210
Introduction ENGR to Thermodynamics
Course announcements
Second Exam:
This Thursday 16 Feb; 8 am; Main Auditorium
Chapters 1-3
Third Exam:
Thursday, 1 Mar; 8 am; Main
Final Exam
21 Mar 2012; 0800-1000;
Lecture 11 201125
ENGR 210 Week 6
Recitation Problems covered this week
Chap 4: 34, 38, 90;
Xtra Problems
Chap 4: 32, 42, 46, 63, 126;
Lecture 11 201125
Energy balance - general case
Heat, work, mass
Ein Eout = Esystem
Ein Eout = Esystem
Q
Lecture 11 201125
Mass flow across a boundary ~ Flow Work
Model with a piston-cylinder device
F=PA
Wflow = F L = P A L = P V
w flow = P v
Total energy of flowing fluid
methalpy,
= P v + e = P v + (u + ke + pe)
= P v + u + (ke + pe)
= h + ke + pe
V2
=h+
+gz
2
Lecture 11 201125
General Control Volume Analysis
Conservation of mass:
- msys = in mi out mo , over a process
- dmsys/dt = in((t)V(t,A)A)iout((t)V(t,A)A)o
Conservation of energy:
Esys=[U+KE+PE]sys = inEin - outEout
= (Qin-Qout)+(Win-Wout)
+(min*inmout*out)
Lecture 11 201125
Steady flow systems
Fluid flows through the control
volume steadily
No properties within the control
volume change with time
mCV = msys = const, Esys = Ecv =
const
No properties change at the
boundaries of CV with time (use
average properties)
Heat and work interactions between
the system and surroundings do not
change with time
Lecture 11 201125
Steady flow
Continuity equation, Energy balance
i = inlet, e = exit
msys = Esys = 0
m m
i
i
e
=0
e
Qin + Win + #mi i % Qout Wout #me e % = 0
$ &
$ &
i
e
Qin Qout + Win Wout + #mi i % #me e % = 0
$ &
$ &
i
e
#'
#'
*%
*%
V2
V2
)
)
Qnet Wnet + -mi ) hi + i + g zi ,. -me ) he + e + g ze ,. = 0
,
,
2
2
-(
.
.
+& e $ (
+&
i$
Lecture 11 201125
Single-stream Steady flow
Continuity equation, Energy balance
mi = me
i
e
i Vi Ai = e Ve Ae = i V = e V
1
1
1 1
Vi Ai = Ve Ae = V = V
vi
ve
vi i ve e
"
Ve2 Vi 2
Q W = m$ he hi +
+ g ze zi
$
2
#
(
If ke~0, pe~0:
q - w = h
Lecture 11 201125
)
%
'
'
&
Nozzles and Diffusers
Nozzle - velocity increases, pressure decreases
Diffuser - velocity decreases, pressure increases
(here: subsonic)
Typically
dQ/dt ~ 0 most of the time
dW/dt = 0
ke 0
pe ~ 0
Lecture 11 201125
Nozzle Example
Air at 600 kPa and 500 K enters an
adiabatic nozzle that has an inlet-to-exit
area ratio of 2:1 with a velocity of 120 m/s
and leaves with a velocity of 380 m/s.
Determine the
(a) Exit temperature
(b) Exit pressure
Lecture 11 201125
1
AIR
2
Nozzle (cont)
System: We take nozzle as the system, which is a
control volume since mass crosses the boundary.
Assumptions
1 This is a steady-flow process since there is no change
with time.
2 Air is an ideal gas with variable specific heats.
3 Potential energy changes are negligible.
4 The device is adiabatic and thus heat transfer is
negligible.
5 There are no work interactions.
Analysis (a) There is only one inlet and one exit, and
thus,
m1 = m2 = m
Lecture 11 201125
Nozzle (cont)
The energy balance for this steady-flow system can be
expressed in the rate form as
i
o
E ut
n E
Rate of net energy transfer
by heat, work, and mass
=
E system (steady)
=0
Rate of change in internal, kinetic,
potential, etc. energies
E in = E out
m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0)
V22 V12
0 = h2 h1 +
2
Lecture 11 201125
Nozzle (cont)
Properties The enthalpy of air at the inlet
temperature of 500 K is h1 = 503.02 kJ/kg
(Table A-17), v1 = 120 m/s; v2 = 380 m/s
therefore,
h2 = h1
V22 V12
2
(380 m/s)
= 503.02 kJ/kg
2
2
(120 m/s) " 1 kJ/kg %
$
'
# 1000 m 2 /s2 &
2
= 438.02 kJ/kg
Then from Table A-17 we interpolate T2 = 436 K
Lecture 11 201125
Nozzle (cont)
The exit pressure is determined from the
conservation of mass relation,
1
1
A2V2 =
A1V1
v2
v1
1
1
A2V2 =
A1V1
RT2 / P2
RT1 / P
1
P2 =
A1T2V1
2 (436.5 K )(120 m/s)
P=
(600 kPa )
1
A2T1V2
1 (500 K )(380 m/s)
= 330.8 kPa = 331 kPa
Lecture 11 201125
Turbines and Compressors
Turbine - produces shaft work
Compressor uses shaft work increase P
Typically
dQ/dt ~ 0 most of the time
dW/dt 0
ke ~ 0
pe ~ 0
Lecture 11 201125
Example Steam Turbine
Steam flows steadily through a turbine at a rate of 45,000
lbm/hr, entering at 1000 psia and 900F and leaving at 5
psia as saturated vapor. If the power generated by the
turbine is 4 MW,
Determine the rate of heat loss from the steam
System: The internal walls of the turbine, which is a
control volume since mass crosses the boundary.
1
Assumptions
1 This is a steady-flow process
since there is no change with time.
2 Kinetic and potential energy
H2O
changes are negligible.
2
Lecture 11 201125
Steam Turbine (cont)
Analysis: There is only one inlet and one exit, so
m1 = m2 = m
We take the turbine as the system, which is a control
volume since mass crosses the boundary. The energy
balance for this steady-flow system can be expressed
in the rate form as
Ein Eout
Rate of net energy transfer
by heat, work, and mass
= Esystem ( steady ) = 0
Rate of change in internal, kinetic,
potential, etc. energies
Ein = Eout
Lecture 11 201125
Steam Turbine (cont)
Expressing the terms that are in and out
mh1 = Qout + W out + mh2
(since ke pe 0)
Qout = m( h2 h1 ) W out
Lecture 11 201125
Steam Turbine (cont)
Properties From the steam tables (Tables A-4E - 6E)
P1 = 1000 psia !
" h1 = 1448.6 Btu/lbm
T1 = 900F
#
P2 = 5 psia
h2 = 1130.7 Btu/lbm
sat .vapor
Lecture 11 201125
Steam Turbine (cont)
Substituting values
Qout = ( 45000/3600 lbm/s)(1130.7 1448.6) Btu/lbm
# 1 Btu &
4000 kJ/s%
(
$1.055 kJ '
= 182.0 Btu/s = 182 Btu/s
Lecture 201125
Example 11 - Compressor
He is compressed from 120 kPa and 310 K to 700 kPa and 430 K.
A heat loss of 20 kJ/kg occurs during the compression.
Determine the power input if the mass flow rate is 90 kg/min.
P2 = 700 kPa
T2 = 430 K
Q
He
90 kg/min
W
P1 = 120 kPa
T1 = 310 K
Lecture 11 201125
Compressor (cont)
Properties The constant pressure specific heat of helium is
cp = 5.1926 kJ/kgK (Table A-2a)
Analysis There is only one inlet and one exit, so m1=m2=m. We
take the compressor as the system, which is a control volume
since mass crosses the boundary. The energy balance for this
steady-flow system can be expressed in the rate form as
i
o
E ut
n E
Rate of net energy transfer
by heat, work, and mass
=
E system (steady)
Rate of change in internal, kinetic,
potential, etc. energies
E in = E out
Lecture 11 201125
=0
Compressor (cont)
Expressing the various terms
W in + mh1 = Qout + mh2 (since ke pe 0)
W in = Qout + m( h2 h1 )
= Qout + mcp (T2 T1 )
Thus
W in = Qout + mcp (T2 T1)
= (90/60 kg/s)(20 kJ/kg)
+ (90/60 kg/s)(5.1926 kJ/kg K)(430 310)K
= 965 kW
Lecture 11 201125
Throttling Valve - flow restrictor
Significant pressure drop
Typically
dQ/dt ~ 0
dW/dt = 0
ke ~ 0
pe ~ 0
Thus
h2 = h1
Lecture 11 201125
Ideal Gases and Throttling
h2 = h1
h = h(T)
T2 = T1
Lecture 11 201125
Mixing Chamber
Typically
dQ/dt ~ 0
dW/dt = 0
ke ~ 0
pe ~ 0
m1 + m2 = m3
m1 h1 + m2 h2 = m3 h3 = ( m1 + m2 ) h3
Lecture 11 201125
Example Mixing Chamber
Water at 50F and 50 psia is heated in a chamber by mixing
with saturated water vapor at 50 psia. If both streams enter
the mixing chamber at the same mass flow rate, find the
temperature and quality of the exiting stream.
T1 = 50F
H2O
(P = 50 psia)
T3, x3
Sat. vapor
m2 = m1
Lecture 11 201125
Mixing Chamber (cont)
System: The mixing chamber is the system, which is a
control volume since mass crosses the boundary.
Assumptions
1 This is a steady-flow process since there is no change
with time.
2 Kinetic and potential energy changes are negligible.
3 There are no work interactions.
4 The device is adiabatic and thus heat transfer is
negligible
Lecture 11 201125
Mixing Chamber (cont)
Analysis The mass and energy balances for this
steady-flow system can be expressed in the rate
form
Mass balance:
min mout = msystem ( steady ) = 0
min = mout
m1 + m2 = m3 = 2 m
m1 = m2 = m
Lecture 11 201125
Mixing Chamber (cont)
Energy balance
E ut
i
o
n E
Rate of net energy transfer
by heat, work, and mass
=
E system ( steady )
=0
Rate of change in internal, kinetic,
potential, etc. energies
E in = E out
m1h1 + m2 h2 = m3 h3 (since Q = W = ke pe 0)
Combining mass and energy gives
mh1 + mh2 = 2mh3 or h3 = (h1 + h2 ) / 2
Lecture 11 201125
Mixing Chamber (cont)
Properties
From steam tables (Tables A-5E - A-6E),
h1 hf @ 50F = 18.07 Btu/lbm
h2 = hg @ 50 psia = 1174.2 Btu/lbm
Lecture 11 201125
Mixing Chamber (cont)
Substituting,
h3 = (18.07 + 1174.2)/2 = 596.16 Btu/lbm
At 50 psia
hf = 250.21 Btu/lbm and hg = 1174.2 Btu/lbm
saturated mixture since hf < h3 < hg
Therefore,
T3 = Tsat @ 50 psia = 280.99F
x3 =
h3 h f
h fg
=
596.16 250.21
= 0.374
924.03
Lecture 11 201125
Heat Exchanger
What is the system?
m1 = m2 = m A
Q = m A ( h2 h1 )
m3 = m4 = mB
m A ( h2 h1 ) = mB ( h3 h4 )
Lecture 11 201125
Heat Exchanger - Example
Steam enters the condenser of a steam power
plant at 20 kPa and a quality of 95% with a
mass flow rate of 2000 kg/h. It is to be cooled
by water from a nearby river by circulating the
water through the tubes within the condenser.
To prevent thermal pollution, the river water
is not allowed to experience a temperature
rise above 10C.
If the steam is to leave the condenser as
saturated liquid at 20 kPa
DETERMINE THE MASS FLOW RATE OF THE
COOLING WATER REQUIRED
Lecture 11 201125
HE - Example
State Pt 3: Steam @ 20 kPa, x=0.95
State Pt 2
T<10C
State Pt 1
Water
State Pt 4: Steam @ 20 kPa, sat liq
Lecture 11 201125
HE (cont)
System: Inner walls of heat exchanger
which includes both streams
Assumptions
This is a steady-flow process since there is no
change with time.
Kinetic and potential energy changes are
negligible.
There are no work interactions.
Heat loss from the device to the surroundings is
negligible and thus heat transfer from the hot
fluid is equal to the heat transfer to the cold
fluid.
Liquid water is an incompressible substance
with constant specific heats at room
temperature.
Lecture 11 201125
HE
The mass balances for this steady-flow
system can be expressed in the rate
form as
Mass balance (for each fluid stream):
(steady )
min mout = msystem
=0
min = mout
m1 = m2 = mw
and
m3 = m4 = ms
Lecture 11 201125
HE
The energy balances for this steadyflow system can be expressed in the
rate form as
Ein Eout
Rate of net energy transfer
by heat, work, and mass
=
e
E(ssytstady)
em
Rate of change in internal, kinetic,
potential, etc. energies
Ein = Eout
m1h1 + m3h3 = m2h2 + m4h4
(NOTE : Q = W = ke pe 0)
Lecture 11 201125
=0
HE Ex
Solving the two equations together
mw (h2 h1) = ms (h3 h4 )
Solving for m w :
w = h3 h4 ms h3 h4 ms
m
h h
cp ( T2 T1)
2
1
Lecture 11 201125
HE Ex
Properties
Steam:
P3 = 20 kPa
x 3 = 0.95
P4 = 20 kPa
sat. liquid
!
" h3 = hf + x 3hfg = 251.42 + 0.95 2357.5
#
= 2491.1 kJ/kg
!
" h4 hf @ 20 kPa = 251.42 kJ/kg
#
Water: c = 4.18 kJ/kgC
Lecture 11 201125
HE Ex
Substituting
(2491.1 251.42) kJ/kg
mw =
(20,000/3600 kg/s)
(4.18 kJ/kg C)(10C)
= 297.7 kg/s = 298 kg/s
Lecture 11 201125
Next
Lecture 11 201125