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Chapter 1 •^ Introduction
1.1 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
1.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0.6 kg/m^3 (see Table A-6). Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth.
Solution: Let R (^) e be the earth’s radius ≈ 6377 km. Then the total mass of air in the atmosphere is
2 t avg avg e 3 2
m dVol (Air Vol) 4 R (Air thickness)
(0.6 kg/m )4 (6.377E6 m) (20E3 m) Ans.
6.1E18 kg
Dividing by the mass of one molecule ≈ 4.8E−23 g (see Prob. 1.1 above), we obtain the total number of molecules in the earth’s atmosphere:
molecules
m(atmosphere) 6.1E21 grams N m(one molecule) 4.8E 23 gm/molecule
= = ≈ Ans. −
1.3E44 molecules
2 Solutions Manual • Fluid Mechanics, Fifth Edition
1.3 For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact with an atmosphere at pressure p (^) a , 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.
1.4 The quantities viscosity μ, velocity V , and surface tension Y may be combined into
a dimensionless group. Find the combination which is proportional to μ. This group has a
customary name, which begins with C. Can you guess its name?
Solution: The dimensions of these variables are { μ} = {M/LT}, {V} = {L/T}, and {Y} =
{M/T^2 }. We must divide μ by Y to cancel mass {M}, then work the velocity into the
group:
2
Y /
M LT T L
hence multiply by V M T L^ T
finally obtain Ans
μ^ ^ =^ =^ = (^) μ V dimensionless. Y
This dimensionless parameter is commonly called the Capillary Number.
1.5 A formula for estimating the mean free path of a perfect gas is:
1.26 1.26 (RT)
(RT) p
μ μ ρ
4 Solutions Manual • Fluid Mechanics, Fifth Edition
Similarly, 1850 m 3 = 1.85E6 liters. Then a metric unit for this water usage is:
L 1 day Q 1.85E6. (b) day 86400 sec
Ans
^
L
s
1.8 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 = 2900 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)/(6894.8) =
10880 lbf/in^2. Substitute the given data into the proposed formula:
2 4
lbf My (2900 lbf in)(1.5 in) 10880 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.
Chapter 1 • Introduction 5
1.9 The dimensionless Galileo number, Ga , expresses the ratio of gravitational effect to
viscous effects in a flow. It combines the quantities density ρ, acceleration of gravity g ,
length scale L , and viscosity μ. Without peeking into another textbook, find the form of
the Galileo number if it contains g in the numerator.
Solution: The dimensions of these variables are { ρ} = {M/L^3 }, { g } = {L/T^2 }, { L } =
{L}, and { μ} = {M/LT}. Divide ρ by μ to eliminate mass {M} and then combine with g
and L to eliminate length {L} and time {T}, making sure that g appears only to the first power:
3 2
M L T
M LT L
^
=^ =
^
while only { g } contains { T }. To keep { g } to the 1st power, we need to multiply it by
{ ρ/μ}^2. Thus { ρ/μ}^2 { g } = { T^2 / L^4 }{ L / T^2 } = { L −^3 }.
We then make the combination dimensionless by multiplying the group by L 3. Thus we obtain:
(^2 2 ) 3 ( )( ) 2. gL Galileo number Ga g L Ans
3 2
gL
1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:
F 3 DV 9 V D^2
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
2 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, π,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****.
Chapter 1 • Introduction 7
1.13 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 5040 ; Power 16(550) 8800 s s (^) in ft s
(1.41 ft^3 s)(5040 lbf ft )^2 0.81 or 8800 ft lbf s
η Ans.
Similarly, one could convert to SI units: Q = 0.04 m^3 /s, ∆p = 241300 Pa, and input power = 16(745.7) = 11930 W, thus h = (0.04)(241300)/(11930) = 0.81. Ans.
1.14 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 = Bg 1/2^ fcn(H), or: {L^3 /T} = {L}{L1/2^ /T}{fcn(H)}, or: {fcn(H)} = { L 3/2 }
8 Solutions Manual • Fluid Mechanics, Fifth Edition
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 = CBg1/2^ H 3/2^ , where C is a dimensionless constant Ans.
1.15 As a practical application of Fig. P1.14, often termed a sharp-crested weir, civil engineers use the following formula for flow rate: Q ≈ 3.3 BH3/2^ , with Q in ft 3 /s and B and H in feet. Is this formula dimensionally homogeneous? If not, try to explain the difficulty and how it might be converted to a more homogeneous form.
Solution: Clearly the formula cannot be dimensionally homogeneous, because B and H do not contain the dimension time. The formula would be invalid for anything except English units (ft, sec). By comparing with the answer to Prob. 1.14 just above, we see that the constant “3.3” hides the square root of the acceleration of gravity.
1.16 Test the dimensional homogeneity of the boundary-layer x -momentum equation:
x
u u p u v g x y x y
Solution: This equation, like all theoretical partial differential equations in mechanics, is dimensionally homogeneous. Test each term in sequence:
^
=^ =^ =^ =^ =
2 3
u u M L L/T p M/LT u v ; x y L T L^2 2 x L^2
M M
L T L T
2 x (^3 )
M L M/LT
{ g } ; L T x L
= = ^ ^ ^ ^ = =^
^2 2 ^2 2
M M
L T L T
All terms have dimension {ML–2^ T –2^ }. This equation may use any consistent units.
1.17 Investigate the consistency of the Hazen-Williams formula from hydraulics:
Q 61.9D2.63 p L
What are the dimensions of the constant “61.9”? Can this equation be used with confidence for a variety of liquids and gases?
10 Solutions Manual • Fluid Mechanics, Fifth Edition
Solution: List the dimensions: { α} = {L^2 /T}, { L } = {L}, { μ} = {M/LT}, { δY} = {M/T^2 }.
We divide δ Y by μ to get rid of mass dimensions, then divide by α to eliminate time:
Y Y 1 1
, then
M LT L L T
T M^ T^ T^ L L
=^ =^ =^ =
Multiply by L and we obtain the Marangoni number: Ans.
L
M =
Y
1.20C (“C” means computer-oriented, although this one can be done analytically.) A baseball, with m = 145 g, is thrown directly upward from the initial position z = 0 and V (^) o = 45 m/s. The air drag on the ball is CV 2 , where C ≈ 0.0010 N ⋅ s 2 /m 2. Set up a differential equation for the ball motion and solve for the instantaneous velocity V(t) and position z(t). Find the maximum height zmax reached by the ball and compare your results with the elementary-physics case of zero air drag.
Solution: For this problem, we include the weight of the ball, for upward motion z :
o +
V t 2 z z (^2) V 0
dV dV F ma , or: CV mg m , solve dt t dt (^) g CV /m
mg Cg m cos( t (gC/m) Thus V tan t and z ln C m C cos
where φ = tan –1[Vo √ (C/mg)]. This is cumbersome, so one might also expect some
students simply to program the differential equation, m(dV/dt) + CV^2 = −mg, with a numerical method such as Runge-Kutta. For the given data m = 0.145 kg, Vo = 45 m/s, and C = 0.0010 N⋅s 2 /m 2 , we compute
0.8732 radians, mg^ 37.72 m^ , Cg^ 0.2601 s 1 , m 145 m C s m C
Hence the final analytical formulas are:
=^ − − =
m V in 37.72 tan(0.8732 .2601t) s cos(0.8732 0.2601t) and z(in meters) 145 ln cos(0.8732)
The velocity equals zero when t = 0.8732/0.2601 ≈ 3.36 s , whence we evaluate the maximum height of the baseball as zmax = 145 ln[sec(0.8734)] ≈ 64.2 meters. Ans.
Chapter 1 • Introduction 11
For zero drag, from elementary physics formulas, V = Vo − gt and z = Vo t − gt 2 /2, we calculate that 2 2 o o max height max
V 45 V (45)
t and z g 9.81 2g 2(9.81)
= = ≈ 4.59 s = = ≈ 103.2 m
Thus drag on the baseball reduces the maximum height by 38%. [For this problem I assumed a baseball of diameter 7.62 cm, with a drag coefficient CD ≈ 0.36.]
1.21 The dimensionless Grashof number , Gr , is a combination of density ρ, viscosity μ,
temperature difference ∆ T , length scale L , the acceleration of gravity g , and the
coefficient of volume expansion β, defined as β = (−1/ ρ)( ∂ρ/ ∂ T )p. If Gr contains both g
and β in the numerator, what is its proper form?
Solution: Recall that { μ/ ρ} = {L^2 /T} and eliminates mass dimensions. To eliminate tem-
perature, we need the product { β∆Τ} = {1}. Then {g} eliminates {T}, and L^3 cleans it all up:
Thus the dimensionless Gr = ρ 2 g β ∆ TL^3^ / μ^2 Ans.
1.22* According to the theory of Chap. 8, as a uniform stream approaches a cylinder of radius R along the line AB shown in Fig. P1.22, –∞ < x < –R, the velocities are
u U (1 R /x );^2 2 v w 0 = (^) ∞ − = = Fig. P1.
Using the concepts from Ex. 1.5, find (a) the maximum flow deceleration along AB; and (b) its location.
Solution: We see that u slows down monotonically from U∞ at A to zero at point B, x = −R, which is a flow “stagnation point.” From Example 1.5, the acceleration (du/dt) is
2 2 2 2 3 3 5
du u u R 2R U 2 2 x u 0 U 1 U , dt t x x x R R
∞ ∞ ∞
This acceleration is negative, as expected, and reaches a minimum near point B, which is found by differentiating the acceleration with respect to x :
2 max decel.
min
d du 5 x 0 if , or. (b) dx dt 3 R du Substituting 1.291 into (du/dt) gives. (a) dt
Ans
Ans
=^ =^ ≈
|
| ∞
2
U
R
Chapter 1 • Introduction 13
1.25 A tank contains 0.9 m 3 of helium at 200 kPa and 20°C. Estimate the total mass of this gas, in kg, (a) on earth; and (b) on the moon. Also, (c) how much heat transfer, in MJ, is required to expand this gas at constant temperature to a new volume of 1.5 m 3?
Solution: First find the density of helium for this condition, given R = 2077 m 2 /(s 2 ⋅K) from Table A-4. Change 20°C to 293 K:
2 3 He He
p 200000 N/m 0.3286 kg/m R T (2077 J/kg K)(293 K)
Now mass is mass , no matter where you are. Therefore, on the moon or wherever,
3 3
mHe = ρ Heυ= (0.3286 kg/m )(0.9 m ) ≈ 0.296 kg Ans. (a,b)
For part (c), we expand a constant mass isothermally from 0.9 to 1.5 m^3. The first law of thermodynamics gives
dQadded − dWby gas = dE = mc (^) v ∆T = 0 since T 2 =T (isothermal) 1
Then the heat added equals the work of expansion. Estimate the work done:
2 2 2 1-2 2 1 1 1 1
m d W p d RT d mRT mRT ln( / ),
= (^) = (^) = (^) =
or: W1-2 = (0.296 kg)(2077 J/kg K)(293 K)ln(1.5/0.9)⋅ = Q1-2 ≈ 92000 J Ans. (c)
1.26 A tire has a volume of 3.0 ft 3 and a ‘gage’ 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 + 2116 lbf/ft^2 = 4608 + 2116 ≈6724 lbf/ft^2
From this compute the density of the air in the tire:
2 3 air
p 6724 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 υg = (0.00732 slug/ft )(32.2 ft/s )(3.0 ft ) ≈ 0.707 lbf Ans.
14 Solutions Manual • Fluid Mechanics, Fifth Edition
1.27 Given temperature and specific volume data for steam at 40 psia [Ref. 13]:
T, °F: 400 500 600 700 800 v, ft 3 /lbm: 12.624 14.165 15.685 17.195 18.
Is the ideal gas law reasonable for this data? If so, find a least-squares value for the gas constant R in m^2 /(s 2 ⋅K) and compare with Table A-4.
Solution: The units are awkward but we can compute R from the data. At 400°F,
2 2 2 3 400 F
p (40 lbf/in )(144 in /ft )(12.624 ft /lbm)(32.2 lbm/slug) ft lbf “R” 2721 ° T (400 459.6) R slug R
V
The metric conversion factor, from the inside cover of the text, is “5.9798”: R (^) metric = 2721/5.9798 = 455.1 m^2 /(s^2 ⋅K). Not bad! This is only 1.3% less than the ideal-gas approxi- mation for steam in Table A-4: 461 m 2 /(s 2 ⋅K). Let’s try all the five data points:
T, °F: 400 500 600 700 800 R, m^2 /(s^2 ⋅K): 455 457 459 460 460
The total variation in the data is only ±0.6%. Therefore steam is nearly an ideal gas in this (high) temperature range and for this (low) pressure. We can take an average value:
5 i i=
p 40 psia, 400 F T 800 F: R. 5
= ° ≤ ≤ ° (^) steam ≈ (^) ≈ Ans
J
R 458 0.6%
kg K
With such a small uncertainty, we don’t really need to perform a least-squares analysis, but if we wanted to, it would go like this: We wish to minimize, for all data, the sum of the squares of the deviations from the perfect-gas law:
5 2 5 i i i 1 i i 1 i
p E p Minimize E R by differentiating 0 2 R T R T
V V
5 i least-squares i 1 i
p 40(144) 12.624 18. Thus R (32.2) (^5) = T 5 860 R 1260 R
(^) ° °
V
For this example, then, least-squares amounts to summing the (V/T) values and converting the units. The English result shown above gives R (^) least-squares ≈ 2739 ft⋅lbf/slug⋅°R. Convert this to metric units for our (highly accurate) least-squares estimate:
R (^) steam≈ 2739/5.9798 ≈ 458 ± 0.6% J/kg K ⋅ Ans.
16 Solutions Manual • Fluid Mechanics, Fifth Edition
1.30 Repeat Prob. 1.29 if the tank is filled with compressed water rather than air. Why is the result thousands of times less than the result of 215,000 ft⋅lbf in Prob. 1.29?
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. 3001 3000, solve 1.940753 slug/ft , p 14.7 1.
= ≈ ^ − ρ≈
3
Hence m water= ρυ= (1.940753)(5 ft ) ≈9.704 slug
The density change is extremely small. Now the work done, as in Prob. 1.29 above, is
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
Ans.
+ ^ ^
≈ × ≈
21 ft lbf ⋅
[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.
1.31 The density of water for 0°C < T < 100°C is given in Table A-1. Fit this data to a
least-squares parabola, ρ = a + bT + cT^2 , and test its accuracy vis-a-vis Table A-1.
Finally, compute ρ at T = 45°C and compare your result with the accepted value of ρ ≈
990.1 kg/m 3.
Solution: The least-squares parabola which fits the data of Table A-1 is:
ρ (kg/m 3 ) ≈ 1000.6 – 0.06986T – 0.0036014T^2 , T in °C Ans.
When compared with the data, the accuracy is less than ±1%. When evaluated at the particular temperature of 45°C, we obtain
ρ45°C ≈ 1000.6 – 0.06986(45) – 0.003601(45)^2 ≈ 990.2 kg/m^3 Ans.
This is excellent accuracya good fit to good smooth data. The data and the parabolic curve-fit are shown plotted on the next page. The curve-fit
does not display the known fact that ρ for fresh water is a maximum at T = +4°C.
Chapter 1 • Introduction 17
1.32 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) 42412 m 3 3 3
Estimate, from the ideal-gas law, the respective densities of helium and air:
He helium (^3) He
p 1.1(101350) kg (a) 0.1832 ; R T 2077(293) (^) m
air air (^3) air
p 101350 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 (42412 m ) (a) m s
ρ υ Ans.
76000 N
W air = ρ airg υ= (1.205)(9.81)(42412) ≈ 501000 N Ans. (b)
The difference between these two, 425000 N , is the buoyancy , or lifting ability, of the blimp. [See Section 2.8 for the principles of buoyancy.]
Chapter 1 • Introduction 19
For EES or the Steam Tables, just program the properties for steam or look it up:
3 3
EES real steam: ρ 1 = 5.23 kg/m Ans. (a), ρ 2 = 2.16 kg/m Ans. (b)
The ideal-gas error is only about 3%, even though the expansion approached the saturation line.
1.35 In Table A-4, most common gases (air, nitrogen, oxygen, hydrogen, CO, NO) have a specific heat ratio k = 1.40. Why do argon and helium have such high values? Why does NH 3 have such a low value? What is the lowest k for any gas that you know?
Solution: In elementary kinetic theory of gases [8], k is related to the number of “degrees of freedom” of the gas: k ≈ 1 + 2/ N , where N is the number of different modes of translation, rotation, and vibration possible for the gas molecule.
Example: Monotomic gas, N = 3 (translation only), thus k ≈ 5/
This explains why helium and argon, which are monatomic gases, have k ≈ 1.67.
Example: Diatomic gas, N = 5 (translation plus 2 rotations), thus k ≈ 7/
This explains why air, nitrogen, oxygen, NO, CO and hydrogen have k ≈ 1.40. But NH 3 has four atoms and therefore more than 5 degrees of freedom, hence k will be less than 1.40 (the theory is not too clear what “N” is for such complex molecules). The lowest k known to this writer is for uranium hexafluoride , 238 UF 6 , which is a very complex, heavy molecule with many degrees of freedom. The estimated value of k for this heavy gas is k ≈ 1..
1.36 The bulk modulus of a fluid is defined as B = ρ( ∂ p/ ∂ρ)S. What are the dimensions
of B? Estimate B (in Pa) for (a) N 2 O, and (b) water, at 20°C and 1 atm.
Solution: The density units cancel in the definition of B and thus its dimensions are the same as pressure or stress:
{B} = {p} = {F/L }^2 = ^ Ans. ^2
M
LT
(a) For an ideal gas, p = C ρ k^ for an isentropic process, thus the bulk modulus is:
Ideal gas: B d (C k^ ) kC k^1 kC k d
= = − = = kp
For N O, from Table A-4, k 2 ≈ 1.31, so BN O 2 = 1.31 atm = 1.33E5 Pa Ans. (a)
20 Solutions Manual • Fluid Mechanics, Fifth Edition
For water at 20°C, we could just look it up in Table A-3, but we more usefully try to estimate B from the state relation (1-22). Thus, for a liquid, approximately,
n n o o o o o
d B [p {(B 1)( / ) B}] n(B 1)p ( / ) n(B 1)p at 1 atm d
For water, B ≈ 3000 and n ≈ 7, so our estimate is
Bwater ≈ 7(3001)p (^) o= 21007 atm≈ 2.13E9 Pa Ans. (b)
This is 2.7% less than the value B = 2.19E9 Pa listed in Table A-3.
1.37 A near-ideal gas has M = 44 and cv = 610 J/(kg⋅K). At 100°C, what are (a) its specific heat ratio, and (b) its speed of sound?
Solution: The gas constant is R = Λ/Μ = 8314/44 ≈ 189 J/(kg⋅K). Then
c (^) v = R/(k − 1), or: k = 1 + R/cv = 1 + 189/610 ≈ 1.31 Ans. (a) [It is probably N O] 2
With k and R known, the speed of sound at 100ºC = 373 K is estimated by
a = kRT = 1.31[189 m /(s^2 2 ⋅ K)](373 K)≈ 304 m/s Ans. (b)
1.38 In Fig. P1.38, if the fluid is glycerin at 20°C and the width between plates is 6 mm, what shear stress (in Pa) is required to move the upper plate at V = 5.5 m/s? What is the flow Reynolds number if “L” is taken to be the distance between plates? Fig. P1.
Solution: (a) For glycerin at 20°C, from Table 1.4, μ ≈ 1.5 N · s/m^2. The shear stress is
found from Eq. (1) of Ex. 1.8:
V (1.5 Pa s)(5.5 m/s)
. (a) h (0.006 m)
Ans
= = ≈ 1380 Pa
The density of glycerin at 20°C is 1264 kg/m 3. Then the Reynolds number is defined by Eq. (1.24), with L = h, and is found to be decidedly laminar, Re < 1500:
3 L
VL (1264 kg/m )(5.5 m/s)(0.006 m) Re. (b) 1.5 kg/m s
Ans
