Qz10: 10 kg/s of ideal gas at 300 K & 1 bar
Insulated compressor
480 K & 4
bar. Negligible KE & PE.
isentropic
compression
&
Wc,rev wc,rev 150 kJ/kg
c = & =
=
= 0.667
Wc
wc
225 kJ/kg
c = 66.7 %
wc,rev
b) isentropic compressor efficiency
150 kJ/kg
c
wc=225
Qz09: 2 kg/s of air at 600 K & 1 bar
compressor
480 K & 9 bar. Surface temperature 500 K, 50 kW heat
to surrounding. Negligible KE & PE, cp=1.287 kJ/kg/K, R= 0.287 kJ/kg/K.
144 2444 144 2444
4
3
4
3
&
&
mhi
mhe
&
&
&
&
&
Wcv = Qcv + m(hi he ) = Qcv + mc p
Qz08a: Change in specific entropy in kJ/kgK for the following cases
T2= 300K p2=800 kPa
c) T1= 300K p1=100 kPa
a) CO2 T1=250K p1=100 kPa
M=32 T2=450K p2=800 kPa
Liquid mercury, cp=0.14 kJ/kgK incompressible
with const cp
Assume CO2 is an ideal gas and use
Qz07: Refrigerator @ SS net power input of 500 W, 3 kW heat from 10oC, deliver heat at 25oC.
KNOWN:
TC=263K
TH=298K
&
Win = 0.5 kW
&
Q = 3 kW
C
FIND: If this is a valid claim
Determine maximum possible coefficient of
perfomance for a refrigerator operati
Qz05: insulated gas turbine at steady state
g = 10
m
s2
i
100m
FIND: vi = ? m/s
&
Wt
e
pe=200 kPa
Te=400 K
Ae=0.01m2
Ai v i
v i = 20
&
Wt = ? kW
ve = ? m/s
&
&
&
me = mi = m
&
&
mi
m RT
i = i i
Ai
Ai pi
i
kJ 1200 K kPa m 3
2kg/s
0.25
vi =
0.01m 2
k
p=const
State 1
3 f (8 C)
0.0317 0.7884 10 3
T (oC)
24
x3 =
=
= 59.8%
o
o
3
p (kPa)
645.66
g (8 C) f (8 C) 0.0525 0.7884 10
(m3/kg) 0.8257 103
u3 = u f (8o C) + x3 u g (8o C) u f (8o C) = 162.67
u (kJ/kg)
82.37
0
x
Alternatively, 1st law for process 2
Qz03a:
1
W=trapezoid area
V2 = 4 m
400 4
64 kPa V
78
1200 kJ
678
44
1
1200 kJ
W12 = pdV = ( p1 + p2 )(V2 V1 )
V =
= 3 m3
2
400 kPa
b) Process 23 linear p(V), work done= trapezoid area
3
V=const
a) Process 12, work is known, linear p(V)
Lin
ear
3
)
2
L
1
Qz02a:
V2 = V1 = 0.5 m
Process 2 is pV=constant p2V2 = p3V3
p3 =
V2
0.5
(1500 kPa )
p2 =
V3
1.5
p3 = 500 kPa
Process 3 linear p(V), but p3=p1
p=constant
Part b) Since V=constant between states (1) and (2)
W12 = 0
Process 2 is pV=constant
W23 = pdV = c ln
Qz01a:
Part a) Force balance for the piston along vertical
direction just at the moment that piston to rise
F
patm
y
mg
=?
For piston
m=50 kg
g=9.81 m/s2
For gas
mg=10g
H=0.1m
mg
A
0.01 m
2
1 N 1 bar
1 kg m/s 2 105 N/m 2
10 5 Pa 1 psi
p = 1.49 bar
4/12/2011
Q01: 10 kg ideal gas with R=0.2 kJ/kg/K and constant cp=1.2 kJ/kg/K in a pistoncylinder assembly at T1=500K and
reversibleisothermal expansion
p2=1 MPa
irreversible adiabatic expansion
p3= 0.1
p1=10 MPa.
MPa and T3=400 K.
isothermal
Alternativ
Q01a: m=10 kg @1200 kPa, 1m3,10m/s 2m3, 20 m/s
with pV2=constant, Q12=600 kJ, PE= 6 kJ
a) final pressure, p2=?
2
V2
W12 =
V1
pdV
cfw_=
pV n = vonst
p2 = 300 kPa
p2V2 p1V1
1 n
(300 2) (1200 1) (kPa m3 )
W12 = 600 kJ
1 2
600 kJ work is done by the system
20.1 SOLUTIONS
1339
CHAPTER TWENTY
Solutions for Section 20.1
Exercises
1. The first vector field appears to be diverging more at the origin, since both fields are zero at the origin and the vectors near the origin are larger in field (I) than they are in
19.1 SOLUTIONS
1307
CHAPTER NINETEEN
Solutions for Section 19.1
Exercises
1. (a) The flux is positive, since F points in direction of positive xaxis, the same direction as the normal vector. (b) The flux is negative, since below the xyplane F points tow
18.1 SOLUTIONS
1257
CHAPTER EIGHTEEN
Solutions for Section 18.1
Exercises
1. Positive, because the vectors are longer on the portion of the path that goes in the same direction as the vector field. 2. Negative because the vector field points in the opposi
17.1 SOLUTIONS
1197
CHAPTER SEVENTEEN
Solutions for Section 17.1
Exercises
1. One possible parameterization is x = 3 + t, 2. One possible parameterization is x = 1 + 3t, 3. One possible parameterization is x = 3 + 2t, 4. One possible parameterization is
16.1 SOLUTIONS
1105
CHAPTER SIXTEEN
Solutions for Section 16.1
Exercises
1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest and largest values the function takes on each small recta
15.1 SOLUTIONS
1039
CHAPTER FIFTEEN
Solutions for Section 15.1
Exercises
1. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point and is a local maximum; the point C is a saddle point. 2. To fi
14.1 SOLUTIONS
951
CHAPTER FOURTEEN
Solutions for Section 14.1
Exercises
1. If h is small, then With h = 0.01, we find fx (3, 2) f (3 + h, 2)  f (3, 2) . h
3.012 (2+1)
f (3.01, 2)  f (3, 2) = fx (3, 2) 0.01 With h = 0.0001, we get fx (3, 2)
0.01
2 2
