Boyle's law

Boyle's law, sometimes referred to as the Boyle–Mariotte law, or Mariotte's law,, is an experimental gas law that describes how the pressure of a gas tends to increase as the volume of the container decreases. A modern statement of Boyle's law is

The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remaiunchanged within a closed system. 

Mathematically, Boyle's law can be stated as

Pressure is inversely proportional to the volume.

or

Pressure multiplied by volume equals some constant .

where P is the pressure of the gas, V is the volume of the gas, and k is a constant. 

The equation states that the product of pressure and volume is a constant for a given mass of confined gas and this holds as long as the temperature is constant. For comparing the same substance under two different sets of conditions, the law can be usefully expressed as

The equation shows that, as volume increases, the pressure of the gas decreases in proportion. Similarly, as volume decreases, the pressure of the gas increases. The law was named after chemist and physicist Robert Boyle, who published the original law in 1662.

SECOND LAW OF THERMODYNAMICS

1. Isobaric ( P is constant)

Constant pressure on a T- s

Work done:

W = P(V2 - V1)

or

W= mR(T2-T1)

Unit : kJ

Heat flow:

Q = mCp ln (T2-T1)

Unit : kJ

The change of entropy is

s2 - s1 = mCp ln (T2/T1)

Unit : kJ/K

s2 - s1 = Cp ln (T2/T1)

Unit : kJ/kgK

2. Isometric (V is constant)

Constant volume on a T-s diagram

Heat flow =

Q = mC(T2-T1)

Unit : kJ

Change of entropy:

s2 - s1 = mCv ln (T2/T1)

Unit : kJ/K

s2 - s1 = Cv ln (T2/T1)

Unit : kJ/kgK

3. Isothermal (T is constant)

Constant temperature process on a T-s diagram

(the shaded area represents the heat supplied during the process)

Change of entropy:

s2 - s1 = R ln (v2/v1) = R ln (p1/p2)

Unit : kJ/kgK

s2 - s1 = mR ln (v2/v1) = mR ln (p1/p2)

Unit : kJ/K

4. Adiabatic 

Reversible adiabatic process on T-s diagram

from the non-flow equation:

W = mCv(T1 - T2)

or since, Cv = R/ γ-1

 W = mR(T1-T2)/γ-1

Or since PV = mRT

W= P1V1 - P2V2/ γ-1

Temperature, Pressure  and Volume for perfect gasses:
T2/T1 = (P2/P1)^γ-1/γ = (V1/V2)^γ-1

5. Polytropic Process

Reverse poly tropic process on a T-s diagram

Work done:

W = P1V1 - P2V2/ n-1

or since PV = mRT

W = mR(T1- T2)/ n-1

Change of entropy:

U2 - U1 = mCv(T2-T1)

Heat Flow:

Q = W + U2 - U1

Written by: Cahaya Athirah binti Mohammad Taufik

Ist LAW OF THERMODYNAMICS

1st LAW OF THERMODYNAMIC

~Energy Conseruatrun~

1.Sound/vibrations 

   • frequency

2.  Chemical energy 

   • food

3.  Nuclear 

   • electric 

4.   Light

   • sun

5.  Mechanical energy 

   • engine 

~Work and Heat Transfer~

example.

~Internal Energy, u~

•isothermal

•isometric 

•isobaric            =    u=Q-W

•polytropic 

•adiabatic

~Enthalpy, H~

 =u+pv

u: internal energy 

p: pressure 

v: volume 

~Continuity Equation~

~Formula~

g: gravity 

Z: height 

p: pressure 

v: volume 

C: velocity 

m: mass flowrate 

~Steam Power Plant~

~Questions & Solutions~

2nd LAW OF THERMODYNAMICS

The Second Law of Thermodynamics

Although the net heat supplied in a cycle equal to the net work done, the gross heat supplied must be greater than the work done; some heat must always be rejected by the system.

        In symbols, 

Q1 - Q2 = W

↓

Qin < Qout

Heat Engine


                            QH - QL = W 
                                     ↓
                                QH  > W

The work producing device that best fits into the definition of a heat engine are:

• The close cycle gas turbine
• The steam power plant

Thermal efficiency

Example

Heat is transferred to heat engine from a furnace at a rate of 252 GJ/hr. If rate of waste heat rejection to a nearby river is 162 GJ/hr, determine the net work done and the thermal efficiency for this heat engine. 

Solution :

                        Thankyou 😊


By : Nurul Faessza Binti Misbah
     (13DEM18F1028)

EXAMPLE QUESTION FIRST LAW OF THERMODYNAMICS PART 2

3. A nozzle is supplied with steam having a specific enthalphy of 2780KJ/KG at the rate of
    9.1KG/MIN. At outlet from the nozzle the velocity of the steam is 1070M/S. assuming that the
    inlet velocity of the steam is negligible and that the process is adiabatic , determine :
 
  a) the specific enthalphy of the steam at the nozzle exit.
  b) the outlet area required if the final specific volume of the steam is 18.75 m3/KG.

         h1 = 2780 KJ/KG.            A2, v2 = 18.75 m3/KG
          m = 9.1 KG/MIN
         c2 = 1070 M/S
         c1 = 0
          Q = 0

SOLUTION :

a)  Q-w = [(h2-h1) + (c2^2-c1^2/2)+ g (z2-z1)]m

           0 = [h2-h1+ c2^2/2] m
           0 = h2 - 2780 (10^3) + 1070^2/2
           0 = h2 - 2207550
          h2 = - 2207550 - 0
               = - 2207.55 KJ/KG.

b)  m = c2A2/v2
   
          A2 = mv2/c2
          A2 = (0.152) (18.75) / 1070
                = 2.66 (10^-3) m2.
                                         

EXAMPLE QUESTION FIRST LAW OF THERMODYNAMIC

1) Steam flows through a turbine stage at the rate of 4500kJ/h. The steam velocity at inlet and outlet are 15m/s and 180m/s respectively. The rate of heat energy flow from the turbine casing to the surroundings is 23kJ/kg of steam flowing. If the specific enthalpy of the steam decrease by 420kJ/kg in passing through the turbine stage. Calculate the power developed.

 ḿ = 4500 kg/h
 c1 = 15 m/s
 c2 =180 m/s
 Q = 23 kJ/s
 ⃤h= -420 kJ/kg

SOLUTION:

 Q-W = [(h2 - h1) + (C2² - C1² )] ḿ

                                                2                                   
         
           (23 x 10³) - w [(-420 x 10³) + (180² - 15²)] (1.25)
                                                                   2

           -w = -(420000) + (16087.5)(1.25) - 23 x 10³ 
           -w = -527.89kw
            w = 527.89kw


2) A rotary pump draws 600kg/hour of atmospheric air and dalivers it at a higer pressure. The specific enthalpy of air at the pump inlet is 300kJ/kg and that at the exit is 509kJ/kg. The heat lost from the pump casing is 5000w. Neglecting the changes in kinectic and potential energy determine the power required to drive the pump.

 ḿ = 600 kg/h
 h1 = 300 kJ/kg
 h2 = 509 kJ/kg
 Q = 5000 w
 w = ?

SOLUTION:

           Q-W = [(h2-h1)] ḿ
           5000 - w = [(509 x 10³ - 300 x 10³)] (0.166)
                      w = 34694 - 5000
                      w = -29.694w

APPLICATION OF STEADY FLOW EQUATION

Let's get started!

The steady flow energy equation is written as

or,in the easy way the equations becomes

and with the flow rate,the equation may be written as

• In Boiler 

In a boiler operating under steady conditions,water is pumped into the boiler along the feed line at the same rate as which  steam leaves the boiler along the steam main, and heat energy is supplied from the furnace at a steady rate.

• In Condensers

If  the condenser is in a steady state then the amount of liquid,usually called condensate,leaving the condenser must be equal to the amount of vapour entering the condenser.

• In Turbine

A turbine is a device which uses a pressure drop to produce work energy which is used to drive an external load.

• In Nozzle

A nozzle utilises a pressure drop to produce an increase in the kinetic energy of the fluid.

In Throttling process 

A throttling process is one in which the fluid is made to flow through a restriction, e.g. a partially opened valve or orifice,causing a considerable drop in the pressure of the fluid.

• In Pump

In applying the steady flow energy equation to a pump,exactly the same arguments are used as for turbine.

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Farah Sakinah bt Norazman

13DEM18F1034