CE 3500 Fluid Mechanics – Fall 2014
Exercises
1 Tire pressure
A tire having a volume of 3 ft3 (0.085 m3) contains air at a gage pressure of 26 psi (179 kPa) and
a temperature of 70 ºF (21ºC). The atmospheric pressure is p atm =14.7 psi (101 kPa). The ideal
gas constant is R=1716
ft lb f
−1 o
=289.6 J kg
−1
. The gravitational acceleration is
slug R
g=32.2 ft s−2 =9.81 m s−2 . Temperature conversions: 0 ºF = 459.67 ºR, 0 ºC = 273.15 ºK).
Determine the density of the air and the weight of the air contained in the tire. Carry out the
calculations in both English and SI units.
o
K
Ideal gas law
[ ]
in.2
( 26 [ psi ] +14.7 [ psi ] ) 144 2
ft
p
slug
ρ=
→ ρ=
=6.448×10−3
RT
ft lbf
ft 3
o
1716
529.67
[
R
]
slug o R
[
ρ=
]
[ ]
[ ]
179 [ kPa ] +101 [ kPa ]
p
kg
→ ρ=
=3.287 3
RT
J
m
289.6
294.15 o K
o
kg K
[ ]
The weight of the air is
[ ] [ ][
[ ] [ ] [ ]
W =ρ g V =6.448×10−3
W =ρ g V =3.287
slug
ft
32.2 2 3 ft 3 ]=0.6229 [ lb f ]
3
ft
s
kg
m
9.81 2 0.085 m3 =2.741 [ N ]
3
m
s
1
CE 3500 Fluid Mechanics – Fall 2014
Exercises
2 Lubricated shaft
A 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in figure below. The
lubricant that fills the 0.3 mm gap between the shafts and bearing is an oil having a kinematic
μ
viscosity ( ν= ρ ) of 8.0 × 10-4 m2 s-1 and a specific gravity SG of 0.91. The density of the water
is ρ H O =1000 kg m−3 . Determine the force F required to pull the shaft at a velocity of 3 m s-1.
2
Assume the velocity distribution in the gap is linear.
25 mm
Bearing
F
Shaft
Bearing
Lubricant
0.5 m
The velocity difference ( du ) between the bearing and the shaft is 3 m s-1 resulting in linearly
(
changing velocity over the 0.3 mm gap ( dy ). Therefore shearing stress τ=μ
)
du
du
=νρ
is
dy
dy
3 [ m s−1 ]
equal to τ=8.0×10 [ m s ] 0.91×10 [ kg m ]
→ τ=7280 [ kg m−1 s−2 ] acting on
−4
3×10 [ m ]
the shaft's surface A=2 π R l = 3.14 × 0.025 m × 0.5 m = 0.03925 m2.
−4
2 −1
3
−3
Therefore F=τ A = 7280 kg m-1 s-2 × 0.03925 m2 = 286 N
2
CE 3500 Fluid Mechanics – Fall 2014
Exercises
3 U-tube manometer
An inverted U-tube manometer containing oil (SG = 0.8 ) is located between two reservoirs as
shown in figure below. The reservoirs on the left, which contains carbon tetrachloride, is closed
and pressurized to 8 psi. The reservoir on the right contains water and is open to the atmosphere.
With the given data, determine the depth of water, h, in the right reservoir.
p=8 [ psi ] , SG oil=0.8 , SG ctc=1.58 , γ w =62.4 [ lb ft −3] ,
8 [ psi ]+ SG ctc γ w 0.3 [ ft ] +SG oil γ w 0.7 [ ft ]+ γ w 2 [ ft ]=γ w h
8 [ psi ]
γ w + SG ctc 0.3 [ ft ]+ SGoil 0.7 [ ft ]+2 [ ft ]=h
h=
1152 [ lb ft −2 ]
62.4 [ lb ft −3 ]
+1.58×0.3 [ ft ] +0.8×0.7 [ ft ] +2 [ ft ]=21.5 [ ft ]
3
CE 3500 Fluid Mechanics – Fall 2014
Exercises
4 Pressure on a piston
A 6 in.-diameter (15.24 cm) piston is located within a cylinder that is connected to a ½ in.diameter (1.27 cm) inclined-tube manometer The fluid in the cylinder and the manometer is oil
(specific weight γ oil =59
lb f
3
=9.27 kN m−3 ).When a weight, W , is placed on the top of the
ft
cylinder, the fluid level in the manometer tube rises 6 in. (15.24 cm) from point (1) to (2). How
heavy is the weight? Assume that the change in position of the piston is negligible. Solve the
problem using both English and SI units.
W = p A , Δ H =Δ L sin θ , p=γ oil Δ H
p=59
[ ]
lb f
ft
3
6 [ in. ]
[ ]
lb
1 [ ft ]
sin 30 o=14.75 2f
12 [ in. ]
ft
p=9.27 [ kN m−3 ] 0.1524 [ m ] sin 30o =706 [ Pa ]
A=
2
r2
π= ( 3 [ in. ] ) π=28.26 [ in.2 ]=0.196 [ ft 2 ] → W =2.89 [ lb f ]
4
(
)
2
0.1524 [ m ]
r2
A= π=
π=0.018 [ m 2 ] → W =12.87 [ N ]
4
2
4
CE 3500 Fluid Mechanics – Fall 2014
Exercises
5 Atmospheric pressure
(
p= p 0 1−
βz
T0
)
g
Rβ
provides the relationship between pressure and elevation in the atmosphere for
those regions in which the temperature varies linearly with elevation ( T =T 0−β z ). Calculate the
value for elevation of 5 km if the lapse rate is β=0.0065 [ ˚ K m−1 ] . The pressure at sea level is
p 0=101.33 [ kPa ] and the sea level temperature T 0=15 [ ˚ C ] . The gas constant is
R=289.6 [ J kg−1 K −1 ] and the gravitational acceleration is g=9.81 [ m s−2] .
(
βz
p= p 0 1−
T0
)
g
Rβ
(
0.0065 [ ˚ C ] 5000 [ m ]
=101.33 [ kPa ] 1−
15 [ ˚ C ] +273.15 [ ˚ K ]
)
9.81 [ m s−2 ]
289.6 [ J kg K
−1
−1
] 0.0065 [ ˚ K m−1 ] =54.311 [ kPa ]
5