Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition

Name: Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition
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Price: 20 USD
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Author: Julian Cooper

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

Julian Cooper

Contributor

4.9

4 months ago

Preview (16 of 1955)

Chapter 1
1-1 (a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K)
(b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K
(c) 0.04 Ibm/(ft-hr)
3600 sec/hr x1.488 = 16.5 2
Ns
m
μ
(d) 1050 Btu
Ibm x 4
1
9.48x10−
J
Btu x 2.20462 Ibm
kg = 2.44 MJ
kg
(e) 12,000 Btu
Ibm x 1
3.412 = 3.52 kW
(f) 14.7 2
Ibf
in x 6894.76 = 101 kPa
1-2 (a) 120 kPa x
2
lbf / in
6.89476kPa = 17.4 lbf/in2
(b) 100 W
m K− x 0.5778 = 57.8 Btu/hr-ft-F
(c) 0.8 2
W
m K− x 0.1761 = 0.14 Btu/hr-ft2-F
(d) 10-6 N-s/m2 x 1
1.488 = 6.7 x 10-7 lbm
ft sec−
(e) 1200 kW x 3412 = 4.1 x 10-6 Btu/hr

2
(f) 1000 kJ
kg x 1 Btu
1.055 kJ x 1 kg
2.2046 lbm = 430 Btu
lbm
1-3 Hp = 50 (ft) x 0.3048 ( m
ft ) = 15.2 m
∆ P = 15.2 m
1000 Pa/kPa x 9.807
1 ( N
kg ) x 1000 (kg/m3) = 149 kPa
1-4 P =∆ 4
12 (ft) x 0.3048 ( m
ft ) x 9.807
1 ( N
kg ) x 1000 ( 3
kg
m )
∆ P = 996 Pa 1.0 kPa≈
1-5
TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE
+ METER CHARGE
( ) ( ) ( ) ( )96,000 kw - hrs 0.045 $ / kw hr + 624 kw 11 50 $ / kw− −
+ $68 = $4,320 + $7,176 + $68 = $11,564
1-6 7 AM to 6 PM 11 hrs/day, 5 days/wk
hrs days
(11) (22) 242 hrs / month
day months =
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

3
( )
( )
( )
624 kw
ratio = 1.57
96,000 kw hr
242 hr
=
⎛ ⎞−
⎜ ⎟
⎝ ⎠
1-7 This is a trial and error solution since eq. 1-1 cannot be solved
explicitly for i.
Answer converges at just over 4.2% using eq. 1-1
1-8 Determine present worth of savings using eq. 1-1
( ) ( )( )12 12
0.012
$1000 1- 1+ 12
P = 0.012
12
P $134,000
−⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
⎛ ⎞
⎜ ⎟
⎝ ⎠
=
1-9 (a) Q VA= = 2 x 3.08 x 10-3 = 6.16 x 10-3m3/s
m 6.16 x 10Q
ρ= = -3 x 998 = 6.15 kg/s
(b) A= 4
π (0.3)2 = 7.07 x 10-2 m2
Q 7.07x10= -2 x 4 = 0.283 m3 / s;
ρ = 1.255 kq/m3
m = 1.225 x 0.283 = 0.347 kg/s
1-10 V = 3x10x20 = 600m3
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

4
= 600 xiQ 1
4 x 1
3600 = 4.17 x 10-2 m3/s
1-11
p p
3
q = mc T c = 4.183 kJ/(kg-K)
= 983.2 kg/m
ρ
∆
1-11 (cont’d)
( ) ( ) ( ) ( )
3 c
3
m kg kJ
q = 1 983.2 4.183 5 20,564
s kg Km
q = 20,564 kw
=
−

kJ
s
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
1-12 =watq airq−
11,200(1)(10) =
= 25000x60x14.7x144x0.24(t 50)
(53.35x510)
−
11,200 = 5601.5 (t2-50); t2 = (11,200/5601.5) + 50 = 70 F
1-13 Diagram as in 1-12 above.
q wat = - q air
1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000)
6279(90-t2) = 29,400

- l purposes only to
students enrolled in courses for which the textbook has been adopted. this work beyond that permitted
5
t2 = 90 - 29,400
6279 = 85.3 C
1-14 q hA(t= s- )t∞
A=
π (1/12) x 10 = 2.618 ft2
st = tsur 212 F≈
= 10x2.618x(212-50) = 4241 Btu/hrq
1-15 A=
π x 0.25x4 = 3.14 16 m2
hA(tq = s- )t∞
h=
s
q
A(t -t )∞
= 1250
3.1416(100 10)− ; h = 4.42 W/(m2 – C)
1-16 (tpq mc= 2-t1) ; m Q x
ρ= ρ = P/RT = 14.7x144/53.35(76+460)
ρ = 0.074 lbm/ft3
m = 5000x0.074x60 = 22,208 lbm/hr
= 0.24 Btu/lbm-Fpc
q = 22,208x0.24(58-76) = -95,939 Btu/hr
Negative sign indicates cooling
1-17 (t1 pm c 3-t1) +
for-profit basis for testing or instructiona
Any other reproduction or translation of

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Problem 1.1
The force,F , of the wind blowing against a building is given by2 2 DF C V A , whereV is
the wind speed, the density of the air,A the cross-sectional area of the building, andDC is
a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Solution 1.12
2
 D
A
F C V
or2
2
 
D
F
C V A
, where2
F MLT
,3
 ML ,1
V LT ,2
A L
Thus, 
    
-2
0 0 0
2-2 -1 2

 
  
D
MLT
C M L T
ML LT L
Hence,DC is dimensionless.

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Problem 1.2
The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed of
sound in the fluid. For an airplane flying at velocityV in air at absolute temperatureT , the
Mach numberMa is,Ma  V
kRT
,
wherek is a dimensionless constant andR is the specific gas constant for air. Show thatMa is
dimensionless.
Solution 1.2
We denote the dimension of temperature by and use Newton’s second law to get2
 ML
F T .
Then 
 
2
2 2
1


   
   
    
   
   
   
L L
T T
M FL ML L
M T F T
or   
1M
.

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Problem 1.3
Verify the dimensions, in both theFLT andMLT systems, of the following quantities which
appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Volume, (b) acceleration, (c) mass, (d) moment of inertia (area), and (e) work.
Solution 1.3
a)3
volume L
b)1 2
acceleration time rate of change of velocity
 
 LT LT
T
c)mass M2
1 2
or with
mass


F MLT
FL T
d) 
  
2 2 4
moment of inertia area second moment of area L L L
e)work force distance  FL2
2 2
or with
work


F MLT
ML T

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Problem 1.4
Verify the dimensions, in both theFLT andMLT systems, of the following quantities which
appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Angular velocity, (b) energy, (c) moment of inertia (area), (d) power, and (e) pressure.
Solution 1.4
a)1angular displacement
angular velocity time

 T
b)energy ~ capacity of body to do worksingle work force distance 
energy FL2
or with 
F MLT
 
 
2 2 2
energy  
MLT L ML T
c) 
  
2 2 4
second mommoment of ine ent of arertia a e ar a  L L L
d) 
 
 
1 2 1 2 3
power rate of doing work    
 FL FLT MLT L T ML T
T
e)  
2 2 2 1 2
2
force
pressure area
F FL MLT L ML T
L
    


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Problem 1.5
Verify the dimensions, in both theFLT system and theMLT system, of the following quanti-
ties which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Frequency, (b) stress, (c) strain, (d) torque, and (e) work.
Solution 1.5
a)-1cycles
frequency= T
time
b)-2
2
-2
-2 -1 -2
2
force F
stress= FL
area L
Since F MLT ,
MLT
stress ML T
L
c) 
0change in length L
strain= L dimensionless
length L
d) 
 
-2 2 -2
torque=force distance FL MLT L ML T
e) 
 
-2 2 -2
work=force distance FL MLT L ML T

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Problem 1.7
Verify the dimensions, in both theFLT system and theMLT system, of the following quantities
which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Acceleration, (b) stress, (c) moment of a force, (d) volume, and (e) work.
Solution 1.7
a)2
2
velocity
acceleration time

 L LT
T
b)2
2
2
2 1 2
2
force
stress area
Since
stress
F FL
L
F MLT
MLT ML T
L


  

c) 
2 2 2
moment of a force force distance  
  FL MLT L ML T
d)3 3
volume (length) L
e) 
2 2 2
work force distance  
  FL MLT L ML T

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Problem 1.8
Ifp is a pressure,V a velocity, and a fluid density, what are the dimensions (in theMLT
system) of (a)/

p , (b)
pV , and (c)2
/

p V ?
Solution 1.8
a)2 2 2 1 2 2 2
3 3 3
     
  
 
p FL MLT L ML T L T
ML ML ML

b)   
1 2 1 3 2 3 3

     
pV ML T LT ML M L T
c)  
 
1 2 0 0 0
2 23 1
dimensionless

 
 

p ML T M L T
V ML LT

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Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition

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