Explain with schematics the operating principle of solid state lasers.

Answer:

Otto cycle for 4 stroke engine:

Assumptions:

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]

[tex]\eta=\dfrac{W_{net}}{Q_{supply}}[/tex]

Answer:

Yes ,at low temperature ductile material behaves like brittle.

Explanation:

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.

Answer:

Explanation:

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)

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]

Answer:

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

Explanation:

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

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]

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]

[tex]w=\dfrac{4200}{20}[/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=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}[/tex]

[tex]s_2-s_1=0.00103[/tex]KJ/kg_k

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

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

Answer:

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

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

Answer:

n=2.32

w= -213.9 KW

Explanation:

[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]Tv^{n-1}=C[/tex]

  [tex]T_1v^{n-1}_1=T_2v^{n-1}_2[/tex]

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

n=2.32

Work in polytropic process given as

       w=[tex]\dfrac{P_1V_1-P_2V_2}{n-1}[/tex]

      w=[tex]mR\dfrac{T_1-T_2}{n-1}[/tex]

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.

Answer:

Force on the bolt = 24.525 kN

Force on the 1st hinge = 8.35 kN

Force on the 2nd hinge = 16.17 kN

Explanation:

Given:

height = 2 m

width =1 m

depth of the door from the water surface = 1.5 m

Therefore,

[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}=\frac{F}{2}[/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.

Answer:

a) True

Explanation:

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.

Answer:

Mass does not affect oscillation frequency.                                                    

Explanation:

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,

sinθ[tex]\simeq[/tex]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

Therefore

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.