Friday, March 29, 2019

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 - P2V2γ-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


Tuesday, March 26, 2019

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~
IMG_1142.jpeg



example.
IMG_1143.jpg



~Internal Energy, u~

•isothermal
•isometric 
•isobaric            =    u=Q-W
•polytropic 
•adiabatic


~Enthalpy, H~
 =u+pv
u: internal energy 
p: pressure 
v: volume 


~Continuity Equation~
IMG_1144.jpg


~Formula~
IMG_1145.jpg

g: gravity 
Z: height 
p: pressure 
v: volume 
C: velocity 
m: mass flowrate 

IMG_1146.jpg


~Steam Power Plant~

IMG_1147.jpg
IMG_1148.jpg
IMG_1149.jpg


~Questions & Solutions~

IMG_1150.jpg
IMG_1151.jpg
IMG_1152.jpg
IMG_1153.jpg

Monday, March 25, 2019

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 
                                     ↓
                                Q> 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.
                                         

Sunday, March 24, 2019

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 = [(h- 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.






    Farah Sakinah bt Norazman
    13DEM18F1034




    Monday, March 18, 2019

    LAB



    Assalamualaikum everyone.
    I want to share about lab about thermofluids that have been do at 17March2019.  





    The lab is:
    PHYSICAL PROPERTIES OF FLUIDS
    Objective for this lab is to define properties of fluids.

    Fluids properties are intimately related to fluid behavior. It is obvious that different fluid can have grossly different characteristic. For example, gasses are light and compressible, whereas liquid are heavy and relatively incompressible.

    To quantify the fluid behaviour deferences, certain fluid properties are used. The fluid properties are mass density, specific weight, specific gravity and specific volume.


    Mass density, ρ is defined as the mass per unit per volume.(SI units, kg/m^3)
    Image result for mass density formula
    Specific weight, ω is defined as the weight per unit volume(SI units,N/M^30
    Image result for specific weight formula(where g=9.81m/s^2)


    Specific gravity or relative density,is the ratio of the weight of the substance to the weight of an equal volume of water at 4 celcius

    Image result for specific weight formula


    Specific volume, v is defined as the reciprocal of mass density. It is used to mean volume per unit mass.(SI unit, m^3/kg)


    Image result for specific volume formula


    Apparatus


    Image result for digital weight scale
    Digital weight scale

    Image result for beaker

    beaker

    we used 4 diferrent type of fluids


    Image result for 2t

    2t oil

    Image result for 4t oil
    4t oil

    Image result for sunlight soap
    sunlight

    Image result for minyak masak
    cooking oil


    Experimental procedure

    1. Measure weight of the empty container.write down the weight
    2. pour 200ml fluids 1 to the container. write down the weight
    3. repeat step from the start with 75ml, and 100ml of fluid 1 volume
    4. repeat the step again with fluid 2.






    Written by :MUHAMMAD DENIAL BIN ANANG SYAFRIZAL ANWAR
    (13DEM18F1027)








    CHAPTER 4 -1st LAW OF THERMODYNAMIC

    chapter 4 :1st Law of The Thermodynamics 
    March 18, 2019
    1st Law of Thermodynamic

    Hi and Assalamualaikum everyone!
    I want to share with u guys some knowledge that I had learnt,so let start's!



    The first thermodynamic law is the formulation of  a more general law of physics (the law of conservation of energy) for thermodynamic processes. The first law of thermodynamics is simply a statement of conservation of energy principle and it asserts that total energy is a thermodynamic property.  Energy can neither be created nor destroyed; it can only change forms.This principle is based on experimental observations and is known as the First Law of Thermodynamics.

    Steady flow process:            ➤ potential  
    (Bernoulli'Theorem)                 ➤ kinetic
                                                    ➤ internal energy 
                                                    ➤ displacement
                                                    ➤ heat
                                                    ➤ work

    Application:                           ➤ boiler
                                                    ➤ condenser 
                                                    ➤ turbine 
                                                    ➤ nozzle
                                                    ➤ throttling
                                                    ➤ pump

    Work(W) and Heat Transfer(Q)

    ∑W=∑Q
    W1+W2=Q1+Q2

    ➣ When work in,means the value of work is negative(-) and when work out,its value is
         positive(+).
    ➣ When heat transfer(Q) in,means the value of Q is positive(+) and when Q out,its
           value is negative(-).

    Internal energy,U

    ΔU=ΔQ-ΔW

    Non-flow process:-
    1.Isothermal,(T constant)
    2.Isometric,(V constant)
    3.Isobaric,(P constant)
    4.Polytropic,(PVn constant)
    5.Adiabatric,(Q=0)

    Entalphy,H

    H=U+PV
    U=internal energy
    P=pressure
    V=volume

    Formula for the continuity equation:-





    Steady Flow Energy Equation





    So that'all i can share with u guys,hope u're understand :))

    DITULIS OLEH:NUR HAZWANI ATIRAH BT ZULKAWI (13DEM18F1036)



    Sunday, March 17, 2019

    Friday, March 15, 2019

    ISOMETRIC, ISOBARIC, POLYTROPIC AND ADABARIC PROCESS


    ISOMETRIC PROCESS 
    -An isochoric process, also called a constant-volume process, an isovolumetric process, or an isometric process, is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant. An isochoric process is exemplified by the heating or the cooling of the contents of a sealed, inelastic container: The thermodynamic process is the addition or removal of heat; the isolation of the contents of the container establishes the closed system; and the inability of the container to deform imposes the constant-volume condition. The isochoric process here should be a quasi-static process.



    ISOBARIC PROCESS 
    -An isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer.
    The term isobaric comes from Greek iso, meaning equal, and baros, meaning weight.
    POLYTEOPIC PROCESS 
    -An ideal isothermal process must occur very slowly to keep the gas temperature constant. An ideal adiabatic process must occur very rapidly without any flow of energy in or out of the system. In practice most expansion and compression processes are somewhere in between, or said to be polytropic.

    ADABATIC PROCESS
    -An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. This puts a constraint on the heat engine process leading to the adiabatic conditionshown below. This condition can be used to derive the expression for the work done during an adiabatic process.





    (VIDEO FOR THIS TOPIC)

     part 1: Work and isobaric processes 



    part 2: Isothermal, isometric, adiabatic processes


    MUHAMMAD NURAMMAR BIN AZAMEE
    (13DEM18F1025)