SOLUTIONS MANUAL
For Fluid Mechanics 8th Edition White
(Author) Graded A+ Latest Update.
, chapter 1 • pressure distribution in a fluid
p2.1 for the two-dimensional stress field in
fig. p2.1, let
find the shear and normal stresses on plane
aa cutting through at 30°.
solution: make cut —aa‖ so that it just hits
the bottom right corner of the element. this
Fig. P2.1
gives the freebody shown at right. now sum
forces normal and tangential to side aa.
denote side length aa as —l.‖
p2.2 for the stress field of fig. p2.1, change the known data to σxx = 2000 psf, σyy
= 3000 psf, and σn(aa) = 2500 psf. compute σxy and the shear stress on plane aa.
solution: sum forces normal to and tangential to aa in the element freebody above, with
σn(aa) known and σxy unknown:
,2-2 2-2
in like manner, solve for the shear stress on plane aa, using our result for σxy:
this problem and prob. p2.1 can also be solved using mohr‘s circle.
p2.3 a vertical clean glass piezometer tube has an inside diameter of 1 mm. when a
pressure is applied, water at 20°c rises into the tube to a height of 25 cm. after correcting
for surface tension, estimate the applied pressure in pa.
solution: for water, let y = 0.073 n/m, contact angle θ = 0°, and γ = 9790 n/m3. the
capillary rise in the tube, from example 1.9 of the text, is
then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m =
0.22 m. the applied pressure is estimated to be p = γhpress = (9790 n/m3)(0.22 m) ≈
2160 pa ans.
θ? Bourdon
p2.4 pressure gages, such as the bourdon gage W gage
in fig. p2.4, are calibrated with a deadweight piston.
if the bourdon gage is designed to rotate the pointer
10 degrees for every 2 psig of internal pressure, how 2 cm Oil
diameter
many degrees does the pointer rotate if the piston and
weight together total 44 newtons?
Fig. P2.4
solution: the deadweight, divided by the piston area, should equal the pressure applied to
the bourdon gage. stay in si units for the moment:
2-3
at 10 degrees for every 2 psig, the pointer should move approximately 100 degrees. ans.
, p2.5 quito, ecuador has an average altitude of 9,350 ft. on a standard day, pressure gage
a in a laboratory experiment reads 63 kpa and gage b reads 105 kpa. express these
readings in gage pressure or vacuum pressure, whichever is appropriate.
solution: convert 9,350 ft x 0.3048 = 2,850 m. we can interpolate in the standard
Altitude Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
good interpolating! then pa = 71500-63000 = 8500 pa (vacuum pressure) ans.(a),
and pb = 105000 - 71500 = 33500 pa (gage pressure) ans.(b)
p2.6 express standard atmospheric pressure as a head, h = p/ρg, in (a) feet of glycerin;
(b) inches of mercury; (c) meters of water; and (d) mm of ethanol.
solution: take the specific weights, γ = ρg, from table a.3, divide patm by γ :
(a) glycerin: h = (2116 lbf/ft2)/(78.7 lbf/ft3) ≈ 26.9 ft ans. (a)
(b) mercury: h = (2116 lbf/ft2)/(846 lbf/ft3) = 2.50 ft ≈ 30.0 inches ans. (b)
(c) water: h = (101350 n/m2)/(9790 n/m3) ≈ 10.35 m ans. (c)
(d) ethanol: h = (101350 n/m2)/(7740 n/m3) = 13.1 m ≈ 13100 mm ans. (d)
p2.7 la paz, bolivia is at an altitude of approximately 12,000 ft. assume a standard
atmosphere. how high would the liquid rise in a methanol barometer, assumed at
20°c? [hint: don‘t forget the vapor pressure.]
solution: convert 12,000 ft to 3658 meters, and table a.6, or eq. (2.20), give
For Fluid Mechanics 8th Edition White
(Author) Graded A+ Latest Update.
, chapter 1 • pressure distribution in a fluid
p2.1 for the two-dimensional stress field in
fig. p2.1, let
find the shear and normal stresses on plane
aa cutting through at 30°.
solution: make cut —aa‖ so that it just hits
the bottom right corner of the element. this
Fig. P2.1
gives the freebody shown at right. now sum
forces normal and tangential to side aa.
denote side length aa as —l.‖
p2.2 for the stress field of fig. p2.1, change the known data to σxx = 2000 psf, σyy
= 3000 psf, and σn(aa) = 2500 psf. compute σxy and the shear stress on plane aa.
solution: sum forces normal to and tangential to aa in the element freebody above, with
σn(aa) known and σxy unknown:
,2-2 2-2
in like manner, solve for the shear stress on plane aa, using our result for σxy:
this problem and prob. p2.1 can also be solved using mohr‘s circle.
p2.3 a vertical clean glass piezometer tube has an inside diameter of 1 mm. when a
pressure is applied, water at 20°c rises into the tube to a height of 25 cm. after correcting
for surface tension, estimate the applied pressure in pa.
solution: for water, let y = 0.073 n/m, contact angle θ = 0°, and γ = 9790 n/m3. the
capillary rise in the tube, from example 1.9 of the text, is
then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m =
0.22 m. the applied pressure is estimated to be p = γhpress = (9790 n/m3)(0.22 m) ≈
2160 pa ans.
θ? Bourdon
p2.4 pressure gages, such as the bourdon gage W gage
in fig. p2.4, are calibrated with a deadweight piston.
if the bourdon gage is designed to rotate the pointer
10 degrees for every 2 psig of internal pressure, how 2 cm Oil
diameter
many degrees does the pointer rotate if the piston and
weight together total 44 newtons?
Fig. P2.4
solution: the deadweight, divided by the piston area, should equal the pressure applied to
the bourdon gage. stay in si units for the moment:
2-3
at 10 degrees for every 2 psig, the pointer should move approximately 100 degrees. ans.
, p2.5 quito, ecuador has an average altitude of 9,350 ft. on a standard day, pressure gage
a in a laboratory experiment reads 63 kpa and gage b reads 105 kpa. express these
readings in gage pressure or vacuum pressure, whichever is appropriate.
solution: convert 9,350 ft x 0.3048 = 2,850 m. we can interpolate in the standard
Altitude Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
good interpolating! then pa = 71500-63000 = 8500 pa (vacuum pressure) ans.(a),
and pb = 105000 - 71500 = 33500 pa (gage pressure) ans.(b)
p2.6 express standard atmospheric pressure as a head, h = p/ρg, in (a) feet of glycerin;
(b) inches of mercury; (c) meters of water; and (d) mm of ethanol.
solution: take the specific weights, γ = ρg, from table a.3, divide patm by γ :
(a) glycerin: h = (2116 lbf/ft2)/(78.7 lbf/ft3) ≈ 26.9 ft ans. (a)
(b) mercury: h = (2116 lbf/ft2)/(846 lbf/ft3) = 2.50 ft ≈ 30.0 inches ans. (b)
(c) water: h = (101350 n/m2)/(9790 n/m3) ≈ 10.35 m ans. (c)
(d) ethanol: h = (101350 n/m2)/(7740 n/m3) = 13.1 m ≈ 13100 mm ans. (d)
p2.7 la paz, bolivia is at an altitude of approximately 12,000 ft. assume a standard
atmosphere. how high would the liquid rise in a methanol barometer, assumed at
20°c? [hint: don‘t forget the vapor pressure.]
solution: convert 12,000 ft to 3658 meters, and table a.6, or eq. (2.20), give
