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.