Thursday, November 30, 2017

Significance of Critical Pressure Ratio

*Consider a nozzle which connects vessel VA and VB. The vessel VA contains steam at high pressure P1. The vessel VB contains steam but its pressure P2 can be varied based on requirements.




*Initially the pressure P2 be equal to the pressure P1 in vessel VA. By that time, there is no flow through nozzle.

*If the pressure P2 is gradually reduced in vessel VB, the discharge through the nozzle increases gradually.

*The pressure P2, at which the discharge is maximum is called critical pressure.

*If the pressure P2 is reduced further, the discharge through the nozzle will not increase and it remains at constant level.




Hence it is confirmed that, at critical pressure ratio, the flow will be maximum.

*The velocity of steam at critical pressure is equal to sonic velocity. This condition will exist at throat cross section of the Convergent divergent nozzle.


Note;

*For convergent – Divergent nozzle, the flow of steam in the convergent portion of the nozzle is subsonic. i.e., the velocity of steam in the convergent section is less than the sound velocity.


*The steam flows in the divergent part of the nozzle is supersonic, i.e., the velocity of the steam in divergent portion is greater than velocity of sound.

Wednesday, November 29, 2017

Point of Contraflexure

# If the end portion of a beam extends beyond the support, the beam is known as overhanging beam.


Overhanging Beam

# In overhanging beams, the Bending Moment* is positive between the supports, whereas the Bending Moment is negative for overhanging portion.

# At some point on the beam, the B.M is zero after changing its sign from positive to negative or vice-versa.

# That point is known as the point of inflexion or point of contraflexure.

Note:


* Bending Moment (B.M): Algebraic sum of the moments of all the forces acting to the left or right of the section gives bending moment.

Saturday, November 25, 2017

Two Reversible Adiabatic Paths Cannot Intersect Each Other

# Let us consider two reversible adiabatic paths 1-3 and 2-3 intersect each other at point 3. 



# Also a reversible constant temperature processes 1-2 be drawn in such a way that it intersects the reversible adiabatic paths at 1 and 2.

# These three reversible processes 1-2, 2-3, 3-1 constitutes a reversible cycle.

# We know that the area under the p-v plot represents the net work output in a cycle. Therefore the area under the three reversible paths represents the net work output in a cycle.

# But such a cycle is not possible, since net work is being produced ia a cycle by a heat engine by exchanging heat with a single reservoir in the processes 1-2, violates the Law of Thermodynamics (Kelvin-Plank). Therefore the consideration of the intersection of the two reversible adiabatic process is wrong.

# Through one point, only one reversible adiabatic can pass.

# As two constant property lines can never intersect each other, it is observed that a reversible adiabatic path must denote some property, which is found later that it is entropy.

Sunday, November 19, 2017

Mach Number

# It is defined as the ratio of local velocity of fluid to the acoustic or sound velocity.
It is a dimensionless number designated by M.

It is expressed as, M = C/a Where, C = Velocity of fluid (air),
   a = Velocity of sound.
# Mach number can also be expressed as the square root of the ratio of inertia force
of flow to the elastic force.

M2 = (Inertia force of flow)/(Elastic force of fluid)
Different Regimes of Compressible flow
Subsonic Flow:
If the flow of fluid with a mach number less than 1, (i.e., C<a) then the flow is called
a subsonic flow. It is characterized by smooth streamlines.


Sonic Flow:
It the fluid flows with Mach number is equal to unity, (i.e., C=a) then the flow is
considered as sonic flow.


Supersonic flow:

If the Mach number is greater than 1 (i.e., C>a) then the flow is called
supersonic flow. It is observed with an oblique shock.


Hypersonic flow:

It the mach number of the flowing fluid is very high, (i.e., M>5) then the flow
is a hypersonic flow. This flow is observed with severe shock in the flow field. 


Note :
M > 5 --> Hypersonic flow

M > 1 --> Supersonic flow

M = 1 --> Sonic flow

M < 1  --> Subsonic flow



Saturday, November 18, 2017

COMBINED FIRST AND SECOND LAW OF THERMODYNAMICS

Combined 1st and 2nd Law of Thermodynamics.


By first law of thermodynamics, ΔQ = dU + ΔW

Since ΔW = pdV

ΔQ = dU + pdV ----------------- 1.

From second law of thermodynamics, (Entropy concept)

ΔQ = T. dS ----------------------- 2.

Put equation 2. in 1.

We get, TdS = dU + pdv ---------------- 3.

We know that, enthalpy h = u + pv

On differentiating, we get

dh = du + pdv + vdp,

From equation 3. , dh = Tds + vdp

Tds = dh - vdp -------------- 4.


The equations 3. and 4. are the thermodynamic equations relating the properties of system.


The following are the relations obtained from the first and second laws.

1. dQ = dE + dW: Holds good for all process, reversible or irreversible and for all systems.

2. dQ = dU + dW: Holds good for any process undergone by a closed system.

3. dQ = dU + pdV: It is good for a closed system , where pdV work is present. This relation true only for quasi-static* process.

4. dQ = TdS: This equation is true only for a reversible processes.

5. TdS = dH –Vdp: This relation hold good for any process, since there is no path function term in the equation. 

6. TdS = dU + pdV: It is good for any reversible or irreversible process, undergone by a closed system. Since the properties in the relation which are independent of the path. 


Note:

* A quasi-static process is a thermodynamic process that happens very slowly  for the system to be in equilibrium. It is reversible.

Friday, November 17, 2017

Nusselt Number (Nu)

Physical Significance of Nusselt Number


# Consider a layer of fluid of thickness L, difference in temperature  ΔT = T2-T1

Heat Transfer through a fluid layer

# Heat transfer through the fluid layer is convection when the fluid involves some motion.

# Heat transfer for this case is QConv = h . A . ΔT ---------1.


# On the contrary, heat transfer through the fluid layer is conduction when the fluid is not in motion.


Heat transfer for this case is QCond = k . A.  ΔT / (L ) --------2.


# By taking their ratio of equation 1. & 2.,


QConv / QCond = (h .  ΔT .  L) / (k .  ΔT)

                         = (h .  L) / (k )


                         = Nu, Nusselt number.



# Nusselt number represents the improvement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid.

# The larger the Nusselt number, the more efficient the convection.


# Nusselt number, Nu is unity for fluid layer represents heat flux across the layer by pure conduction.



Note;


k - Thermal Conductivity, W/m K

h - Convection heat transfer coefficient , W/m2 K






Thursday, November 16, 2017

Comparison of Otto, Diesel and Dual cycles

Based on Same Compression Ratio



# The comparison of Otto, Diesel and Dual cycle for the same compression ratio in p-v plot is shown below.
P-v plot

From p-v plot,

1247➡Otto cycle 


1267 ➨Diesel cycle


12357 ➡Dual cycle



# Here the compression ratio ‘r’ is equal for all cycles.

Compression ratio, r = V2/V1


Heat rejected is also equal for all cycles.


# We know that the efficiency of cyclic process is given by

η = 1- QR/ QS


QS - Heat supplied or absorbed


QR - Heat rejected


For constant heat rejection QR , the cycle efficiency increases with increases in the value of heat supplied, QS.


# In T-s plot, 


T-s Plot

The area under 1247 represents QS for Otto cycle


The area under 1267 represents QS for Diesel cycle

The area under 12357 represents QS for Dual cycle

Since,  QS)Otto > QS)Dual  > QS)Diesel ,

ηOtto > ηDual  > ηDiesel  , for same compression ratio



Note;


Area under p-v diagram represents Work done during the processes.


Area under T-s diagram represents heat transfer during the processes.


p = C ➡ Constant pressure processes.

v = C ➡ Constant volume processes.


PvƔ = C ➡ Isentropic processes.









Wednesday, November 15, 2017

Slope of constant volume and constant pressure processes in T-s plot

# For constant pressure processes(p=C),

ds = m . Cp . dT/T ---------------------- 1.



T-s Plot



# The slope of the curve in T-s plot is dT/dS

Therefore from equation 1.,  dT/dS = T/(m . Cp) ---------------- 2.


For 1 kg of perfect gas,  

Equation 2. becomes

dT/dS = T/Cp    {Since m = 1 kg} -------------------- 3.


# Similarly for constant volume processes(v=C),  

dT/dS = T/Cv --------------------- 4.



# For perfect gas, we know that

Cp > Cv 

or

1/Cp < 1/Cv  --------------------- 5.


# Compare equation 5. with equation 2. & 4.

T/Cp  <  T/Cv

or

{dT/dS} of v=C > {dT/dS} of p=C ----------------6.

Here {dT/dS} is slope of the curve in T-s plot.


# Therefore the slope of the curve for constant volume process (a-b) is higher than that of constant pressure processes (a-b’). 




Note ;

Constant volume process , v=C.

Constant pressure process , p=C.

Cp, Cv - Specific heats at constant pressure and constant volume.

ds - Change in entropy.


Tuesday, November 14, 2017

Prandtl Number (Pr)


It is the dimensionless parameter which gives the relation between the relative thickness of the velocity boundary layer and thermal boundary layer.

It is denoted by Pr.

Pr =  (Momentum diffusivity)/(Thermal diffusivity)

     = V

     = (µ x ρ x Cp )/(ρ x k )

Pr = (µ x Cp )/( k )


Here,

V - Kinematic viscosity, Stroke

μ - Dynamic Viscosity, kg/m-s

α - Thermal diffusivity, k/(ρ x Cp)

Cp - Specific Heat, J/kg K

k - Thermal conductivity, W/mK


Its value is more than 100,000 for heavy oils and less than 0.01 for liquid metals.

It is in the order of 10 for water. 

For gases, it varies from 0.004 to 1, which depicts that both momentum and heat dissipate through the fluid medium at same rate.

It is clear that, heat diffuses very quickly in liquid metals (Pr < 1) relative to momentum transfer. At the same time, as in case of oils, it is very slow. (Pr > 1)

Also, the thermal boundary layer is thicker for liquid metals and thinner for oils comparative to velocity boundary layer.





Note :

The dimensionless number is named after Ludwig Prandtl , who initially set up the concept of boundary layer in  1904 and give major contributions to boundary layer theory.














Monday, November 13, 2017

Nipping of Leaf Springs

In Leaf spring, the stresses in extra full-length leaves are 50 % more than that of  in graduated leaves.

It is important to equalize the stresses in all the leaves on leaf spring.

One of the method of equalizing the stresses is to pre-stress the spring.

This can be done by bending the leaves to different radii of curvature before the assembly.

The full length leaf is given with greater radius of curvature than the adjacent graduated leaves. The shorter leaves are given with smaller radius of curvature than the previous leaves.

The initial gap formed between the extra full length leaf and graduated leaf before the assembly called a ‘nip’.




A difference in radius of curvature is made for such pre-stressing is known as ‘nipping’.




Note;

The leaves are held together by using two U bolts and a center clip.

To keep the leaves in alignment, rebound clips are also provided in order to prevent lateral shifting of leaves.

These springs are used in railway wagons, cars, trucks etc..


Sunday, November 12, 2017

Unit of Refrigeration


Tonne of Refrigeration (TR) 

The amount of heat removed (i.e., refrigeration effect) from 1 tonne* of water from 0 °C to make 1 tonne of ice at 0 °C in 24 hours.

It can also be explained as the amount of heat extracted to make 1 tonne of ice from and at 0 °C in 24 hrs.

To make ice from water at 0 °C, it is important to remove latent heat**.

For water, latent heat = 335 kJ/kg.

    Therefore, one tonne of refrigeration = 1000 X 335 kJ in 24 hrs.

   =  (1000 X 335 )/24 kJ/hr
     
   = 13,958.3 kJ/hr

   =  (1000 X 335 )/(24 X 60) kJ/min

   = 232.6 kJ/min

In practice, 1 tonne of refrigeration is equivalent to 210 kJ/min or 3.5 kJ/Sec of heat.



Note:

*1 tonne is equivalent to weight of 1000 kg.

**Amount of heat transfer required to cause a change of phase in mass of a substance at constant temperature and pressure.