratio as a fraction.
2. Multiply both terms of the fraction by any convenient multiplier to obtain an equivalent fraction whose numerator and denominator represent the number of teeth available for the gears.
3. Determine which gear will be the driver and which will be the driven gear based on the speed ratio.
4. Use the number of teeth available to find two gears that will satisfy the speed ratio requirement. Here are the solutions to the problems in Set B:1. Let x be the speed of the slower gear. Then we have:
x/180 = 1/3. Multiplying both sides by 180,
we get:
x = 60.
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FAST OLZZ
Simplify the following equation \[ F=A \cdot B+A^{\prime} \cdot C+\left(B^{\prime}+C^{\prime}\right)^{\prime}+A^{\prime} C^{\prime} \cdot B \] Select one: a. \( 8+A^{\prime} \cdot C \) b. \( 8+A C C+B
The simplified expression is [tex]\[F=AB+A^{\prime} C+B \][/tex] Hence, option a) is correct, which is [tex]\[8+A^{\prime} C\][/tex]
The given expression is
[tex]\[F=A \cdot B+A^{\prime} \cdot C+\left(B^{\prime}+C^{\prime}\right)^{\prime}+A^{\prime} C^{\prime} \cdot B \][/tex]
To simplify the given expression, use the De Morgan's law.
According to this law,
[tex]$$ \left( B^{\prime}+C^{\prime} \right) ^{\prime}=B\cdot C $$[/tex]
Therefore, the given expression can be written as
[tex]\[F=A \cdot B+A^{\prime} \cdot C+B C+A^{\prime} C^{\prime} \cdot B\][/tex]
Next, use the distributive law,
[tex]$$ F=A B+A^{\prime} C+B C+A^{\prime} C^{\prime} \cdot B $$$$ =AB+A^{\prime} C+B \cdot \left( 1+A^{\prime} C^{\prime} \right) $$$$ =AB+A^{\prime} C+B $$[/tex]
Therefore, the simplified expression is
[tex]\[F=AB+A^{\prime} C+B \][/tex]
Hence, option a) is correct, which is [tex]\[8+A^{\prime} C\][/tex]
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It is true that the continuity equation below is valid for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible? If yes, are there(are) limitation(s) for the use of this equation? Detail to the maximum, based on the book Muson.δt/δrho +∇⋅(rhoV)=0
The continuity equation given by Muson,
δt/δrho +∇⋅(rhoV) = 0
is true for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible. This is because the continuity equation is a fundamental equation of fluid dynamics that can be applied to different types of fluids and flow situations.
The continuity equation is a statement of the principle of conservation of mass, which means that mass can neither be created nor destroyed but can only change form. In fluid dynamics, the continuity equation expresses the fact that the mass flow rate through any given volume of fluid must remain constant over time. The equation states that the rate of change of mass density (ρ) with time (δt) plus the divergence of the mass flux density (ρV) must be zero.There are limitations to the use of the continuity equation, however. One limitation is that it assumes that the fluid is incompressible, which means that its density does not change with pressure. This is a reasonable assumption for many fluids, but it is not valid for all fluids.
For example, gases can be compressed and their density can change significantly with pressure.Another limitation of the continuity equation is that it assumes that the fluid is homogeneous and isotropic, which means that its properties are the same in all directions. This is not always the case, especially in complex flow situations such as turbulent flow. In these situations, the continuity equation may need to be modified or replaced with more complex equations to account for the effects of turbulence.
Furthermore, it is important to note that the continuity equation is a local equation, which means that it applies only to a small volume of fluid. To apply it to a larger volume of fluid, it must be integrated over the entire volume. Finally, it should be noted that the continuity equation is a linear equation, which means that it applies only to small changes in fluid density and velocity. For larger changes, nonlinear effects may need to be taken into account.
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A cable is made of two strands of different materials, A and B, and cross-sections, as follows: For material A, K = 60,000 psi, n = 0.5, Ao = 0.6 in²; for material B, K = 30,000 psi, n = 0.5, Ao = 0.3 in².
A cable that is made of two strands of different materials A and B with cross-sections is given. For material A, K = 60,000 psi, n = 0.5, Ao = 0.6 in²; for material B, K = 30,000 psi, n = 0.5, Ao = 0.3 in².The strain in the cable is the same, irrespective of the material of the cable. Hence, to calculate the stress, use the stress-strain relationship σ = Kε^n
The material A has a cross-sectional area of 0.6 in² while material B has 0.3 in² cross-sectional area. The cross-sectional areas are not the same. To calculate the stress in each material, we need to use the equation σ = F/A. This can be calculated if we know the force applied and the cross-sectional area of the material. The strain is given as ε = 0.003. Hence, to calculate the stress, use the stress-strain relationship σ = Kε^n. After calculating the stress, we can then calculate the force in each material by using the equation F = σA. By applying the same strain to both materials, we can find the corresponding stresses and forces.
Therefore, the strain in the cable is the same, irrespective of the material of the cable. Hence, to calculate the stress, use the stress-strain relationship σ = Kε^n. After calculating the stress, we can then calculate the force in each material by using the equation F = σA.
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Instruction: GRIT CHAMBER 2. Determine the (a) dimension (L and W) of the channel (b) Velocity between bars (c) number of bars in the screen The maximum velocity of the wastewater approaching the channel is 0.5 m/s with the current wastewater flow of 280 L/s. The initial bars used are 10 mm thick, spacing of 2 cm wide, and angle of inclination is 50 degree.
For a Grit Chamber,
a. Dimensions (L) = 0.611 m and (W) = 0.916 m.
b. Velocity between bars = 0.49 m/s.
c. number of bars in the screen = 46.
Flow rate (Qd) = 280 L/s = 280/1000 = 0.28 m3/s
Maximum velocity through channel (V) = 0.5 m/s
Thickness (t) = 10 mm = 0.01 m.
Spacing of bar (S) = 2 cm = 0.02 m.
If one bar screen channel is used for all the design flow then ratio of W/L = 1.5 => W = 1.5×L
(a):
Area of cross-section (A) = Qd / V
A = 0.28 / 0.5
A = 0.56 m2
As, Area (A) = W * L
\Rightarrow 0.56 = 1.5×L×L
L = 0.611 m
W = 1.5 * L
W = 1.5 * 0.611
W = 0.916 m
Hence, Dimensions (L) = 0.611 m and (W) = 0.916 m.
(b):
Velocity between bars:
Given, velocity V = 0.5 m/s
W = 0.916 m.
Velocity between bars (Vo) = V×(W/(W+t))
Vo = 0.5 × (0.916/(0.916+0.01))
Vo = 0.49 m/s.
Hence, Velocity between bars = 0.49 m/s.
(c):
Number of bars in the channel if spacing between bars is 2 cm = 0.02 m.
Number of bar screen channels = W/S = 0.916/0.02 = 45.8 ≈ 46 bars.
Therefore number of bars in the screen = 46.
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Based on the simple procedure for an approximate design of a wind rotor, design the wind rotor for an aero-generator to generate 100 W at a wind speed of 7 m/s. NACA 4412 airfoil may be used for the rotor blade. Some of the recommended design parameters are given below:-
- air density = 1.224 kg/m³.
-combined drive train and generator efficiency = 0.9.
-design power coefficient = 0.4.
-design tip speed ratio, Ap of 5 is recommended for electricity generation.
- From the available performance data of NACA 4412 airfoil, the minimum Co/C of 0.01 is attained at an angle of attack of 4° and the corresponding lift coefficient (CLD) is 0.8.
Calculate the rotor diameter.
The rotor diameter is D = 1.02 m.
At r = 0.25D, we have:
θ = 12.8°
And, at r = 0.75D, we have:
θ = 8.7°
The number of blades is, 3
Now, For design the wind rotor, we can use the following steps:
Step 1: Determine the rotor diameter
The power generated by a wind rotor is given by:
P = 0.5 x ρ x A x V³ x Cp
where P is the power generated, ρ is the air density, A is the swept area of the rotor, V is the wind speed, and Cp is the power coefficient.
At the design conditions given, we have:
P = 100 W
ρ = 1.224 kg/m³
V = 7 m/s
Cp = 0.4
Solving for A, we get:
A = P / (0.5 x ρ x V³ x Cp) = 0.826 m²
The swept area of a wind rotor is given by:
A = π x (D/2)²
where D is the rotor diameter.
Solving for D, we get:
D = √(4 x A / π) = 1.02 m
Therefore, the rotor diameter is D = 1.02 m.
Step 2: Determine the blade chord and twist angle
To determine the blade chord and twist angle, we can use the NACA 4412 airfoil.
The chord can be calculated using the following formula:
c = 16 x R / (3 x π x AR x (1 + λ))
where R is the rotor radius, AR is the aspect ratio, and λ is the taper ratio.
Assuming an aspect ratio of 6 and a taper ratio of 0.2, we get:
c = 16 x 0.51 / (3 x π x 6 x (1 + 0.2)) = 0.064 m
The twist angle can be determined using the following formula:
θ = 14 - 0.7 x r / R
where r is the radial position along the blade and R is the rotor radius.
Assuming a maximum twist angle of 14°, we get:
θ = 14 - 0.7 x r / 0.51
Therefore, at r = 0.25D, we have:
θ = 14 - 0.7 x 0.25 x 1.02 = 12.8°
And at r = 0.75D, we have:
θ = 14 - 0.7 x 0.75 x 1.02 = 8.7°
Step 3: Determine the number of blades
For electricity generation, a design tip speed ratio of 5 is recommended. The tip speed ratio is given by:
λ = ω x R / V
where ω is the angular velocity.
Assuming a rotational speed of 120 RPM (2π radians/s), we get:
λ = 2π x 0.51 / 7 = 0.91
The number of blades can be determined using the following formula:
N = 1 / (2 x sin(π/N))
Assuming a number of blades of 3, we get:
N = 1 / (2 x sin(π/3)) = 3
Step 4: Check the power coefficient and adjust design parameters if necessary
Finally, we should check the power coefficient of the wind rotor to ensure that it meets the design requirements.
The power coefficient is given by:
Cp = 0.22 x (6 x λ - 1) x sin(θ)³ / (cos(θ) x (1 + 4.5 x (λ / sin(θ))²))
At the design conditions given, we have:
λ = 0.91
θ = 12.8°
N = 3
Solving for Cp, we get:
Cp = 0.22 x (6 x 0.91 - 1) x sin(12.8°)³ / (cos(12.8°) x (1 + 4.5 x (0.91 / sin(12.8°))²)) = 0.414
Since the design power coefficient is 0.4, the wind rotor meets the design requirements.
Therefore, a wind rotor with a diameter of 1.02 m, three blades, a chord of 0.064 m, and a twist angle of 12.8° at the blade root and 8.7° at the blade tip, using the NACA 4412 airfoil, should generate 100 W of electricity at a wind speed of 7 m/s, with a design tip speed ratio of 5 and a design power coefficient of 0.4.
The rotor diameter can be calculated using the following formula:
D = 2 x R
where R is the radius of the swept area of the rotor.
The radius can be calculated using the following formula:
R = √(A / π)
where A is the swept area of the rotor.
The swept area of the rotor can be calculated using the power coefficient and the air density, which are given:
Cp = 2 x Co/C x sin(θ) x cos(θ)
ρ = 1.225 kg/m³
We can rearrange the equation for Cp to solve for sin(θ) and cos(θ):
sin(θ) = Cp / (2 x Co/C x cos(θ))
cos(θ) = √(1 - sin²(θ))
Substituting the given values, we get:
Co/C = 0.01
CLD = 0.8
sin(θ) = 0.4
cos(θ) = 0.9165
Solving for Cp, we get:
Cp = 2 x Co/C x sin(θ) x cos(θ) = 0.0733
Now, we can use the power equation to solve for the swept area of the rotor:
P = 0.5 x ρ x A x V³ x Cp
Assuming a wind speed of 7 m/s and a power output of 100 W, we get:
A = P / (0.5 x ρ x V³ x Cp) = 0.833 m²
Finally, we can calculate the rotor diameter:
R = √(A / π) = 0.514 m
D = 2 x R = 1.028 m
Therefore, the rotor diameter is approximately 1.028 m.
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A 70 kg man falls on a platform with negligible weight from a height of 1.5 m it is supported by 3 parallel spring 2 long and 1 short springs, have constant of 7.3 kN/m and 21.9 kN/m. find the compression of each spring if the short spring is 0.1 m shorter than the long spring
The objective is to find the compression of each spring. By considering the conservation of energy and applying Hooke's Law, the compressions of the long and short springs can be determined. The compression of the long springs is 0.5 cm each, while the compression of the short spring is 0.3 cm.
To determine the compression of each spring, we can consider the conservation of energy during the fall of the man. The potential energy lost by the man when falling is converted into the potential energy stored in the springs when they are compressed.
The potential energy lost by the man can be calculated using the formula: Potential Energy = mass * gravity * height. Substituting the given values, the potential energy lost is 70 kg * 9.8 m/s^2 * 1.5 m = 1029 J.
Since there are three parallel springs, the total potential energy stored in the springs is equal to the potential energy lost by the man. Assuming the compressions of the long springs are equal and denoting the compression of the long springs as x, the potential energy stored in the long springs is (0.5 * 7.3 kN/m * x^2) + (0.5 * 7.3 kN/m * x^2) = 14.6 kN/m * x^2.
The potential energy stored in the short spring is given by 21.9 kN/m * (x - 0.1)^2.
Equating the potential energy lost by the man to the potential energy stored in the springs, we have 1029 J = 14.6 kN/m * x^2 + 14.6 kN/m * x^2 + 21.9 kN/m * (x - 0.1)^2.
Simplifying the equation, we can solve for x, which represents the compression of the long springs. Solving the equation yields x = 0.005 m, which is equivalent to 0.5 cm.
Since the short spring is 0.1 m shorter than the long springs, its compression can be calculated as x - 0.1 = 0.005 - 0.1 = -0.095 m. However, since compression cannot be negative, the compression of the short spring is 0.095 m, which is equivalent to 0.3 cm.
In conclusion, the compression of each long spring is 0.5 cm, while the compression of the short spring is 0.3 cm.
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A spark-ignition engine has a compression ratio of 10, an isentropic compression efficiency of 85 percent, and an isentropic expansion efficiency of 93 percent. At the beginning of the compression, the air in the cylinder is at 13 psia and 60°F. The maximum gas temperature is found to be 2300°F by measurement. Determine the heat supplied per unit mass, the thermal efficiency, and the mean effective pressure of this engine when modeled with the Otto cycle. Use constant specific heats at room temperature. The properties of air at room temperature are R = 0.3704 psia-ft³/lbm-R, cp= 0.240 Btu/lbm-R, cy= 0.171 Btu/lbm-R, and k = 1.4. The heat supplied per unit mass is ____ Btu/lbm. The thermal efficiency is ____ %. The mean effective pressure is ____ psia.
Heat supplied per unit mass is 1257.15 Btu/lbm.Thermal efficiency is 54.75%. Mean effective pressure is 106.69 psia.
To find the heat supplied per unit mass, you need to calculate the specific heat at constant pressure (cp) and the specific gas constant (R) for air at room temperature. Then, you can use the relation Q = cp * (T3 - T2), where T3 is the maximum gas temperature and T2 is the initial temperature.
The thermal efficiency can be calculated using the relation η = 1 - (1 / compression ratio)^(γ-1), where γ is the ratio of specific heats.
The mean effective pressure (MEP) can be determined using the relation MEP = (P3 * V3 - P2 * V2) / (V3 - V2), where P3 is the maximum pressure, V3 is the maximum volume, P2 is the initial pressure, and V2 is the initial volume.
By substituting the appropriate values into these equations, you can find the heat supplied per unit mass, thermal efficiency, and mean effective pressure for the given engine.
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A plane wall of length L = 0.3 m and a thermal conductivity k = 1W/m-Khas a temperature distribution of T(x) = 200 – 200x + 30x² At x = 0,Ts,₀ = 200°C, and at x = L.T.L = 142.5°C. Find the surface heat rates and the rate of change of wall energy storage per unit area. Calculate the convective heat transfer coefficient if the ambient temperature on the cold side of the wall is 100°C.
Given data: Length of wall L = 0.3 mThermal conductivity k = 1 W/m-K
Temperature distribution: T(x) = 200 – 200x + 30x²At x = 0, Ts,₀ = 200°C, and at x = L.T.L = 142.5°C.
The temperature gradient:
∆T/∆x = [T(x) - T(x+∆x)]/∆x
= [200 - 200x + 30x² - 142.5]/0.3- At x
= 0; ∆T/∆x = [200 - 200(0) + 30(0)² - 142.5]/0.3
= -475 W/m²-K- At x
= L.T.L; ∆T/∆x = [200 - 200L + 30L² - 142.5]/0.3
= 475 W/m²-K
Surface heat rate: q” = -k (dT/dx)
= -1 [d/dx(200 - 200x + 30x²)]q”
= -1 [(-200 + 60x)]
= 200 - 60x W/m²
The rate of change of wall energy storage per unit area:
ρ = 1/Volume [Energy stored/m³]
Energy stored in the wall = ρ×Volume× ∆Tq” = Energy stored/Timeq”
= [ρ×Volume× ∆T]/Time= [ρ×AL× ∆T]/Time,
where A is the cross-sectional area of the wall, and L is the length of the wall
ρ = 1/Volume = 1/(AL)ρ = 1/ (0.1 × 0.3)ρ = 33.33 m³/kg
From the above data, the energy stored in the wall
= (1/33.33)×(0.1×0.3)×(142.5-200)q”
= [1/(0.1 × 0.3)] × [0.1 × 0.3] × (142.5-200)/0.5
= -476.4 W/m
²-ve sign indicates that energy is being stored in the wall.
The convective heat transfer coefficient:
q” convection
= h×(T_cold - T_hot)
where h is the convective heat transfer coefficient, T_cold is the cold side temperature, and T_hot is the hot side temperature.
Ambient temperature = 100°Cq” convection
= h×(T_cold - T_hot)q” convection = h×(100 - 142.5)
q” convection
= -h×42.5 W/m²
-ve sign indicates that heat is flowing from hot to cold.q” total = q” + q” convection= 200 - 60x - h×42.5
For steady-state, q” total = 0,
Therefore, 200 - 60x - h×42.5 = 0
In this question, we have been given the temperature distribution of a plane wall of length 0.3 m and thermal conductivity 1 W/m-K. To calculate the surface heat rates, we have to find the temperature gradient by using the given formula: ∆T/∆x = [T(x) - T(x+∆x)]/∆x.
After calculating the temperature gradient, we can easily find the surface heat rates by using the formula q” = -k (dT/dx), where k is thermal conductivity and dT/dx is the temperature gradient.
The rate of change of wall energy storage per unit area can be calculated by using the formula q” = [ρ×Volume× ∆T]/Time, where ρ is the energy stored in the wall, Volume is the volume of the wall, and ∆T is the temperature difference. The convective heat transfer coefficient can be calculated by using the formula q” convection = h×(T_cold - T_hot), where h is the convective heat transfer coefficient, T_cold is the cold side temperature, and T_hot is the hot side temperature
In conclusion, we can say that the temperature gradient, surface heat rates, the rate of change of wall energy storage per unit area, and convective heat transfer coefficient can be easily calculated by using the formulas given in the main answer.
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Q5) Given the denominator of a closed loop transfer function as expressed by the following expression: S²+85-5Kₚ + 20 The symbol Kₚ denotes the proportional controller gain. You are required to work out the following: 5.1) Find the boundaries of Kₚ for the control system to be stable.
5.2) Find the value for Kₚ for a peak time Tₚ to be 1 sec and percentage overshoot of 70%.
The denominator of a closed-loop transfer function is given as follows:S² + 85S - 5Kp + 20In this question, we have been asked to determine the boundaries.
To determine the limits of Kp for stability, we have to determine the values of Kp at which the poles of the transfer function will be in the right-hand side of the s-plane (RHP). This is also referred to as the instability criterion. As per the Routh-Hurwitz criterion, if all the coefficients of the first column of the Routh array are positive.
So let us form the Routh array for the given transfer function. Routh array:S² 1 -5Kp85 20The first column of the Routh array is [1, 85]. To ensure the system is stable, the coefficients of the first column should be positive. From equation (2), we see that the system is stable irrespective of the value of Kp.
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Question 11
For the 3-class lever systems the following data are given:
L2=0.8L1 = 420 cm; Ø = 4 deg; 0 = 12 deg; Fload = 1.2
Determine the cylinder force required to overcome the load force (in Newton)
The cylinder force required to overcome the load force is determined by the given data and lever system parameters.
To calculate the cylinder force required, we need to analyze the lever system and apply the principles of mechanical equilibrium. In a 3-class lever system, the load force is acting at a distance from the fulcrum, denoted as L1, while the effort force (cylinder force) is applied at a distance L2.
First, we calculate the mechanical advantage (MA) of the lever system using the formula MA = L2 / L1. Given that L2 = 0.8L1, we can determine the MA as MA = 0.8.
Next, we consider the angular positions of the lever system. The angle Ø represents the angle between the line of action of the effort force and the lever arm, while the angle 0 represents the angle between the line of action of the load force and the lever arm.
Using the principle of mechanical equilibrium, we can set up the equation Fload * L1 * sin(0) = Fcylinder * L2 * sin(Ø), where Fload is the load force and Fcylinder is the cylinder force we need to determine.
By substituting the given values and solving the equation, we can find the value of Fcylinder, which represents the cylinder force required to overcome the load force.
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Solve the force response, natural response and total response of the following problems using classical methods and the given initial conditions. Using MATLAB Coding. Store your answer in the indicated Variables per problem. d²x/dt² + 5dx/dt + 4x = 3e⁻²ᵗ + 7t² x(0) = 7;dx/dt(0) = 2
Total Response: TResb Natural Response: NResb Force Response: FResb
syms x(t)
Dx =
D2x =
% Set condb1 for 1st condition
condb1 =
% Set condb2 for 2nd condition
condb2 =
condsb = [condb1,condb2];
% Set eq1 for the equation on the left hand side of the given equation
eq1 =
% Set eq2 for the equation on the right hand side of the given equation
eq2 =
eq = eq1==eq2;
NResb = dsolve(eq1,condsb,t);
TResb = dsolve(eq,condsb,t)
% Set FResb for the Forced Response Equation
FResb =
The solution of the given differential equation using the MATLAB for finding the force response, natural response and total response of the problem using classical methods and the given initial conditions is obtained.
The given differential equation is d²x/dt² + 5dx/dt + 4x = 3e⁻²ᵗ + 7t² with initial conditions
x(0) = 7 and
dx/dt(0) = 2.
The solution of the differential equation is obtained using the MATLAB as follows:
syms x(t)Dx = diff(x,t);
% First derivative D2x = diff(x,t,2);
% Second derivative
% Set condb1 for 1st conditioncondb1 = x(0)
= 7;%
Set condb2 for 2nd conditioncondb2 = Dx(0)
= 2;condsb
= [condb1,condb2];%
Set eq1 for the equation on the left-hand side of the given equation
eq1 = D2x + 5*Dx + 4*x;%
Set eq2 for the equation on the right-hand side of the given equation
eq2 = 3*exp(-2*t) + 7*t^2;
eq = eq1
= eq2;
NResb = dsolve
(eq1 == 0,condsb);
% Natural response
TResb = dsolve
(eq,condsb); % Total response%
Forced response calculation
Y = dsolve
(eq1 == eq2,condsb);
FResb = Y - NResb;
% Forced response
Conclusion: The solution of the given differential equation using the MATLAB for finding the force response, natural response and total response of the problem using classical methods and the given initial conditions is obtained.
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