selectionmec6 (1)
TRANSCRIPT
Problem 6.38 [Difficulty: 1]
Given: Water at speed 25 ft/s
Find: Dynamic pressure in in. Hg
Solution:
Basic equations pdynamic12
ρ⋅ V2⋅= p ρHg g⋅ ∆h⋅= SGHg ρ⋅ g⋅ ∆h⋅=
Hence ∆hρ V2⋅
2 SGHg⋅ ρ⋅ g⋅=
V2
2 SGHg⋅ g⋅=
∆h12
25fts⋅⎛⎜
⎝⎞⎠
2×
113.6
×s2
32.2 ft⋅×
12 in⋅1 ft⋅
×= ∆h 8.56 in⋅=
Problem 6.39 [Difficulty: 1]
Given: Air speed of 100 km/hr
Find: Dynamic pressure in mm water
Solution:
Basic equations pdynamic12
ρair⋅ V2⋅= p ρw g⋅ ∆h⋅=
Hence ∆hρairρw
V2
2 g⋅⋅=
∆h
1.23kg
m3⋅
999kg
m3⋅
12
× 100kmhr
⋅⎛⎜⎝
⎞⎠
2×
1000 m⋅1 km⋅
⎛⎜⎝
⎞⎠
2×
1 hr⋅
3600 s⋅⎛⎜⎝
⎞⎠
2×
s2
9.81 m⋅×= ∆h 48.4 mm⋅=
Problem 6.42 [Difficulty: 2]
Given: Air jet hitting wall generating pressures
Find: Speed of air at two locations
Solution:
Basic equations pρair
V2
2+ g z⋅+ const= ρair
pRair T⋅
= ∆p ρHg g⋅ ∆h⋅= SGHg ρ⋅ g⋅ ∆h⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data R 287J
kg K⋅⋅= T 10− °C= ρ 999
kg
m3⋅= p 200 kPa⋅= SGHg 13.6=
For the air ρairp
R T⋅= ρair 2.65
kg
m3=
Hence, applying Bernoulli between the jet and where it hits the wall directly
patmρair
Vj2
2+
pwallρair
= pwallρair Vj
2⋅
2= (working in gage pressures)
Hence pwall SGHg ρ⋅ g⋅ ∆h⋅=ρair Vj
2⋅
2= so Vj
2 SGHg⋅ ρ⋅ g⋅ ∆h⋅
ρair=
∆h 25 mm⋅= hence Vj 2 13.6× 999×kg
m3⋅
12.65
×m3
kg⋅ 9.81×
m
s2⋅ 25× mm⋅
1 m⋅1000 mm⋅
×= Vj 50.1ms
=
Repeating the analysis for the second point
∆h 5 mm⋅=patmρair
Vj2
2+
pwallρair
V2
2+= V Vj
2 2 pwall⋅
ρair−= Vj
2 2 SGHg⋅ ρ⋅ g⋅ ∆h⋅
ρair−=
Hence V 50.1ms
⋅⎛⎜⎝
⎞⎠
22 13.6× 999×
kg
m3⋅
12.65
×m3
kg⋅ 9.81×
m
s2⋅ 5× mm⋅
1 m⋅1000 mm⋅
×−= V 44.8ms
=
Problem 6.43 [Difficulty: 2]
Problem 6.44 [Difficulty: 2]
Given: Wind tunnel with inlet section
Find: Dynamic and static pressures on centerline; compare with Speed of air at two locations
Solution:
Basic equations pdyn12
ρair⋅ U2⋅= p0 ps pdyn+= ρair
pRair T⋅
= ∆p ρw g⋅ ∆h⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data T 5− °C= U 50ms⋅= R 287
Jkg K⋅⋅= patm 101 kPa⋅= h0 10− mm⋅= ρw 999
kg
m3⋅=
For air ρairpatmR T⋅
= ρair 1.31kg
m3=
pdyn12
ρair⋅ U2⋅= pdyn 1.64 kPa⋅=
Also p0 ρw g⋅ h0⋅= p0 98.0− Pa= (gage)
and p0 ps pdyn+= so ps p0 pdyn−= ps 1.738− kPa= hsps
ρw g⋅= hs 177− mm=
(gage)
Streamlines in the test section are straight son
p∂
∂0= and pw pcenterline=
In the curved sectionn
p∂
∂ρair
V2
R⋅= so pw pcenterline<
Problem 6.45 [Difficulty: 2]
Problem 6.46 [Difficulty: 2]
Problem 6.49 [Difficulty: 2]
Problem 6.50 [Difficulty: 2]
Given: Siphoning of gasoline
Find: Flow rate
Solution:
Basic equation pρgas
V2
2+ g z⋅+ const=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the gas tank free surface and the siphon exit
patmρgas
patmρgas
V2
2+ g h⋅−= where we assume the tank free surface is slowly changing so Vtank <<,
and h is the difference in levels
Hence V 2 g⋅ h⋅=
The flow rate is then Q V A⋅=π D2⋅
42 g⋅ h⋅⋅=
Qπ
4.5 in⋅( )2
×1 ft2⋅
144 in2⋅
× 2 32.2×ft
s21× ft⋅×= Q 0.0109
ft3
s⋅= Q 4.91
galmin⋅=
Problem 6.54 [Difficulty: 3]
Given: Flow rate through siphon
Find: Maximum height h to avoid cavitation
Solution:
Basic equation pρ
V2
2+ g z⋅+ const= Q V A⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data Q 5Ls⋅= Q 5 10 3−
×m3
s= D 25 mm⋅= ρ 999
kg
m3⋅= patm 101 kPa⋅=
From continuity VQA
=4 Q⋅
π D2⋅
= V4π
0.005×m3
s⋅
1.025 m⋅
⎛⎜⎝
⎞⎠
2×= V 10.2
ms
=
Hence, applying Bernoulli between the free surface and point A
patmρ
pAρ
g h⋅+V2
2+= where we assume VSurface <<
Hence pA patm ρ g⋅ h⋅− ρV2
2⋅−=
From the steam tables, at 20oC the vapor pressure is pv 2.358 kPa⋅=
This is the lowest permissible value of pA
Hence pA pv= patm ρ g⋅ h⋅− ρV2
2⋅−= or h
patm pv−
ρ g⋅V2
2 g⋅−=
Hence h 101 2.358−( ) 103×
N
m2⋅
1999
×m3
kg⋅
s2
9.81 m⋅×
kg m⋅
N s2⋅
×12
10.2ms
⎛⎜⎝
⎞⎠
2×
s2
9.81 m⋅×−= h 4.76 m=
Problem 6.55 [Difficulty: 2]
Problem 6.56 [Difficulty: 2]
Given: Flow through tank-pipe system
Find: Velocity in pipe; Rate of discharge
Solution:
Basic equations pρ
V2
2+ g z⋅+ const= ∆p ρ g⋅ ∆h⋅= Q V A⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the free surface and the manometer location
patmρ
pρ
g H⋅−V2
2+= where we assume VSurface <<, and H = 4 m
Hence p patm ρ g⋅ H⋅+ ρV2
2⋅−=
For the manometer p patm− SGHg ρ⋅ g⋅ h2⋅ ρ g⋅ h1⋅−= Note that we have water on one side and mercury onthe other of the manometer
Combining equations ρ g⋅ H⋅ ρV2
2⋅− SGHg ρ⋅ g⋅ h2⋅ ρ g⋅ h1⋅−= or V 2 g⋅ H SGHg h2⋅− h2+( )⋅=
Hence V 2 9.81×m
s2⋅ 4 13.6 0.15×− 0.75+( )× m⋅= V 7.29
ms
=
The flow rate is Q Vπ D2⋅
4⋅= Q
π
47.29×
ms
⋅ 0.05 m⋅( )2×= Q 0.0143
m3
s=
Problem 6.68 [Difficulty: 3]
Problem 6.69 [Difficulty: 3]
Given: Flow through reducing elbow
Find: Gage pressure at location 1; x component of force
Solution:
Basic equations: pρ
V2
2+ g z⋅+ const= Q V A⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Ignore elevation change 6) p2 = patm
Available data: Q 2.5Ls⋅= Q 2.5 10 3−
×m3
s= D 45 mm⋅= d 25 mm⋅= ρ 999
kg
m3⋅=
From contnuity V1Q
π D2⋅
4
⎛⎜⎝
⎞
⎠
= V1 1.57ms
= V2Q
π d2⋅
4
⎛⎜⎝
⎞
⎠
= V2 5.09ms
=
Hence, applying Bernoulli between the inlet (1) and exit (2) p1ρ
V12
2+
p2ρ
V22
2+=
or, in gage pressures p1gρ
2V2
2 V12
−⎛⎝
⎞⎠⋅= p1g 11.7 kPa⋅=
From x-momentum Rx p1g A1⋅+ u1 mrate−( )⋅ u2 mrate( )⋅+= mrate− V1⋅= ρ− Q⋅ V1⋅= because u1 V1= u2 0=
Rx p1g−π D2⋅
4⋅ ρ Q⋅ V1⋅−= Rx 22.6− N=
The force on the supply pipe is then Kx Rx−= Kx 22.6 N= on the pipe to the right
Problem 6.70 [Difficulty: 3]
Given: Flow nozzle
Find: Mass flow rate in terms of ∆p, T1 and D1 and D2
Solution:
Basic equation pρ
V2
2+ g z⋅+ const= Q V A⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1ρ
V12
2+
p2ρ
V22
2+= where we ignore gravity effects
But we have Q V1 A1⋅= V1π D1
2⋅
4⋅= V2
π D22
⋅
4⋅= so V1 V2
D2D1
⎛⎜⎝
⎞
⎠
2
⋅=
Note that we assume the flow at D2 is at the same pressure as the entire section 2; this will be true if there is turbulent mixing
Hence V22 V2
2 D2D1
⎛⎜⎝
⎞
⎠
4
⋅−2 p2 p1−( )⋅
ρ= or V2
2 p1 p2−( )⋅
ρ 1D2D1
⎛⎜⎝
⎞
⎠
4
−⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
=
Then the mass flow rate is mflow ρ V2⋅ A2⋅= ρπ D2
2⋅
4⋅
2 p1 p2−( )⋅
ρ 1D2D1
⎛⎜⎝
⎞
⎠
4
−⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
⋅=π D2
2⋅
2 2⋅
∆p ρ⋅
1D2D1
⎛⎜⎝
⎞
⎠
4
−⎡⎢⎢⎣
⎤⎥⎥⎦
⋅=
Using p ρ R⋅ T⋅= mflowπ D2
2⋅
2 2⋅
∆p p1⋅
R T1⋅ 1D2D1
⎛⎜⎝
⎞
⎠
4
−⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
⋅=
For a flow nozzle mflow k ∆p⋅= where kπ D2
2⋅
2 2⋅
p1
R T1⋅ 1D2D1
⎛⎜⎝
⎞
⎠
4
−⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
⋅=
We can expect the actual flow will be less because there is actually significant loss in the device. Also the flow will experience avena contracta so that the minimum diameter is actually smaller than D2. We will discuss this device in Chapter 8.