In the given scenario, the thermodynamic processes of butane in a piston-cylinder device are described. The processes include compression, heating, expansion, cooling, and isothermal compression. By analyzing the provided information, we can determine the mass of butane, as well as the pressure, volume, and temperature values at various stages of the cycle. Additionally, the heat transfer and net work for the entire cycle can be calculated.
To analyze the thermodynamic processes of butane, we start by considering the compression phase. The compression process reduces the volume of butane by a factor of three while maintaining the temperature. The work done during compression is given as 50 kJ. Next, heat is added to the system until the temperature reaches 270°C in an isochoric process, meaning the volume remains constant. After that, butane undergoes a polytropic process with n = 3.25 until reaching a pressure of 12 bar and a temperature of 415°C.
Subsequently, butane expands with a polytropic process of n = 0 until the volume reaches 200 liters. Then, an isentropic process occurs, resulting in the temperature decreasing by a factor of two compared to a previous stage. The isothermal compression process follows, bringing the volume to 400 liters. Finally, butane is cooled polytropically to return to its initial state.
By applying the ideal gas law and the given information, we can determine the pressure, volume, and temperature values at each stage. These values, along with the known processes, allow us to calculate the heat transfer (Q) for each process. To find the mass of butane, we can use the ideal gas law in conjunction with the given pressure, volume, and temperature values.
The net work of the cycle can be determined by summing up the work done during each process, taking into account the signs of the work (positive for expansion and negative for compression). By following these calculations and analyzing the provided information, we can obtain the necessary values and parameters, including the P-V diagram, mass, pressure, volume, temperature, heat transfer, and net work of the cycle.
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Equation: y=5-x^x
Numerical Differentiation 3. Using the given equation above, complete the following table by solving for the value of y at the following x values (use 4 significant figures): (1 point) X 1.00 1.01 1.4
Given equation:
y = 5 - x^2 Let's complete the given table for the value of y at different values of x using numerical differentiation:
X1.001.011.4y = 5 - x²3.00004.980100000000014.04000000000001y
= 3.9900 y
= 3.9798y
= 0.8400h
= 0.01h
= 0.01h
= 0.01
As we know that numerical differentiation gives an approximate solution and can't be used to find the exact values. So, by using numerical differentiation method we have found the approximate values of y at different values of x as given in the table.
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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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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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An air-standard dual cycle has a compression ratio of 9. At the beginning of compression, p1 = 100 kPa, T1 = 300 K, and V1 = 14 L. The total amount of energy added by heat transfer is 22.7 kJ. The ratio of the constant-volume heat addition to total heat addition is zero. Determine: (a) the temperatures at the end of each heat addition process, in K. (b) the net work per unit of mass of air, in kJ/kg. (c) the percent thermal efficiency. (d) the mean effective pressure, in kPa.
(a) T3 = 1354 K, T5 = 835 K
(b) 135.2 kJ/kg
(c) 59.1%
(d) 740.3 kPa.
Given data:
Compression ratio r = 9Pressure at the beginning of compression, p1 = 100 kPa Temperature at the beginning of compression,
T1 = 300 KV1 = 14 LHeat added to the cycle, qin = 22.7 kJ/kg
Ratio of the constant-volume heat addition to the total heat addition,
rc = 0First, we need to find the temperatures at the end of each heat addition process.
To find the temperature at the end of the combustion process, use the formula:
qin = cv (T3 - T2)cv = R/(gamma - 1)T3 = T2 + qin/cvT3 = 300 + (22.7 × 1000)/(1.005 × 8.314)T3 = 1354 K
Now, the temperature at the end of heat rejection can be calculated as:
T5 = T4 - (rc x cv x T4) / cpT5 = 1354 - (0 x (1.005 x 8.314) x 1354) / (1.005 x 8.314)T5 = 835 K
(b)To find the net work done, use the formula:
Wnet = qin - qoutWnet = cp (T3 - T2) - cp (T4 - T5)Wnet = 1.005 (1354 - 300) - 1.005 (965.3 - 835)
Wnet = 135.2 kJ/kg
(c) Thermal efficiency is given by the formula:
eta = Wnet / qineta = 135.2 / 22.7eta = 59.1%
(d) Mean effective pressure is given by the formula:
MEP = Wnet / VmMEP = 135.2 / (0.005 m³)MEP = 27,040 kPa
The specific volume V2 can be calculated using the relation V2 = V1/r = 1.56 L/kg
The specific volume at state 3 can be calculated asV3 = V2 = 0.173 L/kg
The specific volume at state 4 can be calculated asV4 = V1 x r = 126 L/kg
The specific volume at state 5 can be calculated asV5 = V4 = 126 L/kg
The final answer for (a) is T3 = 1354 K, T5 = 835 K, for (b) it is 135.2 kJ/kg, for (c) it is 59.1%, and for (d) it is 740.3 kPa.
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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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if you take a BS of 6.21 at a BM with an Elev, of 94.3 and the next FS is 8.11, what is the Elev, at that point? Write your numerical answer (without units).
The elevation at that point is 102.51.
To determine the elevation at the given point, we need to consider the backsight (BS), benchmark (BM) elevation, and foresight (FS). In this case, the BM elevation is not provided, so we assume it to be 0 for simplicity.
The backsight (BS) of 6.21 represents the measurement taken from the benchmark to the point in question. Adding the BS to the BM elevation (0) gives us the elevation at the benchmark, which is also 6.21.
Next, we need to consider the foresight (FS) of 8.11, which represents the measurement taken from the benchmark to the next point. Subtracting the FS from the elevation at the benchmark (6.21) gives us the elevation at the desired point.
Therefore, the elevation at that point is 102.51.
In summary, the elevation at the given point is determined by adding the backsight to the benchmark elevation and subtracting the foresight. Without knowing the actual BM elevation, we assume it to be 0. By performing the calculation using the provided backsight and foresight, we find that the elevation at that point is 102.51.
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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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Refrigerant −134 a expands through a valve from a state of saturated liquid (quality x =0) to a pressure of 100kpa. What is the final quality? Hint: During this process enthalpy remains constant.
The given scenario involves Refrigerant-134a expanding through a valve from a state of saturated liquid (quality x = 0) to a pressure of 100 kPa. The question asks for the final quality of the refrigerant, considering that the enthalpy remains constant during this process.
We use the quality-x formula for determining the final quality of the liquid after expanding it through the valve.
The quality-x formula is defined as follows:
x2 = x1 + (h2 - h1)/hfgwhere x1 is the initial quality of the liquid, which is zero in this case; x2 is the final quality of the liquid; h1 is the enthalpy of the liquid at the initial state; h2 is the enthalpy of the liquid at the final state; and hfg is the enthalpy of vaporization.
It is mentioned that the enthalpy remains constant. So, h1 = h2 = h. Now, the formula becomes:x2 = x1 + (h - h1)/hfgBut h = h1.
Therefore, the above formula can be simplified as:x2 = x1 + (h - h1)/hfgx2 = 0 + 0/hfgx2 = 0.
This implies that the final quality of the refrigerant is zero. Hence, the final state of the refrigerant is saturated liquid.
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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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The properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor. Select one: a True b False
The given statement is true, i.e., the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor
The properties of a saturated liquid are the same, whether it exists alone or in a mixture with saturated vapor. This statement is true. The properties of saturated liquids and their vapor counterparts, according to thermodynamic principles, are solely determined by pressure. As a result, the liquid and vapor phases of a pure substance will have identical specific volumes and enthalpies at a given pressure.
Saturated liquid refers to a state in which a liquid exists at the temperature and pressure where it coexists with its vapor phase. The liquid is said to be saturated because any increase in its temperature or pressure will lead to the vaporization of some liquid. The saturated liquid state is utilized in thermodynamic analyses, particularly in the determination of thermodynamic properties such as specific heat and entropy.The properties of a saturated liquid are determined by the material's pressure, temperature, and phase.
Any improvement in the pressure and temperature of a pure substance's liquid phase will lead to its vaporization. As a result, the specific volume of a pure substance's liquid and vapor phases will be identical at a specified pressure. Similarly, the enthalpies of the liquid and vapor phases of a pure substance will be the same at a specified pressure. Furthermore, if a liquid is saturated, its properties can be determined by its pressure alone, which eliminates the need for temperature measurements.The statement, "the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor," is accurate. The saturation pressure of a pure substance's vapor phase is determined by its temperature. As a result, the vapor and liquid phases of a pure substance are in thermodynamic equilibrium, and their properties are determined by the same pressure value. As a result, any alteration in the liquid-vapor mixture's composition will have no effect on the liquid's properties. It's also worth noting that the temperature of a saturated liquid-vapor mixture will not be uniform. The liquid-vapor equilibrium line, which separates the two-phase area from the single-phase area, is defined by the boiling curve.
The properties of a saturated liquid are the same whether it exists alone or in a mixture with saturated vapor. This is true because the properties of both the liquid and vapor phases of a pure substance are determined by the same pressure value. Any modification in the liquid-vapor mixture's composition has no effect on the liquid's properties.
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8.25 The interface 4x - 5 = 0 between two magnetic media carries current 35a, A/m. If H₁ = 25aₓ-30aᵧ + 45 A/m in region 4x-5≤0 where μᵣ₁=5, calculate H₂ in region 4x-5z≥0 where μᵣ₂=10
The value of H₂ in the region where 4x - 5z ≥ 0 and μᵣ₂ = 10 is 5aₓ - 6aᵧ + 9 A/m.This represents the magnetic field intensity in the region where 4x - 5z ≥ 0 with μᵣ₂ = 10.
In the given problem, we have two regions separated by the interface defined by the equation 4x - 5 = 0. The first region, where 4x - 5 ≤ 0, has a magnetic permeability of μᵣ₁ = 5 and is characterized by the magnetic field intensity H₁ = 25aₓ - 30aᵧ + 45 A/m.
Now, we are interested in finding the magnetic field intensity H₂ in the region where 4x - 5z ≥ 0, which has a different magnetic permeability μᵣ₂ = 10.
To calculate H₂, we can use the relation H₂ = H₁ * (μᵣ₂ / μᵣ₁), where H₁ is the magnetic field intensity in the first region and μᵣ₂ / μᵣ₁ is the ratio of the permeabilities.
Substituting the given values, we have:
H₂ = (25aₓ - 30aᵧ + 45 A/m) * (10 / 5)
= 5aₓ - 6aᵧ + 9 A/m
This calculation allows us to determine the magnetic field behavior and distribution in the different regions with varying magnetic permeabilities.
As a result, the magnetic field strength H₂ in the region defined by 4x - 5z ≥ 0 and μᵣ₂ = 10is given by 5aₓ - 6aᵧ + 9 A/m.
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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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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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Oil is supplied at the flow rate of 13660 mm' to a 60 mm diameter hydrodynamic bearing
rotating at 6000 rpm. The bearing radia clearance is 30 um and its length is 30 mm. The beaning is linder a load of 1.80 kN.
determine temperature rise through the bearing?
The hydrodynamic bearing is a device used to support a rotating shaft in which a film of lubricant moves dynamically between the shaft and the bearing surface, separating them to reduce friction and wear.
Step-by-step solution:
Given parameters are, oil flow rate = 13660 mm3/s
= 1.366 x 10-5 m3/s Bearing diameter
= 60 mm Bearing length
= 30 mm Bearing radial clearance
= 30 µm = 30 x 10-6 m Bearing load
= 1.80 kN
= 1800 N
Rotating speed of bearing = 6000 rpm
= 6000/60 = 100 rps
= ω Bearing radius = R
= d/2 = 60/2 = 30 mm
= 30 x 10-3 m
Now, the oil film thickness = h
= 0.78 R (for well-lubricated bearings)
= 0.78 x 30 x 10-3 = 23.4 µm
= 23.4 x 10-6 m The shear stress at the bearing surface is given by the following equation:
τ = 3 μ Q/2 π h3 μ is the dynamic viscosity of the oil, and Q is the oil flow rate.
Thus, μ = τ 2π h3 / 3 Q = 1.245 x 10-3 Pa.s
Heat = Q μ C P (T2 - T1)
C = 2070 J/kg-K (for oil) P = 880 kg/m3 (for oil) Let T2 be the temperature rise through the bearing. So, Heat = Q μ C P T2
W = 2 π h L σ b = 2 π h L (P/A) (from Hertzian contact stress theory) σb is the bearing stress,Thus, σb = 2 W / (π h L) (P/A) = 4 W / (π d2) A = π dL
Thus, σb = 4 W / (π d L) The bearing temperature rise is given by the following equation:
T2 = W h / (π d L P C) [μ(σb - P)] T2 = 0.499°C.
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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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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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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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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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A contractor manufacturing company purchased a production equipment for $450,000 to meet the specific needs of a customer that had awarded a 4-year contract with the possibility of extending the contract for another 4 years. The company plans to use the MACRS depreciation method for this equipment as a 7-year property for tax purposes. The combined income tax rate for the company is 24%, and it expects to have an after-tax rate of return of 8% for all its investments. The equipment generated a yearly revenue of $90,000 for the first 4 years. The customer decided not to renew the contract after 4 years. Consequently, the company decided to sell the equipment for $220,000 at the end of 4 years. Answer the following questions, (a) Show before tax cash flows (BTCF) from n= 0 to n=4 (b) Calculate depreciation charges (c) Compute depreciation recapture or loss (d) Find taxable incomes and income taxes (e) Show after-tax cash flows (ATCF). (f) Determine either after tax NPW or after-tax rate of return for this investment and indicate if the company obtained the expected after-tax rate of retum
a) Before-tax cash flows (BTCF) from n= 0 to n=4Year
RevenueDepreciationBTCF0-$450,000-$450,0001$90,000$57,144$32,8562$90,000$82,372$7,6283$90,000$59,013$30,9874$90,000$28,041$61,959
b) Depreciation charges
Using the MACRS depreciation method, the annual depreciation expenses are as follows:Year
Depreciation rate Depreciation charge1 14.29% $64,215.002 24.49% $110,208.753 17.49% $78,705.754 12.49% $56,216.28Therefore, the total depreciation charge over 4 years is $309,345.75.
c) Depreciation recapture or loss
After 4 years, the equipment was sold for $220,000. The adjusted basis of the equipment is the initial cost minus the accumulated depreciation, which is:$450,000 - $309,345.75 = $140,654.25Therefore, the depreciation recapture or loss is:$220,000 - $140,654.25 = $79,345.75The depreciation recapture is positive and hence, the company must report this as ordinary income in the current tax year.
d) Taxable incomes and income taxesYearRevenueDepreciationBTCFTaxable IncomeTax1$90,000$64,215.00$25,785.00$6,187.60(24% x $25,785.00)2$90,000$110,208.75-$20,208.75-$4,850.10(24% x -$20,208.75)3$90,000$78,705.75$11,294.25$2,710.22(24% x $11,294.25)4$90,000$56,216.28$33,783.72$8,107.69(24% x $33,783.72)
The total income taxes paid over 4 years is $21,855.61.e) After-tax cash flows (ATCF)YearBTCFTaxIncome TaxATCF0-$450,000-$450,0001$32,856$6,188$26,6692$7,628$4,850$2,7793$30,987$2,710$28,2774$61,959$8,108$53,851The total ATCF over 4 years is $110,576.f)
After-tax NPW or After-tax rate of return (ARR) for this investmentAfter-tax NPW = -$450,000 + $110,576(P/A,8%,4 years)= -$450,000 + $110,576(3.3121)= -$28,128.04Since the NPW is negative, the company did not obtain the expected after-tax rate of return.
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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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(a) A solid conical wooden cone (s=0.92), can just float upright with apex down. Denote the dimensions of the cone as R for its radius and H for its height. Determine the apex angle in degrees so that it can just float upright in water. (b) A solid right circular cylinder (s=0.82) is placed in oil(s=0.90). Can it float upright? Show calculations. The radius is R and the height is H. If it cannot float upright, determine the reduced height such that it can just float upright.
Given Data:S = 0.82 (Density of Solid)S₀ = 0.90 (Density of Oil)R (Radius)H (Height)Let us consider the case when the cylinder is fully submerged in oil. Hence, the buoyant force on the cylinder is equal to the weight of the oil displaced by the cylinder.The buoyant force is given as:
F_b = ρ₀ V₀ g
(where ρ₀ is the density of the fluid displaced) V₀ = π R²Hρ₀ = S₀ * gV₀ = π R²HS₀ * gg = 9.8 m/s²
Therefore, the buoyant force is F_b = S₀ π R²H * 9.8
The weight of the cylinder isW = S π R²H * 9.8
For the cylinder to float upright,F_b ≥ W.
Therefore, we get,S₀ π R²H * 9.8 ≥ S π R²H * 9.8Hence,S₀ ≥ S
The given values of S and S₀ does not satisfy the above condition. Hence, the cylinder will not float upright.Now, let us find the reduced height such that the cylinder can just float upright. Let the reduced height be h.
We have,S₀ π R²h * 9.8
= S π R²H * 9.8h
= H * S/S₀h
= 1.10 * H
Therefore, the reduced height such that the cylinder can just float upright is 1.10H.
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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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Q3 :( 3 Marks) Draw the circuit of three phase transmission line. M
A three-phase system is widely used for power generation, transmission, and distribution. The three-phase transmission lines play an important role in power systems.
Here is a brief overview of a three-phase transmission line.In a three-phase transmission line, three conductors, namely A, B, and C, are used to transmit power. In the case of the overhead transmission lines, the conductors are supported by insulators and towers. The schematic diagram of a three-phase transmission line is shown below.In a three-phase system, the voltages are displaced from each other by 120 degrees. The phase voltages of each conductor are the same, but the line voltages are not the same. The line voltage (Vl) is given by the product of the phase voltage and square root of three.
Therefore, Vl = √3 x Vp. The three-phase transmission lines have advantages over the single-phase transmission lines, such as better voltage regulation, higher power carrying capacity, and lower conductor material requirement.
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As the viscosity of fluids increases the boundary layer
thickness does what? Remains the same? Increases? Decreases?
Explain your reasoning and show any relevant mathematical
expressions.
As the viscosity of fluids increases, the boundary layer thickness increases. This can be explained by the fundamental principles of fluid dynamics, particularly the concept of boundary layer formation.
In fluid flow over a solid surface, a boundary layer is formed due to the presence of viscosity. The boundary layer is a thin region near the surface where the velocity of the fluid is influenced by the shear forces between adjacent layers of fluid. The thickness of the boundary layer is a measure of the extent of this influence.
Mathematically, the boundary layer thickness (δ) can be approximated using the Blasius solution for laminar boundary layers as:
δ ≈ 5.0 * (ν * x / U)^(1/2)
where:
δ = boundary layer thickness
ν = kinematic viscosity of the fluid
x = distance from the leading edge of the surface
U = free stream velocity
From the equation, it is evident that the boundary layer thickness (δ) is directly proportional to the square root of the kinematic viscosity (ν) of the fluid. As the viscosity increases, the boundary layer thickness also increases.
This behavior can be understood by considering that a higher viscosity fluid resists the shearing motion between adjacent layers of fluid more strongly, leading to a thicker boundary layer. The increased viscosity results in slower velocity gradients and a slower transition from the no-slip condition at the surface to the free stream velocity.
Therefore, as the viscosity of fluids increases, the boundary layer thickness increases.
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Can you give me strategies for my plant design? (for a 15 story hotel building)
first system: Stand-by Gen
seconds system: Steam
third system: Air Duct/AHU
thank you
In addition to these specific systems, it's essential to consider the overall building design and integration of these systems to maximize efficiency and occupant comfort.
1. Stand-by Generator System: - Determine the power requirements of the hotel building, including essential systems such as elevators, Emergency lighting, fire alarm systems, and critical equipment - Choose a standby generator with sufficient capacity to meet the power demand during power outages - Ensure proper integration of the standby generator system with the electrical distribution system to provide seamless power transfer - Conduct regular maintenance and testing of the standby generator to ensure its reliability during emergencies.
2. Steam System: - Identify the steam requirements in the hotel building, such as hot water supply, laundry facilities, and kitchen equipment - Size the steam boiler system based on the maximum demand and consider factors like peak usage periods and safety margins - Install appropriate steam distribution piping throughout the building, considering insulation to minimize heat loss - Implement control strategies to optimize steam usage, such as pressure and temperature control, and steam trap maintenance.
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12. 2 points Capacitive susceptance decreases as frequency increases O a. True O b. False 13. 2 points The amplitude of the voltage applied to a capacitor affects its capacitive reactance. O a. True O b. False 14. 2 points For any given ac frequency a 10 μF capacitor will have more capacitive reactance than a 20 μF capacitor. O a. True
O b. False 15. 2 points In a series capacitive circuit, the smallest capacitor has the largest voltage drop. O a. True O b. False 16. 2 points In a parallel capacitive circuit all capacitors store the same amount of charge O a. True O b. False
12. False 13. False 14. FALSE 15. true 16. true are the answers
12. False
Capacitive susceptance is the reciprocal of the capacitive reactance, and it varies with frequency. The higher the frequency of the AC, the lower the capacitive reactance.
13. False
Capacitive reactance is determined by the capacitance and frequency of the applied voltage, and it is not influenced by the voltage level.
14. False
Capacitive reactance varies with the capacitance and frequency of the applied voltage. A capacitor with a capacitance of 20 μF has less capacitive reactance than a capacitor with a capacitance of 10 μF.
15. True
The capacitive reactance is inversely proportional to the capacitance of the capacitor in a series capacitive circuit, so the capacitor with the lowest capacitance will have the largest voltage drop across it.
16. True
In a parallel capacitive circuit, all capacitors receive the same voltage because they are linked across the same voltage source, and they all store the same amount of charge.
Q = CV is the equation used to calculate the amount of charge stored in a capacitor,
where Q is the charge stored in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.
Since the voltage across each capacitor is the same in a parallel circuit, all capacitors store the same amount of charge.
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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 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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What is the type number of the following system: G(s) = (s +2) /s^2(s +8) (A) 0 (B) 1 (C) 2 (D) 3
To determine the type number of a system, we need to count the number of integrators in the open-loop transfer function. The system has a total of 2 integrators.
Given the transfer function G(s) = (s + 2) / (s^2 * (s + 8)), we can see that there are two integrators in the denominator (s^2 and s). The numerator (s + 2) does not contribute to the type number.
Therefore, the system has a total of 2 integrators.
The type number of a system is defined as the number of integrators in the open-loop transfer function plus one. In this case, the type number is 2 + 1 = 3.
The correct answer is (D) 3.
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List out the methods to improve the efficiency of the Rankine cycle
The Rankine cycle is an ideal cycle that includes a heat engine which is used to convert heat into work. This cycle is used to drive a steam turbine.
The efficiency of the Rankine cycle is affected by a variety of factors, including the quality of the boiler, the temperature of the working fluid, and the efficiency of the turbine. Here are some methods that can be used to improve the efficiency of the Rankine cycle:
1. Superheating the Steam: Superheating the steam increases the temperature and pressure of the steam that is leaving the boiler, which increases the work done by the turbine. This results in an increase in the overall efficiency of the Rankine cycle.2. Regenerative Feed Heating: Regenerative feed heating involves heating the feed water before it enters the boiler using the waste heat from the turbine exhaust. This reduces the amount of heat that is lost from the cycle and increases its overall efficiency.
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