TRC2200:  Thermo  Fluids  &  Power  Systems  

LECTURE  10

Introduc)on  &  Basic  Concepts  of  Thermodynamics  

Lecturer:  Alpha  Agape  Gopalai

• Basic concepts to form a sound foundation for the development of the principles of thermodynamics.

•  Explain the basic concepts of thermodynamics such as system,

state, state postulate, equilibrium, process, and cycle.

•  Review concepts of temperature, temperature scales, pressure,

and absolute and gage pressure.

•  Energy •  Introduce an intuitive systematic problem-solving technique.

Lecture  Outline

• Many  form  of  Energy  

Energy

•  In  thermodynamic  analysis,  total  energy  of  a  system  can  be  classified  into  two  groups:  

•  Macroscopic  forms  of  energy    •  Microscopic  forms  of  energy    

Energy

Energy:  Macroscopic Macroscopic  energy:  Forms  of  energy  that  are  possessed  by  a  system  with  respect  to  an    outside  reference  frame.  

Kine)c  Energy  (KE)    

Poten)al  Energy  (KE)    

𝐾𝐸= 1/2 𝑚𝑉2  where      :  m  =  mass  of  a  body  (kg)    

       V  =  velocity  of  the  body  (m/s)  

𝑃𝐸=𝑚𝑔𝑧  where      :  m  =  mass  of  a  body  (kg)    

       g  =  gravita)onal  accelera)on  (m/s2)          z  =  eleva)on  (m)  

Energy:  Microscopic • Microscopic  energy:  Forms  of  energy  that  are  related  to  the  molecular  structure  of  a  system  and  the  degree  of  molecular  ac)vity.  They  are  independent  of  outside  reference  frames.    

 

Sensible  energy:  Kine)c  energy  of  the  molecules.    

Transla)onal  energy     Rota)onal    kine)c  energy    

Vibra)onal  kine)c  energy    

Spin  energy    

Sum  of  all  microscopic  forms  of  energy  is  referred  as  the  internal  energy  of  a  system  (U)  

Energy:  Microscopic •  Latent  energy:  Internal  energy  associated  with  the  phase  of  a  system.    

•  Chemical  energy:  Internal  energy  associated  with  the  atomic  bonds  in  a  molecule.  

•  Nuclear  energy:  Internal  energy  associated  with  the  strong  bonds  within  the  nucleus  of  the  atom.  

Solid   Liquid   Gas  

• We  define  the  total  energy  in  a  system  as:  

• We  can  also  represent  total  energy  as  specific  energy  (per  unit  mass)      

Total  Energy  of  a  System

E =U +KE +PE =U +mV 2

2+mgz

E =me =m u+ ke+ pe( )

E =m(u+V2

2+ gz)

Where    e  =  total  energy  of  a  system  per  unit  mass  (kJ/kg)        ke  =  Kine)c  energy  per  unit  mass  (kJ/kg)    

   pe  =  Poten)al  energy  per  unit  mass  (kJ/kg)        u  =  internal  energy  per  unit  mass  (kJ/kg)  

Total  Energy  of  a  System

Closed  system  Most  closed  systems  remains  sta)onary  during  a  process  à  No  change  in  their  kine)c  and  poten)al  energies  

E =me =m u+ ke+ pe( )

•  Open  system    Open  system  involves  fluid  flow  for  long  periods  of  )me.  It  is  commonly  expressed  in  term  of  energy  flow  rate  (𝐸 )  à  incorpora)ng  mass  flow  rate  ( 𝑚 )  

m = ρV = ρAcVavg    

 ρ  =  fluid  density        V  =  volume  flow  rate    

   Ac  =  Cross-­‐sec1onal  area  of  flow          Vavg  =  the  average  flow  velocity  normal  to  Ac  

•  Energy  can  be  contained  or  transferred  

•  In  a  closed  system  Energy  can  be  transferred  by  •  Heat  • Work  

•  In  a  control  volume  Energy  can  be  transferred  by  •  Heat  • Work  • Mass  

Total  Energy  of  a  System

Mechanical  Energy Mechanical  energy  systems  are  designed  to  transport  fluid  from  one  loca)on  to  another  at  a  specified  flow  rate,  velocity  and  eleva)on  difference.  

Such  systems  may  generate  mechanical  work  in  a  turbine  or  it  may  consume  mechanical  work  in  a  pump/  fan  during  this  process.  

We  can  analyze  these  systems  by  considering  the  mechanical  forms  of  energy  only  and  the  fric)onal  effect  that  cause  the  mechanical  energy  to  be  lost.  

Mechanical  Energy Mechanical energy: The form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine. Kinetic and potential energies: The familiar forms of mechanical energy.

•  A  pump  transfers  mechanical  energy  to  a  fluid  by  raising  its  pressure  •  A  turbine  extracts  mechanical  energy  from  a  fluid  by  dropping  its  pressure  

•  Therefore,  the  pressure  of  a  flowing  fluid  is  also  associated  with  mechanical  energy  

 •  Pressure  on  its  own  is  not  a  form  of  energy  

•  Pressure  ac)ng  on  a  fluid  through  a  distance  produces  work  •  Flow  work  ,  which  can  be  calculated  as    

•  Therefore  

Mechanical  Energy

Mechanical energy of a flowing fluid per unit mass

Mechanical  Energy

Rate of mechanical energy of a flowing fluid

For  a  system  with  fluids  flowing,  we  know  that  

Therefore,  

Mechanical  Energy

Mechanical  Energy

•  16

Mechanical energy change of a fluid during incompressible flow per unit mass

Rate of mechanical energy change of a fluid during incompressible flow

Example

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AssumpIon:  Let’s  assume  wind  flows  steadily  –  at  a  constant  speed  of  8.5  m/s  

Understanding:  A  site  with  a  specified  wind  speed  is  considered.  Wind  energy  per  unit  mass,  for  a  specified  mass,  and  for  a  given  mass  flow  rate  of  air  is  to  be  determined.  

Analysis:  Wind  flow  (energy)  is  being  converted  into  mechanical  energy.  The  only  harvestable  form  of  energy  of  atmospheric  air  is  kine)c  energy,  which  is  captured  by  the  wind  turbine.  

Wind  energy  per  unit  mass  𝑒=𝑘𝑒=   𝑉↑2 /2 = 8.5↑2 /2 =𝟑𝟔.𝟏  𝑱/𝒌𝒈    

Example

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Wind  energy  for  an  air  mass  of  10kg    𝑒=𝑘𝑒=   𝑉↑2 /2 = 8.5↑2 /2 =36.1  𝐽/𝑘𝑔    𝑬=𝒎𝒆=𝟑𝟔𝟏  𝑱    

Wind  energy  for  a  mass  flow  rate  of  1154  kg/s    𝑒=𝑘𝑒=   𝑉↑2 /2 = 8.5↑2 /2 =36.1  𝐽/𝑘𝑔    𝑬 = 𝒎 𝒆=𝟒𝟏.𝟕  𝒌𝑾    

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3-­‐3  ENERGY  TRANSFER  BY  HEAT Heat: The form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference.

Energy is recognized as heat transfer only as it crosses the system boundary.

During an adiabatic process, a system

exchanges no heat with its surroundings.

Heat transfer per unit mass

Amount of heat transfer when heat transfer rate changes with time

Amount of heat transfer when heat transfer rate is constant

Energy  Transfer  by  Heat • AdiabaIc  process:  A  process  during  which  there  is  no  heat  transfer  

•  It  takes  place  when    •  The  system  is  well  insulated    •  System  and  surroundings  are  at  the  same  temperature,  thus  there  is  no  temperature  difference  for  heat  transfer  

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Energy  Transfer  by  Heat Heat  is  transferred  by  three  mechanisms:    •  Conduction: The transfer of energy from

the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles.

•  Convection: The transfer of energy between a solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion.

•  Radiation: The transfer of energy due to the emission of electromagnetic waves (or photons).

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Energy  Transfer

Heat  and  work  are  direc)onal  quan))es.    •  Heat  transfer  to  a  system  and  work  done  on  a  system  are  posiIve  •  Heat  transfer  from  a  system  and  work  done  by    a  system  are  negaIve  

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Energy  Transfer  by  Work •  Work:  The  energy  transfer  associated  with  a  force  acIng  through  a  distance.    

•  If  the  energy  crossing  the  boundary  of  a  closed  system  is  not  heat,  it  must  be  work.                  

•  Work  per  unit  )me  is  called  power  and  is  denoted  𝑊        𝑊 = 𝑑𝑊/𝑑𝑡   

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Work done per unit mass

Heat vs. Work •  Both are recognized at the boundaries of a

system as they cross the boundaries. That is, both heat and work are boundary phenomena.

•  Systems possess energy, but not heat or

work. •  Both are associated with a process, not a

state.

•  Unlike properties, heat or work has no meaning at a state.

•  Both are path functions (i.e., their magnitudes depend on the path followed during a process as well as the end states).

Checkpoint

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The  interior  surfaces  of  the  oven  forms  the  system  boundary.    There  will  definitely  be  a  rise  of  temperature  in  the  oven.      -­‐  That  means  energy  is  being  supplied  to  the  system    What  causes  this  energy  transfer?  Heat  or  Work?    What  is  crossing  the  boundaries?  Heat  or  Work?    

Various  Forms  of  Work

𝑊↓𝑒 =∫𝑡↓1 ↑𝑡↓2 ▒𝑉𝐼𝑑𝑡   𝑊↓𝑒 =𝑉𝐼∆𝑡    Where     𝑊 ↓𝑒   =  Electrical  power  (W)  

 We  =  Electrical  work  (kJ)    ∆𝑡  =  t2  –  t1  =  )me  interval  (s)  

Electrical  Work  𝑊 ↓𝑒 =𝑉𝐼  

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Various  Forms  of  Work ShaR  Work    

Force  acts  through  a  distance  s,  𝑠=2𝜋𝑟𝑛    Shag  work  Wsh:    𝑊↓𝑠ℎ =𝐹𝑠= 𝑇/𝑟 2𝜋𝑟𝑛=2𝜋𝑛𝑇    Where    n  =  number  of  revolu)on      

𝑇=𝐹𝑟  

Power  transmihed  by  the  shag  𝑊 ↓𝑠ℎ :      

  𝑊 ↓𝑠ℎ =2𝜋𝑛 𝑇      

Where     𝑛   =  number  of  revolu)on  per  unit  )me        

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Various  Forms  of  Work Spring  Work  

Rela)onship  between  displacement  x  and  force  F  of  a  spring  is  expressed  as:  

 F  =  kx      

Where    k  =  spring  constant  (kN/m)    

Work  done  by  the  spring,  Wspring    

𝑊↓𝑠𝑝𝑟𝑖𝑛𝑔 = 1/2 𝑘( 𝑥2↓↑2 − 𝑥1↓↑2 )    Where    x1  =  ini)al  displacement  of  the  spring  (m)  

               x2  =  final  displacement  of  the  spring  (m)  

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Various  Forms  of  Work

•  The  work  transfer  needed  to  raise  a  body  is  equal  to  the  change  in  the  potenIal  energy  of  the  body    

•  The  work  transfer  needed  to  accelerate  a  body  is  equal  to  the  change  in  the  kineIc  energy  of  the  body.    

•  It  plays  important  roles  in:  (Considering  the  fric1on  and  other  energy  losses)    

•  The  design  of  elevators    •  The  design  of  automo)ve  and  aircrag  engine  •  Determina)on  of  hydroelectric  power  produced  by  a  dam  

 

Work  Done  to  Raise  or  to  Accelerate  a  Body  

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Example  2 Consider  a  1200  kg  car  cruising  steadily  at  90  km/h.  Now  the  car  starts  climbing  a  hill  that  is  sloped  30o  from  the  horizontal.  If  the  velocity  of  the  car  is  to  remain  constant  during  climbing.  Determine  the  addi)onal  power  that  must  be  delivered  by  the  engine.  

Analysis:  The  addi)onal  power  required  is  simply  the  work  that  needs  to  be  done  per  unit  )me  to  raise  the  eleva)on  of  the  car,  which  is  equal  to  the  change  in  the  poten)al  energy  of  the  car  per  unit  )me  

=  mg  Δz/  Δt  =  mg  Vver)cal    =  1200  kg  x  9.81  m/s2  x  90  km/h  x  sin  30o    =  147  kJ/s  =  147  kW    

W&

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Example  3 Determine  the  average  power  required  to  accelerate  a  900  kg  from  rest  to  velocity  of  80  km/h  in  20  s  on  a  level  road.    

 Analysis:    The  work  needed  to  accelerate  a  body  is  simply  the  change  in  the  kine)c  energy  of  the  body.    

22 2 22 1

1 1 80000( ) 900 02 2 3600a

mW m V V kgs

⎡ ⎤⎛ ⎞= − = × × −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

=222  kJ  

The  average  power  is  determined  from  

22220

aaW kJWt s

= =Δ

& =  11.1  kW  

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Energy  Transfer

Iden)fy  heat  and  work  interac)on  in  the  following  cases  

ü No heat

ü No work

ü Heat input if potato is system

ü No work

ü No heat

ü Work input (electrical)

ü No work ü Heat transfer if

element is turn on.

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Problem  Solving  Technique •  Step  1:  Problem  Statement  

•  Determine  the  interac)on  between  system  and  its  surrounding.    •  Determine  whether  it  is  a  closed  system,  open  system  or  an  isolated  system.    

InteracIon  with  surrounding   Mass    exchange  

No  mass  exchange  

Energy  exchange    

No  energy  exchange   X  

Open  sys.    (Control  volume)  

Closed  sys.    (Control  mass)  

Isolated  system  

Problem  Solving  Technique

•  Step  2:  Schema)c  •  Draw  a  schema)c  diagram  of  the  system  •  Define  your  system  boundary    

Q  is  heat  supplied  to/leaving  the  system  

Problem  Solving  Technique

•  Step  3:  Assump)ons  and  Approxima)ons  •  Is  it  a  quasista)c  or  non-­‐quasista)c  equilibrium  process?    •  Is  it  a  steady  flow  process  or  transient  flow  process?    •  Is  there  any  heat  loss?    

Problem  Solving  Technique •  Step  4:  Physical  Laws  

Determine  physical  laws  that  govern  the  system.  

∑↑▒𝑭 =𝒎𝒂    

Problem  Solving  Technique •  Step  5:  Proper)es  

•  Determine  the  proper)es  of  the  system:  extensive  or  intensive  proper)es.    

•  Does  it  involve  compressed  liquid  or  saturated  liquid,  saturated  vapor  or  superheated  vapor?    

Problem  Solving  Technique •  Step  6:  Calcula)ons  •  Step  7:  Reasoning,  Verifica)on,  and  Discussion  

•  Are  the  results  prac)cally  viable?    •  What  are  the  implica)ons  of  the  results  to  the  problem?      

•  Y.  A.  Cengel,  J.  M.  Cimbala,  R.  H.  Turner,  “Fundamentals  of  Thermal-­‐fluid  Sciences”,  4th  ed.  Chapter  2  &  Chapter  3.  

Reading  Material

AVendance  Code

41  TRC  2200:  Thermo-­‐Fluids  &  Power  Systems