Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

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SOLUTIONS MANUAL
Fluid Mechanics
for Engineers
First Edition
David A. Chin
January 2018

Chapter 1
Properties of Fluids
1.1. (a) Find the dimensions of CL as follows,
CL = FL
1
2 ρV 2 A → CL = MLT−2
(ML−3 )(LT−1 )2 (L2 ) = MLT−2
MLT−2 = 1
Therefore CL is dimensionless and does not depend on the system of units used.
(b) No adjustment factor is necessary when USCS units are used instead of SI units.
1.2. (a) Inserting the dimensions of the variables in the given equation yields
ρ d2 z
dt2 + a dz
dt + bz = c
┌ M
L3
L
T2 + a L
T + b[L] = c
┐ ┌ ┐ ┌ ┐
┌ M
L2 T2 + a L
T + b[L] = c
┐ ┌ ┐
Therefore, the required dimensions of the parameters a, b, and c are,
a = M
L3 T , b = M
L3 T2 , c = M
L2 T2
┌ ┐ ┌ ┐ ┌ ┐
(b) If ρ∗, z∗, and t∗ are the given variables in nonstandard units, then the conversion factors
are:
ρ
ρ∗ = 103 , z
z∗ = 10−3 , t
t∗ = 3600
1.3. (a) Inserting the dimensions of the variables in the given equation yields
Q = 1
n
A 5
3
P 2
3
S 1
2
0 → L3 T−1 = 1
( )
(L2 ) 5
3
L 2
3
(−) 1
2 → L3 T−1 = L 8
3
−
Since the dimension of the left-had side of the equation is not equal to the dimension on
the right-hand side of the equation, the given equation is not dimensionally homogeneous .

(b) If the length units are changed from m to ft and 1 m = 3.281 ft, then inserting this
conversion factor into the given equations requires that
(3.281)−3 Q' = 1
n
(3.281)−2 A' 5
3
[(3.281)−1 P '] 2
3
S 1
2
0
┌ ┐
where the primed quantities have length units of ft. Simplifying the above equation and
removing the primes gives
Q = 3.281 1
3
n
A 5
3
P 2
3
S 1
2
0 → Q = 1.486
n
A 5
3
P 2
3
S 1
2
0
Therefore, the conversion factor to be added is 1.486 .
1.4.
−
−
−
·
·
−
Quantity Dimension Typical SI Unit
energy 2 T−2FL = ML J
force F = MLT−2 N
heat 2 T 2FL = ML J
moment 2 T−2FL = ML N m
momentum MV = MLT−1 kg m/s
power 2FLT−1 T−3= ML W
pressure T−2FL−2 = ML−1 Pa
strain LL−1 =
stress T−2FL−2 = ML−1 Pa
work 2 T 2FL = ML J
1.5.
Given With Prefix
5.63 × 107 N 56.3 MN
8.27 × 105 Pa 827 kPa
23.86 × 10−4 m 2
0.0386 km
7.88 × 105 m 788 km
1.6.
Quantity USCS Abbreviation In SI Units
12 gallons per minute 12 gpm 45.4 L/min
55 miles per hour 55 mph 88.5 km/h
5 feet per second 5 ft/s 1.5 m/s
125 cubic feet per minute 125 cfm 3.54 m3 /min
1000 gallons 1000 gal 3785 L
25 acres 25 ac 10.1 ha
500 horsepower 500 hp 373 kW

1.7. (a)
1 hp = 550 ft·lb
s × 0.3048 m
ft × 4.448 N
lb = 745.7 W
(b)
1 lb/in2 = 1 lb
in2 × 4.448 N
lb × 1
0.02542
in2
m2 = 6.894 × 103 Pa = 6.894 kPa
1.8. From the given data: ρ0 = 1000 kg/m3 , and the density deviates most from 1000 kg/m3 at
T = 100◦C, where ρ = 958.4 kg/m3 . Hence, the maximum error in assuming a density of
1000 kg/m3 is
error = 1000 − 958.4
958.4 × 100 = 4.3%
1.9. From the given data: V1 = 3 L, ρ1 = 1030 kg/m3 , V2 = 5 L, and ρ = 920 kg/m3 . The density
of the mixture, ρm , is given by
ρm = ρ1V1 + ρ2V2
V1 + V2
= (1030)(3) + (920)(5)
3 + 5 = 961 kg/m3
Note: The volumes do not need to be converted from L to m3 since the conversion factor
would cancel out.
1.10. (a) The specific weight, γ, is derived from the density, ρ, using the relation: γ = ρg = 9.807ρ.
Obtaining ρ from Appendix B.1 gives:
Temperature ρ
(◦C) (kg/m3 ) (N/m3 )
γ
0 999.8 9805
20 998.2 9789
100 958.4 9399
(b) The specific gravity, SG, is derived from the density, ρ, using the relation: SG = ρ/ρ4◦ C.
Obtaining ρ from Appendix B.1 gives:
Temperature ρ SG
(◦C) (kg/m3 ) (–)
0 999.8 1.000
20 998.2 0.998
100 958.4 0.958
V1 = 400 L, T1 = 15◦C, and T2 = 90◦C.1.11. From the given data: The densities of water
corresponding to T1 and T2 (from Appendix B.1) are: ρ1 = 999.1 kg/m3 and ρ2 = 965.3 kg/m3 .
(a) The initial mass, m1, in the tank is given by
m1 = ρ1V1 = (999.1)(0.4) = 399.6 kg

The volume of water after heating to 90◦C is given by
V2 = m1
ρ2
= 399.6
965.3 = 0.4140 m3 = 414.0 L
Therefore, the spilled volume, ΔV , is given by
ΔV = V2 V1 = 414 400 = 14 L− −
(b) The spilled mass, Δm is given by
Δm = ρ2ΔV = (965.3)(14 10−3 ) = 13.5 kg×
The percent change in the mass (= percent change in weight) is 13.5/399.6 × 100 =
3.4%
1.12. From the given data: γ = 12 kN/m3 = 12 000 N/m3 . For water at 4◦C: ρw = 1000 kg/m3 .
According to the definitions of density and specific gravity,
ρ = γ
g = (12000)(9.807) = 1224 kg/m3
SG = ρ
ρw
= 1224
1000 = 1.224
1.13. From the given data: SG = 1.5. For water at 4 C:◦ ρw = 1000 kg/m3 . According to the
definitions of density and specific weight,
ρ = SG · ρw = (1.5)(1000) = 1500 kg/m3
γ = ρg = (1500)(9.807) = 14 710 N/m3 14.7 kN m3≃ ·
1.14. For any given volume, V , containing a mixture, let Cm = mass ratio, ρf = density of the pure
fluid, ρm = density of the mixture, mf = mass of pure fluid, mm = mass of mixture, ms =
mass of solids in the mixture, SGf = specific gravity of pure fluid, and SGm = specific gravity
of mixture. Therefore,
mf = ρf V, mm = ρmV, ms = mm mf = (ρm ρf )V− −
Using these relationships yields,
Cm = ms
mm
= (ρm − ρf )//V
ρm//V = 1 − ρf
ρm
→ Cm = 1 − SGf
SGm
1.35. From the given data: ρ = 800 kg/m3 . For water at 4◦C: ρw = 1000 kg/m3 . According to the
definitions given in Equations 1.9 and 1.10,
γ = ρg = (800)(9.807) = 7846 N/m3 7.85 kN/m3≃
SG = ρ
ρw
= 800
1000 = 0.80

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1.16. From the given data: M = 200 kg, and SG = 1.5. At 4 C the density of water is◦ ρw =
1000 kg/m3 . The volume, V , of the reservoir is given by
V = M
ρw SG = 200
(1000)(1.5) = 0.133 m3
·
1.17. From the given data: Wc = 10 N, and Wt = 50 N. For kerosene at 20◦C, ρ = 808 kg/m3 and
γ = 7924 N/m3 (Appendix B.4). Using these data gives the following,
weight of kerosene, Wk = Wt Wc = 50 10 = 40 N− −
volume of kerosene, Vk = Wk
γ = 40
7924 = 5.048 × 10−3 m3 = 5.05 L
mass of kerosene = Wk
g = 40
9.807 = 4.08 kg
1.18. The bulk modulus, Ev, is defined by Equation 1.12 as
Ev = − dp
dV /V (1)
where the density of the fluid, ρ, is defined by
ρ = M
V (2)
where M is the (constant) mass of fluid and V is the volume of fluid that is compressed by
the fluid pressure. Differentiating Equation 2 with respect to V gives
dρ
dV = − M
V 2 (3)
Combining Equations 2 and 3 to eliminate M yields
dρ
dV = − ρ
V
or dV
V = − dρ
ρ (4)
Finally, combining Equations 1 and 4 gives
Ev = dp
dρ/ρ
1.19. Equation 1.13 can be approximated by
Δρ
ρ = Δp
Ev
(1)
and at 20◦C, Ev = 2.18 106 kPa (Table B.1). For× Δρ/ρ = 0.01, Equation 1 becomes
0.01 = Δp
2.18 106 → Δp = 2.18 × 104 kPa = 21.8 MPa
×

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1.20. From the given data: T = 20 C, D = 3 m, R = D/2 = 1.5 m, and◦ Δp = 9 MPa. For water
at 20 C,◦ ρ0 = 998.2 kg/m3 and Ev = 2.18 106 kPa× (from Appendix B.1). Using these data,
the volume, V , of the tank, and the initial mass, m0 is the tank are calculated as
V = 4
3 πR3 = 4
3 π(1.5)3 = 14.14 m3 , m0 = ρ0V = (998.2)(14.14) = 1.411 104 kg×
From the definition of the bulk modulus, Ev,
Ev ≈ Δp
Δρ/ρ0
→ Δm
m0
= Δp
Ev
→ Δm = m0
Δp
Ev
Substituting the given and derived data into this relationship yields,
Δm = (1.411 × 104 ) 9 × 103
2.18 106 = 58.3 kg
×
1.21. From the given data: p1 = 100 kPa, p2 = 20 000 kPa, V1 = 1.700 m3 , and V2 = 1.650 m3 .
Using the definition of the bulk modulus given by Equation 1.12,
Ev ≈ − Δp
ΔV /V = − 20000 − 100
1.650 − 1.700
1.700
= 6.766 × 105 kPa = 677 MPa
1.22. From the given data: V1 = 10 m3 , and Δp = 10 MPa. For benzene, Ev = 1030 MPa (Appendix
B.4). Using the definition of the bulk modulus given by Equation 1.12,
Ev ≈ − Δp
ΔV /V1
→ 1030 ≈ − 10
ΔV
10
→ ΔV ≈ 0.0971 m3
1.23. From the given data: T1 = 10◦C, and T2 = 100◦C. The average coefficient of volume
expansion, β, between T1 and T2 is derived from Appendix B.1 as ¯β = 0.418 10−3 K−1 .×
(a) Applying Equation 1.19

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1.24. from the given data: β = 5.7 10−4 K−1 ,× T1 = 10◦C, and T2 = 90◦C. Applying Equation
1.19 (with Δp = 0) gives
Δρ
ρ ≈ −βΔT = −5.7 × 10−4 (90 − 10) = −0.0432 = −4.32%
1.25. From the given data: T1 = 15◦C, ΔV /V1 = 0.01, and β = 9.5 10−4 K−1× . Using the given
data,
V2 − V1
V = 0.01 → V2
V − 1 = 0.01 → V2
V = 1.01 (1)
1 1 1
Δρ
ρ1
= //m/V2 − //m/V1
//m/V1
→ Δρ
ρ1
= V1
V2
− 1 (2)
Δρ
ρ1
= −5.7 × 10−4 ΔT (3)
where ΔT is the maximum allowable temperature rise. Combining Equations 1 to 3 yields
ΔT = 10.4◦C .
1.26. From the given data: T = 20◦ C = 293 K. Noting that the speed of sound, c, is given by
c = √
vE /ρ , the calculation of the speed of sound in water and mercury are summarized in
the following table:
Medium Ev
(×106 Pa)
ρ
(kg/m3 )
c
(m/s)
Water 2171 998 1475
Mercury 26200 13550 1390
Therefore, the speed of sound in water at 20◦C is 1475 m/s , and the speed of sound in
mercury at 20◦C is 1390 m/s .
1.27. From the given data: c = 1700 m/s, and SG = 1.8. For water at 4◦C, ρw = 1000 kg/m3 .
Using Equation 1.15, the bulk modulus, Ev, is calculated as follows:
c = Ev
SG ρw
→ 1700 = Ev
(1.8)(1000) → Ev = 3.06 × 106 Pa = 3.06 MPa
/
·
/

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(b) The term used to describe a gas in which the continuum approximation is not valid is a
rarefied gas .
1.29. From the given data: p = 101 kPa, and T = 25◦C = 298 K. For He, RHe = 8314/4.003 =
2077 J/(kg K), and for air,· Rair = 8314/28.96 = 287.1 J/(kg K) (from Appendix B.5). Using·
the ideal gas law,
ρHe = p
RHeT = 101 × 103
(2077)(298) = 0.1632 kg/m3
ρair = p
RairT = 101 × 103
(287.1)(298) = 1.181 kg/m3
The specific volume is defined by Equation 1.11 as the volume per unit mass, hence
vHe = 1
ρHe
= 1
0.1632 = 6.13 m3 /kg
vair = 1
ρair
= 1
1.181 = 0.847 m3 /kg
1.30. For air at standard atmospheric pressure, p = 101 kPa and R = 287.1 J/(kg K). Taking· ρ1 =
density from Appendix B.2, and ρ2 = density from ideal gas law, gives:
T
(◦C)
ρ1
(kg/m3 )
T
( K)
ρ2
(kg/m3 )
Δ
(%)
−40 1.514 233 1.5098 −0.27
−20 1.394 253 1.3905 −0.25
0 1.292 273 1.2886 −0.26
5 1.269 278 1.2654 −0.28
10 1.246 283 1.2431 −0.23
15 1.225 288 1.2215 −0.29
20 1.204 293 1.2007 −0.28
25 1.184 298 1.1805 −0.29
30 1.164 303 1.1610 −0.25
40 1.127 313 1.1239 −0.27
50 1.092 323 1.0891 −0.26
60 1.059 333 1.0564 −0.24
70 1.028 343 1.0256 −0.23
80 0.9994 353 0.9966 −0.28
90 0.9718 363 0.9691 −0.27
100 0.9458

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Based on these results, the ideal gas law gives quite accurate estimates with errors less than 0.31% .
1.31. From the given data: ρ = 5 kg/m3 , and p = 450 kPa. Properties of O2 from Appendix B.5:
cp = 909 J/kg K, cv = 649 J/kg K, and R =· · cp cv− = 909 649 = 260 J/kg K. The ideal− ·
gas law, Equation 1.24, gives
ρ = p
RT → 5 = 450 × 103
(260)T → T = 346 K = 73◦C
1.32. From the given data: V = 2 m3 , T = 15◦C = 288 K, and p = 500 kPa. The molar mass of
helium is 4.003 g/mol, and hence the gas constant for helium can be taken as R = 8314/4.003
= 2077 J/kg K. The density, mass, and weight of helium in the tank are given by·
ρ = p
RT = 500 × 103
(2077)(288) = 0.08610 kg/m3
M = ρV = (0.08610)(2) = 0.1722 kg
W = M g = (0.1722)(9.807) = 1.689 N
1.33. From the given data: m = 10 kg, T = 15 C = 288 K, p = 10 MPa, and L = 3D. For pure◦
oxygen, R = Ru/M = 8314/32 = 259.8 J/(kg·K). Using the given data and the ideal gas law,
V = π
4 D 2 L = π
4 D 2 (3D) = 3π
4 D 3 , m = pV
RT → V = mRT
p
Combining these equations gives
3
4 πD 3 = mRT
p → D = 4mRT
3πp
1
3
= 4(10)(259.8)(288)
3π(10 106 )
1
3
= 0.317 m
┌ ┐ ┌
×
┐
Since the length must be three times the diameter, L = 3(0.317) = 0.950 m. The required
dimensions of the tank are a diameter of 317 mm and a length of 0.950 m .
1.34. From the given data: M = 10 kg, T = 60◦C = 333 K, and p = 200 kPa. For air,

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1.36. From the given data: V = 0.1 m3 , T = 20 C = 293 K, and p = 400 kPa. The gas constant◦
for air can be taken as R = 287.1 J/kg·K. The density of air in the tank can be calculated
using Equation 1.24, which gives
ρ = p
RT = 400 × 103
(287.1)(293) = 0.4898 kg/m3
Hence the weight of air in the tank , W , is given by
W = ρV g = (0.4898)(0.1)(9.807) = 0.4803 N 0.48 N≃
The weight (0.48 N) has been rounded to two significant digits to be consistent with the
accuracy of the given data.
1.37. From the given data: V = 10 m 12 m 4 m = 480 m3 ,× × p = 101.3 kPa, T1 = 20◦C = 293.15 K,
and T = 10 C = 283.15 K.2 ◦ For air, R = 287.1 J/(kg K).·
(a) Using the given data at a temperature of 20◦C gives
ρ1 = p
RT1
= 101.3 × 103
(287.1)(293.15) = 1.204 kg/m3 , m1 = ρ1V = (1.204)(480) = 577.7 kg
W1 = m1g = (577.7)(9.807) = 5.666 103 N = 1274 lb×
(b) When the temperature in the room is reduced to 10◦C, then
ρ2 = p
RT2
= 101.3 × 103
(287.1)(283.15) = 1.246 kg/m3 , m2 = ρ2V = (1.246)(480) = 598.1 kg
change = m2 − m1
m1
× 100 = 598.1 − 577.7
577.7 × 100 = 3.53%
1.xx From the given data: V = 600 mL = 6 10−4 m3 ,× p1 = 101.3 kPa,

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1.39. From the given data: p1 = 130 kPa, p2 = 210 kPa, V = 15 L = 0.015 m3 , and T =
30◦C. Required constants: R = 8.314 kJ/kmol K, molar mass of air, M = 28.97 kg/kmol,·
patm = 101 kPa. Using these data with the ideal gas law,
n1 = p1V1
RT1
= (101 + 130)(0.015)
(8.314)(273.15 + 30) = 0.00137 kmol
n2 = p2V2
RT = (101 + 210)(0.015)
(8.314)(273.15 + 30) = 0.00185 kmol
2
mass added = (n2 − n1)M = (0.00185 − 0.00137)28.97 = 0.0139 kg = 13.9 g
1.40. Consider three states of the tire: State 1 is the initial state, State 2 is the heated-up state,
and State 3 is the cooled-down state. From the given data: patm = 101.3 kPa, V = 15 L, p1
(gauge) = 207 kPa, p1 (absolute) = 207 kPa + 101.3 kPa = 308.3 kPa, T1 = 20◦C = 293 K,
T2 = 60◦C = 333 K, and T3 = T1 = 293 K.
(a) Between the initial state and the heated-up state, the air density in the tire remains
constant, so the ideal-gas law gives
ρR = p
T = constant → p1
T1
= p2
T2
→ p2 = T2
T1
p1 = 333
293 (207) = 350 kPa
Hence, the resulting gauge pressure is 350 kPa 101.3 kPa = 249 kPa =− 36 lb/in2 .
(b) Between the heated-up state and the cooled-down state, the air density in the tire
remains constant and the air pressure at the heated-up state is equal to 308.3 kPa. The
ideal-gas law gives
ρR = p
T = constant →

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(a) From the ideal-gas law,
ρ = p
RT → p1
T1
= p2
T2
→ T2
T1
= p2
p1
= 241
207 = 1.164
Using this relationship, the percentage change in temperature is calculated as follows,
% change in T = T2 − T1
T1
× 100 = T2
T1
− 1 × 100 = [1.164 − 1] × 100 = 16.4%
┌ ┐
Note that this percentage change applies to a temperature given in degrees kelvin.
(b) From the given data: T1 = 25◦C = 298 K.
1.164(298) = 347 K = 74◦C .
Hence, from the result in Part (a), T2 =
Note that thhe percentage change in ◦C is much higher
than the percentage change in K.
1.43. (a) From the given data: D = 6 m, T = 20◦C = 293 K, p = 200 kPa, Ru = 8312 J/(kmol K),·
and V = πD 3 /6 = 113.1 m3 . For an ideal gas:
pV = nRT → (200 × 103 )(113.1) = n(8312)(293) → n = 9.29 kmol
Since the relative molecular mass of He is 4.003, the mass of 9.29 kmol of He is (9.29)(4.003)kg
= 37.2 kg .
(b) From the given data: T1 = 25◦C = 298 K, p1 = 210 kPa + 101 kPa = 311 kPa (assuming
atmospheric pressure is 101 kPa), V = 0.025 m3 , and T2 = 50◦C = 323 K. From the
ideal gas law,
V = nRT1
p1
= nRT2
p2
→ p2 = p1
T2
T1
= 311 323
298 = 337 kPa
( ) ( )
This corresponds to a pressure increase of 337 kPa 311 kPa = 26 kPa .−
1.44. From given data: Wrat = 1.5 N, Wbal = 0.5 N, ρair = 1.17 kg/m3 , p = 100 kPa, T = 25◦C =
298 K. Take Ru = 8.314 J/mol K· and molar mass of He, mHe, is 4.003 g/mol. To lift the rat:
Wrat + Wbal + ρHegVbal
' '' '
Weight lifted
= ρairgVbal
' '' '
buoyant

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1.45. Form the given data: z = 10 m, D1 = 5.0 mm, T1 = 20◦C, and patm = 101.3 kPa. At the
bubble release location,
p1 = patm + γz = 101.3 + 9.79(10) = 199.2 kPa, T1 = 20◦C
At the surface of the lake, p2 = 101.3 kPa and for an isothermal process,
p1V1 = p2V2 → V2 = p1
p2
V1 → 4
3 πr 3
2 = p1
p2
4
3 πr 3
1 → r2
r1
= D2
D1
= p1
p2
1
3
→ D2 =
( p1
p2
) 1
3
D1 =
( 199.2
101.3
) 1
3
(5) = 6.3 mm
( ) ( ) ( )
Therefore, the diameter of the bubble when it reaches the surface of the lake is 6.3 mm .
1.46. From the given data: V1 = 1.0 m3 , V2 = 0.4 m3 , and p1 = 101 kPa. For air, the ratio of
specific heats is given by k = 1.40.
(a) Under isentropic conditions, the pressure of the compressed volume, p2, is gi ven by
Equations 1.33 and 1.34 as
p1v k
1 = p2v k
2 → (101)(1.0)1.40 = p2(0.4)1.40 → p2 = 364 kPa
(b) Under isothermal conditions, the pressure of the compressed volume, p2, is g ivenby
Equations 1.33 and 1.34 as
p1v1 = p2v2 → (101)(1.0) = p2(0.4) → p2 = 253 kPa
1.xx From the given data: V0 = 1 L = 10−3 m3 , P0

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The (partial) pressure, PO, of the other gas in the space above the water surface is given by
PO = (0.5P0) V0
Vg
= (0.5 × 101.3) 1
0.25 = 202. 6 kPa
Therefore, the total pressure, P , in the space above the liquid surface is equal to the sum of
the partial pressures of the two gasses plus the vapor pressure of water (2.337 kPa), which
gives
P = PCO2 + PO + pvap = 58.1 + 202.6 + 2.337 = 263.0 kPa
1.47. From the given data: T = 20◦C = 293 K, p = 101.3 Pa, f1 = 0.20, and f2 = 0.8. For O2,
R1 = 259.8 J/(kg K),· and for N2, R2 = 296.7 J/(kg K).· Under the given conditions, the
densities of O2 and N2, represented by ρ1 and ρ2, are given by
ρ1 = p
R1T = 101.3 × 103
(259.8)(293) = 1.331 kg/m3 , ρ2 = p
R2T = 101.3 × 103
(296.7)(293) = 1.166 kg/m3
(a) For each 1 m3 of air there is f1 m3 of O2 and f2 m3 of N2. Therefore, the partial pressures
of O2 and N2, denoted by p1 and p2, are given by
p1 = f1p = (0.2)(293) = 20.3 kPa , p2 = f2p = (0.8)(293) = 81.0 kPa
(b) The density, ρm of the mixture is given by
ρm = f1ρ1 + f2

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Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

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