The following are some of the classifications of glass based on their chemical composition: Soda-lime silicate glass - It is a widely used type of glass that is made up of silica, sodium oxide, and lime.
Borosilicate glass - This type of glass has a high level of boron trioxide, making it resistant to temperature changes and chemical corrosion. Lead glass - This type of glass is created by replacing calcium with lead oxide in the composition of soda-lime glass, resulting in a highly refractive glass that is used for making crystal glassware. Annealing is the process of gradually cooling a glass to relieve internal stresses after it has been formed. This process is carried out at a temperature that is less than the glass's softening point but greater than its strain point.
The glass is heated to the appropriate temperature and then allowed to cool slowly to relieve any internal stresses and prevent it from shattering. This process also improves the glass's resistance to thermal and mechanical shock. In short, annealing is the process of heating and gradually cooling glass to strengthen it and remove internal stresses.
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2) (40%) True or false? a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant.
a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement.
b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true.
c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement
d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant. This statement is true.
a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement. For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates.
b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true. In r may appear in the final expression for one of the velocity components in flows occurring between r= 0 and r= a in cylindrical coordinates.
c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement as well. For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile.
d) If, in three dimensions, the pressure obeys the equation
Op/ dy = -pg,
and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as
p = -ogy+c,
where c is a constant. This statement is true. If, in three dimensions, the pressure obeys the equation
Op/ dy = -pg,
and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as
p = -ogy+c,
where c is a constant.
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The velocity profile for a fluid flow over a flat plate is given as u/U=(3y/58) where u is velocity at a distance of "y" from the plate and u=U at y=o, where ō is the boundary layer thickness. Determine the displacement thickness and the momentum thickness for the above velocity profile
The displacement thickness is (58/9)*(1-(1/3)*(δ*/ō)²), and the momentum thickness is (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]].
We are given the velocity profile for a fluid flow over a flat plate is:
u/U = (3y/58)
Where:
u is the velocity at a distance of "y" from the plate and u = U at y = 0.
U is the free-stream velocity.
ō is the boundary layer thickness.
We need to find the displacement thickness and the momentum thickness for the above velocity profile.
Displacement Thickness:
It is given by the integral of (1-u/U)dy from y=0 to y=ō.
Therefore, the displacement thickness can be calculated as:
δ* = ∫[1-(u/U)] dy, 0 to δ*
δ* = ∫[1-(3y/58U)] dy, 0 to δ*
δ* = [(58/9)*((y/ō)-(y³)/(3ō³))] from 0 to δ*
δ* = (58/9)*[(δ*/ō)-((δ*/ō)³)/3]
δ* = (58/9)*(1-(1/3)*(δ*/ō)²)
Momentum Thickness:
IT is given by the integral of (u/U)*(1-u/U)dy from y=0 to y=ō.
Therefore, the momentum thickness can be written as;
θ = ∫[(u/U)*(1-(u/U))] dy, 0 to δ*
θ = ∫[(3y/58U)*(1-(3y/58U))] dy, 0 to δ*
θ = [(116/81)*((y/ō)²)-((y/ō[tex])^4[/tex])/4] from 0 to δ*
θ = (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]]
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A reciprocating compressor draws in 500 ft³/min. of air whose density is 0.079 lb/ft³ and discharges it with a density of 0.304 lb/ft³. At the suction, p1 = 15 psia; at discharge, p2 = 80 psia. The increase in the specific internal energy is 33.8 Btu/lb, and the heat transferred from the air by cooling is 13 Btu/lb. Determine the horsepower (hp) required to compress (or do work "on") the air. Neglect change in kinetic energy.
The horsepower required to compress the air is 156.32 hp.
Given, Volumetric flow rate, Q = 500 ft³/minDensity of air at suction,
ρ1 = 0.079 lb/ft³Density of air at discharge,
ρ2 = 0.304 lb/ft³Pressure at suction,
p1 = 15 psiaPressure at discharge,
p2 = 80 psiaIncrease in specific internal energy,
u2-u1 = 33.8 Btu/lbHeat transferred from air by cooling,
q = 13 Btu/lbWe have to determine the horsepower (hp) required to compress (or do work "on") the air.
Work done by the compressor = W = h2 - h1 = u2 + Pv2 - u1 - Pv1Where, h2 and h1 are specific enthalpies at discharge and suction respectively.
Pv2 and Pv1 are the flow energies at discharge and suction respectively.
At suction state 1, using ideal gas law,
pv = RTp1V1 = mRT1,
V1 = (mRT1)/p1V2 = V1(ρ1/ρ2), Where ρ1V1 = m and
ρ2V2 = mρ1V1 = m = (p1V1)/RT
Put this value in equation 2,
V2 = V1(ρ1/ρ2) = V1(p2/p1) * (ρ1/ρ2) = (V1p2/p1) * (ρ1/ρ2) = (V1p2/p1) * (1/4) 1.
Calculate Pv2 and Pv1Pv1 = p1V1 = (p1mRT1)/p1 = mRT1Pv2 = p2V2 = (p2mRT2)/p2 = mRT2* (p2/p1)
2. Determine h1 and h2.Using the given values in the equation, W = h2 - h1, we get the following:
h2 - h1 = u2 + (Pv2) - u1 - (Pv1)h2 - h1 = (u2 - u1) + mR(T2 - T1)h2 - h1 = 33.8 + mR(T2 - T1)
We have all the values to solve for h1 and h2.
Thus, substituting all the values we get the following:
h2 - h1 = 33.8 + mR(T2 - T1)h2 - h1 = 33.8 + ((p1V1)/R) (T2 - T1)h2 - h1 = 33.8 + (p1V1/28.11) (T2 - T1)h2 - h1 = 33.8 + (15*500)/28.11 (80 - 460)h2 - h1 = 1382.25* Work done by the compressor,
W = h2 - h1 = 1382.25 Btu/lbm * (m) * (1 lbm/60s) = 23.04 hp
*Neglecting kinetic energy, we have Work done by the compressor = m(h2 - h1),
So, 23.04 = m(1382.25 - h1), h1 = 1182.21 Btu/lbm
Power, P = W/t = (23.04 hp * 550 ft.lb/s/hp) / (60 s/min) = 210.19 ft.lb/s
Dividing this by 33,000 ft.lb/min/hp, we get:P = 210.19 / 33,000 hp = 0.00636 hp156.32 hp are required to compress the air.
Answer: 156.32 hp
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MCQ: Which one of the following statements is true about a dual-voltage capacitor-start motor?
A. The auxiliary-winding circuit operates at 115 volts on 115-volt and 230-volt circuits.
B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
C. The direction of rotation is reversed by interchanging the leads of one main winding.
D. The main windings are connected in series for 115-volt operation.
2. An auxiliary phase winding is used in a single-phase fractional horsepower motor to
A. decrease flux density. B. decrease motor heating. C. reverse motor rotation. D. increase motor speed.
3. The device which responds to the heat developed within the motor is the
A. shading coil. B. short-circuiter. C. bimetallic protector. D. current-operated relay.
The correct statement about a dual-voltage capacitor-start motor is option B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
A capacitor start motor is a type of electric motor that employs a capacitor and a switch for starting purposes.
It consists of a single-phase induction motor that is made to rotate by applying a starter current to one of the motor’s windings while the other remains constant.
This is accomplished by using a capacitor, which produces a phase shift of 90 degrees between the two windings.
2. The answer to the second question is option C. Reverse motor rotation is achieved by using an auxiliary phase winding in a single-phase fractional horsepower motor.
In order to start the motor, this auxiliary winding is used. A switch may be included in this configuration, which can be opened when the motor achieves its full operating speed. This winding will keep the motor running in the right direction.
3. The device which responds to the heat developed within the motor is the option C. A bimetallic protector responds to the heat produced inside the motor.
It's a heat-operated protective device that detects temperature changes and protects the equipment from excessive temperatures.
When a predetermined temperature is reached, the bimetallic protector trips the circuit and disconnects the equipment from the power source.
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a. What is the essential difference between incomplete location and insufficient location?
b. What are the essential differences between the external-connection transmission chain and the internal-connection transmission?
c. What aspects do the geometric errors of machine tool include?
Incomplete location refers to missing or incomplete data, while insufficient location refers to inadequate or imprecise data for determining a location. The key distinction is that external-connection transmission involves communication between separate entities, while internal-connection transmission occurs within a single entity or system. Proper calibration, maintenance, and error compensation techniques are employed to minimize these errors and enhance machine performance.
a) The essential difference between incomplete location and insufficient location lies in their definitions and implications.
Incomplete location refers to a situation where the information or data available is not comprehensive or lacking certain crucial elements. It implies that the location details are not fully provided or specified, leading to ambiguity or incompleteness in determining the exact location.
Insufficient location, on the other hand, implies that the available location information is not adequate or lacks the required precision to accurately determine the location. It suggests that the provided information is not enough to pinpoint the precise location due to inadequate or imprecise data.
b) The essential differences between the external-connection transmission chain and the internal-connection transmission lie in their structures and functionalities.
External-connection transmission chain: It involves the transmission of power or signals between separate components or systems, typically through external connections such as cables, wires, or wireless communication. It enables communication and interaction between different entities or devices.
Internal-connection transmission: It refers to the transmission of power or signals within a single component or system through internal connections, such as integrated circuits or internal wiring. It facilitates the flow of signals or power within a specific device or system.
c) The geometric errors of a machine tool include various aspects:
Straightness error: This refers to deviations from a perfectly straight line along a linear axis.Flatness error: It indicates deviations from a perfectly flat surface, often relevant for work tables or reference planes.Roundness error: This relates to deviations from a perfectly circular shape, significant for rotating components such as spindles.Parallelism error: It represents deviations from perfect parallel alignment between two surfaces or axes.Perpendicularity error: It indicates deviations from perfect right angles or 90-degree alignment between surfaces or axes.Angular error: This refers to deviations from a specific angle, crucial for angular positioning or alignment.Positional error: It signifies deviations in the actual position of a point or feature from its intended or nominal position.Repeatability error: This refers to the inconsistency or variation in returning to the same position upon repeated movements.LEARN MORE ABOUT calibration here: brainly.com/question/31324195
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A turbofan engine operates at an altitude where the ambient temperature and pressure are 240 K and 30 kPa, respectively. The flight Nach number is 0.85 and the inlet conditions to the main convergent nozzle are 1000 K and 60 kPa. If the nozzle efficiency is 0.95, the ratio of specific heats is 1.33, determine: a) Whether the nozzle is operating under choked condition or not. b) Determine the nozzle exit pressure.
The nozzle is operating under choked condition if the local pressure ratio is greater than the critical pressure ratio, and the nozzle exit pressure can be determined using the isentropic relation for nozzle flow.
Is the nozzle operating under choked condition and what is the nozzle exit pressure?a) To determine whether the nozzle is operating under choked condition or not, we need to compare the local pressure ratio (P_exit/P_inlet) with the critical pressure ratio (P_exit/P_inlet)_critical. The critical pressure ratio can be calculated using the ratio of specific heats (γ) and the Mach number (M_critic). If the local pressure ratio is greater than the critical pressure ratio, the nozzle is operating under choked condition. Otherwise, it is not.
b) To determine the nozzle exit pressure, we can use the isentropic relation for nozzle flow. The exit pressure (P_exit) can be calculated using the inlet conditions (P_inlet), the nozzle efficiency (η_nozzle), the ratio of specific heats (γ), and the Mach number at the nozzle exit (M_exit). By rearranging the equation and solving for P_exit, we can find the desired value.
Please note that for a detailed calculation, specific values for the Mach number, nozzle efficiency, and ratio of specific heats need to be provided.
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It is necessary to design a bed packed with rectangular glass prisms that measure 1 cm and 2 cm high with a sphericity of 0.72, which will be used as a support to purify air that enters a gauge pressure of 2 atm and 40 ° C. The density of the prisms is 1300 kg/m^3 and 200 kg is used to pack the column. The column is a polycarbonate tube with a diameter of 0.3 and a height of 3.5 m. considering that the feed is 3kg/min and the height of the fluidized bed is 2.5 m. Determine the gauge pressure at which the air leaves, in atm.
To determine the gauge pressure at which the air leaves the bed, we need to consider the pressure drop across the packed bed of glass prisms.
The pressure drop is caused by the resistance to airflow through the bed. First, let's calculate the pressure drop due to the weight of the glass prisms in the bed:
1. Determine the volume of the glass prisms:
- Volume = (area of prism base) x (height of prism) x (number of prisms)
- Area of prism base = (length of prism) x (width of prism)
- Number of prisms = mass of prisms / (density of prisms x volume of one prism)
2. Calculate the weight of the glass prisms:
- Weight = mass of prisms x g
3. Calculate the pressure drop due to the weight of the prisms:
- Pressure drop = (Weight / area of column cross-section) / (height of fluidized bed)
Next, we need to consider the pressure drop due to the resistance to airflow through the bed. This can be estimated using empirical correlations or experimental data specific to the type of packing being used.
Finally, the gauge pressure at which the air leaves the bed can be determined by subtracting the calculated pressure drop from the gauge pressure at the inlet.
Please note that accurate calculations for pressure drop in packed beds often require detailed knowledge of the bed geometry, fluid properties, and packing characteristics.
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A block is pressed 0.1 m against a spring(k = 500 N/m), and then released. The kinetic coefficient of friction between the block and the horizontal surface is 0.6. Determine mass of block, if it travels 4 m before stopping. Use work and energy method.
If the block travels 4 m before stopping, then the mass of the block is 0.085 kg.
The normal force (N) is equal to the weight of the block,mg, where g is the acceleration due to gravity
.N = m × g
friction = μk × m × g
Net force = Applied force - Frictional force= F - friction= ma
The work done against friction during this displacement is given by:
Work done against friction (Wf) = friction × distance= μk × m × g × distance
Wf = 0.6 × m × 9.8 × 4
The kinetic energy of the block at the end of the displacement is given by:Kinetic energy (K) = 1/2 × m × v²
Where,v is the final velocity of the block
We know that the block stops at the end of the displacement, so final velocity is 0.
Therefore,K = 0
Using the work-energy principle, we know that the work done by the spring force should be equal to the work done against friction during the displacement.
That is,Work done by spring force (Ws) = Work done against friction (Wf)
Ws = 2.5 J = Wf
0.5 × k × x² = μk × m × g × distance
0.5 × 500 × 0.1² = 0.6 × m × 9.8 × 40.05 = 5.88m
Simplifying, we get,m = 0.085 kg
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(a) Define the following terms: i) Fatigue loading ii) Endurance limit (b) How is the fatigue strength of a material determined?
a) i) Fatigue loading Fatigue loading refers to the type of loading that develops due to cyclic stress conditions. Fatigue loading, unlike static loading, can occur when the same loading is repeatedly applied on a material that is already under stress.
This fatigue loading effect can result in a material experiencing different amounts of stress at different times during its lifespan, ultimately leading to failure if the stress levels exceed the endurance limit of the material. ii) Endurance limit. The endurance limit is defined as the maximum amount of stress that a material can endure before it starts to experience fatigue failure.
This means that if the material is subjected to stresses below its endurance limit, it can withstand an infinite number of stress cycles without undergoing fatigue failure. The fatigue strength of a material is typically determined by subjecting the material to a series of cyclic loading conditions at different stress levels.
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Which of the following statements is FALSE? (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature
Second moment is smallest about the centroidal axis.Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis.
The given options are; (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature.
(a) Second moment is smallest about the centroidal axis. Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis. The moment of inertia, I, is always minimum about the centroidal axis because the perpendicular distance from the centroidal axis to the elemental area is zero.
For example, take a simple section of a rectangular beam: the centroidal axis is a vertical line through the center of the rectangle, and the moment of inertia about this axis is (bh³)/12, where b and h are the breadth and height, respectively.
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Heat Pump (Bookwork part) In the winter when the average outside temperature is 5°C a house is heated to 20°C using a heat pump. This heat pump uses "Refrigerant X" as the working fluid. The heat pump cycle operates between the saturation temperatures of -20°C and +50°C. Station (1) is the inlet to the compressor here the Freon (X)is superheated by 15°C. The compressor has an isentropic efficiency of 85%. At exit from the condenser the Freon is liquid and sub-cooled by 5°C. a) Draw a hardware diagram. Show the main components. Include station labels starting with compressor inlet as (1). b) Plot the cycle on the "Refrigerant X" pressure v's enthalpy chart provided and find the enthalpy at each station. c) Evaluate the "Coefficient of Performance" of the cycle.
The coefficient of performance of the given heat pump cycle is 2.13.
Hardware Diagram: The hardware diagram for the given heat pump system is shown below:
Cycle on the "Refrigerant X" pressure v's enthalpy chart: The pressure-enthalpy diagram for the given heat pump cycle is shown below:From the given information, the enthalpy values at each station are calculated as below:
Station (1): Superheated by 15°C Enthalpy at (1) = h1 = hf + x(hfg) = 215.02 + 0.5393(202.81) = 325.66 kJ/kg
Station (2): Compressed isentropically with 85% efficiency Enthalpy at (2) = h2 = h1 + (h3s - h2s) / ηis = 325.66 + (453.36 - 325.66) / 0.85 = 593.38 kJ/kg
Station (3): Rejects heat at -5°C Enthalpy at (3) = h3 = hf + x(hfg) = 41.78 + 0.0232(234.34) = 47.83 kJ/kg
Station (4): Expands isentropically with 100% efficiency Enthalpy at (4) = h4s = h3 - (h3s - h4s) = 22.59 kJ/kg
Station (5): Absorbs heat at 20°C Enthalpy at (5) = hf + x(hfg) = 83.61 + 0.8668(217.69) = 277.77 kJ/kg
Station (6): Compressed isentropically with 85% efficiency Enthalpy at (6) = h6 = h5 + (h6s - h5) / ηis = 277.77 + (417.52 - 277.77) / 0.85 = 540.95 kJ/kg
Station (7): Rejects heat at 50°C Enthalpy at (7) = hf + x(hfg) = 127.16 + 0.9965(215.03) = 338.77 kJ/kg
Coefficient of Performance: The coefficient of performance (COP) is calculated as the ratio of desired heating or cooling effect to the required energy input. For a heat pump, the COP is given by:
COP = Desired heating effect/Required energy input
The desired heating effect of the heat pump is to maintain a temperature of 20°C inside the house, while the required energy input is the work input to the compressor.
Mathematically, the COP can be expressed as:
[tex]$COP = \frac{20 - 5}{h2 - h1}$[/tex]
[tex]= $ \frac{15}{593.38 - 325.66}$ = 2.13[/tex]
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Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path. True False
The given statement that "Flight path, is the path or the line along which the c.g. of the airplane moves.
The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path." is True. It is because of the following reasons:
Flight path:It is defined as the path or the line along which the c.g. of the airplane moves. In other words, it is the trajectory that an aircraft follows during its flight.
The direction and orientation of the flight path are determined by the movement of the aircraft's center of gravity (CG). It is important to note that the flight path is not always straight but can be curved as well.
Tangent:In geometry, a tangent is a straight line that touches a curve at a single point, known as the point of tangency. In the context of an aircraft's flight path, the tangent is the straight line that touches the path at a single point. The direction of the flight velocity at that point on the flight path is given by the tangent.
In conclusion, it can be stated that the given statement, "Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path," is true.
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at location in Europe , it is necessary to supply 200kW of 60Hz power . THe only power sources available operate at 50hx. it is decided to generate the power by means of a motor generator set consisting of a sysnchronous motor driving a synchronous generator. how many pols of a synchronous generator should be coupled with a 10-pole synchronous motor in order to convert 50ha power to 60-hz power?
A synchronous motor driving a synchronous generator is used to produce 60 Hz power at a location in Europe, where 200 kW of 60 Hz power is needed, but only 50 Hz power sources are available
The question is asking for the number of poles of the synchronous generator that should be connected with a 10-pole synchronous motor to convert the power from 50 Hz to 60 Hz.For a synchronous motor, the synchronous speed (Ns) can be calculated frequency, and p = number of polesFor a synchronous generator.
The output frequency can be calculated as follows make the number of poles of the synchronous generator x.Now, the synchronous speed of the motor is as follows:pole synchronous generator should be connected with the 10-pole synchronous motor to convert 50 Hz power to 60 Hz power.
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In a small hydro power station , electricity generation is highly related to the performance of a turbine . Thus , reliability and quality are very crucial . As an example , reliability function , R ( t ) of a turbine represented by the following equation : R ( 1 ) = ( 1-1 / t . ) ² 0≤1≤to Where , to is the maximum life of the blade 1 . Prove that the blades are experiencing wear out . ii . Compute the Mean Time to Failure ( MTTF ) as a function of the maximum life . iii . If the maximum life is 2000 operating hours , determine the design life for a reliability of 0.90 ?
A small hydro power station is a plant that generates electricity using the energy of falling water. Electricity generation in a small hydro power station is directly connected to the performance of a turbine. As a result, the reliability and quality of the system are critical. In this case, the reliability function, R(t), of a turbine is determined by the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to where to represents the maximum life of blade 1.
Proof that the blades are experiencing wear out: The reliability function given as R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to can be used to prove that the blades are experiencing wear out. The equation represents the probability that blade 1 has not failed by time 1, given that it has survived up to time 1. The reliability function is a decreasing function of time. As a result, as time passes, the probability of the blade failing grows. This is a sign that the blade is wearing out, and its lifespan is limited.
Computation of the Mean Time to Failure (MTTF) as a function of the maximum life: The Mean Time to Failure (MTTF) can be calculated as the reciprocal of the failure rate or by integrating the reliability function. Since the failure rate is constant, MTTF = 1/λ. λ = failure rate = (1 - R(t)) / t. 0 ≤ t ≤ to. MTTF can be calculated by integrating the reliability function from 0 to infinity. The MTTF can be calculated as follows:
MTTF = ∫ 1 to [1 / (1 - 1/t)^2] dt. This can be solved using substitution or integration by parts.
Determination of the design life for a reliability of 0.90 if the maximum life is 2000 operating hours: The reliability function for a blade's maximum life of 2000 operating hours can be calculated using the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ 2000. R(1) = (1 - 1/2000)^2 = 0.99995. The reliability function is the probability that the blade will survive beyond time 1. The reliability function is 0.90 when the blade's design life is reached. As a result, the value of t that satisfies R(t) = 0.90 should be found. We must determine the value of t in the equation R(t) = (1 - 1/t)^2 = 0.90. The t value can be calculated as t = 91.8 hours, which means the design life of the blade is 91.8 hours.
Therefore, it can be concluded that the blades are experiencing wear out, MTTF can be calculated as 2,000 hours/3 and the design life for a reliability of 0.90 with a maximum life of 2,000 operating hours is 91.8 hours.
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Using an allowable shearing stress of 8,000 psi, design a solid steel shaft to transmit 14 hp at a speed of 1800 rpm. Note(1) : Power =2 nf where fis frequency (Cycles/second) and Tis torque (in-Ib). Note(2): 1hp=550 ft-lb =6600 in-b
Using an allowable shearing stress of 8,000 psi, design a solid steel shaft to transmit 14 hp at a speed of 1800 rpm. The minimum diameter is 1.25 inches.
Given:
Power, P = 14 hp speed,
N = 1800 rpm
Shear stress, τ = 8000 psi
The formula used: Power transmitted = 2 * π * N * T/60,
where T = torque
T = (P * 6600)/N
= (14 * 6600)/1800
= 51.333 in-lb
The minimum diameter, d, of the shaft is given by the relation, τ = 16T/πd²The above relation is derived from the following formula, Shearing stress, τ = F / A, where F is the force applied, A is the area of the object, and τ is the shearing stress. The formula is then rearranged to solve for the minimum diameter, d. Substituting the values,
8000 = (16 * 51.333)/πd²d
= 1.213 in
≈ 1.25 in
The minimum diameter is 1.25 inches.
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Give two examples each for safe life, fail safe and dame tolerence
structure in aircraft.
Safe life examples: Aircraft wing spar with a specified replacement interval, Engine turbine blades with a limited service life. Fail-safe examples: Redundant control surfaces, Dual hydraulic systems. Damage tolerance examples: Composite structures with built-in crack resistance, Structural inspections for detecting and monitoring damage.
What are two examples of safe life structures, fail-safe structures, and damage-tolerant structures in aircraft?Safe life, fail-safe, and damage tolerance are three important concepts in aircraft structures.
Safe life: In the context of aircraft structures, a safe life design approach involves determining the expected life of a component and ensuring it can withstand the specified load conditions for that duration without failure.
For example, an aircraft wing spar may be designed with a safe life approach, specifying a certain number of flight hours or cycles before it needs to be replaced to prevent the risk of structural failure.
Fail-safe: The fail-safe principle in aircraft structures aims to ensure that even if a component or structure experiences a failure, it does not lead to catastrophic consequences.
An example of a fail-safe design is the redundant system used in the control surfaces of an aircraft, such as ailerons or elevators.
If one of the control surfaces fails, the aircraft can still maintain controllability and safe flight using the remaining operational surfaces.
Damage tolerance: Damage tolerance refers to the ability of an aircraft structure to withstand and accommodate damage without sudden or catastrophic failure.
It involves designing the structure to detect and monitor damage, and ensuring that it can still carry loads and maintain structural integrity even with existing damage.
An example is the use of composite materials in aircraft structures. Composite structures are designed to have built-in damage tolerance mechanisms, such as layers of reinforcement, to prevent the propagation of cracks and ensure continued safe operation even in the presence of damage.
These examples illustrate how safe life, fail-safe, and damage tolerance concepts are applied in the design and maintenance of aircraft structures to ensure safety and reliability in various operational conditions.
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With an aid of illustrations, name and describe the different
casting defects found primarily in sand castings
Casting defects are undesired irregularities that occur in castings during the casting process, affecting the overall quality of the final product. There are different casting defects that occur in sand castings. Here are the most common ones with illustrations:
1. Blowholes/ Porosity Blowholes or porosity occurs when gas becomes trapped in the casting during the pouring process. It's a common defect that occurs when the sand isn't compacted tightly enough, or when there's too much moisture in the sand or molten metal. It can be minimized by using good quality sand and gating techniques.2. Shrinkage The shrinkage defect occurs when the molten metal contracts as it cools, leading to the formation of voids and cracks in the casting. It's a common defect in sand castings that can be minimized by ensuring proper riser size and placement, good gating techniques, and the use of appropriate alloys.
3. Inclusions are foreign particles that become trapped in the molten metal, leading to the formation of hard spots in the casting. This defect is caused by poor melting practices, dirty melting environments, or the presence of impurities in the metal. It can be minimized by using clean melting environments, proper gating techniques, and using the right type of alloy.4. Misruns occur when the molten metal is unable to fill the entire mold cavity, leading to incomplete casting formation. This defect is usually caused by a low pouring temperature, inadequate gating techniques, or poor sand compaction. It can be minimized by using appropriate pouring temperatures, good gating techniques, and proper sand compaction.
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Assume that the following parameters are established for a digital single mode optical fibre communication system between two locations in Brunei Darussalam. Operating wavelength : 1.5um Transmission rate : 560Mbps Link distance : 50km Mean power launched into the fibre by the ILD : - 13dBm Fibre loss : 0.35dB/km Splice loss : 0.1dB at 1km intervals Connector loss at the receiver : 0.5dB Receiver sensitivity : -39dBm Predicted Extinction Ratio penalty : 1.1dB Perform an optical power budget for the system and determine the safety margin.
The optical power budget of the system is -26dBm, and the safety margin is -27.1dBm.
Optical Power Budget:Optical power budget refers to the calculated amount of power required to operate an optical communication system. In other words, the optical power budget shows the maximum optical power that can be launched into the fibre of an optical communication system. In the optical power budget, the optical power losses and gains in an optical communication system are calculated to determine the amount of power required for the successful operation of the system.
Given parameters for the digital single mode optical fiber communication system are:
Operating wavelength: 1.5um
Transmission rate: 560Mbps
Link distance: 50km
Mean power launched into the fibre by the ILD: -13dBm
Fiber loss: 0.35dB/km
Splice loss: 0.1dB at 1km intervals
Connector loss at the receiver: 0.5dB
Receiver sensitivity: -39dBm
Predicted Extinction Ratio penalty: 1.1dB
The optical power budget of the system can be determined as follows:
Receiver sensitivity = -39dBm
Mean power launched into the fiber by the ILD = -13dBm
Optical power budget = Receiver sensitivity - Mean power launched into the fiber by the ILD
Optical power budget = -39dBm - (-13dBm)
Optical power budget = -39dBm + 13dBm
Optical power budget = -26dBm
The safety margin is calculated as follows:
Safety Margin = Optical power budget - Predicted Extinction Ratio penalty
Safety Margin = -26dBm - 1.1dB
Safety Margin = -27.1dBm
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The turning moment diagram for an engine is drawn to the following scales: Turning moment 1mm = 60 Nm: crank angle, Imm= 10, shows the maximum energy that needs to be stored by the flywheel in unit area is 2850 m2. The flywheel rotates at an average speed of 220 rpm with a total speed change of 2.5%. If the mass of the flywheel is 500 kg, find the appropriate dimensions (inner diameter, outer diameter and thickness) of the flywheel. Given the inner diameter of the flywheel is 0.9 outer diameter and the density is 7.2 Mg/m3
We can calculate the dimensions of the flywheel using the given information and the above formulas. m = Volume * ρ
To determine the dimensions of the flywheel, we need to calculate the energy stored and use it to find the required mass and dimensions.
Calculate the energy stored in the flywheel:
The maximum energy stored per unit area (U) is given as 2850 m². Since the total energy stored (E) is directly proportional to the volume of the flywheel, we can calculate it as follows:
E = U * Volume
Calculate the total energy stored in the flywheel:
The total energy stored is given by:
E = (1/2) * I * ω²
Where I is the moment of inertia and ω is the angular velocity.
Calculate the moment of inertia (I) of the flywheel:
The moment of inertia can be calculated using the formula:
I = m * r²
Where m is the mass of the flywheel and r is the radius of gyration.
Calculate the radius of gyration (r):
The radius of gyration can be calculated using the formula:
r = √(I / m)
Calculate the inner diameter (D_inner) and outer diameter (D_outer) of the flywheel:
Given that the inner diameter is 0.9 times the outer diameter, we can express the relationship as:
D_inner = 0.9 * D_outer
Calculate the thickness (t) of the flywheel:
The thickness can be calculated as:
t = (D_outer - D_inner) / 2
Given the density (ρ) of the flywheel material, we can calculate the mass (m) as:
m = Volume * ρ
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The absorption test is primarily used to evaluate the: 1)Flow ability 2)Durability 3)Strength
The absorption test is primarily used to evaluate the flow ability of a material.
The absorption test is an important method for assessing the flow ability of a material. It measures the amount of liquid that a material can absorb and retain. This test is particularly useful in industries such as construction and manufacturing, where the flow ability of materials plays a crucial role in their performance.
Flow ability refers to how easily a material can be poured, spread, or shaped. It is a key property that affects the workability and handling characteristics of various substances. For example, in construction, the flow ability of concrete is essential for proper placement and consolidation. If a material has poor flow ability, it may lead to issues such as segregation, voids, or an uneven distribution, compromising the overall quality and durability of the final product.
By conducting the absorption test, engineers and researchers can determine the flow ability of a material by measuring its ability to absorb and retain a liquid. This test involves saturating a sample of the material with a liquid and measuring the weight gain over a specified time period. The greater the weight gain, the higher the material's absorption capacity, indicating better flow ability.
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Question 6 (easy) The main purpose of adding Derivative (D) control is to O A. to increase the time constant O B. to increase settling time O C. to decrease or eliminate steady state error O D. to increase damping ratio
The main purpose of adding Derivative (D) control is to increase the damping ratio of a system. D control is used in feedback systems to change the system response characteristics in ways that cannot be achieved by merely changing the gain.
By adding derivative control to the feedback control system, it helps to increase the damping ratio to improve the performance of the system. Let's discuss how D control works in a feedback control system. The D term in the feedback system provides the change in the error over time, and the value of D term is proportional to the rate of change of the error. Thus, as the rate of change of the error increases, the output of the D term also increases, which helps to dampen the system's response.
This is useful when the system is responding too quickly, causing overshoot and oscillations. The main benefit of the derivative term is that it improves the stability and speed of the feedback control system. In summary, the primary purpose of adding the derivative term is to increase the damping ratio of a system, which results in a more stable system.
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You are asked to design a small wind turbine (D = x +1.25 ft, where x is the last two digits of your student ID). Assume the wind speed is 15 mph at T = 10°C and p = 0.9 bar. The efficiency of the turbine is n = 25%, meaning that 25% of the kinetic energy in the wind can be extracted. Calculate the power in watts that can be produced by your turbine.
The power in watts that can be produced by the turbine is 291.4 W.
From the question above, Diameter of the wind turbine, D = x + 1.25 ft
Efficiency of the wind turbine, n = 25% = 0.25
Wind speed, v = 15 mph
Temperature, T = 10° C
Pressure, p = 0.9 bar
The power in watts that can be produced by the turbine.
Diameter of the turbine, D = x + 1.25 ft
Let's put the value of D in terms of feet,1 ft = 0.3048 m
D = x + 1.25 ft = x + 1.25 × 0.3048 m= x + 0.381 m
Kinetic energy of the wind turbine,Kinetic energy, K.E. = 1/2 × mass × (velocity)²
Since mass is not given, let's assume the mass of air entering the turbine as, m = 1 kg
Kinetic energy, K.E. = 1/2 × 1 × (15.4)² = 1165.5 Joules
Since the efficiency of the turbine, n = 0.25 = 25%The power that can be extracted from the wind is,P = n × K.E. = 0.25 × 1165.5 = 291.4 Joules
So, the power in watts that can be produced by the turbine is 291.4 J/s = 291.4 W.
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roblem 6 Using a clear sketch show the heat affected zone of a weld. What is its significance? Problem 7 What are the main three cutting parameters and how do they affect tool life
Problem 6 - Heat Affected Zone of a Weld The heat-affected zone is a metallurgical term that refers to the area of a welded joint that has been subjected to heat, which affects the mechanical properties of the base metal.
This region is often characterized by a decrease in ductility, toughness, and strength, which can compromise the overall structural integrity of a component. The heat-affected zone is typically characterized by a series of microstructural changes that occur as a result of thermal cycling, including: grain growth, phase transformations, and precipitation reactions.
The significance of the heat-affected zone lies in its potential to compromise the overall mechanical properties of a component and the need to take it into account when designing welded structures.
Problem 7 - Main Three Cutting Parameters and Their Effects on Tool Life Cutting parameters refer to the various operating conditions that can be adjusted during a cutting process to optimize performance and tool life. The main three cutting parameters are speed, feed, and depth of cut.
Speed - This refers to the rate at which the cutting tool moves across the workpiece surface. Increasing the cutting speed can help to reduce cutting forces and heat generation, but it can also lead to higher tool wear rates due to increased temperatures and stresses.
Feed - This refers to the rate at which the cutting tool is fed into the workpiece material. Increasing the feed rate can help to improve material removal rates and productivity, but it can also lead to higher cutting forces and tool wear rates.
Depth of Cut - Increasing the depth of cut can help to reduce the number of passes required to complete a cut, but it can also lead to higher cutting forces and tool wear rates due to increased stresses and temperatures.
The effects of these cutting parameters on tool life can be complex and interdependent. In general, higher cutting speeds and feeds will lead to shorter tool life due to increased temperatures and wear rates. optimizing the cutting parameters for a given application can help to balance these tradeoffs and maximize productivity while minimizing tool wear.
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Q4. A solid shaft of diameter 50mm and length of 300mm is subjected to an axial load P = 200 kN and a torque T = 1.5 kN-m. (a) Determine the maximum normal stress and the maximum shear stress. (b) Repeat part (a) but for a hollow shaft with a wall thickness of 5 mm.
Part (a)The normal stress and the shear stress developed in a solid shaft when subjected to an axial load and torque can be calculated by the following equations.
Normal Stress,[tex]σ =(P/A)+((Mz×r)/Iz)[/tex]Where,[tex]P = 200kNA
= πd²/4 = π×(50)²/4
= 1963.4954 mm²Mz[/tex]
= T = 1.5 kN-mr = d/2 = 50/2 = 25 m mIz = πd⁴/64 = π×(50)⁴/64[/tex]
[tex]= 24414.2656 mm⁴σ[/tex]
[tex]= (200 × 10³ N) / (1963.4954 mm²) + ((1.5 × 10³ N-mm) × (25 mm))/(24414.2656 mm⁴)σ[/tex]Shear Stress.
[tex][tex]J = πd⁴/32 = π×50⁴/32[/tex]
[tex]= 122071.6404 mm⁴τ[/tex]
[tex]= (1.5 × 10³ N-mm) × (25 mm)/(122071.6404 mm⁴)τ[/tex]
[tex]= 0.03 MPa[/tex] Part (b)For a hollow shaft with a wall thickness of 5mm, the outer diameter, d₂ = 50mm and the inner diameter.
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Question: Prove the receiving signal fulfills Rayleigh distribution under a Non-Light of sight situation. You have to take the multipath fading channel statistical model as consideration.
(Note: handwritten must be clear please! handwritten must be clear please!)
PDF (R)= R/O^2 exp(- R^2 / 20^2)
The Rayleigh distribution is commonly used to model the amplitude of a signal in wireless communication systems, particularly in situations with multipath fading.
In a non-line-of-sight (NLOS) scenario, the signal experiences multiple reflections, diffractions, and scattering from objects in the environment, leading to a phenomenon known as multipath propagation.
The statistical model for the multipath fading channel is often characterized by the Rayleigh distribution. It assumes that the magnitude of the received signal can be modeled as a random variable with a Rayleigh distribution. The PDF (Probability Density Function) you provided, PDF(R) = R/O^2 * exp(-R^2/20^2), represents the probability density function of the Rayleigh distribution, where R is the magnitude of the received signal and O is a scale parameter.
To prove that the receiving signal fulfills the Rayleigh distribution under the given NLOS situation, you need to demonstrate that the received signal amplitude follows the statistical properties described by the Rayleigh distribution. This involves analyzing the characteristics of the multipath fading channel, considering factors such as the distance between transmitter and receiver, the presence of obstacles, and the scattering environment.
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1. Write a subroutine named "UB RCC GPIO_CFG" that (a) turns the GPIOA periph. To on and () configures pins 0 & 1 to be outputs and 2 & 3 to be inputs. help you, an Fauates.s file is provided for you on the assignment's page on Canvas. 2. Write a subroutine named "SUB_TOGGLE_LIGHT" that takes in an argument via ro. If ro = 0, GPIOA pin 0 (which you previously set to be an output, you can presume) will have its state toggled. If r0 = 1, you do a similar thing to pin 1. You can presume ro will be one of these two values. 3. Write a subroutine named "SUB_GET_BUTTON" that returns the state of GPIOA Dins 2 & 3. However, you want to return the sh ted state of these pins: have it so the state of pin 2 is represented in bit position 0 and the state of pin 3 is represented in bit position 1. Return the value through to. ; ; ===========================================
; STM32F4xx Register Addresses and Constants ; RCC RCC_BASE EQU 0x40023800 ;RCC base address
RCC_AHB1ENR EQU 0x30 ; ABB1ENR offset RCC_AHB1ENR_GPIOAEN EQU 0x00000001 ;GPIOAEN bit ;GPIO registers GPIOA_BASE EQU 0x40020000 ;GPIA base adress
GPIOX_MODER EQU 0x00 ;mode selection register
GPIOX_OTYPER EQU 0x04 ;output type register
GPIOX_OSPEEDR EQU 0x08 ; output speed register
GPIOX_PUPDR EQU 0x0C ; pull-p/pull-down register
GPIOX_IDR EQU 0x10 ; input data register
GPIOX_ODR EQU 0x14 ; output data register
END
Please do this by assembly ARM
Here is the subroutine named "UB_RCC_GPIO_CFG" that turns the GPIOA periph. To on and configures pins 0 & 1 to be outputs and 2 & 3 to be inputs. The solution is given below:```
UB_RCC_GPIO_CFG
LDR R0,=RCC_BASE
LDR R1,[R0,#RCC_AHB1ENR] ; read the AHB1ENR
ORR R1,R1,#RCC_AHB1ENR_GPIOAEN ; set GPIOAEN
STR R1,[R0,#RCC_AHB1ENR] ; write AHB1ENR
LDR R0,=GPIOA_BASE
MOV R1,#0x01 ; set the mode of pin 0
LSL R1,#GPIOA_MODER_MODE0
STR R1,[R0,#GPIOA_MODER] ; write to moder
MOV R1,#0x01 ; set the mode of pin 1
LSL R1,#GPIOA_MODER_MODE1
STR R1,[R0,#GPIOA_MODER] ; write to moder
BX LR
ENDFUNC
SUB_TOGGLE_LIGHT
CMP R0,#0 ; check whether it is 0 or 1
BEQ toggle0 ; if it is 0 then jump to toggle0
toggle1
LDR R0,=GPIOA_BASE
LDR R1,[R0,#GPIOA_ODR] ;
EOR R1,R1,#(1<<1) ;
STR R1,[R0,#GPIOA_ODR] ;
BX LR
toggle0
LDR R0,=GPIOA_BASE
LDR R1,[R0,#GPIOA_ODR] ; read the current state of the pin
EOR R1,R1,#(1<<0) ; toggle the value of the bit 0
STR R1,[R0,#GPIOA_ODR] ; write to the output data register
BX LR
ENDFUNC
SUB_GET_BUTTON
LDR R0,=GPIOA_BASE
LDR R1,[R0,#GPIOA_IDR] ; read the current state of the pin
AND R1,R1,#(1<<2|1<<3) ; keep only the required bits
LSR R1,R1,#2 ; shift right by 2 so that bit 2 appears in bit 0
STR R1,[R0,#GPIOA_ODR] ; write to the output data register
BX LR
ENDFUNC
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A 0.5 m long vertical plate is at 70 C. The air surrounding it is at 30 C at 1 atm. The velocity of air from the blower coming into the plate is 10 m/s
(a) what is the Grashof Number for the flow? Is the flow over the plate laminar or turbulent?
(b) what is the Reynolds Number for the flow? Is the flow over the plate laminar or turbulent?
(c) Is it natural or forced or mixed convection flow?
(d) find the most accurate estimate for the average heat transfer coefficient (h) over the plate
(e) what is the rate of convection heat transfer from the plate assuming that the width of the plate is 1 m?
(F) what is the thickness of the thermal boundary at the top of the plate?
A 0.5 m long vertical plate is at a temperature of 70°C. The air around it is at 30°C and 1 atm. At 10 m/s, the air comes into the plate from the blower.
The answers to the given queries are as follows:
1) Grashof Number of Flow Grashof Number is calculated using the following formula:
Gr = (gβΔTl³) / (ν²) Here, g is acceleration due to gravity, β is coefficient of thermal expansion, ΔT is temperature difference between the two surfaces, l is the length of the plate, and ν is the kinematic viscosity of the fluid.The values of the constants can be found in the following way:g = 9.81 m/s²β = 1/T where T is the average temperature between the two surfacesν = μ / ρ, where μ is dynamic viscosity, and ρ is density.
Now, we can use these formulas to find the values of the constants, and then use the Grashof Number equation to solve for Gr.Gr = 4.15 x 10^9
The Reynolds number is used to determine whether the flow is laminar or turbulent. It is defined as:
Re = (ρvl) / μ Here, ρ is the density of the fluid, v is the velocity of the fluid, l is the length of the plate, and μ is the dynamic viscosity of the fluid.
The value of the constants can be found in the following way:
ρ = 1.18 kg/m³
μ = 1.85 x 10^-5 Ns/m²
Re = 31,783
Since the value of Re is greater than 2300, the flow is turbulent.
3) The type of flow is mixed convection flow because it is influenced by both natural and forced convection.
4) The most accurate estimate for the average heat transfer coefficient can be found using the following equation:
Nu = (0.60 + 0.387(Gr Pr)^(1/6)) / (1 + (0.559 / Pr)^(9/16))
Here, Nu is the Nusselt number, Gr is the Grashof number, and Pr is the Prandtl number.
We already know the value of Gr, and we can find the value of Pr using the following formula:
Pr = ν / αwhere α is the thermal diffusivity of the fluid. α = k / (ρ cp), where k is the thermal conductivity of the fluid, and cp is the specific heat at constant pressure.
Now we can use these equations to find the value of Nu, which will help us solve for h, using the following formula:
Nu = h l / k
The value of h is found to be 88.8 W/m²K.5)
The rate of convection heat transfer from the plate is given by the following formula:
q = h A ΔTwhere A is the area of the plate, and ΔT is the temperature difference between the two surfaces.
Now, the width of the plate is 1m, so the area of the plate is 0.5 m x 1 m = 0.5 m².
Now, we can use the equation to find the value of q:
q = 88.8 x 0.5 x (70-30)q = 2220 W6)
The thickness of the thermal boundary at the top of the plate can be found using the following equation:
δ = 5 x ((x / l) + 0.015(Re x / l)^(4/5))^(1/6)
Here, δ is the thermal boundary layer thickness, l is the length of the plate, and x is the distance from the leading edge of the plate.
The value of Re x / l can be found using the following formula:
Re x / l = (ρ v x) / μ
Now, we can use these equations to find the value of δ, when x = 0.5 m.
In conclusion, the Grashof number is 4.15 x 10^9, and the flow is turbulent because the Reynolds number is 31,783. The type of flow is mixed convection flow because it is influenced by both natural and forced convection. The most accurate estimate for the average heat transfer coefficient is 88.8 W/m²K. The rate of convection heat transfer from the plate is 2220 W. Finally, the thickness of the thermal boundary at the top of the plate is 0.0063 m.
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1. Explain any one type of DC motor with a neat
diagram.
2. Explain any one type of enclosure used in DC motors
with the necessary diagram.
1. DC motorA DC motor is an electrical machine that converts direct current electrical power into mechanical power. These types of motors function on the basis of magnetic forces. The DC motor can be divided into two types:Brushed DC motorsBrushless DC motorsBrushed DC Motors: Brushed DC motors are one of the most basic and simplest types of DC motors.
They are commonly used in low-power applications. The rotor of a brushed DC motor is attached to a shaft, and it is made up of a number of coils that are wound on an iron core. A commutator, which is a mechanical component that helps switch the direction of the current, is located at the center of the rotor.
Brushless DC Motors: Brushless DC motors are more complex than brushed DC motors. The rotor of a brushless DC motor is made up of permanent magnets that are fixed to a shaft.
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IF an 85% efficient alternator operating at 1800RPM were putting
out 100kW of power how much torque would need tro be delivered by
the prime mover?
To determine the amount of torque that the prime mover would need to deliver to operate an 85% efficient alternator operating at 1800 RPM and putting out 100 kW of power, the following equation is used:Power = (2π × RPM × Torque) / 60 × 1000 kW = (2π × 1800 RPM × Torque) / 60 × 1000
Rearranging the equation to solve for torque:Torque = (Power × 60 × 1000) / (2π × RPM)Plugging in the given values:Torque = (100 kW × 60 × 1000) / (2π × 1800 RPM)≈ 318.3 Nm
Therefore, the prime mover would need to deliver about 318.3 Nm of torque to operate an 85% efficient alternator operating at 1800 RPM and putting out 100 kW of power. This can also be written as 235.2 lb-ft.
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A polymeric cylinder initially exerts a stress with a magnitude (absolute value) of 1.437 MPa when compressed. If the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 x10¹² Pa-s, respectively, what will be the approximate magnitude of the stress, in MPa, exerted by the spring after 1.8 day(s)? Answer Format: X.X Unit: MPa
The stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MP
a. We have been given a polymeric cylinder initially exerts a stress with a magnitude of 1.437 MPa
when compressed and the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 × 10¹² Pa-s respectively.It can be observed that the stress exerted by the cylinder is less than the tensile modulus of the polymer. Therefore, the cylinder behaves elastically.
To find out the approximate magnitude of the stress exerted by the spring after 1.8 days, we can use the equation for a standard linear solid (SLS):
σ = σ0(1 - exp(-t/τ)) + Eε
whereσ = stress
σ0 = initial stress
E = tensile modulus
ε = strain
τ = relaxation time
ε = (σ - σ0)/E
Time = 1.8 days = 1.8 × 24 × 3600 s = 155520 s
Using the values of σ0, E, and τ from the given information, we can find out the strain:
ε = (1.437 - 0)/16.5 × 10⁶ε = 8.71 × 10⁻⁸
From the equation for SLS, we can write:
σ = σ0(1 - exp(-t/τ)) + Eεσ
= 1.437(1 - exp(-155520/2 × 10¹²)) + 16.5 × 10⁶ × 8.71 × 10⁻⁸σ
= 1.437(1 - 0.99999999961) + 1.437 × 10⁻⁴σ ≈ 0.176 MPa
Thus, the stress exerted by the spring after 1.8 days is approximately 0.176 MPa.
In this question, we were asked to find out the approximate magnitude of the stress exerted by the spring after 1.8 days. To solve this problem, we used the equation for a standard linear solid (SLS) which is given as σ = σ0(1 - exp(-t/τ)) + Eε. Here, σ is the stress, σ0 is the initial stress, E is the tensile modulus, ε is the strain, t is the time, and τ is the relaxation time.Using the given values, we first found out the strain. We were given the initial stress and the tensile modulus of the polymer. Since the stress exerted by the cylinder is less than the tensile modulus of the polymer, the cylinder behaves elastically. Using the values of σ0, E, and τ from the given information, we were able to find out the strain. Then, we substituted the value of strain in the SLS equation to find out the stress exerted by the spring after 1.8 days. The answer we obtained was approximately 0.176 MPa.
Therefore, we can conclude that the magnitude of the stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MPa.
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