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Fluid Mechanics, 8th edition
In SI Units
Frank M. White
Solutions Manual
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Chapter 1 •^ Introduction
P1.1 A gas at 20°C may be rarefied if it contains less than 10^12 molecules per mm^3. If Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent?
Solution: The mass of one molecule of air may be computed as
Molecular weight 28.97 mol^1 m 4.81E 23 g Avogadro’s number 6.023E23 molecules/g mol
− = = = − ⋅
Then the density of air containing 10^12 molecules per mm^3 is, in SI units,
12 3
3 3
molecules g 10 4.81E 23 mm molecule g kg 4.81E 11 4.81E 5 mm m
Finally, from the perfect gas law, Eq. (1.13), at 20°C = 293 K, we obtain the pressure:
^
2 3 2
kg m p RT 4.81E 5 287 (293 K). m s K
4.0 Pa ns
P1.3 For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact with an atmosphere at pressure pa, must undergo shear stress and hence begin to flow.
Solution: Assume zero shear. Due to element weight, the pressure along the lower and right sides must vary linearly as shown, to a higher value at point C. Vertical forces are presumably in balance with ele- ment weight included. But horizontal forces are out of balance, with the unbalanced force being to the left, due to the shaded excess-pressure triangle on the right side BC. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result.
Fig. P1.
P1.4 Sand, and other granular materials, definitely flow , that is, you can pour them from a container or a hopper. There are whole textbooks on the “transport” of granular materials [54]. Therefore, is sand a fluid? Explain.
Solution : Granular materials do indeed flow , at a rate that can be measured by “flowmeters”. But they are not true fluids, because they can support a small shear stress without flowing. They may rest at a finite angle without flowing, which is not possible for liquids (see Prob. P1.3). The maximum such angle, above which sand begins to flow, is called the angle of repose. A familiar example is sugar, which pours easily but forms a significant angle of repose on a heaping spoonful. The physics of granular materials are complicated by effects such as particle cohesion, clumping, vibration, and size segregation. See Ref. 54 to learn more.
P1.5 A formula for estimating the mean free path of a perfect gas is:
p
RT
RT
where the latter form follows from the ideal-gas law, ρ = p/RT. What are the dimensions
of the constant “1.26”? Estimate the mean free path of air at 20°C and 7 kPa. Is air rarefied at this condition?
Solution: We know the dimensions of every term except “1.26”:
2 3 2
M M L
{ } {L} { } { } {R} {T} { }
LT L T
^
Therefore the above formula (first form) may be written dimensionally as
3 2 2
{M/L T}
{L} {1.26?} {1.26?}{L}
{M/L } [{L /T }{ }]
Since we have {L} on both sides, {1.26} = {unity}, that is, the constant is dimensionless. The formula is therefore dimensionally homogeneous and should hold for any unit system.
For air at 20°C = 293 K and 7,000 Pa, the density is ρ = p/RT = (7,000)/[(287)(293)] =
0.0832 kg/m^3. From Table A-2, its viscosity is 1.80E−5 N ⋅ s/m^2. Then the formula predicts a mean free path of
1/
(0.0832)[(287)(293)]
Ans.
= ≈ 9.4E 7 m −
This is quite small. We would judge this gas to approximate a continuum if the physical
scales in the flow are greater than about 100 , that is, greater than about 94 μm.
P1.8 Suppose that bending stress σ in a beam depends upon bending moment M and
beam area moment of inertia I and is proportional to the beam half-thickness y. Suppose also that, for the particular case M = 2,900 in⋅lbf, y = 1.5 in, and I = 0.4 in^4 , the predicted
stress is 75 MPa. Find the only possible dimensionally homogeneous formula for σ.
Solution: We are given that σ = y fcn(M,I) and we are not to study up on strength of
materials but only to use dimensional reasoning. For homogeneity, the right hand side must have dimensions of stress, that is,
2
M
{ } {y}{fcn(M,I)}, or: {L}{fcn(M,I)} LT
σ = ^ =
or: the function must have dimensions (^2 )
M
{fcn(M,I)} L T
= ^ ^
Therefore, to achieve dimensional homogeneity, we somehow must combine bending moment, whose dimensions are {ML^2 T–2}, with area moment of inertia, {I} = {L^4 }, and end up with {ML–2T–2}. Well, it is clear that {I} contains neither mass {M} nor time {T} dimensions, but the bending moment contains both mass and time and in exactly the com-
bination we need, {MT–2}. Thus it must be that σ is proportional to M also. Now we
have reduced the problem to:
2 2 2
M ML
yM fcn(I), or {L} {fcn(I)}, or: {fcn(I)} LT T
^
{L − 4 }
We need just enough I ’s to give dimensions of {L–4}: we need the formula to be exactly inverse in I. The correct dimensionally homogeneous beam bending formula is thus:
σ = where {C} ={unity} Ans.
My C , I
The formula admits to an arbitrary dimensionless constant C whose value can only be
obtained from known data. Convert stress into English units: σ = (75 MPa)/(6,894.8) =
10880 lbf/in^2. Substitute the given data into the proposed formula:
2 4
lbf My (2,900 lbf in)(1.5 in) 10,880 C C , or: in I 0.4 in
σ Ans.
= = = C ≈1.
The data show that C = 1, or σ = My/I , our old friend from strength of materials.
P1.9 A hemispherical container, 26 inches in diameter, is filled with a liquid at 20°C and weighed. The liquid weight is found to be 1,617 ounces. ( a ) What is the density of the fluid, in kg/m^3? ( b ) What fluid might this be? Assume standard gravity, g = 9. m/s^2.
Solution : First find the volume of the liquid in m^3 :
Hemisphere volume =
� (26 𝑖𝑖)^3 = 4,601 𝑖𝑖 3 ÷ �61,
Liquid mass = 1,617 𝑜𝑜 ÷ 16 = 101 𝑙𝑙𝑚 �0.
Then the liquid density =
𝒎 𝟑^
From Appendix Table A.3, this could very well be ammonia. Ans .( b )
P1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:
F 3 DV 9 V D^2
p
= pμ + ρ
where D = sphere diameter, μ = viscosity, and ρ = density. Is the formula homogeneous?
Solution: Write this formula in dimensional form, using Table 1-2:
{F} {3 }{ }{D}{V} 9 { }{V} {D}?^2
p
= p μ + ^ ρ
2 2 3 2
ML M L M L
or: {1} {L} {1} {L }? T LT T L T
^ ^
=^ +
where, hoping for homogeneity, we have assumed that all constants (3, p,9,16) are pure ,
i.e., {unity}. Well, yes indeed, all terms have dimensions {ML/T^2 }! Therefore the Stokes- Oseen formula (derived in fact from a theory) is dimensionally homogeneous****.
P1.13 The efficiency η of a pump is defined as
Q p Input Power
where Q is volume flow and ∆p the pressure rise produced by the pump. What is η if
∆p = 35 psi, Q = 40 L/s, and the input power is 16 horsepower?
Solution: The student should perhaps verify that Q∆p has units of power, so that η is a
dimensionless ratio. Then convert everything to consistent units, for example, BG:
2 2 2
L ft lbf lbf ft lbf Q 40 1.41 ; p 35 5,040 ; Power 16(550) 8, s s in ft s
(1.41 ft 3 s)(5,040 lbf ft )^2 0.81 or 8,800 ft lbf s
η Ans.
Similarly, one could convert to SI units: Q = 0.04 m^3 /s, ∆p = 241,300 Pa, and input power = 16(745.7) = 11,930 W, thus h = (0.04)(241,300)/(11,930) = 0.81. Ans.
P1.14 The volume flow Q over a dam is proportional to dam width B and also varies with gravity g and excess water height H upstream, as shown in Fig. P1.14. What is the only possible dimensionally homo-geneous relation for this flow rate?
Solution: So far we know that Q = B fcn(H,g). Write this in dimensional form:
3
2
L
{Q} {B}{f(H,g)} {L}{f(H,g)}, T L or: {f(H,g)} T
Fig. P1.
So the function fcn(H,g) must provide dimensions of {L^2 /T}, but only g contains time. Therefore g must enter in the form g1/2^ to accomplish this. The relation is now
Q = Bg1/2fcn(H), or: {L^3 /T} = {L}{L1/2/T}{fcn(H)}, or: {fcn(H)} = { L3/2 }In order for fcn(H) to provide dimensions of {L3/2}, the function must be a 3/2 power. Thus the final desired homogeneous relation for dam flow is:
Q = C B g1/2^ H3/2, where C is a dimensionless constant Ans.
P1.15 The height H that fluid rises in a liquid barometer tube depends upon the liquid density ρ , the barometric pressure p , and the acceleration of gravity g. (a) Arrange these four variables into a single dimensionless group. (b) Can you deduce (or guess) the numerical value of your group?
Solution : This is a problem in dimensional analysis , covered in detail in Chapter 5. Use the symbols for dimensions suggested with Eq. (1.2): M for mass, L for length, T for time, F for force, { H }= { L }, { ρ } = { M / L^3 }, { g } = { L / T^2 }, { p } = { F / L^2 } = { M/ ( LT^2 )} where the change in pressure dimensions uses Newton’s law, { F } = { ML/T^2 }. We see that we can cancel mass by dividing density by pressure:
�
𝑀𝐿−1^ 𝑇 −^
𝐿^2
We can eliminate time by multiplying by { g }: {( ρ/p )( g )} = {( T^2 / L^2 ) ( L / T^2 )} = { L -1}. Finally, we can eliminate length by multiplying by the height { H }:
�� 𝜌𝜌 𝑝 �^ (𝐻)�^ =^ {𝐿
−1}{𝐿} = {1} dimensionless
Thus the desired dimensionless group is ρgH / p , or its inverse, p / ρgH. Answer (a)
(b) You might remember from physics, or other study, that the barometer formula is p ≈ ρgH. Thus this dimensionless group has a value of approximately 1.0, or unity. Answer ( b )
P1.18 (“” means “difficult”—not just a plug-and-chug, that is) For small particles at low velocities, the first (linear) term in Stokes’ drag law, Prob. 1.10, is dominant, hence F = KV, where K is a constant. Suppose
a particle of mass m is constrained to move horizontally from the initial position x = 0 with initial velocity V = Vo. Show (a) that its velocity will decrease exponentially with time; and (b) that it will stop after travelling a distance x = mVo/K.
Solution: Set up and solve the differential equation for forces in the x -direction:
∑ = − = − = (^) ∫ = −∫ o
V t x x V 0
dV dV m F Drag ma , or: KV m , integrate dt dt V K Solve and (^) ∫ ( ) Ans. (a,b)
t mt K (^) o mt K o 0
mV V V e x V dt 1 e K
= −^ /^ = = − −^ /
Thus, as asked, V drops off exponentially with time, and, as , o
V
t x K m
P1.19 In his study of the circular hydraulic jump formed by a faucet flowing into
a sink, Watson [53] proposes a parameter combining volume flow rate Q , density ρ and
viscosity μ of the fluid, and depth h of the water in the sink. He claims that the grouping
is dimensionless, with Q in the numerator. Can you verify this?
Solution : Check the dimensions of these four variables, from Table 1.2:
Can we make this dimensionless? First eliminate mass { M } by dividing density by viscosity,
that is, ρ/ μ has units {T/L^2 }. (I am pretending that kinematic viscosity is unfamiliar to the
students in this introductory chapter.) Then combine ρ/μ and Q to eliminate time: ( ρ/μ) Q
has units {L}. Finally, divide that by a single depth h to form a dimensionless group:
{ Q } = { L^3^ / T } ; { } ρ = { M / L^3 } ; { }μ = { M / LT } ; { } h ={ } L
{ } {1} dimensionless. Watson is correct. { / }{ }
Q M L L T
Ans h M LT L
P1.20 Books on porous media and atomization claim that the viscosity μ and surface
tension ϒ of a fluid can be combined with a characteristic velocity U to form an
important dimensionless parameter. ( a ) Verify that this is so. ( b ) Evaluate this parameter
for water at 20°C and a velocity of 3.5 cm/s. NOTE: Extra credit if you know the name
of this parameter.
Solution : We know from Table 1.2 that { μ}= {ML-1T-1}, { U } = {LT-1}, and { ϒ }= {FL-1} =
{MT-2}. To eliminate mass {M}, we must divide μ by ϒ , giving { μ/ ϒ } = {TL-1}.
Multiplying by the velocity will thus cancel all dimensions:
The grouping is called the Capillary Number. ( b ) For water at 20°C and a velocity of 3.
cm/s, use Table A.3 to find μ = 0.001 kg/m-s and ϒ = 0.0728 N/m. Evaluate
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P1.21 Aeronautical engineers measure the pitching moment M o of a wing and then write it in the following form for use in other cases: 𝑀o = 𝛽 𝑉 2 𝐴 𝐶 𝜌 where V is the wing velocity, A the wing area, C the wing chord length, and ρ the air density. What are the dimensions of the coefficient β?
Solution : Write out the dimensions of each term in the formula:
{𝑀o} = {𝐹𝐿} = �
𝑀𝐿^2
𝐿^2
� {𝐿^2 }{𝐿} �
𝐿^3
𝑀𝐿^2
Thus { β } = {unity} or dimensionless. It is proportional to the moment coefficient in aerodynamics.
is dimensionless, as is its inverse, .( )
U
Ans a U
2
U kg m s m s Ans b kg s U
P1.24 Air, assumed to be an ideal gas with k = 1.40, flows isentropically through a nozzle. At section 1, conditions are sea level standard (see Table A.6). At section 2, the temperature is –50°C. Estimate ( a ) the pressure, and ( b ) the density of the air at section 2.
Solution : From Table A.6, p 1 = 101,350 Pa, T 1 = 288.16 K, and ρ 1 = 1.2255 kg/m^3.
Convert to absolute temperature, T 2 = -50°C = 223.26 K. Then, for a perfect gas with constant k ,
Alternately, once p 2 was known, we could have simply computed ρ 2 from the ideal-gas law.
ρ 2 = p 2 / RT 2 = (41,400)/[287(223.16)] = 0.647 kg/m^3
P1.25 On a summer day in Narragansett, Rhode Island, the air temperature is 74ºF and the barometric pressure is 14.5 lbf/in^2. Estimate the air density in kg/m^3.
Solution : This is a problem in handling awkward units. Even if we use the BG system, we have to convert. But, since the problem calls for a metric result, better we should convert to SI units: T = 74ºF + 460 = 534ºR x 0.5556 (inside front cover ) = 297 K p = 14.5 lbf/in^2 x 6,894.8 (inside front cover ) = 100,700 Pa The SI gas constant, from Eq. (1.12), is 287 m^2 /(s^2 ∙K). Thus, from the ideal gas law, Eq. (1.10),
𝜌 =
100,700 N/m^2
(
m^2 s^2 ∙ K)(297 K)
N ∙ s^2 m^4
This doesn’t look like a density unit, until we realize that 1 N ≡ 1 kg∙m/s^2. Making this substitution, we find that ρ = 1.18 kg/m^3 Answer Notice that faithful use of SI units will lead to faithful SI results, without further conversion.
2 2 /(^ 1)^ 1.4/(1.4 1)^ 3. 1 1 2 2 2 1/(^ 1)^ 1/(1.4 1)^ 2. 1 1 3 2
Thus (0.4087)(101,350 ) .( )
( ) (0.7744) 0.
Thus (0.5278)(1.2255 / )
k k
k
p T p T p Pa Pa Ans a T T kg m
− −
− −
0.647 kg / m^3 Ans b .( )
P1.26 A tire has a volume of 3.0 ft^3 and a ‘gage’ pressure (above atmospheric pressure)
of 32 psi at 75°F. If the ambient pressure is sea-level standard, what is the weight of air in the tire?
Solution: Convert the temperature from 75°F to 535°R. Convert the pressure to psf:
p = (32 lbf/in )(144 in /ft )^2 2 2 + 2,116 lbf/ft^2 = 4,608 + 2,116 ≈6,724 lbf/ft^2
From this compute the density of the air in the tire: 2 3 air
p 6,724 lbf/ft 0.00732 slug/ft RT (1717 ft lbf/slug R)(535 R)
Then the total weight of air in the tire is 3 2 3
Wair = ρ g u= (0.00732 slug/ft )(32.2 ft/s )(3.0 ft ) ≈ 0.707 lbf Ans.
P1.27 For steam at a pressure of 45 atm, some values of temperature and specific volume are as follows, from Ref. 23: T , ºF 500 600 700 800 900 v , ft^3 /lbm 0.7014 0.8464 0.9653 1.074 1.
Find an average value of the predicted gas constant R in m^2 /(s^2 ∙K). Does this data reasonably approximate an ideal gas? If not, explain.
Solution : If ideal, the calculated gas constant would, from Table A.4, be about 461 m^2 /(s^2 ∙K). Try this for the first value, at T = 500ºF = 960ºR. Change to SI units, using the inside front cover conversions: T = 500ºF = 960ºR ÷ 1.8 = 533 K; p = 45(101,350 Pa) = 4,561 kPa v = 0.7014 ft^3 /lbm ÷ 16.019 = 0.0438 m^3 /kg
𝑝 𝑣 = 𝑅 𝑇 , or: 𝑅 =
𝐴2∙^ 𝐾
This is about 19% lower than the recommended value of R steam = 461 m^2 /(s^2 ∙K). Continue by filling out the rest of the table: T , ºF 500 600 700 800 900 R , m^2 /(s^2 ∙K) 374 409 427 437 443
These are all low, with an average of 418, nine per cent low. The temperature is too low and the pressure too high. We are too near the saturation line for the ideal gas law to be accurate.
Solution: First evaluate the density change of water. At 1 atm, ρ o ≈ 1.94 slug/ft^3. At
120 psi(gage) = 134.7 psia, the density would rise slightly according to Eq. (1.22): 7 3 o
p 134. 3,001 3,000, solve 1.940753 slug/ft , p 14.7 1.
ρ = ≈ ^ − ρ≈ Hence mwate (^) r= ρu= (1.940753)(5 ft )^3 ≈9.704 slug
The density change is extremely small. Now the work done, as in Prob. 1.29 above, is
u
∫ ∫ (^) ∫
2 2 2 1 - 2 2 avg 2 1 1 1 avg
m m d W p d p d p p m for a linear pressure rise
+ ^ ^ ≈ (^) × (^) (^) ≈ (^)
3 1-2 (^2 )
14.7 134.7 lbf 0.000753 ft Hence W 144 (9.704 slug) (^2) ft 1.9404 slug
21 ft lbf ⋅ Ans.
[Exact integration of Eq. (1.22) would give the same numerical result.] Compressing
water (extremely small ∆ρ) takes ten thousand times less energy than compressing air,
which is why it is safe to test high-pressure systems with water but dangerous with air.
P1.31 One cubic foot of argon gas at 10°C and 1 atm is compressed isentropically to a new pressure of 600 kPa. ( a ) What will be its new density and temperature? ( b ) If allowed to cool, at this new volume, back to 10°C, what will be the final pressure? Assume constant specific heats.
Solution : This is an exercise in having students recall their thermodynamics. From Table A.4, for argon gas, R = 208 m^2 /(s^2 -K) and k = 1.67. Note T 1 = 283K. First compute the initial density:
For an isentropic process at constant k ,
( b ) Cooling at constant volume means ρ stays the same and the new temperature is
283K. Thus
2 1 3 (^1 2 ) 1
p N m kg m RT (^) m s K K
2 2 2 1. 2 1 1 2 2 /( 1) 2 1.67 / 0. 2 1 1
5.92 ( ) ( ) , Solve .( ) 101,350 1.
5.92 ( ) ( ) , Solve 305 .( ) 283
k
k k
p Pa Ans a p Pa p T T T C Ans a p T K
−
4.99 kg/m^3
578 K
2 (^3 333) s K K Pa Ans b
m m
kg
p = ρ RT = = = 294 kPa
P1.32 A blimp is approximated by a prolate spheroid 90 m long and 30 m in diameter. Estimate the weight of 20°C gas within the blimp for (a) helium at 1.1 atm; and (b) air at 1.0 atm. What might the difference between these two values represent (Chap. 2)?
Solution: Find a handbook. The volume of a prolate spheroid is, for our data,
(^2) LR 2 2 (90 m)(15 m) (^2) 42,412 m 3 3 3
u = p = p ≈
Estimate, from the ideal-gas law, the respective densities of helium and air:
He helium (^3) He
p 1.1(101,350) kg (a) 0.1832 ; R T 2,077(293) m
air air (^3) air
p 101,350 kg (b) 1.. R T 287(293) m
Then the respective gas weights are
3 He He (^3 )
kg m W g 0.1832 9.81 (42, 412 m ) (a) m s
= ρ u = ^ ^ ≈ Ans.
76, 000 N
W air = ρ airg u= (1.205)(9.81)(42,412) ≈ 501, 000 N Ans. (b)
The difference between these two, 425,000 N , is the buoyancy , or lifting ability, of the blimp. [See Section 2.8 for the principles of buoyancy.]
P1.33 A tank contains 9 kg of CO 2 at 20ºC and 2.0 MPa. Estimate the volume of the tank, in m^3. Solution : All we have to do is find the density. For CO 2 , from Table A.4, R = 189 m^2 /(s^2 ∙K). Then
𝜌 =
([189 m^2 /(s^2 ∙ K](20 + 273 K)
Then the tank volume = m / ρ = (9 kg ) / (36.1 kg / m^3^ ) = 0.25 m^3 Ans.
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