The First Law of
Thermodynamics
Open System
(Control Volume)
1
First low of thermodynamics for
open Systems

Reminder of an open System.
Open
system = Control
volume
It is a properly selected region
in space.
Mass and energy can cross
its boundary.
2
Control volume involves two
main processes
 Steady


flow processes.
Fluid flows through the control volume
steadily.
Its properties are experiencing no
change with time at a fixed position.
 Unsteady

flow processes.
Fluid properties are changing with time.
3
Mass balance for steady flow
processes

We already showed that for steady flow
 i   m e
m

Many engineering devices involve a single
stream (one inlet and one exit only).
m 1  m 2
or 1 A1V1   2 A2V2
4
Energy Balance for SteadyFlow Systems 0
E in  E out
E in  E out  E sys
Qin W in  E mass ,in  Qout W out  E mass ,out
Qin W in   mi i  Qout W out   mee
Q in  Qout W in W out 
1 2
1 2
m
(
Pv

u

V

gz
)

m
(
Pv

u

V  gz )i
 e

e
i
2
2
1 2
1 2
Q W   m e (Pv  u  V  gz )e   m i (Pv  u  V  gz )i
2
2
1 2
1 2
Q W   m e (h  V  gz )e   m i (h  V  gz )i
2
2
5
Let us look at some common steady
flow devices
Only one in and one out
More than one
inlet and exit
6
For single stream steady flow
devices, the 1st low becomes



 
V
V
Q  W  mi hi 
 g  zi   me he 
 g  ze 
2
2











2
i


2
e
Often the change in kinetic energy and potential energy is small.




Q W  m i hi  m e he
q  w  he  hi
Per unit mass
7
1V1 A1  2V2 A2
Nozzles
A1
A2
1 A1
V2  V1
 2 A2
 A1  A2
A nozzle is a device that
increases the velocity of a
fluid at the expense of
pressure
1   2 for liquids
1   2 for low speed gas
V2  V1
8
1V1 A1   2V2 A2
Diffusers
A diffuser is a device that
slows down the velocity of a
fluid causing an increase in its
pressure
A1
A2
1 A1
V2  V1
 2 A2
 A1  A2
1  2 for liquids
1  2 for low speed gas
V2  V1
9
Diffusers
10
Nozzles and Diffusers
(1st low analysis)




2
2



Ve  Vi

Q  W  m he  hi 
 g  ze  zi 


2


Is there work in this system? NO
Is there heat transfer? let us say: NO
In fact, it depends on the problem!
Does the fluid change elevation? NO

2
e

2
i
V V
0  he  hi  
2
?
Q : What happened to the m
Ans: It is divided out
11
which can be rearranged to
2
2
e
Vi
V
hi 
 he 
2
2
In a nozzle, enthalpy is converted into kinetic energy
How can you find the mass flow rate in a nozzle?
m  1V1 A1   2V2 A2
V1 A1 V2 A2
m 

v1
v2
12
Example (4-9): Deceleration of
Air in a Diffuser
Air at 10oC and 80 kpa enters the diffuser of a jet
engine steadily with a velocity of 200 m/s. The inlet
area of the diffuser is 0.4 m2. The air leaves the
diffuser with a velocity that is very small compare to
the inlet velocity.
Determine
(1) The mass flow rate of the air and
(2) The temperature of the air leaving the
diffuser.
13
Example (4-10): Acceleration
of Steam in a Nozzle
Steam at 1.72 Mpa (250 psia) and 371C (700
F) steadily enters a nozzle whose inlet area
is 0.019 m2 (0.2 ft2.) The mass flow rate of
the steam through the nozzle is 4.54 kg/s
(10 lbm/s) . Steam leaves the nozzle at 1.38
Mpa (200 psia) with a velocity of 274.3 m/s
(900 ft/s). The heat losses from the nozzle
per unit mass of the steam are estimated to
be 2.8 KJ/kg (1.2 Btu/lbm).
Determine:
(a) the inlet velocity and
(b) the exit temperature of the steam.
Answers:
a) 41.0 m/s or 134.4 ft/s,
b) 350 C or 661.9 F
14
Solution of Example: Acceleration of Steam in a Nozzle
Note that there is heat transfer (Q).
So, you have to go back to the general form
of the 1st low for single stream devices and
get the following:
2
2
e
Vi
V
q  hi 
 he 
2
2
Turbines
A turbine is a device that produces work at the
expense of temperature and pressure.
As the fluid passes through the turbine, work is done
against the blades, which are attached to a shaft. As a
result, the shaft rotates, and the turbine produces work.
16
17
18
19
Compressors
Inlet
A compressor is a device that increases the
pressure of a fluid by adding work to the system.
Work is supplied from an external source through a
rotating shaft.
Compressor
Win
Exit
20
21
22
23
Turbines and Compressors




2
2

   
Ve  Vi



h

h
(
kJ
/
kg
)


Q 
W wm
h

h


g
z

z
W me ehe i  ihi 2 ( W ) e i 



Is there work in this system?
Is there heat transfer?

Yes!
Negligible because of insulation. Exception: Internal
cooling in some compressors.
Does the fluid change elevation?
NO
Does the kinetic energy change?
Usually it can be ignored
24
Example (4-12): Power
Generation by a Steam Turbine
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 the figure on the right.
a) Compare the magnitude 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.
Answers: a) h = -885.9 kJ/kg,
ke = 14.95 kJ/kg, pe = -0.04 kJ/kg,
b) 871.0 kJ/kg, and c) 5.74 kg/s
25
Throttling Valve
A throttling valve reduces
the fluid pressure.
P1>P2
It is small device and thus
the flow through it may be
assumed adiabatic (q=0)
since there is neither
sufficient time nor large
enough area for any effective
heat transfer to occur.
26
Throttling Valve






2
2
0  h e  hi 

Ve hVei  hi
Q  W  m he  hi 
 g  ze  zi 


2
isenthalpic
device


Is there work in this system?
NO
Is there heat transfer?
Usually it can be ignored
Does the fluid change elevation?
NO
Does the fluid change velocity?
Usually it can be ignored
27
What happens to the fluid temperature
a cross throttling Valves ?
h1  h2
u1  p1v1  u2  p2v2
if p2v2  p1v1  u2  u1  T 
if p2v2  p1v1  u2  u1  T 
28
Throttling Valves
(incompressible substance )
For incompressible substance (like water),  is constant
v1  v2
and
P1  P2
u1  p1v1  u2  p2v2
 p2v2  p1v1  u2  u1
T
will rise
For incompressible substance only!
29
Throttling Valves (compressible
substance: Vapor)
Example (4-13): Expansion of Refrigerant-134a in a
Refrigerator
Refrigerant-134a enters the capillary tube of a
refrigerator as saturated liquid at 0.8 MPa and is
throttled to a pressure of 0.12 MPa.
Determine the quality of the refrigerant at the final state
and the temperature drop during this process.
<Answers: 0.339, -53.69oC>
30
Solution of Example: Expansion of Refrigerant-134a in a Refrigerator
Notice that T2 <T1
Throttling Valves (Compressible
substance: an Ideal Gas)


What happens if the gas is ideal?
For ideal gases



h = Cp  T
But  h = 0
So…  T = 0
The inlet and outlet temperatures are the same!!!
32
Mixing Chamber
Mixing two or more fluids is a
common engineering process
Mixing
Chamber
The mixing chamber does not have to
be a distinct “chamber.” An ordinary Telbow, or a Y-elbow in a shower, for
example, serves as the mixing
chamber for the cold- and hot-water
streams as shown in the figure (Left).
33
Mixing Chamber


 




2
2
 




Ve
Vi
Qnet Wnet   mee he e 2  gze   i mi i hi  2  gzi 









0   m h    m h 

We no longer have only one inlet and one exit stream
Is there any work done?
No
Is there any heat transferred?
No
Is there a velocity change?
No
Is there an elevation change?
No
34
Mixing Chamber

Material Balance


m  m
i

Mixing
Chamber



m m m
e
1
2
3
Energy balance





m h  m h m h  m h  m h
i
i
e e
1
1
2
2
3
35
3
Example(4-14): Mixing of Hot and Cold
Waters in a Shower
Consider an ordinary shower where
hot water at 140oF is mixed with cold
water at 50oF. If it is desired that a
steady stream of warm water at 110oF
be supplied, determine the ratio of
the mass flow rates of the hot to cold
water. Assume the heat losses from
the mixing chamber to be negligible
and the mixing to take place at a
pressure of 20 psia.
<Answer: 2.0>
36
Heat Exchanger
A heat exchanger is a
device where two moving
fluids exchange heat
without mixing.
37
Heat Exchangers
Your analysis approach will depend on how you define your system
38
Heat Exchangers
System (a): entire HX

Mass Balance

Divide into two separate
streams with equal inlet
and outlet flow rates
m1  m 2 , m 3  m 4

Energy balance


Two inlets
Two outlets
m1h1  m 3h3  m 2 h2  m 4 h4
39
Heat Exchangers
System (b): Single stream

Mass Balance

Considering one single
stream with one inlet and
one outlet flow rates
m1  m 2

Energy balance



One inlet
One outlet
Plus heat transfer
Q  m 2 h2  m1h1
40
Example: Cooling of Refrigerant-134a by Water
Refrigerant-134a is to be cooled by
water in a condenser. The refrigerant
enters the condenser with a mass
flow rate of 6 kg/min at 1 MPa and
70oC and leaves at 35oC. The cooling
water enters at 300 kPa and 15oC and
leaves at 25oC. Neglecting any
pressure drop, determine
(a) the mass flow rate of the cooling
water required and (b) the heat
transfer rate from the refrigerant to
water.
<Answers: a) 0.486 kg/s, b) 20.35 kJ/s>
41