8 - McGraw Hill
Fluid Mechanics, 4th Edition, by Frank M. White
NOTES ABOUT ERRATA IN THE FIRST PRINTING (not necessarily in order, as new corrections are added to the end of the document):
Page 51, Problem 1.62, line 2:
“If the hydrogen-water interface...” (not ‘hydrogen-air)
Page 88, Eq. (2.52): When changing to blue, the first denominator vanished:
(Presently, the denominator tan( is missing – it was there in Final Proofs.)
Page 186, Problem 3.18, last line: add data for b and h:
Add data: “...estimate umax in m/s for b = h = 10 cm.”
Page 187, Problem 3.25, next-to-last line:
Change Δu to ΔU: “...Find ΔU as a function of stream speed...”
Page 189, Problem 3.36, line 2: add ‘-in-’ (for inch)
“...through a 3-in-pipe and entrains...” (not ‘ 3-pipe’ )
Page 205, Problem 3.150, the equation “hf ( ...” needs a subscript “1” on V:
Page 409, Prob. 6.58, the very bottom of the right column:
“... and (z = 100 m....” (not (x = 100 m)
Page 458, Table 7.2, bottom, elliptical cylinder. Numbers on the left side of the
figures should have a colon, not a decimal point:
1:1 , 2:1 , 4:1 , and 8:1 (not 1.1, 2.1, 4.1, 8.1)
Page 500, Eq. (8.17): Second integral should have d(, not d(:
Page 551, Eqs. (8.199abc): Compositor failed to restore all denominators:
Page 613, line 12: the second “1” should be a plus sign (+):
fLD1 = fLD2 + DeltafLD , not “= fLD2 1 DeltafLD”
Page 627: Add an EES Icon to the Solution to Example 9.17.
Page 641, Problem 9.9, line 3, change ‘at an exit pressure equal to’:
to: “at Vexit = 1600 m/s to an ambient pressure of 54 kPa.”
Page 647, Problem 9.80, Forgot to specify the tire volume:
“...thousandths of an inch. The tire volume is 2.5 ft2.
Page 762, Problem 11.70, last line, add data:
“...what is the flow rate if d = 8 inches.”
Pages 806-812: Answers to Selected Problems – corrections in red:
2.8 (b) DALR = 9.77 (C/km
2.28 4400 ( 400 ft
2.38 (a) p1gage = ( (m-(a)gh – ((t-(a)gH
2.48 F = 39700 N
2.68 18,040 N
2.102 124 kN
2.116 19100 N/m3
2.126 h/H=Z-(Z2-1+()1/2, (=d/H, Z=(2+(-()/2, (=pa/((gH)
2.130 Slightly unstable, MG = -0.007 m
2.136 MG = L2/(3(R) – 4R/(3()
2.157 77 r/min, minimum pressure halfway between B and C (not 2.156)
3.18 (b) Q = 16bhumax/9
3.62 3100 N
3.70 91 lbf
3.74 Fx = 0 , Fy = - 17 N , Fz = - 126 N
3.88 V = ( + [(2 + 2(Vj]1/2 , ( = (Q/2k
3.138 μ ’ πρgd4(H+L)/(128LQ) – (2(Q/(16(L)
3.162 Q = 166 ft3/min, (p = 0.0204 lbf/in2
3.164 (a) 5.25 kg/s ; (b) 91 cm
3.178 h2 = 2.03 ft (subcritical) or 0.74 ft (supercritical)
4.70 (= (cos(/(r2), ( = 2am
4.76 (a) 0.106 m from A; (b) 0.333 m above the wall
4.78 (a) Vwall,max = m/L; (b) pmin at x = L
5.14 (/x = fcn((Ux/()
5.36 (a) QlossR/(A(T) = constant
5.50 (a) F/((UL) = constant
5.68 (a) F/((U) = constant; (b) No, not plausible
6.56 (a) 188 km; (b) 27 MW
6.104 (p = 0.0305 lbf/in2
7.18 ( = 0.16(; Fdrag = 0.024 N
7.34 725 N
7.52 CD(ReL)1/2 = 2.67 (by numerical integration)
7.64 (t1000-2000m = 202 s
7.118 (a) 21 m/s; (b) 360 m
8.82 (a) 4.5 m/s; (b) 1.13; (c) 1.26 hp
9.2 (a) V2 = 450 m/s, (s = 515 J/kg-K; (b) V2 = 453 m/s, (s = 512 J/kg-K
9.48 (a) 313 m/s; (b) 0.124 m/s; (c) 0.00331 kg/s
9.72 Mass flow = 0.5 kg/s, pe = 208 kPa, Mae = 0.407
9.78 Case A: 0.0071 kg/s; B: 0.0068 kg/s
11.14 1870 W
11.66 1240 ft3/min
Fluid Mechanics, 4th Ed., by Frank M. White NEW 4/8/99
FURTHER ERRATA IN THE 1st PRINTING, discovered by Christoffer Norberg
p. 67, line 1 below Fig. 2.6: ... and p2 = pa at z2 = 0: (subscript a , not z)
p. 138, line 10: “...double-integral flux terms required...” (not triple-integral)
p. 169, Fig. E3.18, near top arrow: should be (s = 700 hp, (not (s).
p. 180, Eqs. (2) and (3) need negative signs:
p. 181, equation below Eq. (1) and the sentence “We relate...”:
A1V1 = A2V2
V1 = (2V2 (not (2V2 )
p. 232, Eq. (4.47); Minus sign before the square bracket: (v = - [......
p. 233, 4 lines below Eq. (4.50); ...using the linear-momentum equation (4.32)
p. 240, line 8: ...as follows. From Eq. (1.41) the definition ...
p. 256, Fig. 4.15, left side: just plain U ( not U( ).
p. 293, 3 lines above Eq. (5.24a): ...variables from Eqs. (5.23) into Eqs. (5.21) and (5.22)
p. 294, last term of Eq. (5.25) should have a minus sign:
p. 330, line 19: ... no analyses which can simulate the fine scale random ...
p. 339, Eq. (6.26): reverse the subscripts on V: ... = ((V2 – V1) = 0
p. 353, center of “Iterative Solution”, two places, change 6.54 to 6.64:
“... compute fnew ( 0.0201 [Eq. (6.64)]. (two places)
p. 373, Eq. (6.104), power of 4: “... = 1 – d14/d24 – Cp (6.104) ”
p. 385, Fig. 6.28a, and p. 386, Fig. 6.28b, if possible without excessive expense,
change Mt = 0.2 to Mat = 0.2 . (Mach number should be Ma)
p. 393, Eq. (6.120): drop the letter L: “ f = (const)(U) (6.120) ”
p. 415, Problem P6.118: add the flow rate as data:
“... compute the overall pressure drop p1-p2 in lbf/ft2 if Q = 20 ft3/s. ”
p. 480, Problem P7.43, last line: change to 5 mm instead of cm:
“... in m/s, at 5 mm above the element. ”
p. 486, Prob. P7.96, line 5: the W should be ( (cap omega):
“... for the rotation rate ( as a function of U, D, ...”
p. 504, if not too expensive to change a label in Figure 8.7: The line from
the ith vortex to the point (x,y) should be labeled ri , not rt.
p. 531, Fig. 8.22b: If inexpensive, change the signs on y:
y = 1/2 b in upper right of the wing, y = - ½ b in the lower left.
p. 533, Eq. (8.81), “L = ...”: Add a denominator (1+2/AR), shilling style:
p. 563, Prob. P8.94: change 8.84 to 8.86 and change 8.85 to 8.87:
“... streamlines from Eq. (8.86) are everywhere ... lines from Eq. (8.87) ...”
p. 582, Ans. (c): add subscript 1 : “ (o1 = ... ” (not just (o = )
p. 645, Prob. P9.62: Add a data sentence at the end: The exit area is 10 cm2.
p. 649, Prob. P9.115: Add data at the end: Assume To1 = 305 K.
p. 651, Prob. P9.137: “... all combinations of ( and h in ...” (not ()
p. 697, Prob. P10.25: Delete the last a: “ ... range 0 < y/a < 0.866.” (not 0.866a)
p. 698, Prob. P10.27: “... wall shear stress is 0.15 lbf/ft2, and compare...” (not 0.18)
p. 700, Prob. P10.68: “... the maximum Froude number is 1.47. ” (not 1.715)
p. 703, Prob. P10.103: Modify the last two sentences: “Is this a mild or steep slope,
of type 1, 2, or 3? Take R = 50 cm. ”
p. 704, Prob. P10.108: Add data at the end: “ Let So = 0.005. ”
p. 765, Prob. 11.94: Add data at end: “ ... power specific speed if n = 200 rev/min.”
p. 767, Design Project: Should be labeled D11.1 (not P11.1)
p. 811, Answer to 9.72: Mass flow = 0.5 kg/s, pe = 185 kPa, Mae = 0.407
NOTE: If further errors are detected, please notify Frank M. White, Dept. of Mechanical Engineering, University of Rhode Island, 92 Upper College Road, Kingston RI 02881. Phone 401-789-1733, Fax 401-874-2355, E-mail WhiteF@egr.uri.edu. Thanks.
-----------------------
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
EES
[pic]
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- mcgraw hill instructor log in
- mcgraw hill online for students
- mcgraw hill lausd
- studysync connected mcgraw hill lausd
- mcgraw hill student resources website
- lausd mcgraw hill connected ed
- connected ed mcgraw hill school
- connected mcgraw hill com lausd
- mcgraw hill instructor resources
- mcgraw hill textbook website
- mcgraw hill education connected ed
- mcgraw hill connected ed math