Reaktion: Kyanit = Sillimanit (∆ r G = ∆ f G Sil - ∆ f G Ky )
Fluid MechanicsFluid Mechanics SOLUTIONS TO PROBLEMS Section 15.1 Pressure P15.2 Let 4 F g be its...
Transcript of Fluid MechanicsFluid Mechanics SOLUTIONS TO PROBLEMS Section 15.1 Pressure P15.2 Let 4 F g be its...
Chapter 15 1 Fluid Mechanics
SOLUTIONS TO PROBLEMS Section 15.1 Pressure
P15.2 Let Fg be its weight. Then each tire supports
Fg
4,
so P =
FA=
Fg
4A
yielding Fg = 4AP = 4 0.024 0 m2( ) 200 ! 103 N m2( ) = 1.92 ! 104 N
P15.3
P =FA=
50.0 9.80( )! 0.500 " 10#2( )2
= 6.24 " 106 N m2
P15.4 The Earth’s surface area is 4!R2 . The force pushing inward over this area amounts to
F = P0A = P0 4!R2( ) . This force is the weight of the air:
Fg = mg = P0 4!R2( ) so the mass of the air is
m =
P0 4!R2( )g
=1.013 " 105 N m2( ) 4! 6.37 " 106 m( )2#
$%&'(
9.80 m s2 = 5.27 " 1018 kg .
Section 15.2 Variation of Pressure with Depth P15.10 The pressure on the bottom due to the water is Pb = ! gz = 1.96 " 104 Pa
So, Fb = PbA = 5.88 ! 106 N
On each end, F = PA = 9.80 ! 103 Pa 20.0 m2( ) = 196 kN
On the side, F = PA = 9.80 ! 103 Pa 60.0 m2( ) = 588 kN
2 Fluid Mechanics
*P15.11 The fluid in the hydraulic jack is originally exerting the same pressure as the air outside. This pressure P0 results in zero net force on either piston. For the equilibrium of piston 2 we require
500 lb = P ! P0( )A = P ! P0( )" 1.5 in.
2#$%
&'(
2
Let F1 represent the force the lever bar exerts on piston 1. Then similarly
F1 = P ! P0( )" 0.25 in.
2#$%
&'(
2
We ignore the weights of the pistons, sliding friction, and the slight difference in fluid pressure P
due to the height difference between points 1 and 2. By division
F1500 lb
=0.25 in.1.5 in.
!"#
$%&
2
F1 =500 lb
36
We say the hydraulic lift has an ideal mechanical advantage of 36. Next for the lever bar we ignore
weight and friction, assume equilibrium, and take torques about the fixed hinge.
!" = 0 F1 2 in.( ) ! F 12 in.( ) = 0 F =
F16
The lever has an ideal mechanical advantage of 6. By substitution,
F =
500 lb36 !6
= 2.31 lb
Section 15.3 Pressure Measurements P15.12 (a) We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the
water surface in the basin and point 2 at the water surface in the straw:
P1 + ! gy1 = P2 + ! gy2
1.013 ! 105 N m2 + 0 = 0 + 1 000 kg m3( ) 9.80 m s2( ) y2 y2 = 10.3 m (b) No atmosphere can lift the water in the straw through
zero height difference.
P15.15 !P0 = "g!h = #2.66 $ 103 Pa : P = P0 + !P0 = 1.013 " 0.026 6( ) # 105 Pa = 0.986 # 105 Pa
Chapter 15 3
Section 15.4 Buoyant Forces and Archimedes’s Principle P15.18 Fg = m + !sV( ) g must be equal to Fb = !wVg
Since V = Ah , m + !sAh = !wAh
and A =
m!w " !s( )h
FIG. P15.18
P15.19 At equilibrium F! = 0 or Fapp + mg = B where B is the buoyant force.
The applied force, Fapp = B ! mg
where B = Vol !water( ) g
and m = Vol( )!ball .
So, Fapp = Vol( ) g !water " !ball( ) = 4
3#r3 g !water " !ball( )
FIG. P15.19
Fapp =
43! 1.90 " 10#2 m( )3 9.80 m s2( ) 103 kg m3 # 84.0 kg m3( ) = 0.258 N
P15.21 (a) P = P0 + ! gh
Taking P0 = 1.013 ! 105 N m2 and h = 5.00 cm
we find Ptop = 1.017 9 ! 105 N m2
For h = 17.0 cm , we get Pbot = 1.029 7 ! 105 N m2
Since the areas of the top and bottom are A = 0.100 m( )2 = 10!2 m2 we find
Ftop = PtopA = 1.017 9 ! 103 N
and Fbot = 1.029 7 ! 103 N
FIG. P15.21
(b) T + B ! Mg = 0
where B = !wVg = 103 kg m3( ) 1.20 " 10#3 m3( ) 9.80 m s2( ) = 11.8 N
and Mg = 10.0 9.80( ) = 98.0 N
Therefore, T = Mg ! B = 98.0 ! 11.8 = 86.2 N
(c) Fbot ! Ftop = 1.029 7 ! 1.017 9( ) " 103 N = 11.8 N
which is equal to B found in part (b).
4 Fluid Mechanics
P15.23 (a) According to Archimedes, B = !waterVwater g = 1.00 g cm3( ) 20.0 " 20.0 " 20.0 # h( )[ ] g
But B = Weight of block = mg = !woodVwood g = 0.650 g cm3( ) 20.0 cm( )3 g
0.650 20.0( )3 g = 1.00 20.0( ) 20.0( ) 20.0 ! h( ) g
20.0 ! h = 20.0 0.650( ) so h = 20.0 1 ! 0.650( ) = 7.00 cm (b) B = Fg + Mg where M = mass of lead
1.00 20.0( )3 g = 0.650 20.0( )3 g + MgM = 1.00 ! 0.650( ) 20.0( )3 = 0.350 20.0( )3 = 2 800 g = 2.80 kg
P15.25 The balloon stops rising when !air " !He( ) gV = Mg and !air " !He( )V = M ,
Therefore, V =
M!air " !He
=400
1.25e"1 " 0.180 V = 1 430 m3
Section 15.7 Bernoulli’s Equation P15.30 Volume flow rate = A1v1 = A2v2
20.0 L60.0 s
1 000 cm3
1 L!
"#$
%&= ' 1.00 cm( )2 vhose = ' 0.500 cm( )2 vnozzle
(a) vhose =
333 cm3 s3.14 cm2 = 106 cm s
(b) vnozzle =
333 cm3 s0.785 cm2 = 424 cm s
P15.33 Apply Bernoulli’s equation between the top surface and the
exiting stream.
P0 + 0 + ! gh = P0 +
12! vx
2 + ! gh
vx2 = 2g h0 ! h( ) ! vx = 2g h0 " h( )
x = vxt : y = h = 1
2g t2
! t = 2h
g
and x = vx
2hg= 2g h0 ! h( ) 2h
g: x = 2 h h0 ! h( )
FIG. P15.33
Chapter 15 5 *P15.34 Assuming the top is open to the atmosphere, then
P1 = P0 .
Note P2 = P0 .
Flow rate = 2.50 ! 10"3 m3 min = 4.17 ! 10"5 m3 s . (a) A1 >> A2 so v1 << v2
Assuming v1 = 0 ,
P1 +! v1
2
2+ ! gy1 = P2 +
! v22
2+ ! gy2
v2 = 2gy1( )1 2 = 2 9.80( ) 16.0( )[ ]1 2 = 17.7 m s
(b) Flow rate = A2v2 =
! d2
4"
#$%
&'17.7( ) = 4.17 ( 10)5 m3 s
d = 1.73 ! 10"3 m = 1.73 mm