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HOMEWORK SET #1
| Problem |
Hints |
Selected
Answers |
| 1.4-Case1 |
|
b. Never bankrupt.
Minimum balance is $2960
c. -$1580
d. $263.33 |
| 1.4-Case
2 |
The
answer shown assumes that the deposit and expenses all take place at
the beginning of the month. If you make a different assumption, your
answer will be different. |
b. Never bankrupt.
Minimum balance is $3035.52
c. -$1470.34
d. not given |
HOMEWORK SET #2
| Problem |
Hints |
Answers |
| B.2 |
Show all your work. |
a) m=9.32 slugs, m=300 lbm
b) 300 lbm, W = 300 lbf
c) |
| B.3 |
Show all your work. |
|
| 3.2 |
Clearly define your systems. |
a) -40 kg/s, -130 kg/s, -160 kg/s, 100
kg/s
b) 5 D.O.F. |
| 3.1 |
You'll need the rate form of mass conservation, and you get to integrate it too!
Only selected results are shown to the right. |
a) At t=4 min, dm/dt = -1.26
lbm/min, mtank =992.7 lbm. At t = 10 min, dm/dt = 0.642 lbm/min,
mtank = 998.07 lbm.
Net change is -0.07 lbm after 20 min. |
| 3.5 |
Remember the definition of volumetric flow rate, and
its relation to mass flow rate for incompressible stuff. |
a) (2/3)Uhb
b) 0.174 kg/s |
| 3.6 |
|
a) 4.99 mm
b) 0.25 m3/s
c) 0.0125 m/s |
HOMEWORK SET #3
| Problem |
Hints |
Answers |
| 3.8 |
What is the exit area? |
a) 471 kg/s
b) 0.471 m3/s
c) For q =
30o V = 1.44 m/s |
| 3.10 |
|
a) Vdot = 3
m3/min, V = 6.37 m/s
b) dhlower/dt = 0.06
m/min (increasing) |
| 3.11 |
|
9.58 s |
| 3.13 |
Selected
answers |
a) O2 -
16.9%, CO2 - 5.7% ...
b) mass O2 - 0.87 g, mass N2 - 3.78 g
c) 28.59
d) 0.0047 m3 |
| 3.14 |
oR =
oF + 460 |
a) CH4 - 48.4%,
C2H6 - 34.9%, CO2 - 10.2%, N2 - 6.5%
b) 21.5 lbm/lbmol
c) 771 lbm/min |
HOMEWORK SET #4
| 3.16 |
Be careful of the independence of you various
equations. |
mdot4 = 2302.5 kg/hr,
mdot5 = 265.9 kg/hr, mdot6 =2036.6 kg/hr, mfF4 = 6.08%, mfS4 = 53.1%, mfP4=
0.11%, mfW4 = 40.7%, mfF6 = 6.87%, mfS6 =
60.0%, mfP6 = 0.12% |
| P1 |
If you need to, you can assume that the distance
between the community pipeline and the ground is zero ft. |
a)
b) 80 ft
c) 48.1 min |
| 3.17 |
You'll need more than one system. As usual, be
careful of the independence of you various equations. |
mdot2 = 96.51 kg/hr,
mdot3 = 91.40 kg/hr, mdot4 =135.1 kg/hr, mdot5 = 30.0 kg/hr, mdot6 = 126.5
kg/hr, mfS1 = 0.85, mfA2 = 0.9901, mfS3 = 0.93,
mfA4 = 0.9091, mfA5 = 0.9091, mfA6 = 0.9709 |
| 3.18 |
|
|
HOMEWORK SET #5
| Problem |
Hints |
Answers |
| 4.9 |
|
i1=4.57 A
i2 = 2.29 A
i3 = 1.14 A |
| 4.10 |
|
Ix = 5.38 A, Vx
= 27.69 V |
| P2 |
|
60 W |
| 4.12 |
|
Ix = -3.24 A, Vx
= 753.9 V |
| 4.13 |
|
Ix = -6.40 A |
| 4.14 |
|
Ix
= 12.6 A |
HOMEWORK SET #6
| Problem |
Hints |
Answers |
| 5.13 |
|
0.108 kg |
| 5.14 |
The answer for P
is for 6.09. Express your answer in terms of mdot. |
a = 6.09
or 52.9
P = 26.8*mdot |
| 5.15 |
You will need to use the chain rule to integrate, that is, dv/dt =
(dv/dt)*(dx/dx)=vdv/dx. |
565.7
m/s |
| 5.16 |
The maximum velocity will occur when the acceleration is zero.
You should be able to find the time from the plot. |
a)
5 s
b) 49.9 ft/s
c)17.88 s |
HOMEWORK SET #7
| Problem |
Hints |
Answers |
| 5.17 |
What is the
relationship of the
accelerations of the boxes if they stay together? What is the normal force between the boxes if they separate? |
a) aA
= 0.997 ft/s2 (up the incline)
aB = 1.62 ft/s2 (up the incline)
b) 1.87 ft |
| 5.18 |
Calculate the
friction force and compare to the maximum allowable friction force to see if it slides. |
a) aA
= aB = 0.667 m/s2
b) aA = aB = 1.0 m/s2
c) aA = 5.1 m/s2, aB = 0.98 m/s2 |
| 5.19 |
The
angle the package is heading down is 30 degrees with respect to
horizontal. |
a)
0.742 m/s
b) 23.9 N-s (38.9 up from horizontal left) |
| 5.20 |
|
a) A was speeding
b) vB = 50 km/hr, vA = 192 km/hr |
| 5.21 |
|
a) 8.51 km/hr
b) 6.67 N |
HOMEWORK SET #8
| Problem |
Hints |
Answers |
| 5.22 |
|
a)
Fx = 161 lbf (right),
Fy = 122.3 lbf (down) |
| 5.23 |
|
a)
0.86 m/s (right)
b) aA = 0.97 m/s2 (left)
aC = 2.94 m/s2 (right)
c) t = 0.22 s |
| 6.10 |
|
Ex
= 90 kN (left)
Ey = 200 kN (up)
ME = 180 kN-m (CCW) |
| 6.11 |
|
a)
47.2
b) Ox = 27.7 N (right)
Oy = 4.55 N (down) |
HOMEWORK SET #9
| Problem |
Hints |
Answers |
| 6.12 |
|
6
in/s |
| 6.13 |
You
will need to find the velocity of the sand hitting the belt. |
Cx
= 90 N (right)
Cy = 2362 N (up)
Dy = 2900 N (up) |
| 6.14 |
Assume
that all surfaces are "smooth", that is, frictionless. |
Fc =
3.43 N (20 up from right)
B = 24.4 N (73.4 up from left) |
| 6.15 |
From LM
and AM all we can solve for is acceleration. Use kinematics to find the distance. |
a)
9.66 ft/s2
b) 2.88 ft/s2
c) 17.34 ft |
HOMEWORK SET #10
| Problem |
Hints |
Answers |
| 7.5 |
|
a)
1.15 m/s
b) 1.75 m/s |
| 7.42 |
|
4.05
m/s |
| 7.43 |
|
20.3
ft |
| 7.44 |
Does
the answer to c) make sense to you? This would obviously not be a
very good design for a bungee jumping tower. |
a)
k = 533 lbf/ft
b) d = 37 ft
c) 710 ft/s2 |
| 7.45 |
Be
careful with units. Be sure to consider the impact first. When
you eventually use conservation of energy take note that the springs will
be initially compressed some due to the 20 lbf object sitting on them. |
0.271
ft |
| 7.46 |
The
spring will be initially compressed. Also be careful when
determining the distance the 0.5 kN force will travel. |
3.47
m/s |
HOMEWORK SET #11
| Problem |
Hints |
Answers |
| 7.9 |
Only
selected answers are provided to the right. The process should look something like
a pie wedge for part c). Note that V1 = V3. |
b)
process 1-2: 128.8 kJ
process 2-3: -320 kJ
process 3-1: 0
d) -191.2 kJ |
| 7.10 |
|
a)
186.5 ft-lbf
b) -745.9 ft-lbf
c) 359.5 ft-lbf, 0.462 BTU |
| 7.11 |
|
b) 2
kW at t=1.2 h
0 kW at t=2.4 h
c) 5 kW-h, 18000 kJ |
| 7.13 |
For
part b) be sure to convert q to
rad. |
a) Wgas
= 0.116 kJ, 31% spring, 69% atm.
b) 0.427 kJ
c) 0.312 kJ |
HOMEWORK SET #12
| Problem |
Hints |
Answers |
| 7.15 |
Solve
for the volumetric flow rates in terms of the mass flow rates and the
densities. |
a)
0.02 kg/s
b) 4.2
c) 2.74 kW or 3.67 hp |
| 7.16 |
|
a)
0.022 kg/s
b) -264 kW or 264 kW (out of system)
c) 0.0136 m2 |
| 7.17 |
|
a)
194,000 Btu/h (out)
b) 258 A
c) 111.8 ft-lbf |
| 7.25 |
Only
selective answer are given to the right. This is the first year I
have used this problem, so the answers have not been checked. If you
get different numbers and can't find an error in your work, please let me
know. |
a)
mixer shaft: 100 rpm
b) 5236 ft-lbf/s, 9.52 hp
d) Q = 581.8 ft-lbf/s = 1.06hp
f) 30.9 ft-lbf
g) 0.23 R |
HOMEWORK SET #13
| Problem |
Hints |
Answers |
| 7.20 |
This
is the first year I have used this problem (and the following 3), so the
answers have not been checked. If you get different numbers and
can't find an error in your work, please let me know. |
a)
703.5 kg/min
b) 187.6 kW
c) 2990 N-m |
| 7.21 |
Selected
answers given. For part c), what do you know about the change
in temperature if PV = constant for an ideal gas? |
a)
1.169 x 10-4 kg
b) 20.8 J
|
| 7.30 |
Selected
answers given. There is an error in the problem statement. At
the beginning of the problem it talks about using oil and and the end it
talks about water. Assume the impressible substance for the problem
is oil. You will need to look up all the appropriate properties in
the back of the book. |
a)
553.8 R
b) 5.83 ft2
c) 0.0029 ft2 |
| 7.34 |
|
a)
20.03 m/s
b) 17.82 m/s |
HOMEWORK SET #14
| Problem |
Hints |
Answers |
| 7.35 |
Note the
importance of
sketching the system correctly. Also note the difference between the CONSERVATION
PRINCIPLE where Qdot appears
and the RELATIONSHIP for Qdot
in part d). |
a)
b) i) i = 1.2 mA
ii) Wdot
= 0.0144 W
iii) Qdot
= -Wdot = -0.0144 W
d) Tsurface
= 316 K
d) -103.7 J
|
| 7.36M |
|
a)
1.16 A
b) 41.9 W (out)
c) 164.7 C |
| P3 |
|
Not
available |
| 7.37 |
Selected
answers given |
a)
Process 1-2: W = 150 kJ, Q = -374.64 kJ
Process 2-3: W = -160.94 kJ
Process 3-4: Q = 86.11 kJ
Process 4-1: W = 138.53 kJ
b) not available
c) not available |
| 7.47 |
Use
0.5 kg of air, not 2 kg as the problem is stated in the text. Only
selected answers are given. Use the ideal gas model to get
temperatures. |
a)
DU12
= 620 kJ
W23 = 320 kJ
DU31
= -120 kJ
b) refrigeration |
HOMEWORK SET #15
| Problem |
Hints |
Answers |
| 7.40 |
|
a)
3.02
b) 69.8 kW-hr/day
c) $167.52/mo
d) $506.3/mo |
| 8.1 |
Be sure and
correctly identify your system and corresponding heat transfer rates and surface
temperatures, e.g. what leaves the gear box enters the air layer. |
a) -254.4 Btu/hr
b) 0.450 Btu/(hr-R)
c) 0.030 Btu/(hr-R)
d) 0.480 Btu/(hr-R) |
| 8.2 |
Be sure to
explain why this device is impossible. |
a) 700 kW
b) -2.5 kW/K
c) impossible |
| 8.15 |
Remember what
s2-s1
(little s) is for an incompressible substance, and don't forget that you
need absolute temperature in the equation. (Come on and sing it
with me now: When in doubt, use absolute temperature!) |
a) 0, 30 kW
b) 0.0695 kW/K, 50.8 kW
c) |
| 8.16 |
What term goes
away in the S-accounting equation for internally reversible processes? |
a)
475oC, 686
m/s
b) 0.201
kW/K, 606 m/s
c) Entropy
generation
is a bad, bad dog, and it will always hurt you. |
HOMEWORK SET #16
| Problem |
Hints |
Answers |
| 8.17 |
Choose you systems
the way we've indicated, and all is well. Also note that you want the finite
time forms of your accounting equations. |
a)17,
500 kJ
b) 8.9 kJ/K
c) 56.2 kJ/K
d) |
| 8.21 |
Remember that
absolute temperature thing. What are acceptable ranges for Sgen? |
987.8
kW |
| 8.9 |
Don't be a xenophobe!
Embrace English units with good humour. |
a)
13.1 HP, 66,667 B/hr, 51.9 B/(hr-oR)
b) 12.67
(Oh my, that's big!), 7895 B/hr (and that's little!)
c) You
bet it works! (Would we steer you wrong?) |
| 8.11 |
Why is the answer
for c) < the answer for b)? Where are the extra losses hidden? |
a) 22 MW, 22%
b) 33.5%
c) 31.5% |
|
