Explain with schematics the operating principle of solid state lasers.

Explain with schematics the operating principle of solid state lasers.

2 months ago

Solution 1

Guest Guest #9791789
2 months ago


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.  

📚 Related Questions

Explain the Otto cycle of a 4 stroke engine.
Solution 1


Otto cycle for 4 stroke engine:


1.Air is a working fluid it will behave like ideal gas.

2.Mass of air is constant(close system)

3.All process is reversible process.

4.Specific heat of air does not depends on temperature.

4 stroke engine is an internal combustion engine.It works on 4 processes like intake ,compression,power and heat exhaust.To complete one cycle ,piston moves from top dead center to bottom dead center two times.

From the Otto cycle

Process 1-2 is isentropic  compression.

Process 2-3 is heat addition.

Process 3-4 is isentropic expansion.

Process 4-1 is heat rejection.

Petrol engine works on Otto cycle.  

Efficiency of cycle [tex]\eta[/tex]


As temperature decreases a ductile material can become brittle - ductile-to-brittle transition
Solution 1


Yes ,at low temperature ductile material behaves like brittle.


Yes,as temperature decreases a ductile material can become brittle.

In metals ductile to brittle transition temperature is around 0.1 to 0.2 Tm(melting point temperature) and in ceramics  ductile to brittle transition temperature is around 0.5 to 0.7 Tm(melting point temperature) .

We can easily see that from graph between fracture toughness and  temperature.In the graph when temperature is low then the ductile material is behaving like brittle material.But when temperature is above a particular value then material behaves like ductile.


An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat rejection from the condenser) c. 3-4 Irreversible expansion d. 4-1 Evaporation (or) heat addition to the evaporator Sketch the processes on T-S diagram.
Solution 1



The deatailed diagram of VCRS is given below such

1-2=Isentropic compression in which temperature increases at constant entropy

2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)

3-4=Irreversible expansion or throttling in which enthalpy remains constant

4-1=Isobaric heat addition(Evaporation)

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.
Solution 1




[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



527.58 =[tex]m\times 0.71(200-50)[/tex]



[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])              


Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.
Solution 1


a)[tex]T_2=868.24 K[/tex] ,[tex]P_2=2.32 bar[/tex]

b) [tex]s_2-s_1=0.0206[/tex]KW/K


P=4200  KW ,mass flow rate=20 kg/s.

Inlet of turbine

 [tex]T_1[/tex]=807°C,[tex]P_1=5 bar,V_1=100 m/s[/tex]

Exits of turbine

 [tex]V_2=125 m/s[/tex]

Inlet of diffuser

[tex]P_3=1 bar,V_3=15 m/s[/tex]

Given that ,use air as ideal gas

R=0.287 KJ/kg-k,[tex]C_p[/tex]=1.005 KJ/kg-k

Now from first law of thermodynamics for open system at steady state


Here given that turbine is adiabatic so Q=0

Air treat ideal gas   PV=mRT, Δh=[tex]C_p(T_2-T_1)[/tex]

[tex]w=\dfrac{P}{mass \ flow\ rate}[/tex]


w=210 KJ/kg

Now putting the values

[tex]1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210[/tex]

[tex]T_2=868.24 K[/tex]

Now to find pressure

We know that for adiabatic [tex]PV^\gamma =C[/tex] and for ideal gas Pv=mRT

⇒[tex]\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}[/tex]

[tex]\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}[/tex]

[tex]P_2=2.32 bar[/tex]

For entropy generation



[tex]s_2-s_1=0.00103\times 20[/tex]KW/K

[tex]s_2-s_1=0.0206[/tex] KW/K

A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.
Solution 1


w =  -28.8 kJ/kg

q = 723.13 kJ/kg


Given :

Initial properties of piston  cylinder assemblies

Temperature, [tex]T_{1}[/tex] = -20°C

Quality, x = 70%

           = 0.7

Final properties of piston  cylinder assemblies

Temperature, [tex]T_{2}[/tex] = 180°C

Pressure, [tex]P_{2}[/tex] = 6 bar

From saturated ammonia tables at [tex]T_{1}[/tex] = -20°C  we get

[tex]P_{1}[/tex] = [tex]P_{sat}[/tex] = 1.9019 bar

[tex]v_{f}[/tex] = 0.001504 [tex]m^{3}[/tex] / kg

[tex]v_{g}[/tex] = 0.62334 [tex]m^{3}[/tex] / kg

[tex]u_{f}[/tex] = 88.76 kJ/kg

[tex]u_{g}[/tex] = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

[tex]v_{1}[/tex] = [tex]v_{f}[/tex]+x ([tex]v_{g}[/tex]-[tex]v_{f}[/tex]

                       = 0.001504+0.7(0.62334-0.001504)

                       = 0.43678 [tex]m^{3}[/tex] / kg

[tex]u_{1}[/tex] = [tex]u_{f}[/tex]+x ([tex]u_{g}[/tex]-[tex]u_{f}[/tex]

                       = 88.76+0.7(1299.5-88.76)

                       =936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

[tex]v_{2}[/tex] = 0.3639 [tex]m^{3}[/tex] / kg

[tex]u_{2}[/tex] = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

           [tex]w = \left (\frac{P_{1}+P_{2}}{2}  \right )\left ( v_{2}-v_{1} \right )[/tex]

           [tex]w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )[/tex]

           [tex]w = -28794.52[/tex] J/kg

                       = -28.8 kJ/kg

Now for heat transfer

[tex]q = (u_{2}-u_{1})+w[/tex]

[tex]q = (1688.2-936.27)-28.8[/tex]

          = 723.13 kJ/kg

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the piston until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.
Solution 1



w= -213.9 KW


[tex]V_1=0.3m^3,T_1=298 K[/tex]

[tex]V_2=0.1m^3,T_1=1273 K[/tex]

Mass of air=1 kg

For polytropic process  [tex]pv^n=C[/tex] ,n is the polytropic constant.



[tex]298\times .3^{n-1}_1=1273\times .1^{n-1}_2[/tex]


Work in polytropic process given as



Now by putting the values

w=[tex]1\times 0.287\dfrac{289-1273}{2.32-1}[/tex]

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)
Solution 1


Force on the bolt = 24.525 kN

Force on the 1st hinge = 8.35 kN

Force on the 2nd hinge = 16.17 kN



height = 2 m

width =1 m

depth of the door from the water surface = 1.5 m


[tex]\bar{y}[/tex] =1.5+1 = 2.5 m

Hydrostatic force acting on the door is

[tex]F= \rho \times g\times \bar{y}\times A[/tex]

[tex]F= 1000 \times 9.81\times 2.5\times 2\times 1[/tex]

         = 49050 N

         = 49.05 kN

Now finding the Moment of inertia of the door about x axis

[tex]I_{xx}=\frac{1}{12}\times b\times h^{3}[/tex]

[tex]I_{xx}=\frac{1}{12}\times1\times 2^{3}[/tex]

               = 0.67

Now location of force, [tex]y^{*}[/tex]

[tex]y^{*}=\bar{y}+\frac{I_{xx}}{A\times \bar{y}}[/tex]

[tex]y^{*}=2.5+\frac{0.67}{2\times 1\times 2.5}[/tex]

             = 2.634

Therefore, calculating the unknown forces

[tex]F=F_{A}+R_{B}+R_{C} = 49.05[/tex]  ------------------(1)

Now since [tex]\sum M_{R_{A}}=0[/tex]

∴ [tex]R_{B}\times L+R_{C}\times L-F\times \frac{1}{2}=0[/tex]

  [tex]R_{B}+R_{C}-F\times \frac{1}{2}=0[/tex]


  [tex]R_{B}+R_{C}=24.525[/tex]        -----------------------(2)

From (1) and (2), we get

[tex]R_{A} = 49.05-24.525[/tex]

                = 24.525 kN

This is the force on the Sliding bolt

Taking [tex]\sum M_{R_{C}}=0[/tex]

[tex]F\times 0.706-R_{A}\times 0.84-R_{B}\times 1.68 = 0[/tex]

[tex]49.05\times 0.706-24.525\times 0.84-R_{B}\times 1.68 = 0[/tex]

[tex]R_{B}[/tex] =8.35 kN

This is the reaction force on the 1st hinge.

Now from (1), we get

[tex]R_{C}[/tex] =16.17 kN

This is the force on the 2nd hinge.

A roller support acts like a contact boundary condition as it can produce a reaction force as a push response to a body but will not produce a pull force to hold a body from moving away. a)True b)- False
Solution 1


a) True


Roller can provide reaction for push support but it can not provide reaction for pull support.  

From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.

On the other hand hinge support can provide reaction in both the direction.

So we can say that roller can not proved reaction for pull support.

A pendulum has an oscillation frequency (T) which is assumed to depend upon its length (L), load mass (m) and the acceleration of gravity (g). Determine the relationship between oscillation frequency, length, load mass and acceleration of gravity. Differentiate as well which variable does not affect the oscillation frequency.
Solution 1


Mass does not affect oscillation frequency.                                                    


Let the bob of the pendulum makes a small angular displacement θ. When the pendulum is displaced from the equilibrium position, a restoring force tries to act upon it and it tries to bring the pendulum back to its equilibrium position. Let this restoring force be F.

Therefore, F = -mgsinθ  

Now for pendulum, for small angle of θ,


Therefore, F = -mgθ

Now from Newton's 2nd law of motion,

F = m.a = -mgθ

[tex]\Rightarrow m.\frac{d^{2}x}{dt^{2}} = - mg\Theta[/tex]

Now since, x = θ.L

[tex]\Rightarrow L.\frac{d^{2}\Theta }{dt^{2}}= -g\Theta[/tex]

[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}= -\frac{g}{L}.\Theta[/tex]

[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}+\frac{g}{L}.\Theta =0[/tex]

Therefore, angular frequency

 [tex]\omega ^{2}[/tex] = [tex]\frac{g}{L}[/tex]

ω = [tex]\sqrt{\frac{g}{L}}[/tex]

Also we know angular frequency is , ω = 2.π.f

where f is frequency


2πf = [tex]\sqrt{\frac{g}{L}}[/tex]

f = [tex]\frac{1}{2 \pi }\sqrt{\frac{g}{L}}[/tex]

So from here we can see that frequency,f is independent of mass, hence it does not affect frequency.