The First Law of
Thermodynamics
• It is the law that relates the various forms of energies
for system of different types.
• It is simply the expression of the conservation of
energy principle.
• Based on experimental observations, the first law of
thermodynamics states that
– “Energy can be neither created nor destroyed during a
process; it can only change forms.”
• For the system shown above, the conservation of energy principle or
the first law of thermodynamics is expressed as
 Total Energy
  Total Energy
  The Change in Total 


 
=

Entering
the
System
Leaving
the
System

 
  Energy of the System 
• The total energy of the system, Esystem, is given as
Ein  Eout  Esystem
Esystem = Internal Energy + Kinetic Energy + Potential Energy
Esystem = U + KE + PE
• If the system does not move with a velocity and has no change
in elevation, it is called a stationary system, and the
conservation of energy equation reduces to
Ein  Eout  U + KE + PE
Ein  Eout  U
Mechanisms of Energy Transfer, Ein and Eout
• Heat Transfer, (Q) Heat transfer to a system (heat gain) increases
the energy of the molecules and thus the internal energy of the
system and heat transfer from a system (heat loss) decreases.
• Work Transfer, (W) Work transfer to a system (i.e., work done on a
system) increases the energy of the system, and work transfer from a
system (i.e., work done by the system) decreases .
• Mass Flow, (m) When mass enters a system, the energy of the
system increases because mass carries energy with it (in fact, mass is
energy). Likewise, when some mass leaves the system, the energy
contained within the system decreases .
• The energy balance can be written more explicitly as
Ein  Eout  (Qin  Qout )  (Win  Wout )  ( Emass ,in  Emass ,out )  ESystem
• Or on a rate form, as
 ESystem Change in internal, kinetic,  kJ 
Ein  Eout byNetheat,
energy transfer
work and mass
potential, etc..energies

Ein  Eout

Rate of net energy transfer
by heat, work and mass

  E System

Rate change in internal, kinetic,
potential, etc..energies
 kW 
The first law and a closed system
• For the closed system where the mass never crosses the system
boundary, then the energy balance is
Qin -Qout +Win -Wout =E system
• Closed system undergoing a cycle
– For a closed system undergoing a cycle, the initial and final states
are identical, and thus
• The initial and final states are identical
Esystem  E2  E1  0
Ein  Eout  0
Ein  Eout
Wnet ,out  Qnet ,in
• If the total energy is a combination of internal energy, kinetic
energy and potential energy
E  U  KE  PE
m(V22  V12 )
Q12  W12  U 2  U1  
 mg (Z 2  Z1 )
2
• For negligible changes in kinetic and potential energy
Q12  U 2  U1   W12
Internal energy and Enthalpy
• Internal energy
– The internal energy includes some complex forms of energy show up
due to translation, rotation and vibration of molecules.
– It is designated by U and it is extensive property.
– Or per unit mass as, specific internal energy, u  U
m
• If we take two phase as liquid and vapor at a given saturation
pressure or temperature
U  U f Ug
mu  m f u f  mg u g
u  u f  xu fg
Enthalpy
• It is another extensive property which has a unit of energy and it is
denoted by H.
• The enthalpy is a convenient grouping of the internal energy, pressure,
and volume and is given by
H  U  PV
– The enthalpy per unit mass is,
h
H
m
h= u+Pv
Q12  U 2  U1   W12
W12  Pdv  P(V2  V1 )
Q12  U2  U1    PV2  PV1 
Q12  H 2  H1
W12  PV2  PV1
Q12  U2  PV2   U1  PV1 
Specific Heat
• It defined as; the energy required to raise the
temperature of a unit mass of a substance by one
degree.
• It is an intensive property of a substance that will
enable us to compare the energy storage capability
of various substances. The unit is KJ Kg℃ or. KJ KgK
• In thermodynamics, we are interested in two kinds of
specific heats: specific heat at constant volume and
specific heat at constant pressure.
• The specific heat at constant volume can be viewed
as the energy required to raise the temperature of the
unit mass of a substance by one degree as the volume
is maintained constant.
• Here the boundary work is zero because the volume
is constant
• From first law δQ  dU
• Per unit mass  q  du
 q  Cv dT
Cv dT  du
 du 
Cv  

 dT v
• The specific heat at constant pressure Cp can be
viewed as the energy required to raise the
temperature of the unit mass of a substance by one
degree as the pressure is maintained constant.
• From first law δQ  dU  PdV  d  U  PV   dH
 q  C p dT
• Per unit mass  q  dh
C p dT  dh
 dh 
Cp  

dT

p
Internal Energy, Enthalpy, and Specific Heats of Ideal
Gases
• We defined an ideal gas as a gas whose temperature, pressure, and
specific volume are related by
Pv  RT
• From the specific heat relation
du  Cv dT
u2  u1  Cv dT
• Or taking average value of specific heat for narrow temperature
difference
u2  u1  Cave,v (T2  T1 )
dh  C p T  dT
h2  h1  C p dT
h2  h1  Cave, p (T2  T1 )
Relation between CP and CV for Ideal Gases
h  u  RT
dh  du  RdT
• Replacing dh by C p dT and du by Cv dT we have
C p dT  Cv dT  RdT
C p  Cv  R
• At this point, we introduce another ideal-gas property called
the specific heat ratio k, defined as
K
Cp
Cv
C p  KCv
Cv 
R
K 1
Cp 
Cp
K
R
KCv  Cv  R
Cp 
K
R
K 1
Internal Energy, Enthalpy, and Specific Heats of
Solids and Liquids
• A substance whose specific volume (or density) is constant is
called an incompressible substance.
• The specific volumes of solids and liquids essentially remain
constant during a process. Therefore, liquids and solids can be
approximated as incompressible substances.
• It can be mathematically shown that the constant-volume and
constant-pressure specific heats are identical for
incompressible substances.
• The specific heat can be expressed as
C p  Cv  C
Example
1. The initial pressure and volume of a pistoncylinder arrangement is 200kPa and 1m3
respectively. 2000kJ of heat is transferred to
the system and the final volume is 2m3.
Determine the change in the internal energy
of the fluid.
2. A piston–cylinder device initially contains 0.8 m3 of
saturated water vapor at 250 kPa. At this state, the
piston is resting on a set of stops, and the mass of the
piston is such that a pressure of 300 kPa is required to
move it. Heat is now slowly transferred to the steam
until the volume doubles. Show the process on a P-v
diagram with respect to saturation lines and
determine (a) the final temperature, (b) the work done
during this process, and (c) the total heat transfer.
3.A piston-cylinder contains steam initially at 1
Mpa, 450 oC and 2.5m3. Steam is allowed to cool
at constant pressure until it first start condensing.
Show the process on a T-v diagram with respect to
saturation lines and determine.
a) The mass of the steam
b) The final temperature
c) The amount of heat transfer
4. A piston–cylinder device initially contains steam at 200 kPa,
200°C, and 0.4 m3. At this state, a linear spring is touching the
piston but exerts no force on it. Heat is now slowly
transferred to the steam, causing the pressure and the volume
to rise to 300 kPa and 0.6 m3, respectively. Show the process
on a P-v diagram with respect to saturation lines and determine
(a) the final temperature, (b) the work done by the steam, and
(c) the total heat transferred.
The First Law and the Control Volume
• The conservation of mass and the conservation of energy principles
for open systems or control volumes apply to systems having mass
crossing the system boundary or control surface.
• In addition to the heat transfer and work crossing the system
boundaries, mass carries energy with it as it crosses the system
boundaries.
• Hence the conservation of mass principle can be used to relate mass which
entering and leaving a system. It can be expressed as
• The net mass transfer to or from a control volume during a process (a time
interval t) is equal to the net change (increase or decrease) in the total
mass within the control volume during that process (t). That is,
 Total mass entering   Total mass leaving   Net change in mass


-
= 

the
CV
during
Δt
the
CV
during
Δt
within
the
CV
during
Δt

 
 

• Thermodynamic processes involving control volumes can be considered in
two groups: steady-flow processes and unsteady-flow processes.
min  mout  mCV
( kg )
min  mout  dmCV / dt
(kg / s)
m  V
Steady state process
• The flow through a control volume is at steady state if, “the
property of the substance at a given position within or at the
boundaries of the control volume do not change with time”.
• During a steady-flow process, the total amount of mass
contained within a control volume does not change with time
(mCV= constant).
min  mout  dmCV / dt
dmCV
  m CV  0
dt
min  m out
m
in
  mout
in V in  out V out
V in  V out
in  out incompressible assumption
Vin Ain  Vout Aout
Unsteady state process
• The properties within the control volume change with time but
remain uniform at any instant of time.
• Typical example:- filling and empting processes where most of
the cases average value of properties must be used.
dmcv
0
dt
dmcv
  mi   m e
dt
dmcv
 mi  m e
dt
Flow Work and The Energy of a Flowing
Fluid
• Unlike closed systems, control volumes involve mass flow
across their boundaries, and some work is required to push the
mass into or out of the control volume.
• This work is known as the flow work, or flow energy.
W flow
F  PA
 FL  PAL  PV
( kJ )
w flow  Pv
w flow,in  Pv
and w flow,exit  Pe ve
i i
W flow,in  ( Pv
i i ) mi and W flow,exit  ( Pe ve ) me
W flow   Peve  me   Pv
i i  mi
W  W flow  W cv
Development of energy balance
• The general representation of the first law of thermodynamics
Q12  W12  E2  E1
• The first law for open system will also have the same form, but
W12  W flow  Wcv E =U +KE+PE
E = Internal Energy + Kinetic Energy + Potential Energy
• The fluid entering or leaving a control volume possesses
an additional
V2
form of energy, the flow energy Pv
e  u  ke  pe  u 
 gz
2
• Then the total energy of a flowing fluid on a unit-mass basis (denoted
by) becomes
  Pv  e  Pv  (u  ke  pe)
• But the combination Pv + u has been previously defined as the
enthalpy h. So the relation in the above equation reduces to
V2
  h  ke  pe  h 
 gz
2
( kJ / kg )
• For inlet
V 2i
ei  Pv
 gzi  Pv
i i  ui 
i i
2
V 2i
ei  Pv
 gzi
i i  hi 
2
• For outlet
V 2e
ee  Pe ve  he 
 gze
2
• General equation
Ein  Eout  dEsystem / dt
0( steady )
0
Qin  Win   m  Qout  Wout   m
in
out
V2
V2
Qin  Win   m(h 
 gz )  Qout  Wout   m(h 
 gz )
2
2
in
out
• In such cases, it is common practice to assume heat to be
transferred into the system (heat input) at a rate of Q, and work
produced by the system (work output) at a rate of W , and then
solve the problem. The first-law or energy balance relation in
that case for a general steady-flow system becomes


V22  V12
Q W  m h2  h1 
 g ( z2  z1 ) 
2


V22  V12
q  w  h2  h1 
 g ( z2  z1 )
2
• When the fluid experiences negligible changes in its kinetic and
potential energies (that is, ke = 0, pe = 0), the energy balance
equation is reduced further to
q  w  h2  h1
Some Steady-Flow
Engineering Devices
Nozzles and Diffusers
• Nozzles and diffusers are commonly utilized in jet engines,
rockets, spacecraft, and even garden hoses.
• A nozzle is a device that increases the velocity of a fluid at the
expense of pressure.
• A diffuser is a device that increases the pressure of a fluid by
slowing it down.
Ein  Eout
min  m out
m1  m2  m




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit






Vi 2 
Ve2 
mi  hi 
  me  he 

2
2 



Ve  2(hi  he )  Vi 2
Turbines
• In steam, gas, or hydroelectric power plants, the device that
drives the electric generator is the turbine.
• As the fluid passes through the turbine, work is done against
the blades, which are attached to the shaft. As a result, the
shaft rotates, and the turbine produces work.
Ein  Eout
min  m out
m1  m2  m




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit




mi hi  m e h e  W out
W out  m(hi  he )
Compressors
• Compressors, as well as fans, are devices used to increase the
pressure of a fluid.
• Work is supplied to these devices from an external source
through a rotating shaft.




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit




min  m out
 W net  m(he  hi )
W net  m(hi  he )
Pumps
• The work required when pumping an incompressible liquid in
an adiabatic steady-state, steady-flow process is given by


V22  V12
Q W  m  h2  h1 
 g ( z2  z1 ) 
2


• The enthalpy difference can be written as
h2  h1   u2  u1    Pv 2   Pv 1 
• The pumping process for an incompressible liquid is
essentially isothermal, and the internal energy change is
approximately zero. Since v2 = v1 = v the work input to the
pump becomes


V V
 W  m  v  P2  P1  

2
2
2
1
2
 W  m v  P2  P1  
 g ( z2  z1 ) 

W in, pump  m v  P2  P1  
Throttling Valves
• Throttling valves are any kind of flow-restricting devices that
cause a significant pressure drop in the fluid.
• The pressure drop in the fluid is often accompanied by a large
drop in temperature, and for that reason throttling devices are
commonly used in refrigeration and air-conditioning
applications.




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit




min  m out
mi h i  m e h e
hi  h e
Mixing Chambers
• The mixing of two fluids occurs frequently in engineering
applications. The section where the mixing process takes place
is called a mixing chamber.
• The ordinary shower is an example of a mixing chamber.
E in  E out




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit




m1 h1  m 2 h 2  m3 h3
m
in
  mout
m1  m 2  m3
m 2  m3  m1


m1 h1   m3  m1  h 2  m3 h3


m1 (h1  h 2 )  m3 (h3  h 2 )
m1  m3
( h3  h 2 )
(h1  h2 )
Heat Exchangers
• Heat exchangers are normally well-insulated devices that
allow energy exchange between hot and cold fluids without
mixing the fluids.
min  m out
m1  m 2  m w
m3  m 4  m R
E in  E out




Vi 2
Ve2
Qnet   mi  hi 
 gzi   W net   me  he 
 gze 
2
2
inlet
exit




m1 h1  m3 h3  m 2 h 2  m 4 h 4
m w (h1  h 2 )  m R (h 4  h3 )
Example
1.Steam at 0.4 MPa, 300 oC, enters an adiabatic
nozzle with a 1m/sec and leaves at 0.2MPa
with a quality of 90%. Find the exit velocity, in
m/s.
2. Steam flows steadily through an adiabatic turbine. The inlet
conditions of the steam are 4 MPa, 500°C, and 80 m/s, and the
exit conditions are 30 kPa, 92 percent quality, and 50 m/s. The
mass flow rate of the steam is 12 kg/s.
Determine
(a) the change in kinetic energy,
(b) the power output, and
(c) the turbine inlet area.
3. The power output of an adiabatic steam turbine
is 5 MW, and the inlet and the exit conditions of
the steam are as indicated in Figure below.
a) Compare the magnitudes of h, ke, and pe.
b) Determine the work done per unit mass of
the steam flowing through the turbine.
c) Calculate the mass flow rate of the steam.
3. Nitrogen gas is compressed in a steady-state,
steady-flow, adiabatic process from 0.1 Mpa, 25
oC.
During the compression process the
temperature become 125 oC. If the mass flow rate
is 0.2kg/s, determine the work done on the
nitrogen, in kW. (use cp=1.039kJ/kg.K)
4. Saturated steam at 0.4 MPa is throttled to
0.1MPa, 100 oC. Determine the quality of the
steam at 0.4MPa.
5. Steam at 0.2MPa, 300oC, enters a mixing
chamber and is mixed with cold water at 20oC,
0.2MPa, to produce 20kg/s of saturated liquid
at 0.2MPa. What are the required steam and
cold water flow rates?
6. Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A
contains 2-kg steam at 1 MPa and 300°C while Tank B contains 3-kg
saturated liquid–vapor mixture with a vapor mass fraction of 50 percent.
Now the partition is removed and the two sides are allowed to mix until the
mechanical and thermal equilibrium are established. If the pressure at the
final state is 300 kPa, determine (a) the temperature and quality of the
steam (if mixture) at the final state and (b) the amount of heat lost from the
tanks.
7. Air enters an adiabatic nozzle steadily at 300
kPa, 400°C, and 30 m/s and leaves at 250kPa
and 33m/s. The inlet area of the nozzle is 0.08
m2. Determine (a) the mass flow rate through
the nozzle, (b) the exit temperature of the air,
and (c) the exit area of the nozzle.
9. Refrigerant-134a is throttled from the
saturated liquid state at 700 kPa to a pressure
of 160 kPa. Determine the temperature drop
during this process and the final specific
volume of the refrigerant.
10. In steam power plants, open feedwater heaters are frequently
utilized to heat the feedwater by mixing it with steam bled off the
turbine at some intermediate stage. Consider an open feedwater
heater that operates at a pressure of 1000 kPa. Feedwater at 50°C
and 1000 kPa is to be heated with superheated steam at 200°C
and 1000 kPa. In an ideal feedwater heater, the mixture leaves
the heater as saturated liquid at the feedwater pressure.
Determine the ratio of the mass flow rates of the feedwater and
the superheated vapor for this case.
11. An adiabatic air compressor is to be powered by a
direct-coupled adiabatic steam turbine that is also
driving a generator. Steam enters the turbine at 12.5
MPa and 500°C at a rate of 25 kg/s and exits at 10
kPa and a quality of 0.92. Air enters the compressor
at 98 kPa and 295 K at a rate of 10 kg/s and exits at 1
MPa and 620 K. Determine the net power delivered
to the generator by the turbine.