Chapter 4
The First Law of
Thermodynamics
Steam power still reigns on
the Cumbres and Toltec
Scenic Railway in Colorado
and New Mexico.
ThermoNet
Thermodynamics: An Integrated Learning System
P.S. Schmidt, O.A. Ezekoye, J.R. Howell and D.K. Baker
Copyright (c) 2005
by John Wiley & Sons, Inc
4. The First Law of Thermodynamics
• Conservation of Energy
• Energy Balance
EIN  EOUT
dECV

dt
• EIN,OUT = Energy transferred across system
boundary
• ECV = Energy contained within system boundary
Chapter 4 The First Law of Thermodynamics
4.1 Closed Systems
• Mass does not cross system boundary
• Energy crosses system boundary.
• Mass Balance
– dmCV/dt = 0
– mCV = constant
system
boundary
QOUT
• Energy Balance
– ECM = U + KE + PE
– KE = mCMv2/2gC
– PE = mCMzg/gC
QIN
WIN or WOUT
ECM  ECM(t 2 ) - ECM(t1)  QIN  WIN   QOUT +WOUT 
Chapter 4 The First Law of Thermodynamics
4.2 Open (Control Volume) Systems
• Denote with CV subscript (e.g., mCV)
• Mass and energy cross system boundary
• On the following slides,
– Compare combustion in open and closed
systems
– See a gas turbine that is analyzed as an
open system
Chapter 4 The First Law of Thermodynamics
4.2 Open (Control Volume) Systems
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Chapter 4 The First Law of Thermodynamics
4.2 Open (Control Volume) Systems
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Chapter 4 The First Law of Thermodynamics
4.2.1 Conservation of Mass
• Rate Basis
dmCV
mIN  mOUT =
dt
• Time Interval
t2
 [m
IN
(t)-mOUT (t)] dt=mCV (t1 )  mCV (t 2 )
t=t1
• Useful Relations
m
V vA X

v
v
V
= Volumetric flow rate [m3/s or ft3/s]
– AX = cross-sectional flow area [m2 or ft2]
–
Chapter 4 The First Law of Thermodynamics
4.2.2 Flow Work and Enthalpy
• Mass crossing system boundary
– Carries energy u + ke + pe per unit mass flow
– Does flow work Pv per unit mass flow
– Recall enthalpy, h = u + Pv
– Total energy entering/leaving system due to mass
transfer is u + ke + pe + Pv = h + ke + pe per unit mass
flow.
Chapter 4 The First Law of Thermodynamics
4.2.3 The First Law
• Change in energy for open system is sum of
– Shaft work: Present if rotating shaft crosses boundary
– Boundary (PdV) work: Present if dVCV/dt  0
– Heat Transfer
– Energy transfer by mass transfer (u + ke + pe)
EIN  EOUT
dECV

dt
where
ECV  mCV u  ke  pe 
EIN,OUT  Q  W  m h  ke  pe 
or QIN  WIN  mIN,i hi  kei  pei  


 QOUT  WOUT  mOUT,j hj  ke j  pe j  
Chapter 4 The First Law of Thermodynamics
dECV
dt
4.3 Steady-State Steady-Flow Processes
• Steady-State (SS):
d  CV
0
dt
where ( )CV is any property
of the system (e.g., m or E)
• Steady-Flow (SF):
.
d
IN,OUT
dt
0
.
where ( )CV is any. transfer
.
. across the system
boundary (e.g., Q, W or m)
Chapter 4 The First Law of Thermodynamics
4.3 Steady-State Steady-Flow Processes
• Steady-State Steady-Flow (SSSF) = No changes
with time
• Mass Balance
N
M
i1
j1
 mIN,i   mOUT,j
dmCV

dt

0, SS
N
M
i1
j1
 mIN,i   mOUT,j
• If 1 stream (i.e., 1-inlet and 1-outlet)
mIN  mOUT  m
Chapter 4 The First Law of Thermodynamics
4.3 Steady-State Steady-Flow Processes
• SSSF Energy Balance
EIN  EOUT 
dECV
dt
0, SS

EIN  EOUT
N
QIN  WIN   mIN,i hi  kei  pei 
i1
N

 QOUT  WOUT   mOUT,j hj  ke j  pe j
j1

• If 1 stream (i.e., 1-inlet and 1-outlet) and dividing
by mass flow rate
qIN  wIN  h  ke  pe IN  qOUT  wOUT  h  ke  pe OUT
Chapter 4 The First Law of Thermodynamics
4.3.1 Nozzles and Diffusers
A diffuser converts high
speed, low pressure
flow to low speed, high
pressure flow
A nozzle converts high
pressure, low speed
flow to low pressure,
high speed flow
• On next page, see a nozzle in a turbojet engine
Chapter 4 The First Law of Thermodynamics
4.3.1 Nozzles and Diffusers
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Chapter 4 The First Law of Thermodynamics
4.3.1 Nozzles and Diffusers
• Common Assumptions
– SSSF
– No work or heat transfer
– Neglect changes in pe
• Energy Balance: Crossing out terms assumed 0
qIN
0
 wIN
0
 qOUT

 h  ke  pe
0
 wOUT
h  ke IN  h  ke OUT
Chapter 4 The First Law of Thermodynamics
0
0


IN
 h  ke  pe
0

OUT


v2 
v2 
 h 
  h 

2gC 
2gC 


IN
OUT
4.3.2 Throttling
• Throttling: Reduces Pressure
• Common Assumptions:
– SSSF
– No work or heat transfer
– Neglect changes in pe and ke
Throttling
Valve
• Energy Balance:
qIN
0
 wIN
 qOUT
0
0

 h  ke
 wOUT
0
0
 pe

 h  ke
0
0

IN
 pe
0

OUT

• Isenthalpic (h = constant) Process
Chapter 4 The First Law of Thermodynamics
hIN  hOUT
4.3.3 Pumps, Fans, and Blowers
• Pumps: Pressurize or
move liquids
• Fans & Blowers: Move
air
mOUT
TOUT
mIN,TIN,PIN
• Common Assumptions:
– SSSF
– No heat transfer
– Neglect changes in pe and ke
POUT
Pump Schematic
• Energy Balance for fan & blower
wIN  hOUT  hIN
• Energy Balance for pump (assuming ICL)
wIN  v POUT  PIN 
Chapter 4 The First Law of Thermodynamics
WIN
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)
Chapter 4 The First Law of Thermodynamics
4.3.4 Turbines
• Common assumptions for turbine:
– SSSF
– Adiabatic (q = 0)
– Neglect kinetic and potential energies
• Turbine energy balance (Single Stream)
  dE
 Q  W  m h  ke  pe     Q
m h  ke  pe 

W

IN
OUT
 IN

IN   OUT

OUT 
dt


EIN

EOUT
WOUT  m hIN  hOUT 

Per unit mass flow

wOUT  hIN  hOUT
Chapter 4 The First Law of Thermodynamics

0, SS
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)
Chapter 4 The First Law of Thermodynamics
4.3.4 Compressors
• Common assumptions for compressor:
– SSSF
– Adiabatic (q = 0)
– Neglect kinetic and potential energies
• Compressor energy balance
  dE
 Q  W  m h  ke  pe     Q
m h  ke  pe 

W

IN
OUT
 IN

IN   OUT

OUT 
dt


EIN

EOUT
WIN  m hOUT  hIN 

Per unit mass flow

wIN  hOUT  hIN
Chapter 4 The First Law of Thermodynamics

0, SS
4.3.5 Heat Exchangers
• Allows heat transfer from
one fluid to another without
mixing
• Example: Car Radiator
Chapter 4 The First Law of Thermodynamics
4.3.5 Heat Exchangers in Steam Power Plant
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Chapter 4 The First Law of Thermodynamics
4.3.5 Heat Exchangers
• Common Assumptions
– SSSF
– Externally adiabatic
– Neglect kinetic and potential
energies
• Energy Balance
Q  W 
IN
 IN
 

m h  ke  pe  

IN 
  QOUT  WOUT 



  dE
m h  ke  pe 

OUT 
dt
 



0,SS
mCOLD hOUT,COLD  hIN,COLD  mHOT hIN,HOT  hCOLD,HOT
Chapter 4 The First Law of Thermodynamics

4.3.6 Mixing Devices
• Combine 2 or more streams
• Common in industrial processes
• Common assumptions
– SSSF
– Adiabatic
– Neglect kinetic and potential energies
• Energy Balance (Streams 1 & 2 mixing to form 3)
Q  W 
IN
 IN
 

m h  ke  pe  

IN 
  QOUT  WOUT 

  dE
m h  ke  pe 

OUT 
dt
 

Chapter 4 The First Law of Thermodynamics

m1h1  m2h2  m3h3
0,SS
4.3.6 Mixing Devices
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Chapter 4 The First Law of Thermodynamics
4.4 Transient (Unsteady) Analysis
• Typically open system not at steady state
– Tank Filling
– Tank Emptying
• Mass Balance:
t2
 mIN - mOUT  dt  mCV (t2 )  mCV (t1)
t1
t2
• Energy Balance:  EIN  t   EOUT  t   dt  ECV  t 2   ECV  t1 
t1

v2 gz 
EIN,OUT  t   Q  W  m  h 



2gc gc 


1 v2 gz 
ECV  m  u 


2
g
g
c
c

Chapter 4 The First Law of Thermodynamics
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: mIN  mOUT =m(t 2 )  m(t1 )
• Energy Balance:
t2

QIN  WIN  mINhIN  - QOUT  WOUT  tt1 mOUT (t)(hOUT (t)dt 
 
 
v2 gz  
v2 gz  
 ECV (t 2 )  ECV (t1 )= m  u 

 m  u 



2gc gc   CV,2  
2gc gc   CV,1
 
Chapter 4 The First Law of Thermodynamics
4.4.2 Tank Filling
• Simplest USUF analysis:
– No outlet flow
– Assume adiabatic
• Mass Balance: mIN =m2  m1
• Energy Balance:
t
 QIN  WIN  mINhIN  - QOUT  WOUT   2 mOUT (t)(hOUT (t)dt 

 

t  t1
 
 
v2
gz  
v 2 gz  
= m  u 

 m  u 







2gc gc  
2g
g
 

c
c
  CV,1
 CV,2  

Chapter 4 The First Law of Thermodynamics
mINhIN   mu CV,2   mu CV,1