This document contains solutions to 6 fluid mechanics problems involving concepts like Bernoulli's equation, dynamic pressure, and flow rate calculations. Problem 6.70 asks the reader to express the mass flow rate through a nozzle in terms of the pressure difference ∆p between the inlet and exit, the inlet temperature T1, and the diameters D1 and D2 of the inlet and exit. The solution uses Bernoulli's equation between the inlet and exit, along with the relationship between flow velocity and pipe diameter, to derive an expression for the exit velocity V2 in terms of these parameters.

1. Problem 6.38 [Difficulty: 1]
Given: Water at speed 25 ft/s
Find: Dynamic pressure in in. Hg
Solution:
Basic equations pdynamic
1
2
ρ⋅ V
2
⋅= p ρHg g⋅ ∆h⋅= SGHg ρ⋅ g⋅ ∆h⋅=
Hence ∆h
ρ V
2
⋅
2 SGHg⋅ ρ⋅ g⋅
=
V
2
2 SGHg⋅ g⋅
=
∆h
1
2
25
ft
s
⋅⎛
⎜
⎝
⎞
⎠
2
×
1
13.6
×
s
2
32.2 ft⋅
×
12 in⋅
1 ft⋅
×= ∆h 8.56 in⋅=

2. Problem 6.39 [Difficulty: 1]
Given: Air speed of 100 km/hr
Find: Dynamic pressure in mm water
Solution:
Basic equations pdynamic
1
2
ρair⋅ V
2
⋅= p ρw g⋅ ∆h⋅=
Hence ∆h
ρair
ρw
V
2
2 g⋅
⋅=
∆h
1.23
kg
m
3
⋅
999
kg
m
3
⋅
1
2
× 100
km
hr
⋅⎛
⎜
⎝
⎞
⎠
2
×
1000 m⋅
1 km⋅
⎛
⎜
⎝
⎞
⎠
2
×
1 hr⋅
3600 s⋅
⎛
⎜
⎝
⎞
⎠
2
×
s
2
9.81 m⋅
×= ∆h 48.4 mm⋅=

3. Problem 6.42 [Difficulty: 2]
Given: Air jet hitting wall generating pressures
Find: Speed of air at two locations
Solution:
Basic equations
p
ρair
V
2
2
+ g z⋅+ const= ρair
p
Rair T⋅
= ∆p ρHg g⋅ ∆h⋅= SGHg ρ⋅ g⋅ ∆h⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data R 287
J
kg K⋅
⋅= T 10− °C= ρ 999
kg
m
3
⋅= p 200 kPa⋅= SGHg 13.6=
For the air ρair
p
R T⋅
= ρair 2.65
kg
m
3
=
Hence, applying Bernoulli between the jet and where it hits the wall directly
patm
ρair
Vj
2
2
+
pwall
ρair
= pwall
ρair 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
m
3
⋅
1
2.65
×
m
3
kg
⋅ 9.81×
m
s
2
⋅ 25× mm⋅
1 m⋅
1000 mm⋅
×= Vj 50.1
m
s
=
Repeating the analysis for the second point
∆h 5 mm⋅=
patm
ρair
Vj
2
2
+
pwall
ρair
V
2
2
+= V Vj
2
2 pwall⋅
ρair
−= Vj
2
2 SGHg⋅ ρ⋅ g⋅ ∆h⋅
ρair
−=
Hence V 50.1
m
s
⋅⎛
⎜
⎝
⎞
⎠
2
2 13.6× 999×
kg
m
3
⋅
1
2.65
×
m
3
kg
⋅ 9.81×
m
s
2
⋅ 5× mm⋅
1 m⋅
1000 mm⋅
×−= V 44.8
m
s
=

4. Problem 6.43 [Difficulty: 2]

5. 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 pdyn
1
2
ρair⋅ U
2
⋅= p0 ps pdyn+= ρair
p
Rair T⋅
= ∆p ρw g⋅ ∆h⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data T 5− °C= U 50
m
s
⋅= R 287
J
kg K⋅
⋅= patm 101 kPa⋅= h0 10− mm⋅= ρw 999
kg
m
3
⋅=
For air ρair
patm
R T⋅
= ρair 1.31
kg
m
3
=
pdyn
1
2
ρair⋅ U
2
⋅= 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= hs
ps
ρw g⋅
= hs 177− mm=
(gage)
Streamlines in the test section are straight so
n
p
∂
∂
0= and pw pcenterline=
In the curved section
n
p
∂
∂
ρair
V
2
R
⋅= so pw pcenterline<

6. Problem 6.45 [Difficulty: 2]

7. Problem 6.46 [Difficulty: 2]

8. Problem 6.49 [Difficulty: 2]

9. Problem 6.50 [Difficulty: 2]
Given: Siphoning of gasoline
Find: Flow rate
Solution:
Basic equation
p
ρgas
V
2
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
patm
ρgas
patm
ρgas
V
2
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 D
2
⋅
4
2 g⋅ h⋅⋅=
Q
π
4
.5 in⋅( )
2
×
1 ft
2
⋅
144 in
2
⋅
× 2 32.2×
ft
s
2
1× ft⋅×= Q 0.0109
ft
3
s
⋅= Q 4.91
gal
min
⋅=

10. Problem 6.54 [Difficulty: 3]
Given: Flow rate through siphon
Find: Maximum height h to avoid cavitation
Solution:
Basic equation
p
ρ
V
2
2
+ g z⋅+ const= Q V A⋅=
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data Q 5
L
s
⋅= Q 5 10
3−
×
m
3
s
= D 25 mm⋅= ρ 999
kg
m
3
⋅= patm 101 kPa⋅=
From continuity V
Q
A
=
4 Q⋅
π D
2
⋅
= V
4
π
0.005×
m
3
s
⋅
1
.025 m⋅
⎛
⎜
⎝
⎞
⎠
2
×= V 10.2
m
s
=
Hence, applying Bernoulli between the free surface and point A
patm
ρ
pA
ρ
g h⋅+
V
2
2
+= where we assume VSurface <<
Hence pA patm ρ g⋅ h⋅− ρ
V
2
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⋅− ρ
V
2
2
⋅−= or h
patm pv−
ρ g⋅
V
2
2 g⋅
−=
Hence h 101 2.358−( ) 10
3
×
N
m
2
⋅
1
999
×
m
3
kg
⋅
s
2
9.81 m⋅
×
kg m⋅
N s
2
⋅
×
1
2
10.2
m
s
⎛
⎜
⎝
⎞
⎠
2
×
s
2
9.81 m⋅
×−= h 4.76 m=

11. Problem 6.55 [Difficulty: 2]

12. Problem 6.56 [Difficulty: 2]
Given: Flow through tank-pipe system
Find: Velocity in pipe; Rate of discharge
Solution:
Basic equations
p
ρ
V
2
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
patm
ρ
p
ρ
g H⋅−
V
2
2
+= where we assume VSurface <<, and H = 4 m
Hence p patm ρ g⋅ H⋅+ ρ
V
2
2
⋅−=
For the manometer p patm− SGHg ρ⋅ g⋅ h2⋅ ρ g⋅ h1⋅−= Note that we have water on one side and mercury on
the other of the manometer
Combining equations ρ g⋅ H⋅ ρ
V
2
2
⋅− SGHg ρ⋅ g⋅ h2⋅ ρ g⋅ h1⋅−= or V 2 g⋅ H SGHg h2⋅− h2+( )⋅=
Hence V 2 9.81×
m
s
2
⋅ 4 13.6 0.15×− 0.75+( )× m⋅= V 7.29
m
s
=
The flow rate is Q V
π D
2
⋅
4
⋅= Q
π
4
7.29×
m
s
⋅ 0.05 m⋅( )
2
×= Q 0.0143
m
3
s
=

13. Problem 6.68 [Difficulty: 3]

14. Problem 6.69 [Difficulty: 3]
Given: Flow through reducing elbow
Find: Gage pressure at location 1; x component of force
Solution:
Basic equations:
p
ρ
V
2
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 2.5
L
s
⋅= Q 2.5 10
3−
×
m
3
s
= D 45 mm⋅= d 25 mm⋅= ρ 999
kg
m
3
⋅=
From contnuity V1
Q
π D
2
⋅
4
⎛
⎜
⎝
⎞
⎠
= V1 1.57
m
s
= V2
Q
π d
2
⋅
4
⎛
⎜
⎝
⎞
⎠
= V2 5.09
m
s
=
Hence, applying Bernoulli between the inlet (1) and exit (2) p1
ρ
V1
2
2
+
p2
ρ
V2
2
2
+=
or, in gage pressures p1g
ρ
2
V2
2
V1
2
−⎛
⎝
⎞
⎠⋅= 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 D
2
⋅
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

15. Problem 6.70 [Difficulty: 3]
Given: Flow nozzle
Find: Mass flow rate in terms of ∆p, T1 and D1 and D2
Solution:
Basic equation
p
ρ
V
2
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)
p1
ρ
V1
2
2
+
p2
ρ
V2
2
2
+= where we ignore gravity effects
But we have Q V1 A1⋅= V1
π D1
2
⋅
4
⋅= V2
π D2
2
⋅
4
⋅= so V1 V2
D2
D1
⎛
⎜
⎝
⎞
⎠
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 V2
2
V2
2
D2
D1
⎛
⎜
⎝
⎞
⎠
4
⋅−
2 p2 p1−( )⋅
ρ
= or V2
2 p1 p2−( )⋅
ρ 1
D2
D1
⎛
⎜
⎝
⎞
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅
=
Then the mass flow rate is mflow ρ V2⋅ A2⋅= ρ
π D2
2
⋅
4
⋅
2 p1 p2−( )⋅
ρ 1
D2
D1
⎛
⎜
⎝
⎞
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅
⋅=
π D2
2
⋅
2 2⋅
∆p ρ⋅
1
D2
D1
⎛
⎜
⎝
⎞
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅=
Using p ρ R⋅ T⋅= mflow
π D2
2
⋅
2 2⋅
∆p p1⋅
R T1⋅ 1
D2
D1
⎛
⎜
⎝
⎞
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅
⋅=
For a flow nozzle mflow k ∆p⋅= where k
π D2
2
⋅
2 2⋅
p1
R T1⋅ 1
D2
D1
⎛
⎜
⎝
⎞
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅
⋅=
We can expect the actual flow will be less because there is actually significant loss in the device. Also the flow will experience a
vena contracta so that the minimum diameter is actually smaller than D2. We will discuss this device in Chapter 8.