Low hardness is not a characteristic property of ceramic material.
Answer: Option D
Explanation:
One of the most important properties of ceramic material is the hardness that the material displays. The hardness owes to the joining of brittle fracture and plastic flow that makes the material to defend against penetration.
The hardness can be tested with the Vickers test. The hardness of ceramic makes its dominance in the use of construction purpose and manufacture of products. Other general properties are less conductivity, high melting temperature, etc.
Not a characteristic property of ceramic material (d) low hardness.
(a) High temperature stability is a defining feature of ceramics, as they can withstand extreme temperatures without significant degradation.
(b) High mechanical strength is another key characteristic, making ceramics suitable for applications where strength and durability are essential.
(c) Low elongation refers to the minimal deformation under stress, which is typical of ceramics due to their brittle nature.
(d) However, low hardness is not a characteristic property of ceramics. Ceramics are known for their high hardness, making them resistant to wear and abrasion.
A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator absorbs heat.
Answer:
275 Kelvin
Explanation:
Coefficient of Performance=11
[tex]T_H=\text {Absolute Temperature of high temperature reservoir=300 K}[/tex]
[tex]T_L=\text {Absolute Temperature of low temperature reservoir}[/tex]
[tex]\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}[/tex]
What is refrigeration capacity and what is meant by a "ton" of refrigeration?
Answer:
1 ton refrigeration =3.517 kJ/s = 3.517 kW
Explanation:
Refrigeration capacity is defined at the measure of the effective cooling capacity of a refrigerator which is expressed in Btu per hour or in tons.
1 ton capacity is a unit of air conditioning and refrigeration which measure the capacity of air conditioning and refrigeration unit.
One ton is equal to removal of 3025kcal heat per hour
1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP
Horizontal wind turbines have same design for offshore and on shore wind farms. a)-True b)- False
Answer: False
Explanation: Horizontal axis wind turbines are usually used for generation of the electric power on the off-shore. The generation of horizontal-axis wind turbine works well when it is installed away from the shore because it supports large sized wind turbines so that they can generate high amount of electricity.They are usually not preferred for the on-shore wind farms because they can have small sized wind turbines only.Therefore the statement given is false.
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twice of that of loose side, F1= 2F2. Please calculate the F1, Fe, and Fo.
Answer:
F₁ = 1500 N
F₂ = 750 N
[tex]F_{e}[/tex] = 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
[tex]F_{max}[/tex] = 3 x [tex]F_{e}[/tex]
where [tex]F_{e}[/tex] is centrifugal force
[tex]F_{e}[/tex] = [tex]F_{max}[/tex] / 3
= 1500 / 3
= 500 N
Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed Mold c)- Preforming d)- Postforming e)- None of the above f)- All the above
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
What is Euler's equation?
Answer:
[tex]e^{ix} = cosx + i sinx[/tex]
Explanation:
In mathematics, Euler's formula is an equation in complex analysis, that gives a relationship between an exponential factor and the trigonometric functions.
The Euler equation is:
[tex]e^{ix} = cosx + i sinx[/tex]
Here,
e - base of the natural logarithm
i - imaginary unit
x - argument given in radians
sin , cos - trigonometric functions sine and cosine respectively.
Describe the importance of ferrite and austenite stabilizing elements in steels
Answer:
The importance of ferrite and austenite stabilizing elements in steels .
Explanation:
Alloying -
The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .
The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .
Stabilizing austenite -
The process by which temperature is increased , in which Austenite exists .
Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .
Hence, the elements are -
Cobalt , Nickel , Manganese , Copper.
The examples of the Austenitic steels are -
Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.
Stabilizing ferrite –
The process by which temperature is decreased , in which austenite exists .
Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.
Hence, the elements are -
Aluminium , Silicon , Tungsten , Chromium , Molybdenum , Vanadium
The examples of the Ferritic steels are -
F-Cr alloys , transformer sheets steel ( 3% Si ).
The manufacturing department of a factory is the only department that needs to be practicing Lean Manufacturing philosophy. a)- True b)- False
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the above
Answer:
Explanation:
load = 4500lb lift height= 30 ft
time =15 s
velocity=[tex]\frac{30}{15}[/tex] ft/s
velocity=2 ft/s
power = force[tex]\times[/tex] velocity
power=[tex]{4500}\times2[/tex]
power= 9000 lb ft/s
1 hp= 550 lb ft/s
power= [tex]\frac{9000}{550} =16.36[/tex] hp
Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature rise due to friction?
Answer:
The surface temperature increases when two bodies are rubbed against each other due to friction.
Explanation:
No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.
Friction is a resistance experienced by the two bodies when they are moved against each other.
The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.
Thus, resulting in rise in the surface temperature of the two bodies.
Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.
To select the center of a line what option must be selected? A. Ortho B. Dynamic input C. Object snap
Answer:
The correct answer is c. object snap
Explanation:
In Autocad, the object snap is defined as a drawing aid that is used together with other different commands to help to draw accurately. It also allows snapping onto a specific object location when there is a picking point. and thus, it helps to select the center of a line.
What is a truss? What separates a truss from a frame and other forms of rigid bodies?
Answer:
The application of force is the main difference between truss and frame.
Explanation:
Truss:
Truss is a collection of beams,which use to handle the tensile and compression loads . That collection of beams creates rigid structure.
The load on the truss will be acting always at the at the hinge. Truss is widely used in the construction areas.
Frame:
Like truss, it is also a combination of beams and used to handle the loads. The main difference between truss and frame is the application of load. In the frame load can apply at the any point of the member of frame along with hinge.
Truss are connected by pin joint and can not transfer moment ,on the other hand frames are connected by rigid joint like welding so frame can transfer moment.
Truss and frame both forms a rigid structure and is used in the construction areas.
In vibration analysis, can damping always be disregarded?
In vibration analysis, damping cannot always be disregarded. This is especially the case when the system is excited near the resonance frequency.
A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- False
Answer:
The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is
b) False
Explanation:
Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.
Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.
How are fluids distributed and transported in a fluid power system?
Answer and explanation :
Fluid distribution is a new technique to produce and to transmit power from one place to other its play a major role in power distribution it is a process of using fluid (any type of fluid as oil or water ) under pressure to generate to control or to transmit
fluid power system is divided into two types
Hydraulic fluid power systempneumatic fluid power system
What are the qualities required for a refrigerant? What is meant by the term "flash chamber?"
Answer:
A good quality refrigerant should be eco friendly.
Explanation:
A refrigerant is a substance that can extract and transfer heat from body to another body or medium.
The desirable qualities required for a refrigerant are :
1. A refrigerant should not deplete ozone layer.
2. A good quality refrigerant should have a low boiling point.
3. It should also have a low melting point.
4. Thermal conductivity of the refrigerant should be high for fast heat transfer.
5. It should have low specific heat.
6. It should have high latent heat.
7. It should have low vapour density.
8. Refrigerant should have high critical pressure and temperature.
9. It should have high enthalpy of vapourization for maximum heat absorption.
10. Refrigerants should not be toxic in nature and non flammable.
11. It should have high coefficient of performance for the working temperature range.
12. It should be easily available and cheap.
Flash Chamber :
A flash chamber in the refrigeration system is also know as the mixing chamber. It is normally used in multistage refrigeration system and is placed in between the expansion valve and the evaporator.
The flash chamber sends only the liquid refrigerant to the evaporator by seperating the liquid from the vapour refrigerant in order to increase the efficiency.
there are totally (___) slip system in an FCC metal.
Answer: 12
Explanation: In FCC metal lattice there are total four octahedral slip plane and six direction where each are common to two octahedral plane and so that gives total of three slip direction . So the total slip system is multiplication of the slip plane and the slip direction that is twelve.
Slip system=Slip plane×Slip direction
Slip system =4×3=12
Therefore there are total 12 slip system in the FCC metal lattice.
In an adiabatic process, the temperature of the system remains constant. a)- True b)- False
Answer:
A. True
Explanation:
The uniform wall thickness that is usually targeted for plastic injection molded parts is roughly: A. 0.5 mm B. 3 mm C. 7 mm D. 12 + mm
Answer:
B. 3 m
Explanation: For plastic injection moulded the thickness is generally between 2 mm to 3 mm
the wall is not too thick because during cooling process there should be defects so thickness of wall is no too high and there is also a problem if we use thicker wall that we need more material for moulding process so the thickness should be in between 2 to 3 mm which is in option B so option B will be the correct option
Thermosetting polymers are polymers that becomes soft and pliable when heated. ( True , False )
Answer:
Thermosetting polymers are polymers that becomes soft and pliable when heated is false
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 500C at a rate of 0.065 kg/s and leaves at 40°C. The refrigerant enters the evaporator at 20°C with a quality of 23 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine: (a) The degrees of subcooling of the refrigerant in the condenser, b)-The mass flow rate of the refrigerant . (c) The heating load and the COP of the pump, and d)-The theoretical minimum power input to the compressor for the same heating load.
Answer:2.88
Explanation:
What are the major limitations of melt spinning method? a)-good for limited alloys and limited shapes b)-good for limited shapes, high cost c)-good for limited thickness and limited shapes d)-high cost, high energy consumption
Answer:d
Explanation:
Melt spinning is the economical process to manufacture synthetic fiber.
Major limitations of Melt spinning is
only suitable for thermoplastic fiberHigh heat inputRequire periodic maintenance of machinePrecise control of temperature is must.Although it is very efficient method as it does not create any environment Pollution
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) has the following dimensions: ID = 150mm, OD = 250 mm and thickness = 37 mm. What is the rotational speed in rpm that would lead to the flywheel's fracture?
Answer:
N = 38546.82 rpm
Explanation:
[tex]D_{1}[/tex] = 150 mm
[tex]A_{1}= \frac{\pi }{4}\times 150^{2}[/tex]
= 17671.45 [tex]mm^{2}[/tex]
[tex]D_{2}[/tex] = 250 mm
[tex]A_{2}= \frac{\pi }{4}\times 250^{2}[/tex]
= 49087.78 [tex]mm^{2}[/tex]
The centrifugal force acting on the flywheel is fiven by
F = M ( [tex]R_{2}[/tex] - [tex]R_{1}[/tex] ) x [tex]w^{2}[/tex] ------------(1)
Here F = ( -UTS x [tex]A_{1}[/tex] + UCS x [tex]A_{2}[/tex] )
Since density, [tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{M}{A\times t}[/tex]
[tex]M = \rho \times A\times t[/tex][tex]M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t[/tex]
[tex]M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37[/tex]
[tex]M = 8252963901[/tex]
∴ [tex]R_{2}[/tex] - [tex]R_{1}[/tex] = 50 mm
∴ F = [tex]763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}[/tex]
F = 33618968.38 N --------(2)
Now comparing (1) and (2)
[tex]33618968.38 = 8252963901\times 50\times \omega ^{2}[/tex]
∴ ω = 4036.61
We know
[tex]\omega = \frac{2\pi N}{60}[/tex]
[tex]4036.61 = \frac{2\pi N}{60}[/tex]
∴ N = 38546.82 rpm
A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.06 ft*lb b. 19.24 ft*lb ? c)-31.17 ftlb d) 119.07 ftlb e)-9.92 ft*lb
Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
Power: Rate of change of work with respect to time is called power.
We know that P=[tex]Torque\times speed[/tex]
[tex]\omega=\frac{2\pi N}{60}[/tex] rad/sec
So that P=[tex]\dfrac{2\pi NT}{60}[/tex]
So 3281.08=[tex]\dfrac{2\pi \times 2329\times T}{60}[/tex]
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb
Fluid enters a device at 4 m/s and leaves it at 2 m/s. If there is no change in the PE of tihe flow, and there is no heat and (non-flow) work across boundaries of the device, what is the increase in specific enthalpyof the fluid (hg-hi) in kJ/kg? Assume steady state operation of the device.
Answer:
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
Explanation:
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here given Q=0 ,w=0
So [tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]
[tex]V_1=4 m/s,V_2=2 m/s[/tex]
[tex]h_1+\dfrac{4^2}{2000}=h_2+\dfrac{2^2}{2000}[/tex]
[tex]h_2-h_1=6\times 10^{-3}[/tex]
So increase in specific enthalpy
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
Discuss the importance of dust and fluff removal from spinning mills and how it is realised in air conditioning plants
Answer:
Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.
Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc. Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.
The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.
An oscillating mechanism has a maximum displacement of 3.2m and a frequency of 50Hz. At timet-0 the displacement is 150cm. Express the displacement in the general form Asin(wt + α).
Given:
max displacement, A = 3.2 m
f= 50 Hz
at t = 0, displacement, d = 150 cm = 1.5 m
Solution:
Displacement in the general form is represented by:
d = Asin(ωt ± α)
d = 3.2sin(2πft ± α)
d = 3.2sin(100πt ± α)
where,
A = 3.2 m,
ω = 2πf = 100π
Now,
at t = 0,
1.5 = 3.2sin(100π(0) ± α )
1.5 = 3.2sinα
sin α = [tex]\frac{1.5}{3.2}[/tex] = 0.4687
α = [tex]sin^{-1}(0.46875)[/tex] = 27.95° = 0.488 radian
Now, we can express displacement in the form of 'Asin(wt + α)' as:
d = 3.2sin(100πt ± 0.488 )
________is the frequency that an engineered system fails, expressed for example, in failures per hour.
Answer:
Failure Rate
Explanation:
Failure rate is the frequency with which an engineered system or component fails, expressed for example in failures per hour.
It is often denoted by the Greek letter λ (lambda) and is important in reliability theory.
Failure rate is usually time dependent, and an intuitive corollary is that the rate changes over time versus the expected life cycle of a system
A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop of 9.375 V across it, what is the value of R2? (a) 2000Ω (b) 1500Ω (c) 1000Ω (d) 500Ω
Answer:
1500Ω
Explanation:
Given data
voltage = 15 V
total Resistance = 4000Ω
potential drop V = 9.375 V
To find out
R2
Solution
we know R1 +R2 = 4000Ω
So we use here Ohm's law to find out current I
current = voltage / total resistance
I = 15 / 4000 = 3.75 × [tex]10^{-3}[/tex] A
Now we apply Kirchhoffs Voltage Law for find out R2
R2 = ( 15 - V ) / current
R2 = ( 15 - 9.375 ) / 3.75 × [tex]10^{-3}[/tex]
R2 = 1500Ω
Air enters a compressor at 100 kPa, 10°C, and 220 m/s through an inlet area of 2 m2. The air exits at 2 MPa and 240°C through an area of 0.5 m2. Including the change in kinetic energy, determine the power consumed by this compressor, in kW.
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]
[tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
[tex]h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W[/tex]
We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]and we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate= [tex]density\times area\times velocity[/tex]
So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
=1.23×200×2 Kg/s
=541.17 Kg/s
[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]
[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
[tex]V_{2}[/tex]=80.07 m/s
Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]
So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]
[tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]
Now by putting the values
[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW