Chapter 8 A standard household water meter is incorporated into

Problem 8.80

Water is pumped through a

6

0-m-long, 0.3-m-diameter pipe from a lower reservoir to a

higher reservoir whose surface is

1

0 m above the lower one. The sum of the minor loss

coefficients for the system is KL = 14.5. When the pump adds 40 kW to the water the

flowrate is

3

m

0.20 s. Determine the pipe roughness.

Solution 8.80

with

3

2

m

0.2 m

s2.82 s

(0.3 m)

4

Q

V

A

π

== =

Thus, from Eq. (1)

(1)

•

(2)

•

Thus, from the Moody chart (see the figure above) with Re = 7.55 × 105 and f = 0.0560, it

follows that 0.028

D

ε

=, or

ε

= 0.028(0.3 m) = 0.0084 m

Transition range

Laminar ﬂow

Wholly turbulent ﬂow

0.1

0.09

0.08

0.07

0.06

0.05

0.04

0.05

0.04

0.03

0.02

0.015

0.01

0.008

Re = VD

_____

μ

ρ

Problem 8.81

Natural gas −



==×





2

5

3

slugs ft

0.0044 and 5.2 10

s

ft v

ρ

is pumped through a horizontal 6-in.–

diameter cast-iron pipe at a rate of lb

8

00 hr . If the pressure at section (1) is 50 psi (abs),

determine the pressure at section (2)

8

mi downstream if the flow is assumed

incompressible. Is the incompressible assumption reasonable? Explain.

Solution 8.81

22

2

11 2 2

12

222

pV p V V

zzf

ggDg

γγ

++=+++

, where z1 = z2, and V1 = V2 (1)

==







0.00085 ft 0.0017

6ft

12

D

ε

(from Table B.1 Equivalent Roughness for New Pipes [Adapted

from Moody (Ref. 7) and Colebrook (Ref. 8)])

Problem 8.82

As shown in the figure below, a standard household water meter is incorporated into a lawn

irrigation system to measure the volume of water applied to the lawn. Note that these

meters measure volume, not volume flowrate. With an upstream pressure of p1 = 50 psi, the

meter registered that 120 ft3 of water was delivered to the lawn during an “on” cycle.

Estimate the upstream pressure, p1, needed if it is desired to have 150 ft3 delivered during an

“on” cycle. List any assumptions needed to arrive at your answer.

Solution 8.82

The energy equation for this flow is

(1)

Irrigation

system:

pipes, ﬁttings,

nozzles, etc.

WATER

METER

(1)

(2)

Irrigation

system:

pipes, ﬁttings,

nozzles, etc.

•

WATER

METER

f

D

V

2

K

L

ℓ

Σ

Problem 8.83

A fan is to produce a constant air speed of 40 m/s throughout the pipe loop shown in the

figure below. The 3-m-diameter pipes are smooth, and each of the four 90° elbows has a

loss coefficient of 0.30. Determine the power that the fan adds to the air.

Solution 8.83

22

11 2 2

12

22

Ls

pV p V

zhhz

gg

γγ

++ − + = + +

Fan

V

= 40 m/s

D

= 3 m

10 m

20 m

Hence,

2

m

40

(20201010) m s

0.0083 1.2 111 m

m

3 m 29.81

s

h





+++





=+=









Transition range

Laminar ﬂow

Wholly turbulent ﬂow

0.1

0.09

0.08

0.07

0.06

0.05

0.04

103

104

105

106

107

0.05

0.04

0.03

0.02

0.015

0.01

Re = VD

_____

μ

ρ

Problem 8.84

Air flows in a horizontal 100-ft-long, 24-in.× 24-in. duct at the rate of 5000 ft3/min. The air

then flows through an expansion into 200 ft of 36-in.× 36-in. duct. The expansion has a loss

coefficient of 0.80, based on the higher inlet velocity. Calculate the static pressure change

across the expansion and the static pressure loss across the entire duct system. The duct is

made of galvanized sheet metal. Use a constant air density of 0.0024 slug/ft3.

Solution 8.84

gives

Note that there is a static pressure drop since the friction losses give a pressure loss greater

than the static pressure rise due to the decrease in velocity. To find pa–pd, apply the

mechanical energy equation from a to d.

GALVANIZED SHEET METAL

A2 = 36

″

×

36

″

V2

A1 = 24

″

×

24

″

V1

abc

d

2

.

22

aa dd

as d L

V

pV p

gz w gz gh

ρρ

++ −=++ +

Now ws = 0, za = zb. Then

21





The mechanical energy equation becomes

Assuming standard air with v = 1.57 × 10−4 2

f

t/s, the Reynolds numbers are:

()

3

5

11 1

1242

1

min

5000 ft / min (2 ft) 60 s

Re 2.65 10

(2 ft) (1. ft57 10 / s)

VD QD

vAv −







=== =×

×

and

The absolute roughness for galvanized sheet metal (taken as galvanized iron) is found from

Table 8.1 to be

ε

= 0.0005 ft. Then

1

0.0005 ft 0.00025

2ftD

ε

==

and 2

2

0.0005 ft 0.00017

3ftD

ε

==

=– 0.00385 psi

ad

p

Problem 8.85

The turbine shown in the figure below develops 400 kW. Determine the flowrate if (a) head

losses are negligible or (b) head loss due to friction in the pipe is considered. Assume

f = 0.02. Note: There may be more than one solution or there may be no solution to this

problem.

Solution 8.85

120 m of 0.30-m-diameter

cast-iron pipe

1 m

20 m

Diffuser

T

120 m of 0.30-m-diameter

cast-iron pipe

(1)

•

(2)

1 m

20 m

Diffuser

T

•

Determine the roots of this cubic equation. Let 3

22

392 1020

V

VF−+≡

b) Include head loss ( f = 0.02): From Eq. (1)

2

22 2

1m 11.1

0.3 m

VA D

V

VV V

AD



== = =











or

2

22

2

120 m 52.0

2

0 m 1 0.02 (11.1) m

m

0.3 m 29.81

s

V



=+ +



 







Thus,

V

5000

4000

Problem 8.86

Water flows from the nozzle attached to the spray tank shown in the figure below.

Determine the flowrate if the loss coefficient for the nozzle (based on upstream conditions)

is 0.75 and the friction factor for the rough hose is 0.11.

Solution 8.86

Thus, with f = 0.11 and KL = 0.75 Eq. (1) gives

Nozzle diameter

= 7.5 mm

0.80 m

p

= 150 kPa

D

= 15 mm

𝓵

= 1.9 m

40

°

Nozzle diameter

= 7.5 mm

0.80 m

(1)

•

p

= 150 kPa

D

= 15 mm

•

(2)

Problem 8.87

Water flows through the pipe shown in the figure below. Determine the net tension in the

bolts if minor losses are neglected and the wheels on which the pipe rests are frictionless.

Solution 8.87

Application of the momentum equation to the control volume indicated gives

33 2 3

()0

B

p

AF mVV−= − =

, since 23

V

V=

Thus, FB = p3A3 (1)

80 mm

3.0 m

Bolts

Galvanized iron

20 m10 m

80 mm

3.0 m

Bolts

Galvanized iron

(1)

•

(2)

•

(3)

from (3) Re = 1.88 × 5

1

0; from (4) f = 0.0245 ≠ 0.02. Try again.

Assume 0.0245

f

=, or m

2.40 s

V

=, Re = 1.72 × 5

1

0 or f = 0.0245 Checks.

Thus, m

2.40 s

V

=

Problem 8.88

When the pump shown in the figure below adds 0.2 hp to the flowing water, the pressures

indicated by the two gages are equal. Determine the flowrate.

Length of pipe between gages = 60 ft

Pipe diameter = 0.1 ft

Pipe friction factor = 0.03

Filter loss coefficient = 12

Solution 8.88

The pump adds 0.2 hp of power.

ft lb

550 ft lb

s

0.2 hp 110

1 hp s

W

⋅

=× =



Convert to head by:

PumpFilter

(2)

(1)

Problem 8.89

The pump shown in the figure below adds 25 kW to the water and causes a flowrate of

3

m

0.04 s. Determine the flowrate expected if the pump is removed from the system. Assume

f = 0.016 for either case and neglect minor losses.

Solution 8.89

But,

3

Nm

25 10 s63.8 m

Nm

9.80 10 0.04

s

m

P

h

Q

γ

⋅

×

== =



×





40-mm-diameter

nozzle

60-mm-diameter,

30-m-long pipe;

f

= 0.016

Pump

60-mm-diameter,

30-m-long pipe;

(1)

•

Thus,

Problem 8.90

The vented storage tank shown in the figure below is used to refuel race cars at a race track.

A total of 40 ft of steel pipe (I.D. = 0.957 in.), two 90° regular elbows, and a globe valve

make up the system. Calculate the time needed to put 20 gal of fuel in a car tank. The

pressures, p2 and p1, are equal, the connections are threaded, and the fuel has the properties

of 68 °F normal octane (

ν

= 8.31×10−6 ft2/s).

Solution 8.90

Assume constant fluid density and apply the mechanical energy equation from point 1 to

point 2. No work is done.

Since 22

4Q

VD

π

=,

Figure 8.22 gives KLent = 0.5, and Table 8.2 gives KL 90° = 1.5 and KLvalve = 10, the Reynolds

number is

h

= 12 ft

Vent

Racing fuel

1

2

p2

=

patm

Table 8.1 gives

Using the known numerical values in Eq. (1) gives

Equations (2) and (3) are now solved with the Moody chart for Q. Using a trial and error

procedure, guess f = 0.023 = wholly turbulent value. Then

or

t = 101.3 s.