First Law of Thermodynamics: Open System

Contributed by:

Jonathan James Fri, Jan 21, 2022 07:38 PM UTC

Turbines, Compressors, Heat exchangers

1. Chapter 4
The First Law of
Control Volume:
Part 3
Assoc.Prof.Dr.Sommai
Priprem
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2. 4.3.4 Turbines
 Turbine: Enthalpy  Shaft work
 Used in
 Almost all power plants
 Some propulsion systems (e.g.,
turbofan and turbojet engines)
 Working Fluid:
 Liquids (e.g., hydro power plants)
 Vapors (e.g., steam power plants)
 Gases (e.g., gas power plants) w
Turbine
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3. Water Turbine
compressor
Gas Turbine
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4. i
4.3.4 Turbines
w
Turbine
 Common assumptions for turbine:
 SSSF e
 Adiabatic (q = 0)
 Neglect kinetic and potential energies
 Turbine energy balance (Single Stream)
Vi 2 Ve2
q  (hi   zi g ) w  (he   ze g ) J / kg
2 2
w (hi  he ) J / kg
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5. 4.3.4 Compressors
 Compressor: Shaft work  Increase pressure & enthalpy of
vapor or gas
 Often like turbine run in reverse
 Used in
 Gas power plants (e.g., gas turbine engine)
 Turbo propulsion systems (e.g., turbofan and turbojet engines).
 Industry (e.g., supply high pressure gas)
 Working Fluids  Gas, Vapor. Not Liquid (pump used)
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6. Wc
Reciprocating Compressor
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7. 4.3.4 Compressors
 Common assumptions for compressor:
 SSSF
Compressor
 Adiabatic (q = 0) w
 Neglect kinetic and potential energies
 Compressor energy balance
Vi 2 Ve2
q  (hi   zi g ) w  (he   ze g ) J / kg
2 2
w (hi  he ) J / kg
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8. 4.3.5 Heat Exchangers
 Allows heat transfer from
one fluid to another
without mixing
 Example: Car Radiator
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9. 4.3.5 Heat Exchangers
 Common Assumptions
 SSSF
 no work involve, wcv=0
 Externally adiabatic
 Neglect kinetic and
potential energies
Heat exchanger
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10. Heat Exchanger analysis technique 1
Whole Equipment
1. Control Volume over the heat exchanger
1
2. Applied 1st law: 2- inlets & 2- exit
3. No heat transfer out of the heat exchanger
3
4. No work, Change in KE & PE are negligible 4
5. Mass flow rate of each stream (hot stream Heat exchanger
and cold stream) remains unchanged 2
Control Volume
6. Use properties of each fluid
o o Vi 2 o o Ve2
Qcv   m i (hi   zi g ) Wcv   m e (he   ze g )
2 2
o o
 m i hi  m e he Why there is no Q
o o in the equation?
m1 (h1  h2 ) m 3 ( h4  h3 ) It is a heat
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exchanger, isn’t it?

11. Heat Exchanger analysis technique 2
Superimposed
1. Separated cold out of hot stream.
2. Control Volume over each stream then 1 Control Volume 1
applied 1st law Cold Fluid
3. No work, change in KE & PE are negligible
4. Heat transfer out from hot stream =
Heat transfer into cold stream
5. Mass flow rate of each stream (hot stream 2
and cold stream) remains unchanged Heat Transfer, Qcv
6. Use properties of each fluid
o o Vi 2 o o Ve2 3
4
Qcv  m i (hi   zi g ) Wcv  m e (he   ze g )
2 2 Hot Fluid
o o o
Cold : Qcv  m1 h1 m1 h2  Qcv m1 (h2  h1 ) Control Volume 2
o o o
Hot :  Qcv  m3 h3 m3 h4   Qcv m3 (h4  h3 )
o o
m1 (h1  h2 ) m3 (h4  h3 ) รศ.ดร.สมหมาย ปรีเปรม

12. 4.3.6 Mixing Devices
Mixing water:
m3h3
hot water:
m2h2
Mixing
Cold water: Chambe
m1h1 r
 Combine 2 or more streams
 Common in industrial processes
 Common assumptions
 SSSF
 Adiabatic
 Neglect kinetic and potential energies
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13. 4.4 Transient (Unsteady) Analysis
 Typically open system not at steady state
 Tank Filling
 Tank Emptying
t2
 Mass Balance:   m
 IN - m
 OUT  dt  mCV (t2 )  mCV (t1)
t1
t2
  E IN  t   E OUT  t   dt  ECV  t 2   ECV  t1 
t1
 Energy Balance: 2
 v gz 
E IN,OUT  t   Q  W  m  h 
   
 2gc gc 
 1 v2 gz 
ECV m  u   
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เปรมgc gc 

14. 4.4.1 Uniform State Uniform Flow
(USUF)
Uniform State: All properties uniform across system at any instant in time
Uniform Flow: All mass flow properties at each inlet and outlet are uniform
across the stream
Neglect kinetic and potential energies
Mass Balance : ∑mi - ∑me = m 2 – m1
Energy Balance:
Q + ∑mi(hi + ½ Vi2 +zig) = W + ∑me(hi + ½ Ve2 +zeg)
+ [m2(u2+½ V22+zg) – m1(u1+½ V12+z1g)]
Subscripts: i = inlet, e = exit 1 = at time start, 2 = at time end
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15. 4.4.2 Tank Filling
 Simplest USUF analysis:
No outlet flow

 Assume adiabatic, W =0, KE, PE = 0
Mass Balance : ∑mi - ∑me = m2 – m1
mi = m2 – m1
Energy Balance:
Q + ∑mi(hi + ½ Vi2 +zig) = W + ∑me(he + ½ Ve2 +zeg)
+ [m2(u2+½ V22+z2g) – m1(u1+½ V12+z1g)]
mihi = m2u2– m1u1
Subscripts: i = at inlet, e = at exit, 1 = at time start, 2 = at time end
What is the equation, if
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16. Conservation of Examples are follow
Mass and Energy
can be applied
everywhere in the
whole wide world
End of Chapter 4
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