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ME 300 HW Set #2
Date Due: Friday, January 23, 2015
SP3. A supply line, turbine, and tank arrangement is shown below. The steam in the supply line
is at a pressure of 15.0 bar and 400˚C. From this supply line, steam flows through a
turbine. At the exit of the turbine, the steam flows into a large tank with a volume of 40
m3. Initially, the tank is evacuated. Steam flows from the supply line through the turbine
into the tank until the tank pressure is 10.0 bar. The temperature of the steam in the tank at
the final condition is measured to be 320˚C. Determine the work done by the turbine
during the tank filling process. Assume that the process is adiabatic, and neglect potential
and kinetic energy effects.
Given:
Line : TL  400C , pL  15.0 bar
Tank : Initial state 1: evacuated , Vtank  40 m3
Final state 2 : T2  320C , p2  10.0 bar
Find:
turbine work Wt
System Sketch:
Assumptions:
(1) adiabatic process Q CV  0
(2)
(3)
(4)
(5)
neglect potential energy effects
neglect kinetic energy effects
inlet flow to tank only, no exit flow
neglect volume of steam in turbine compared to tank
Basic Equations:
dECV  t
dt

1
Q CV  WCV
3

2
Vi 2

  m i  hi 
 g zi
2
i








e
3

2
Ve2

m e  he 
 g ze
2


4
Solution:
Integrating the first law equation over the tank filling process :
ECV  ECV 2  ECV 1  U 2  mi u2   WCV  mi hL
 WCV  mi  hL  u2 
From Table A  4,
TL  400C , pL  15.0 bar  hL  3255.8
kJ
kg
T2  320C , p2  10.0 bar  v 2  0.2678
m3
kg
mi  m2,tank 
tank

v2
u2  2826.1
40 m3
 149.4 kg
m3
0.2678
kg

kJ 
WCV  mi  hL  u2   149.4 kg   3255.8  2826.1

kg 

WCV  64, 200 kJ
kJ
kg





SP4.
A vertical piston-cylinder device initially contains 0.1 m3 of a saturated liquid-vapor
mixture (SLVM) of R134a at 20°C and an initial quality of 0.8. At this initial condition,
the piston is resting on a stop. The piston-cylinder device is connected to a supply line
with superheated R134a at 100°C and a pressure of 16.0 bar. The valve between the
supply line and the piston-cylinder device is opened and is left open until the pressure in
the piston-cylinder device reaches 12.0 bar. The piston is observed to start moving when
the pressure in the cylinder is 8.0 bar and continues to rise until it reaches a second stop.
At the second stop, the piston-cylinder volume is 0.2 m3. The final temperature and
pressure in the piston-cylinder device are 70°C and 12.0 bar, respectively. Determine the
heat transfer to or from the piston-cylinder device for the filling process.
pL = 16.0 bar
TL = 100ºC
p2 = 12.0 bar
T2 = 70ºC
V2 = 0.3 m3
T1=20ºC x1=0.8
V1=0.1 m3
State 1
Valve
State 2
Given:
Line : TL  100C , pL  16.0 bar
Piston  Cylinder : Initial state 1: T1  20C , x1  0.05, V1  0.1 m3
Final state 2 : T2  70C , p2  12.0 bar , V2  0.3 m3
Find:
heat transfer QCV  Q12
System Sketch:
pL = 16.0 bar
TL = 100ºC
p2 = 12.0 bar
T2 = 70ºC
V2 = 0.3 m3
T1=20ºC x1=0.8
V1=0.1 m3
State 1
State 2
Valve
Assumptions:
(1) hi  hL during the filling process
(2) neglect potential energy effects
(3) neglect kinetic energy effects
(4) inlet flow to piston-cylinder (PC) system only, no exit flow
Basic Equations:
Conservation of mass :
dmCV  t 

dt
dECV  t  
First Law :
 QCV  WCV
dt
3
4
 m   m
i
i
e
3

2
V2

  m i  hi  i  g zi
2
i


2
ECV  U CV  KECV  PECV
dU CV  t  
 QCV  WCV  m i hi
dt
2
WCV   p d 
1
u1  1  x1  u f  x1u g
Solution:
 m i
e
v1  1  x1  v f  x1v g






e
3

2
V2

m e  he  e  g ze
2


4





Integrating the consv of mass equation over the process:
mCV  m2  m1  mi
Integrating the consv of energy equation over the process:
ECV  U 2  U1  m2u2  m1u1  QCV  WCV  mi hL 
QCV  WCV   m2  m1  hL  m2u2  m1u1
Calculating the mass in the PC system at the initial and final states:
From Tables A-10 and A-12,
T1  20C  v f  0.8157 103
m3
m3
, v g  0.0358
kg
kg
v1  1  x1  v f  x1v g  0.2  0.8157 103   0.8  0.0358   0.0288
u f  76.80
m3
kg
kJ
kJ
, u g  237.91
kg
kg
u1  1  x1  u f  x1u g  0.2  76.80   0.8  237.91   205.7
T2  70C , p2  12.0 bar 
TL  70C , pL  16.0 bar 
kJ
kg
m3
kJ
, u2  275.59
kg
kg
kJ
hL  327.46
kg
v 2  0.01947
0.1 m3
V2
0.3 m3
3.47

kg
m


 15.4 kg
2
m3
m3
v2
0.0288
0.01947
kg
kg
During the piston-cylinder expansion process,
kJ
p  constant  8.0 bar  800 3
m
2
kJ 

WCV  W12   p dV  p V2  V1    800 3   0.3  0.1 m3   160 kJ
1
m 

m1 
V1

v1
QCV  WCV   m2  m1  hL  m2u2  m1u1  160 kJ  15.4  3.47 kg  327.46 kJ 
 15.4 kg  275.59 kJ    3.47 kg  205.7 kJ 
QCV  216 kJ