Lecture_3_Heat and t..

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Example Problems
A gas undergoes a thermodynamic cycle consisting of three processes:
Process 1 - 2:
Compression with pV = constant from p1 = 105 Pa,
V1 = 1.6 m3, V2 = 0.2 m3, (U2 –U1) = 0.
Process 2 – 3:
Constant pressure expansion from V2 to V3.
Process 3 – 1:
Constant volume, (U3 –U1) = -3549 kJ.
(a) Sketch the cycle.
(b) Determine the heat transfer and the work for process 2 – 3.
(a)
p
2
3
Notes:
For 1 - 2: Since U12 = 0, Q12 = W12
1
V
For 3 – 1: Since dV31 = 0, W31 = 0.
Ucycle = 0.
Example Problems
(b)
U cycle  U12  U 23  U 31
0
 0  ΔU 23  -3549  0
U 23  3549 J
The work in the process 2 – 3 is given by,
dW23  VV3 pdV  p2 (V3  V2 )
2
In order to determine this work, we need V2 since we know that V3 = V1.
We can get V2 from the relation, pV = constant .
p1V1  constant  p2V2 ; p2  p1V1 / V2  8x105 Pa.
W23  p2 (V3  V2 )  8x105 (1.6  0.2)  1120 kJ
Q23  U 23  W23  3549  1120  4669 kJ.
Also note that since U12  0, Q12  W12 and,
W12   pdV  
V2
V1
V2 dV
constant
dV
 p1V1 
 1.6 x105 ln  0.2 /1.6   332.7 kJ
V1 V
V
Example Problems
Summary in kJ, note that U for the cycle = 0.
U12 = 0
W12 = -332.7
Q12 = -332.7
U23 = 3549
W23 = 1120
Q23 = 4669
U31 = -3549
W31 = 0
Q31 = -3549
Wcycle = Qcycle = 787.3 kJ
A gas in a piston-cylinder assembly undergoes an expansion process
for which the relationship between pressure and volume is given by
PVn = constant = A
The initial pressure is 3 bars, the initial volume is 0.1m3
and the final volume is 0.2 m3. Determine the work for the process in kJ if
(a) n = 1.5, (b) n = 1.0 and (c) n = 0.
In each case the work is given by,
p
 pdV
A
Vn
 V21 n  V11 n 
dV
W  A n  A

1

n
V


V1
V2
The constant A can be evaluated at either end state
 V21 n  V11 n
dV
W  A n  A
 1 n
V1 V
V2
p2V2n  p1V1n  A
  p2V2  p1V1 


1

n
 

This expression is valid for all n except n =1.
(a) To evaluate W, the pressure at state 2 is required.
n
 V1 
 0.1 
p2  p1     3bar  

V
0.2


 2
1.5
 1.06bar
 1.06bar  0.2m3  3bar  0.1m3  105 N / m2
W 

1

1.5

 1bar
 17.6kJ
  1kJ 
 3

10
N
/
m



(b) For n = 1 we need to consider this special case, pV  A
V2
W   p1V1  ln
V1
W  20.79kJ
(c) For n = 0, the pressure volume relation is just p = A
W   pdV  p V2  V1 
W  30kJ
Example Problems
A gas is enclosed in a cylinder with a movable
piston. Under conditions that the walls are
Adiabatic, a quasi-static increase in volume
results in a decrease in pressure according to,
PA, VA
PV 5 / 3  constant for Q = 0.
WAB
A
D
105
P (Pa)
Find the quasi-static work done on the
system and the net heat transfer to the
system in each of the three processes,
ADB, ACB and the linear AB. In the
process ADB the gas is heated at
constant pressure (P = 105 Pa) until its
volume increases from 10-3 to 8 x 10-3
m3. The gas is then cooled at constant
volume until its pressure decreases to
105/32 Pa. The other processes (ACB
and AB) can be similarly interpreted
according to the figure.
105/32
C
B
10-3
8 x 10-3
V (m3)
Example Problems
U = Q – W
From the first Law:
For the adiabatic process:
U = – W
U  U B  U A  WAB   PA 
VB
VA
5/3
 VA 
 V  dV
 
3
= PAVA5 / 3 VB 2 / 3  VA 2 / 3 
2
3
=  25  100   112.5 J
2
For the process ADB:
WADB   PdV PA VD  VA   700 J
but, U B  U A  QADB  WADB
and from the result of the adiabatic process
QADB  U B  U A   WADB  112.5  700  587.5 J
Example Problems
Similarly for the process ACB: we find that WACB = -21.9 J and QACB = -90.6 J
For the process AB: W can also calculated from the area:
WAB  360.9 J and from the first law, QAB = 248.4 J
Note that while we can calculate QADB and QACB , we can not calculate QAD , QDB , QAC
and QCB , separately for we do not know (UD – UA) or (UC – UA).
Given the standard enthalpies of formation, H f ,find the enthalpy for the
following reaction at 298K and 1 atmosphere pressure.
o
H of  kJ / mol 
MnSiO3
MnO
SiO2
-246
-384
-910
MnSiO3 (s, 298)  MnO(s,298) + SiO2 (quartz,298)
H o 

products
H of 

H of
reac tan ts
 910  384  (246)  1048 KJ
Given the specific heat functionality find the heat of reaction at 800K.
Heat capacity Constants (J/mol-K)
MnSiO3
MnO
SiO2
c x 10-5
b x 103
a
110
46.0
46.9
16.2
8.2
34.2
-25.8
-3.7
-11.3
Temp range (K)
298-1500
298-900
298-1000
First bring each of the components in the reaction from 298K to 800K
dH  C p dT

800
298
dH  
800
298
C p dT
H (800)  H (298)  
800
298
H (800)  H of  
800
H (800)  H  
800
298
o
f
298
C p dT
C p dT
 a  bT  cT  dT
2
Then evaluate H for the reaction at 800K
H (800)  H  
o
f
800
298
 a  bT  cT  dT
2
800
b 2 c 

 H of   aT 
T 

2
T  298

a =-17.1
b = 26.2 x 10-3
c = 10.8 x 105
 26.2 x 10-3
2
2 
H (800)  H   17.1 800  298   
800

298


2


1 
5  1
 10.8 x 10  


800
298 

o
f
H (800)  1048kJ  1.0kJ  1049kJ
P,V, T Relations and Thermodynamic Properties
Graphical equation of state
P, v, T surface projected
on the P-T plane.
Phase diagram
P,v, T surface projected
on the P- v plane.
P,v, T surface for a substance
that expands on freezing.
v = V/m, specific volume
P,V, T Relations and Thermodynamic Properties
Graphical equation of state
P,v, T surface for a substance
that contracts on freezing.
Phase diagram and P-v surface
for a substance that contracts on
freezing.
P,V, T Relations and Thermodynamic Properties
Phase change: consider a container of liquid water heated at constant
pressure (1 Atm ~ 105 Pa)
(a) The temperature of the water
rises to 100 C.
(b) At 100 C the heat energy goes
into converting the liquid water to
water vapor. The volume of the
system increases. The temperature
of the two – phase, liquid/vapor
system stays constant (100 C) until
the last drop of liquid disappears.
(c) After all the fluid is converted to vapor, the temperature and volume rise as
heat is added to the vapor.
P,V, T Relations and Thermodynamic Properties
Saturated liquid
Saturated vapor
x
1-x
g
f
vf
v
P,v, T surface projected on
the T- v plane for water.
Lever rule:
x(vg  v)  (1  x)(v  v f )
v  (1  x)v f  xvg rule of mixtures
x
v  vf
vg  v f
vg-v
v-vf
vg
vf
v
vg
x is the mass fraction of the
vapor called the quality.
x
mvap
mliq  mvap
P,V, T Relations and Thermodynamic Properties
Specific internal energy and enthalpy
• u = U/unit mass
• h = H/unit mass
(J/kg)
(J/kg)
u = Q/mass – W/mass
h = u + pv
The simple rule of mixtures can always be used for any of the specific
quantities (v, u, h):
h = (1-x)hf + xhg
u = (1-x)uf + xug
v = (1-x)vf + xvg
h, u, v are the specific enthalpy, internal energy, and volume in the two-phase
fluid/gas region.
Example
Water in a piston cylinder assembly undergoes two processes from an initial state
where the pressure is 106 Pa and the temperature is 400 C.
Process 1-2:
Process 2-3:
The water is cooled as it compressed at constant pressure to the
saturated vapor state at 106 Pa.
The water is cooled at constant volume to 150 ºC.
Sketch the processes on p-v and T-v diagrams.
For the overall process determine the work and heat transfer in kJ/kg.
Water
1
10
2
1
400 C
179.9 C
4.758
3
v (m3/kg)
150 C
T (ºC)
P (105 Pa)
Boundary
106 Pa
400 C
179.9 C 2
4.758 x 105 Pa
150 C
3
v (m3/kg)
The only work done in this process is the constant pressure compression,
process 1-2, since process 2-3 is constant volume.
W   pdV  p V2  V1 
V2
V1
W
V V 
 p  2  1   p  v2  v1 
m
 m m
Since for state 1 we know P and T, we can get the specific volume from a data Table,
(properties of superheated water vapor) v1 = 0.3066 m3/kg. The specific volume
at state 2 is the saturated value at 10 bar v2 = 0.1944 m3/kg.
W/m = 106 (0.1944 – 0.3066) = -112.2 kJ/kg
The minus sign indicates
work done ON the system
From the 1st Law, Q = U + W, and dividing through by the mass,
Q U W
W

   u3  u1  
m
m
m
m
u1 is the specific internal energy in state 1, which we can get also get
from a Table as 2957.3 kJ/kg.
P (105 Pa)
To get u3 note that we are in a 2-phase fluid/vapor region, so u3 will be a linear
combination of uf and ug determined from the quantity of fluid and vapor present
at v3 = v2. The mass fraction of vapor present or “quality” is given by,
Table
2
10
1
400 C
179.9 C
4.758
3
v (m3/kg)
vf
v3 v
g
150 C
v3  v f
0.1994  1.0905 x103
x3 

 0.494
3
vg  v f 0.3928  1.0905 x10
Then using the rule of mixtures we can evaluate
u3.
u3  1  x  u f  xu g  1583.9 kJ/kg
631.68
2559.5
Finally,
Q U W
W

   u3  u1    1583.9  2957.3   112.2   1485.6 kJ/kg
m
m
m
m
The minus sign indicates that heat is
transferred OUT of the system
A two-phase liquid-vapor mixture is initially at 5 bar in a closed container of
volume 0.2 m3 and “quality”, x = 0.10. The system is heated until only saturated
water remains. Determine the mass of the water in the tank and the final pressure.
Given Data
V = 0.2 m3
P1= 5 bar
x1 = 0.10
x2 = 1.0 (saturated water)
H2O
Using the Table:
State 1:
vf1= 1.0296 x 10-3 m3/kg
vg1= 0.3749 m3/kg
The water is in a closed system and the total volume and mass are constant.
P (bar)
v1  xvg1  (1  x)v f 1  0.03842m3 / kg
1 m

v V
V
0.2m3
m 
 5.21kg
v 0.03842m3 / kg

?
P2 =?
2
5.0
P1
1
v
v2  v1
P2  Psat @ v2  51.45bar linearly extrapolated between
vf1
v v
g1
50 and 60 bar.

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