Tuesday, March 3, 2015

Thermodynamics

                                                            Thermodynamics

PERSONAL NOTES::

1 st Law Of Thermodynamics

Concepts::

~Energy conservation
~Energy can not be  created nor destroyed but can change its form.
~Heat is a form of energy
~Absolute  values of internal energies cant be calculated
~If heat is added to a system then it has a + sign convection
~If work is done by the system it has a -  sign convection

Gibbs Phase Rule

It determines minimum no of intensive variables that must be arbitarirly fixed to fix all the other properties iof the system.
The phse ruke variables are the Temperature, Pressure and the Composition.

At F=0 system in invarient, Then maximum no of pahses that can exist in equilibrium in given by P= 2+N

At equilibrium driving forces are zero

Chemical potential causes mass to flow across phases rather concentraction difference

                                  PVT BEHAVIOR OF PURE FLUIDS

When  a fuel is burned under adiabatic conditions in the presence of oxygen or air , the maximum temp. attained by the system is called the adiabatic flame temperature.
Max temp. is attainable when the fuel  is burned in the theoretically amount of pore oxygen.
Combustion in the presence of excess air or the oxygen will have adiabatic flame temp less than the maximum attainable theoretical adiabatic flame temp.
Flame temperature calculation are applicable for the exothermic reactions only.


The terms B/V, B/V2 etc of the virial expansion depends upon molecular interaction in ideal gases such interaction do not exist

When Papproches to 0  volume increses and term like B/v approaches to less contribution
Hence as P approaches 0  Z approaches 1

Internal energy is a function of temperature and pressure.
But for an ideal gas it is the function of Temperature only
For an ideal gas cp, cv, Δu, Δ H are all functions of functions of temperature only

                                           2nd Law Of Thermodynamics

Concepts

Heat is a less useful form of energy in contrast to the forms like electrical, mechanical or work
Work can be converted into other forms of energy and neglecting the frictional losses the                    conversion can approach to the efficiency of 100%
It is impossible by a cyclic process to convert heat absorbed by the system to convert completely        into work done
No engine can have efficiency greater than that of Carnot engine operating between same temperature levels.
The thermal efficiency of the Carnot engine depends only on the temperature levels and not  upon the       working substance of the engine.
Coefficient of performance (COP) of heat pumps can never be infinity.

Entropy
The entropy change of a heat reservoir is always given by Q/T
Entropy change is a state function and change is identical whether the change is bought about by a      reversible or irreversible processes


Appliactions of Thermodynamics
Turbines
Turbines uses superheated steam as working fluid exoands to convert internal energy as shaft work of the turbine.
Work of a turbine is
Ws= H2-H1
Isentropic work of a turbine is the maximum work obtainable from expansion process of a aturbine
Efficiency, n= Ws/w(isentropic)=ΔH/ΔH isentropic
Efficiency of an expander is in the range 0.7 to 0.8
Isentropic expansion results in greater change in enthalpy and hence more work
Isentropic expansion is proceede at constant entropy
Calculations are first made for isentropic expansion and then taking into consideration efficiency for actual expansio calculations

Monday, March 2, 2015

Bracket Support Solved Example::

::Problem~01::

Design a bracket support for a vertical cylinder vessel with the help of following data :

  1. Diameter of vessel = 3 m
  2. Height of vessel = 20 m
  3. Clearance from vessel bottom to foundation = 2.0 m
  4. weight of vessel with contents = 5 Tons
  5. Wind pressure = 120-150 kg/m2
  6. No. of brackets = 4 m
  7. Diameter Of anchor bolt circle = 1.65 m
  8. Height of bracket from foundation = 3.0 m
  9. permissible steel structure :
  • Tension = 14.0 kg/cm2
  • Compression = 1240 kg/cm2
  • Bending = 1500 kg/cm2

  1. Permissible bearing pressure
  • From concrete = 38 kg/cm2
View Solution



Saturday, February 14, 2015

Condenser Thermal Design:: Solved Examples

Problem.1

28000 kg/hr of pure ethyl alcohol at 2 psi is to be condensed by water from 85 degree F to 120.
A dirt factor of 0.003 should be provided. A pressure drop at 2 psi is allowable for vapour and 10 psi for the water. Calculate the required size of a 1-2 horizontal condenser using 1 inch OD, 14 BWG, Tubes 16 ft long on 1 1/4 inch triangular pitch.

View solution::

Problem.2

30000 kg/hr of pure ethyl alcohol at 2 psi is to be condensed by water from 85 degree F to 125 Degree F. A dirt factor of 0.004 should be provided. A pressure drop at 2 psi is allowable for the vapour and 10 psi for the water.

Calculate the required size of 1-2 horizontal condenser using 1 inch OD 14 BWG, tubes 16 ft long on 1 1/4 inch triangular pitch.

View solution::

Problem.3

A horizontal 1-2 condenser is required for the condenser of 27200 kg/hr of substantially pure n-propyl alcohol coming from top of distilling column operating at 15 psig at which pressure it boils at 118 Degree C. Water at 29.5 degree C will be used as the cooling medium. A dirt factor of 0.003 is required with allowable pressure drops of 2.0 psi for the vapour pressure and 10 psi for the water.

Assume that 8 ft tubes are used. Tubes are to be 3/4 Inch OD. 16 BWG on 15/16 Inch triangular pitch. Assume fan tube passes and the maximum baffle spacing. Design the condenser.

View solution::

Rotary Drier Design Example

Problem.1 

A rotary drier using counter-current flow is to be used to dry 12000 kg/hr of wet salt containing 5% water (wet basis). Heated air at 147 Degree C with 50 Degree C wet bulb temp is available. The specific heat of the salt is 0.21. The outlet temp of air and salt are 72 degree C and 93 degree C respectively. Calculate the length and diameter of the drier required.

DATA::

Humidity of entering air = 0.05 kg water/kg dry air
Mass velocity of air = 5000 kg/hr-m2
Spe. Heat of salt = 0.21 kcal/kg. C

View Solution::


Thursday, February 12, 2015

Routh Problem 6 Solution Page


Routh Stability Test :: Solved Examples

Problem.01.

Examine the stability of the characteristics equation :

s5 + 2 s4 +4 s3 +8 s2 +3 s + 1 = 0

using routh's criteria.

View Solution::


Problem.02.

For the system given Below write down the characterictics equation with the help of the routh test, comment on the stability for kc = 4  . Also find the value of kc for which the system is unstable












View Solution::

Problem.03.

The open loop transfer function of the control system is given as ::

           G(s) =    Kc(s + 1)      
                     S (s + 2) (s + 3)
Problem.04.

By the means of Routh test determine the stability of the system when kc = 2.0










Problem.05.

For the control system in Fig, determine the value of K above which the system is unstable.

Determine the value of K for which two of the roots are on the imaginery axis and determine the values of these imaginery roots and the remaining roots.




Problem.06.

A control system is represented by the means of a block diagram as shown in the fig.

Comment on the stability of the system.


Problem.07.

Consider the following feedback control system and make stability analysis with routh criterion, Find the critical stability conditions when Ţi  = 0.1

S3 + 2 s 2(2 + Kc) s + Kci = 0

View Solution::