Answer Explanation:
CASCADE REFRIGERATION SYSTEMS :
cascade refrigeration systems are commonly used in the liquefaction of natural gas and some other gases
cascade refrigeration system is also known as freezing system. It uses two types of refrigerants having different boiling points.this system is employed to obtain temperature of -40 to -80It allows stable ultra low temperature operationexamples of cascade refrigeration system: LNG (Liquefied natural gas ) plants mostly used cascade refrigeration system
ADVANTAGES OF CASCADE REFRIGERATION SYSTEMS:
energy is saved because the system allows the use of refrigerant gasesthe running cost is inexpensive repair is easyA cascade refrigeration system utilizes multiple linked refrigeration cycles to achieve ultra-low temperatures not possible with a single cycle. It features a cascade control mechanism for precise temperature management, using heat exchangers to interconnect and sequentially cool each stage.
Explanation:Understanding Cascade Refrigeration SystemsA cascade refrigeration system is a complex refrigeration strategy that employs multiple refrigeration cycles linked together to achieve cooling. Typically, these systems are utilized in environments requiring very low temperatures, which cannot be efficiently or cost-effectively achieved through a single refrigeration cycle. The essence of a cascade system lies in its deployment of two or more refrigeration units in series, each referred to as a "stage," with each successive stage operating at a lower temperature range than its predecessor.
The interconnection between these stages involves a heat exchanger, where a refrigerant from one cycle cools the condenser of the next lower temperature cycle. This process is crucial for achieving the desired low temperatures. One common application is in industrial refrigeration, where maintaining precise low temperatures is critical for process efficiency and product quality.
A key feature of cascade refrigeration systems is the use of cascade control, which enhances system responsiveness to temperature changes and maintains tight control over the process. This control mechanism consists of a master-slave relationship, where the output of one control loop (master) serves as the setpoint for the next (slave), creating a highly integrated control environment. Such setup enables rapid adjustment to changing operating conditions, ensuring the system swiftly responds to disturbances while maintaining the desired temperature levels.
The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump has 98% efficiency, what would have been the enthalpies at the inlet and outlet if the pump was 100% efficient?
Answer:500,551.02
Explanation:
Given
Initial enthaly of pump \left ( h_1\right )=500KJ/kg
Final enthaly of pump \left ( h_2\right )=550KJ/kg
Final enthaly of pump when efficiency is 100%=[tex]h_2^{'}[/tex]
Now pump efficiency is 98%
[tex]\eta [/tex]=[tex]\frac{h_2-h_1}{h_2^{'}-h_1}[/tex]
0.98=[tex]\frac{550-500}{h_2-500}[/tex]
[tex]h_2=551.02KJ/kg[/tex]
therefore initial and final enthalpy of pump for 100 % efficiency
initial=500KJ/kg
Final=551.02KJ/kg
A Mariner vessel, floating at a draft of 23'6", has a GM of 1.5 feet which does not meet the required GM standard. How far above the keel must 1,400 tons be loaded to increase the GM to 2.0 feet?
Answer:
0.5 feet
Explanation:
it is given that the martin floats at draft of 23'6"
GM=1.5
The load is given as follows
1400 tons is loaded as 2 feet above keel
1400 tons-----kg----2 feet
final kg = [tex]\frac{final moment }{final dispacement}[/tex]
[tex]\frac{weight}{1400 kg}[/tex] = [tex]\frac{kg}{2 feet}[/tex] = [tex]\frac{moment of keel }{2800}[/tex]
final kg = [tex]\frac{2800}{1400}[/tex]=2 feet
final GM =2 feet-1.5 feet
=0.5 feet
If you add 10 J of heat to a system so that the final temperature of the system is 200K, what is the change in entropy of the system? a)-0.05 J/K b)-0.30 J/k c)-1 J/K d)-9 J/K e)-2000 J/K
Answer:
0.05 J/K
Explanation:
Given data in question
heat (Q) = 10 J
temperature (T) = 200 K
to find out
the change in entropy of the system
Solution
we will solve this by the entropy change equation
i.e ΔS = ΔQ/T ...................1
put the value of heat Q and Temperature T in equation 1
ΔS is the enthalpy change and T is the temperature
so ΔS = 10/200
ΔS = 0.05 J/K
What is the principle of operation of a mechanical dynamometer?
The mechanical dynamometer is an instrument used to measure forces or to calculate the weight of objects. The traditional dynamometer, invented by Isaac Newton, bases its operation on the stretching of a spring that follows the law of elasticity of Hooke in the measurement range. Like a scale with elastic spring, it is a spring scale, but it should not be confused with a scale of saucers (instrument used to compare masses).
These instruments consist of a spring, generally contained in a cylinder that in turn can be inserted into another cylinder. The device has two hooks or rings, one at each end. The dynamometers have a scale marked on the hollow cylinder that surrounds the spring. When hanging weights or exerting a force on the outer hook, the cursor of that end moves on the external scale, indicating the value of the force.
The dynamometer works thanks to a spring or spiral that has inside, which can be lengthened when a force is applied on it. A point or indicator usually shows, in parallel, the force.
Internal flow is one in which the flow is not bounded. a) True b) False
False it is External
Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False
Answer: False
Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.
If I add 30J of heat to a system so that the final temperature of the system is 300K, what is the change in entropy of the system? a)-1 J/K b)- 3 J/K c)- 1 J/K d)- 9 J/K e)- 10 J/K
Answer:
0.1 J/K
Explanation:
entropy change equation is as followed:
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
where ΔS=entropy change
Q=Heat transfer
T= temperature
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
[tex]\Delta S=\frac{30}{300}[/tex]
[tex]\Delta S=0.1 J/K[/tex]
hence the change in entropy of system which is [tex]\Delta S[/tex]is equal to 0.1 J/K
A closed system contains propane at 35°c. It produces 35 kW of work while absorbing 35 kW of heat. What is process? the temperature of the system after this process.
Answer:
35°c
Explanation:
Given data in question
heat = 35 kw
work = 35 kw
temperature = 35°c
To find out
temperature of the system after this process
Solution
we know that first law of thermodynamics is Law of Conservation of Energy
i.e energy can neither be created nor destroyed and it can be transferred from one form to another form
first law of thermodynamics is energy (∆E) is sum of heat (q) and work (w)
here we know
35 = 35 + m Cv ( T - t )
35-35 = m Cv ( T-t )
T = t
here T = final temperature
t = initial temperature
it show final temp is equal to initial temp
so we can say temp after process is 35°c
Air is heated from 50 F to 200 F in a rigid container with a heat transfer of 500 Btu. Assume that the air behaves as an ideal gas. Determine the volume of air [ft3] if the initial pressure is 2 atm. Also show the process on a P-v state diagram. Use the following temperature conversion: T[R] = T[F] + 460.
Answer:
[tex]V=68.86ft^3[/tex]
Explanation:
[tex]T_1[/tex] =10°C,[tex]T_2[/tex] =93.33°C
Q=500 btu=527.58 KJ
[tex]P_1= 2atm[/tex]
If we assume that air is ideal gas PV=mRT, ΔU=[tex]mC_v(T_2-T_1)[/tex]
Actually this is closed system so work will be zero.
Now fro first law
Q=ΔU=[tex]mC_v(T_2-T_1)[/tex]+W
⇒Q=[tex]mC_v(T_2-T_1)[/tex]
527.58 =[tex]m\times 0.71(200-50)[/tex]
m=4.9kg
PV=mRT
[tex]200V=4.9\times 0.287\times (10+273)[/tex]
[tex]V=1.95m^3[/tex] ([tex]V=1m^3=35.31ft^3[/tex])
[tex]V=68.86ft^3[/tex]
A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of 300 KNm, as shown. The modulus of rigidity G for the steel is 80 GPa. Determine: (a) The maximum shearing stress in the shaft. (b) The shearing stress on a traverse cross section at the inside surface of the shaft (c) The magnitude of the angle of twist for a (L) -2.5 m length.
Answer:
a. [tex]\tau=51.55 MPa[/tex]
b.[tex]\tau=42.95MPa[/tex]
c.[tex]\theta=7.67\times 10^{-3}[/tex] rad.
Explanation:
Given: [tex]D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa [/tex]
We know that
[tex]\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}[/tex]
J for hollow shaft [tex]J=\dfrac{\pi (D_o^4-D_i^4)}{64}[/tex]
(a)
Maximum shear stress [tex]\tau =\dfrac{16T}{\pi Do^3(1-K^4)}[/tex]
[tex]K=\dfrac{D_i}{D_o}[/tex]⇒K=0.83
[tex]\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}[/tex]
[tex]\tau=51.55 MPa[/tex]
(b)
We know that [tex]\tau \alpha r[/tex]
So [tex]\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}[/tex]
[tex]\dfrac{51.55}{\tau}=\dfrac{210}{175}[/tex]
[tex]\tau=42.95MPa[/tex]
(c)
[tex]\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}[/tex]
[tex]\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}[/tex]
[tex]\theta=7.67\times 10^{-3}[/tex] rad.
A slider of mass 0.25 kg on a string, 0.5 m long is rotating around a pivot on a frictionless table. The velocity of the slider is initially 0.05 m/s. When the string is pulled into a radius of 0.125 m how fast is the mass spinning?
Answer:
0.025 m/sec
Explanation:
we have given m =0.25 kg
velocity=0.05 m/sec
radius =0.5 meter
the centrifugal force produced due to rotational motion
[tex]F_c=\frac{mv^2}{r}=\frac{0.25\times 0.05^2}{0.5}=0.00125 N[/tex]
now again using this equation for finding the final velocity
[tex]0.00125=\frac{mv_{final}^2}{r}=\frac{0.25v_{final}^2}{0.125}[/tex]
[tex]v_{final}=\sqrt{\frac{0.00125\times 0.125}{0.25}}=0.025\ m/sec[/tex]
so the final speed of mass spring will be 0.025 m/sec
0il with a relative density of 0,8 flows in a pipe of diameter 60 mm. A venturi meter having a throat diameter of 35 mm is installed in the pipeline. The pressure difference is measured with a mercury manometer. The levels of the manometer differ by 22 mm. The venturi meter has a discharge coefficient of 0,98. Calculate the flow rate of the oil.
Answer:
the flow rate of the oil is 2.5 m³/s
Explanation:
Given data
relative density (S) = 0.8
diameter (d1) = 60 mm = 0.06 m
diameter (d2) = 35 mm = 0.035 m
height (h) = 22 mm = 0.022 m
discharge coefficient (Cd) = 0.98
To find out
the flow rate of the oil
solution
we know the formula for rate of flow i.e.
flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] / [tex]\sqrt{a1^{2} a2^{2} }[/tex] ...............1
here first we find area a1 and a2 i.e.
a1 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.06² = 0.002827 m²
a2 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.035² = 0.000962 m²
and now we find n = (density of mercury / density of oil) - 1 × h
n = ((13.56 / 0.8) - 1) × 0.022 = 0.3509
put all these value in equation 1
flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] / [tex]\sqrt{a1^{2} a2^{2} }[/tex]
flow rate = 0.98× 0.002827× 0.000962 [tex]\sqrt{2*9.81*0.3509}[/tex] / [tex]\sqrt{0.002827^{2} 0.000962^{2} }[/tex]
flow rate = 2.571386 m³/s
What is the overall transfer function for a closed-loop system having a forward-path transfer function of 5/(s + 3) and a negative feedback-path transfer function of 10?
Answer:
transfer function T(S)=[tex]\frac{5}{S+53}[/tex]Explanation:
NEGATIVE FEEDBACK TRANSFER FUNCTIONnegative feedback control of the amplifier is achieved by applying output voltage signal back to inverting input terminal by feedback
transfer function is T=[tex]\frac{g}{1+Gh}[/tex]
where G=forward Path gain
H=negative feedback gain
here G=[tex]\frac{5}{S+3}[/tex]
H=10
T(S)=[tex]\frac{G}{1+GH}[/tex]
=[tex]\frac{5}{S+53}[/tex]
What is considered a method for inducing heat transfer? (1) -heat power and convection (2)-preseribed temperature (3)-radiation (4)-thermal insulation (5)-prescribed strain
Answer: (1) heat power and convection
(3)radiation
Explanation: Heat can be transferred in many different ways such as conduction,radiation form and convection etc.
Convection is a method of transferring of the heat from a particular surface by the help of fluids .E.g.- air
Radiation is the method of transfer of heat by the emission or absorption process in the other surface.E.g.- earth getting warm due to sun.
Therefore the answer to the question is option (1) and (3).
The Manufacturing sector is the only sector where Lean manufacturing philosophy can be applied. a)- True b)- False
Answer:
b). false
Explanation:
Lean manufacturing
Lean manufacturing, a philosophy developed by Toyota Production System are means to eliminate wastes. They are defined as the techniques or management activities in eliminating wastes and increasing the efficiency inside an organisation.
According to the concept of lean manufacturing, mainly seven types of wastes are identified. They are :
1. Transportation waste
2. Inventory waste
3. Over production
4. Waiting
5. Defects
6. Motion waste
7. Non utilized talent
All these waste affect greatly to the efficiency of an organisation and devalue its services. Lean manufacturing advises to prevent all these waste in order to increase the productivity.
All the management activities and techniques used in lean manufacturing may be different according to the business application but they are all based on the same basic principle of removing wastes and errors and increase efficiency.
The different sectors that are benefiting from lean manufacturing methodology are
Healthcare
Hospitality
Food and Beverage
Government
Manufacturing
Lean manufacturing can be used in different sectors.
How is heat transfer defined in an internally reversible process
Answer:
Heat transfer for a internally reversible process.
Explanation:
Internally reversible means that there is entropy generation ' with in ' the system.
Heat transfer of a process is considered to be reversible if it occurs because of any minute temperature difference between the surrounding and the system .
Let us consider an example ,
Transferring of the heat across the difference in temperature of 0.0001 °C appears as more reversible than for the difference in temperature of 100 °C .
Hence ,
By heating or cooling a system for a number of infinitesimally small steps , we can approximate a reversible process.
A horizontal pipe is fitted with a nozzle. The inlet diameter of the nozzle is 40 mm and the outlet diameter is 20 mm. The flow rate in the pipe is 1.2 m3 /min and water density is 1000 kg/m3 . Determine the force exerted by the nozzle on the water.
Answer:
969.68N
Explanation:
d₁=0.04 m A₁=[tex]\frac{\pi d^2_{1} }{4}[/tex]
[tex]A_{1} =\frac{\pi \times .04^2}{4}= 0.00125m^{2} \[/tex]
d₂=0.02 m A₂=[tex]\frac{\pi d^2_{2} }{4}[/tex]
[tex]A_{2} =\frac{\pi \times .02^2}{4}= 0.00031m^{2} \[/tex]
Q=1.2m³/min Q=1.2/60=0.02m³/s
using continuity equation
Q₁=A₁v₁
v₁=Q₁/A₁=0.02/0.00125=16m/s
Q₂=A₂v₂
v₂=Q₂/A₂=0.02/0.00031=64.5m/s
[tex]F_{inlet}=\rho A_{1}v_1^{2}[/tex]
[tex]F_{inlet}=1000\times 0.00125\times16^{2}=320N[/tex]
[tex]F_{outlet}=\rho A_{2}v_2^{2}[/tex]
[tex]F_{outlet}=1000\times 0.00031\times64.5^{2}=1289.68N[/tex]
Force on the nozzle=F_{outlet}-F_{inlet}
= 1289.68-320
=969.68N
That the larger volume of chimney will enhance natural convection is due to (a) Higher thermal conductivity (c) Larger radiation surface area (b) Increase in the buoyancy force (d) Increase in the volume expansion coefficient
Answer:
Out of the four options provided, the most accurate answer is
option b) increase in the buoyancy force
Explanation:
Natural convection is a process in which thermal expansion of fluid takes place naturally due to natural buoyancy resulting in motion of fluid when it is heated.
Differences in densities result in buoyancy and natural convection depends on buoyancy force. Also higher air temperatures are found at lower densities which is found at the outlets of the channels and the larger the channel size, the larger is the buoyancy force (as the density difference will be higher).
An aluminum electrical cable is 20 mm in diameter is covered by a plastic insulation (k = 1 W/m-k) of critical thickness. This wire is placed in a room with an air flow heat transfer coefficient of 50 W/m^2-K. Compared to the bare aluminum wire, the heat loss from this insulated wire will be a) LESS b) MORE c) SAME d) ZERO
Answer:
the heat loss from this insulated wire is less
Explanation:
Given data in question
diameter of cable (d) = 20 mm
( K ) = 1 W/m-k
heat transfer coefficient (h) = 50 W/m²-K
To find out
the heat loss from this insulated wire
solution
we will find out thickness of wire
heat loss is depend on wire thickness also
we have given dia 20 mm
so radius will be d/2 = 20/ 2 = 10 mm
Now we find the critical thickness i.e.
critical thickness = K / heat transfer coefficient
critical thickness = 1 / 50 = 0.02 m i.e. 20 mm
now we can see that critical thickness is greater than radius 10 mm
so our rate of heat loss will be decreasing
so we can say our correct option is (a) less
A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False
Answer:
True
Explanation:
For point in xz plane the stress tensor is given by[tex]\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right][/tex]
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane
Takt time is the rate at which a factory must produce to satisfy the customer's demand. a)- True b)- False
Answer: a)True
Explanation: Takt time is defined as the average time difference between the production of the two consecutive unit of goods by the manufacturer and this rate is matched with the demand of the customer. This is the time which is calculated to find the acceptable time for which the goods unit must be produced by the factory to meet the needs of the customer. Therefore , the statement is true that takt time is the rate at which a factory must produce to satisfy the customer's demand.
The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03
Answer:
Answer is c 0.500
Explanation:
[tex]SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}[/tex]
We know that [tex]\rho_{water}=62.42lb/ft^{3}[/tex]
Applying values we get
[tex]SpecificGravity=\frac{31.2}{62.4}=0.5[/tex]
A heat engine operates between a hot reservoir at 2000°C and the atmosphere (cold reservoir) at 25°C. it produces 50 MW of power while rejecting 40 MW of waste heat. Determine the maximum possible thermal efficiency of the engine in percent.
Answer:
55.56%
Explanation:
Given data
Temprature of hot reservior =2000°c=2273k
Temprature of Cold reservior=25°c=298k
Power produced by engine=50MW
Heat rejected =40MW
we know that Effeciency(η) of heat engine=[tex]\frac{Work produced}{heat supplied}[/tex]
Also we know that
heat supplied[tex]\left ({Q_s} \right )=work produced{W}+Heat rejected{Q_r}[/tex]
Q_s=50+40=90MW
η=[tex]\frac{W}{Q_s}[/tex]
η=[tex]\frac{50}{90}[/tex]
η=55.55%
What is the most common type of pump?
Answer:
The most common type of pumps are Positive displacement and Non positive displacement pumps.
Explanation:
Pumps are two types:
(A) Positive displacement pump
(a)Gear pump
(1) Ge rotor pumps
(2)Internal gear pumps
(3)Lobe pumps
(4) External gear pumps
(b)Piston pump
(1)Radial piston
(2)Axial piston
(c)vane pump
(B) Non positive displacement pump
(a) Centrifugal pump
Give reasons why the control of dimensional tolerances in manufacturing is important.
Answer:
Tolerance is important for the very fact that providing proper tolerances ensures proper fittings of different parts.
Explanation:
Tolerance and dimensioning is an important link between manufacturing and engineering.
Tolerance optimization leads to high cost of machined part to be produced and also provides good quality product. Whereas loose tolerance means reduction in cost but poor quality product. hence it is very important and critical to provide the right tolerance while designing a product.
Tolerance also influence what type of production processes to be selected by the process planners. The optimization of the tolerances during the design phase has a positive impact on the results coming out of the manufacturing processes.
Providing proper tolerances ensures that the parts will fit properly.
Therefore, providing proper tolerances, the engineers shares the responsibility to manufacture the parts correctly.
Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.
Answer:
Velocity component in x-direction [tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex].
Explanation:
v=3xy+[tex]x^{2}[/tex]y
We know that for incompressible flow
[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]
[tex]\frac{\partial v}{\partial y}=3x+x^{2}[/tex]
So [tex]\frac{\partial u}{\partial x}+3x+x^{2}=0[/tex]
[tex]\frac{\partial u}{\partial x}= -3x-x^{2}[/tex]
By integrate with respect to x,we will find
[tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex]+C
So the velocity component in x-direction [tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex].
In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle temperature is 2031K. Determine the thermal efficiency.
Answer:
[tex]\eta[/tex]=0.60
Explanation:
Given :Take [tex]\gamma[/tex]=1.4 for air
[tex]P_1=100 KPa ,T_1=300K[/tex]
[tex]\frac{V_1}{V_2}[/tex]=r ⇒ r=16
As we know that
[tex]T_2=T_1(r^{\gamma-1})[/tex]
So [tex]T_2=300\times 16^{\gamma-1}[/tex]
[tex]T_2[/tex]=909.42K
Now find the cut off ration [tex]\rho[/tex]
[tex]\rho=\frac{V_3}{V_2}[/tex]
[tex]\frac{V_3}{V_2}=\frac{T_3}{T_2}[/tex]
[tex]\rho=\frac{2031}{909.42}[/tex]
[tex]\rho=2.23[/tex]
So efficiency of diesel engine
[tex]\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}[/tex]
Now by putting the all values
[tex]\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}[/tex]
So [tex]\eta[/tex]=0.60
So the efficiency of diesel engine=0.60
Describe the slip mechanism that enables a metal to be plastically deformed without fracture.
Answer and explanation:
Deformation means change in position plastic deformation mainly cause due to motion of dislocation
THERE ARE MAINLY TWO MECHANISM BY WHICH PLASTIC DEFORMATION TAKES PLACE
SLIPTWINNINGSLIP : slip is a process of sliding of blocks over one another along the planes these planes are called slip planes slip takes place when the shear stress exceeds than the critical value of stress distance between the slip planes are called slip lines the resistance for slip plane is very less as compared to any other planes the slip plane is the plane has very high density
Explain the reasons for abandoning a well.
Answer:
explained below
Explanation:
An Abandoned well is well no longer in use or in such a state of despair that ground water can no longer be pumped out of it in useable quantity.
Following are few reasons for abandoning wells:
1. When the level ground water level falls down the well becomes redundant. And in recent times the ground water level has fallen to appreciable magnitude.
2. Wells represent potential conduits or pathways for surface contaminants to reach ground water supply.The ground water contamination at your well is likely to show up in municipal water supply.
3. Moreover, if the well is unused it can cause physical hazard to people and animals living nearby. As these well grow vegetation around them thus hiding their hole.
Explain with schematics the operating principle of solid state lasers.
Explanation:
A solid state laser contains a cavity like structure fitted with spherical mirrors or plane mirrors at the end filled with a rigidly bonded crystal. It uses solid as the medium. It uses glass or crystalline materials.
It is known that active medium used for this type of laser is a solid material. This lasers are pumped optically by means of a light source which is used as a source of energy for the laser. The solid materials gets excited by absorbing energy in the form of light from the light source. Here the pumping source is light energy.