9.1 SOLUTIONS
633
CHAPTER NINE
Solutions for Section 9.1
Exercises
1. The first term is 21 + 1 = 3. The second term is 22 + 1 = 5. The third term is 23 + 1 = 9, the fourth is 24 + 1 = 17, and the fift
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 piston-cylinder assembly at T1=500K and
reversible-isothermal expansion
p2=1 MPa
irreversible adiabatic expansion
p3=
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
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
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 10-3
u3 = u f (8o C) + x3 u g (8o C) u f (8o C) = 162.67
u
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
A
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 possib
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 incompre
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
&
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
EEEE 281
Fall 2015
Quiz 4
W
Problem 1: 20 points
Use superposition to nd the power dissipated by the SOO-kQ resistor in Fig.
5.50.
gure 5.50
500 kn
60 [LA 2.? M cfw_1 5 MD
. - ., _,J_ o L
E g 7 f,"
RIT Dubai
EEEE 281
MIDTERM 1
Fall 2014
Dr. Tlili B.
Closed book, closed notes
Time allowed: 1 11 30 min
1. Show all your work to receive credit.
2. Give your answer in Engineering notations. Problem
RIT Dubai
EEEE 281
Quiz 2 Solutions
Fall 2015
Use Kirchhoffs and Ohms laws in a step-by-step procedure to evaluate all
the currents and voltages in the circuit of Fig. 3.50. (b) Calculate the power
ab
10.1 SOLUTIONS
689
CHAPTER TEN
Solutions for Section 10.1
Exercises
1. Let f (x) = 1 = (1 - x)-1 . Then f (0) = 1. 1-x f (x) f (x) f (x) f (4) (x) f (5) (x) f (6) (x) f (7) (x) = = = = = = =
1!(1 - x)
11.1 SOLUTIONS
761
CHAPTER ELEVEN
Solutions for Section 11.1
Exercises
1. (a) (III) An island can only sustain the population up to a certain size. The population will grow until it reaches this limit
12.1 SOLUTIONS
861
CHAPTER TWELVE
Solutions for Section 12.1
Exercises
1. The distance of a point P = (x, y, z) from the yz-plane is |x|, from the xz-plane is |y|, and from the xy-plane is |z|. So, B
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
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 an
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 parameterizati
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.
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 x-axis, the same direction as the normal vector. (b) The