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 u to U: “...Find U 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
2.284400 400 ft
2.38(a) p1gage = ( m-a)gh – (t-a)gH
2.48F = 39700 N
2.6818,040 N
2.102124 kN
2.11619100 N/m3
2.126h/H=Z-(Z2-1+)1/2, =d/H, Z=(2+-)/2, =pa/(gH)
2.130Slightly unstable, MG = -0.007 m
2.136MG = L2/(3R) – 4R/(3)
2.15777 r/min, minimum pressure halfway between B and C (not 2.156)
3.18(b) Q = 16bhumax/9
3.623100 N
3.7091 lbf
3.74Fx = 0 , Fy = - 17 N , Fz = - 126 N
3.88V = + [2 + 2Vj]1/2 , = Q/2k
3.138gd4(H+L)/(128LQ) – 2Q/(16L)
3.162Q = 166 ft3/min, p = 0.0204 lbf/in2
3.164(a) 5.25 kg/s ; (b) 91 cm
3.178h2 = 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/(AT) = 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
6.104p = 0.0305 lbf/in2
7.18 = 0.16; Fdrag = 0.024 N
7.34725 N
7.52CD(ReL)1/2 = 2.67 (by numerical integration)
7.64t1000-2000m = 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
9.72Mass flow = 0.5 kg/s, pe = 208 kPa, Mae = 0.407
9.78Case A: 0.0071 kg/s; B: 0.0068 kg/s
11.141870 W
11.661240 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 = 1/2b 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 . Thanks.
1