Solutions Manual for
Fluid Mechanics
Seventh Edition in SI Units
Frank M. White
Chapter 4
Differential Relations
for Fluid Flow
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc.
(“McGraw-Hill”) and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following restrictions, and if the
recipient does not agree to these restrictions, the Manual should be promptly returned
unopened to McGraw-Hill: This Manual is being provided only to authorized
professors and instructors for use in preparing for the classes using the affiliated
textbook. No other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any student or other
third party. No part of this Manual may be reproduced, displayed or distributed in
any form or by any means, electronic or otherwise, without the prior written
permission of the McGraw-Hill.
© 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators
for course preparation. If you are a student using this Manual, you are using it without permission.
,P4.1 An idealized velocity field is given by the formula
Is this flow field steady or unsteady? Is it two- or three-dimensional? At the point (x, y, z) =
(–1, +1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the
acceleration.
Solution: (a) The flow is unsteady because time t appears explicitly in the components. (b)
The flow is three-dimensional because all three velocity components are nonzero.
(c) Evaluate, by laborious differentiation, the acceleration vector at (x, y, z) = (−1, +1, 0).
du ∂ u ∂u ∂u ∂u
= +u +v +w = 4x + 4tx(4t) − 2t 2 y(0) + 4xz(0) = 4x +16t 2 x
dt ∂ t ∂x ∂y ∂z
dv ∂ v ∂v ∂v ∂v
= + u + v + w = −4ty + 4tx(0) − 2t 2 y(−2t 2 ) + 4xz(0) = −4ty + 4t 4 y
dt ∂ t ∂x ∂y ∂z
dw ∂ w ∂w ∂w ∂w
= +u +v +w = 0 + 4tx(4z) − 2t 2 y(0) + 4xz(4x) = 16txz +16x 2 z
dt ∂t ∂x ∂y ∂z
dV
or: = (4x +16t 2 x)i + (−4ty + 4t 4 y)j + (16txz +16x 2 z)k
dt
at (x, y, z) = (−1, +1, 0), we obtain
(d) At (–1, +1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans.
P4.2 A three-dimensional flow field has the velocity given by
V = 4x 2 i + 3xyj + 5t(x + y)zk
Find the local and total accelerations in terms of x, y, z, and t.
Solution
Velocity V = (4x 2 , 3xy, 5t(x + y)z)
∂V
Local acceleration = (0, 0, 5(x + y)z)
∂t
Convective acceleration
∂u ∂u ∂u
ax = u +v +w
∂x ∂y ∂z
= (4x 2 )(8x) + (3xy)(0) + 5t(x + y)z(0)
= 32x 3
,2
∂v ∂u ∂u
ay = u +v +w
∂x ∂y ∂z
= (4x 2 )(3y) + (3xy)(3x) + 5t(x + y)z(0)
= 12x 2 y + 9x 2 y
= 21x 2 y
∂w ∂w ∂w
az = u +v +w
∂x ∂y ∂z
= (4x 2 )(5tz) + (3xy)(5tz) + 5t(x + y)z(5t)(x + y)
= 20tx 2 z +15txyz + 25t 2 (x + y)2 z
Total acceleration = Local accel + convective accel
∂v
= (32x 3 , 21x 2 y, 20tx 2 z + 15txyz + 25t 2 (x + y)2 z + 5(x + y)z)
∂t
P4.3 Flow through the converging nozzle in Fig. P4.3 can be approximated by the one-
dimensional velocity distribution
(a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case
Vo = 3 m/s and L = 15 cm, compute the acceleration, in g’s, at the entrance and at the exit.
Fig. P4.3
Solution: Here we have only the single ‘one-dimensional’ convective acceleration:
du 2(3)2 2x
For L = 0.15 m and Vo = 3 m/s, = 1+ = 120(1+13.3x), with x in m
dt 0.15 0.15
At x = 0, du/dt = 120 m/s2 (12 g’s); at x = L = 0.15 m, du/dt = 360 m/s2 (37 g’s). Ans. (b)
, 3
P4.4 A two-dimensional velocity field is given by
V = (x2 – y2 + x)i – (2xy + y)j
in arbitrary units. At (x, y) = (1, 2), compute (a) the accelerations ax and ay, (b) the velocity
component in the direction θ = 40°, (c) the direction of maximum velocity, and (d) the
direction of maximum acceleration.
Solution: (a) Do each component of acceleration:
du ∂u ∂u
= u + v = (x 2 − y 2 + x)(2x +1) + (−2xy − y)(−2y) = a x
dt ∂x ∂y
dv ∂v ∂v
= u + v = (x 2 − y 2 + x)( −2y) + (−2xy − y)(−2x −1) = a y
dt ∂x ∂y
At (x, y) = (1, 2), we obtain ax = 18i and ay = 26j Ans. (a)
(b) At (x, y) = (1, 2), V = –2i – 6j. A unit vector along a 40° line would be n = cos40°i +
sin40°j. Then the velocity component along a 40° line is
V40° = V⋅n40° = (−2i − 6 j) ⋅ (cos 40°i + sin 40°j) ≈ 5.39 units Ans. (b)
(c) The maximum acceleration is amax = [182 + 262]1/2 = 31.6 units at ∠55.3° Ans. (c, d)
P4.5 A simple flow model for a two-dimensional converging nozzle is the distribution
x y
u = U o (1+ ) v = −U o w=0
L L
(a) Sketch a few streamlines in the region 0<x/L<1 and 0<y/L<1, using the method of
Section 1.11. (b) Find expressions for the horizontal and vertical accelerations.
(c) Where is the largest resultant acceleration and its numerical value?
Solution: The streamlines are in the x-y plane and are found from the velocities:
dx dy dx dy
= or integrate : ∫ = −∫ Cancel U o
u v U o (1+ x / L) Uo y / L
L ln(1+ x / L) = − L ln(y / L) + const , or : ln[(y / L)(1+ x / L)] = constant
y C
Finally the streamlines : = Ans.(a)
L 1+ x / L
Fluid Mechanics
Seventh Edition in SI Units
Frank M. White
Chapter 4
Differential Relations
for Fluid Flow
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc.
(“McGraw-Hill”) and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following restrictions, and if the
recipient does not agree to these restrictions, the Manual should be promptly returned
unopened to McGraw-Hill: This Manual is being provided only to authorized
professors and instructors for use in preparing for the classes using the affiliated
textbook. No other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any student or other
third party. No part of this Manual may be reproduced, displayed or distributed in
any form or by any means, electronic or otherwise, without the prior written
permission of the McGraw-Hill.
© 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators
for course preparation. If you are a student using this Manual, you are using it without permission.
,P4.1 An idealized velocity field is given by the formula
Is this flow field steady or unsteady? Is it two- or three-dimensional? At the point (x, y, z) =
(–1, +1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the
acceleration.
Solution: (a) The flow is unsteady because time t appears explicitly in the components. (b)
The flow is three-dimensional because all three velocity components are nonzero.
(c) Evaluate, by laborious differentiation, the acceleration vector at (x, y, z) = (−1, +1, 0).
du ∂ u ∂u ∂u ∂u
= +u +v +w = 4x + 4tx(4t) − 2t 2 y(0) + 4xz(0) = 4x +16t 2 x
dt ∂ t ∂x ∂y ∂z
dv ∂ v ∂v ∂v ∂v
= + u + v + w = −4ty + 4tx(0) − 2t 2 y(−2t 2 ) + 4xz(0) = −4ty + 4t 4 y
dt ∂ t ∂x ∂y ∂z
dw ∂ w ∂w ∂w ∂w
= +u +v +w = 0 + 4tx(4z) − 2t 2 y(0) + 4xz(4x) = 16txz +16x 2 z
dt ∂t ∂x ∂y ∂z
dV
or: = (4x +16t 2 x)i + (−4ty + 4t 4 y)j + (16txz +16x 2 z)k
dt
at (x, y, z) = (−1, +1, 0), we obtain
(d) At (–1, +1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans.
P4.2 A three-dimensional flow field has the velocity given by
V = 4x 2 i + 3xyj + 5t(x + y)zk
Find the local and total accelerations in terms of x, y, z, and t.
Solution
Velocity V = (4x 2 , 3xy, 5t(x + y)z)
∂V
Local acceleration = (0, 0, 5(x + y)z)
∂t
Convective acceleration
∂u ∂u ∂u
ax = u +v +w
∂x ∂y ∂z
= (4x 2 )(8x) + (3xy)(0) + 5t(x + y)z(0)
= 32x 3
,2
∂v ∂u ∂u
ay = u +v +w
∂x ∂y ∂z
= (4x 2 )(3y) + (3xy)(3x) + 5t(x + y)z(0)
= 12x 2 y + 9x 2 y
= 21x 2 y
∂w ∂w ∂w
az = u +v +w
∂x ∂y ∂z
= (4x 2 )(5tz) + (3xy)(5tz) + 5t(x + y)z(5t)(x + y)
= 20tx 2 z +15txyz + 25t 2 (x + y)2 z
Total acceleration = Local accel + convective accel
∂v
= (32x 3 , 21x 2 y, 20tx 2 z + 15txyz + 25t 2 (x + y)2 z + 5(x + y)z)
∂t
P4.3 Flow through the converging nozzle in Fig. P4.3 can be approximated by the one-
dimensional velocity distribution
(a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case
Vo = 3 m/s and L = 15 cm, compute the acceleration, in g’s, at the entrance and at the exit.
Fig. P4.3
Solution: Here we have only the single ‘one-dimensional’ convective acceleration:
du 2(3)2 2x
For L = 0.15 m and Vo = 3 m/s, = 1+ = 120(1+13.3x), with x in m
dt 0.15 0.15
At x = 0, du/dt = 120 m/s2 (12 g’s); at x = L = 0.15 m, du/dt = 360 m/s2 (37 g’s). Ans. (b)
, 3
P4.4 A two-dimensional velocity field is given by
V = (x2 – y2 + x)i – (2xy + y)j
in arbitrary units. At (x, y) = (1, 2), compute (a) the accelerations ax and ay, (b) the velocity
component in the direction θ = 40°, (c) the direction of maximum velocity, and (d) the
direction of maximum acceleration.
Solution: (a) Do each component of acceleration:
du ∂u ∂u
= u + v = (x 2 − y 2 + x)(2x +1) + (−2xy − y)(−2y) = a x
dt ∂x ∂y
dv ∂v ∂v
= u + v = (x 2 − y 2 + x)( −2y) + (−2xy − y)(−2x −1) = a y
dt ∂x ∂y
At (x, y) = (1, 2), we obtain ax = 18i and ay = 26j Ans. (a)
(b) At (x, y) = (1, 2), V = –2i – 6j. A unit vector along a 40° line would be n = cos40°i +
sin40°j. Then the velocity component along a 40° line is
V40° = V⋅n40° = (−2i − 6 j) ⋅ (cos 40°i + sin 40°j) ≈ 5.39 units Ans. (b)
(c) The maximum acceleration is amax = [182 + 262]1/2 = 31.6 units at ∠55.3° Ans. (c, d)
P4.5 A simple flow model for a two-dimensional converging nozzle is the distribution
x y
u = U o (1+ ) v = −U o w=0
L L
(a) Sketch a few streamlines in the region 0<x/L<1 and 0<y/L<1, using the method of
Section 1.11. (b) Find expressions for the horizontal and vertical accelerations.
(c) Where is the largest resultant acceleration and its numerical value?
Solution: The streamlines are in the x-y plane and are found from the velocities:
dx dy dx dy
= or integrate : ∫ = −∫ Cancel U o
u v U o (1+ x / L) Uo y / L
L ln(1+ x / L) = − L ln(y / L) + const , or : ln[(y / L)(1+ x / L)] = constant
y C
Finally the streamlines : = Ans.(a)
L 1+ x / L
