Problem 6.74
Two reservoirs, which differ in surface elevation by 40 m, are connected by a new commercial
steel pipe of diameter 8 cm. If the desired weight flow rate is 200 N/s of water at 20C, what is
the proper length of the pipe? Neglect minor losses.
Neglect minor losses.
Solution 6.74
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/m-s. For commercial steel,
Problem 6.75
You wish to water your garden with 100 ft of
5
8
–in-diameter hose whose roughness is 0.011 in.
What will be the delivery, in ft3/s, if the gage pressure at the faucet is 60 lbf/in2? If there is no
nozzle (just an open hose exit), what is the maximum horizontal distance the exit jet will carry?
Neglect minor losses.
Solution 6.75
For water, take
= 1.94 slug/ft3 and
= 2.09E−5 slug/fts. We are given
/d = 0.011/(5/8) 0.0176. For constant area hose, V1 = V2 and energy yields
Problem 6.76
The small turbine in Fig. P6.76 extracts 400 W of power from the water flow. Both pipes are
wrought iron. Compute the flow rate Q in m3/h. Why are there two solutions? Which is better?
Neglect minor losses.
Solution 6.76
For water, take
= 998 kg/m3 and
= 0.001 kg/ms. For wrought iron, take
0.046 mm,
hence
/d1 = 0.046/60 0.000767 and
/d2 = 0.046/40 0.00115. The energy equation, with
V1 0 and p1 = p2, gives
Problem 6.77*
Modify Prob. 6.76 into an economic analysis, as follows. Let the 40 m of wrought-iron pipe have
a uniform diameter d. Let the steady water flow available be Q = 30 m3h. The cost of the turbine
is $4 per watt developed, and the cost of the piping is $75 per centimeter of diameter. The power
generated may be sold for $0.08 per kilowatt hour. Find the proper pipe diameter for minimum
payback time, i.e., minimum time for which the power sales will equal the initial cost of the
system.
Neglect minor losses.
Problem 6.76
The small turbine in Fig. P6.76 extracts 400 W of power from the water flow. Both pipes are
wrought iron. Compute the flow rate Q in m3/h. Why are there two solutions? Which is better?
Neglect minor losses.
Solution 6.77
With flow rate known, we need only guess a diameter and compute power from the energy
equation similar to Prob. 6.76:
Problem 6.78
In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in
m3h, if the fluid is water at 20C. Which way is the flow?
Neglect minor losses.
Solution 6.78
For water, take
= 998 kgm3 and
= 0.001 kgms. For commercial steel, take
0.046 mm,
hence
d = 0.04660 0.000767. With p1, V1, and V2 all 0, the energy equation between
Problem 6.79
A garden hose is to be used as the return line in a waterfall display at a mall. In order to select
the proper pump, you need to know the roughness height inside the garden hose. Unfortunately,
roughness information is not supplied by the hose manufacturer. So you devise a simple
experiment to measure the roughness. The hose is attached to the drain of an above-ground
swimming pool, the surface of which is 3.0 m above the hose outlet. You estimate the minor loss
coefficient of the entrance region as 0.5, and the drain valve has a minor loss equivalent length of
200 diameters when fully open. Using a bucket and stopwatch, you open the valve and measure
the flow rate to be 2.0 × 10−4 m3/s for a hose that is 10.0 m long and has an inside diameter of
1.50 cm. Estimate the roughness height in mm inside the hose.
Neglect minor losses.
Solution 6.79
First evaluate the average velocity in the hose and its Reynolds number:
Problem 6.80
The head-versus–flow-rate characteristics of a centrifugal pump are shown in Fig. P6.80. If this
pump drives water at 20C through 120 m of 30-cm-diameter cast-iron pipe, what will be the
resulting flow rate, in m3/s?
Neglect minor losses.
Solution 6.80
For water, take
= 998 kg/m3 and
= 0.001 kg/ms. For cast iron, take
0.26 mm, hence
/d = 0.26/300 0.000867. The head loss must match the pump head:
Problem 6.81
The pump in Fig. P6.80 is used to deliver gasoline at 20C through 350 m of
30–cm-diameter galvanized iron pipe. Estimate the resulting flow rate, in m3/s. (Note that the
pump head is now in meters of gasoline.)
Neglect minor losses
Solution 6.81
For gasoline, take
= 680 kg/m3 and
= 2.92E−4 kg/ms. For galvanized iron, take
0.15
mm, hence
/d = 0.15/300 0.0005. Head loss matches pump head:
Problem 6.82
Fluid at 20C flows through a horizontal galvanized-iron pipe 20 m long and 8 cm in diameter.
The wall shear stress is 90 Pa. Calculate the flow rate in m3/h if the fluid is (a) glycerin; and
(b) water.
Neglect minor losses
Solution 6.82
(a) For glycerin, take
= 1260 kg/m3 and
= 1.49 kg/ms. For galvanized iron, take
0.15 mm, hence
/D = 0.15/80 0.001875. But we are guessing this flow is laminar:
Problem 6.83
For the system of Fig. P6.55, let z = 80 m and L = 185 m of cast-iron pipe. What is the pipe
diameter for which the flow rate will be 7 m3/h?
Neglect minor losses
Solution 6.83
For water, take
= 998 kg/m3 and
= 0.001 kg/ms. For cast iron, take
0.26 mm, but d is
unknown. The energy equation is simply
Problem 6.84
It is desired to deliver 60 m3/h of water at 20C through a horizontal asphalted cast-iron pipe.
Estimate the pipe diameter which will cause the pressure drop to be exactly 40 kPa per 100 m of
pipe length.
Neglect minor losses
Solution 6.84
Write out the relation between p and friction factor, taking “L” = 100 m:
Problem 6.85
For the system in Prob. P6.53, a pump, which delivers 15,000 W to the water, is used at night to
refill the upper reservoir. The pipe diameter is increased from 12 cm to provide more flow. If
the resultant flow rate is 90 m3/h, estimate the new pipe size.
Neglect minor losses
Problem 6.53
Water at 20C flows by gravity through a smooth pipe from one reservoir to a lower one. The
elevation difference is 60 m. The pipe is 360 m long, with a diameter of 12 cm. Calculate the
expected flow rate in m3/h.
Neglect minor losses.
Solution 6.85
For water at 20C, Table A.3,
= 998 kg/m3 and
= 0.001 kg/m-s. Recall that z = 60 m and
Problem 6.86
SAE 10 oil at 20C flows at an average velocity of 2 m/s between two smooth parallel horizontal
plates 3 cm apart. Estimate (a) the centerline velocity, (b) the head loss per meter, and (c) the
pressure drop per meter.
Neglect minor losses
Solution 6.86
For SAE 10 oil, take
= 870 kg/m3 and
= 0.104 kg/ms. The half-distance between plates is
called “h” (see Fig. 6.37). Check Dh and Re:
Problem 6.87
A commercial-steel annulus 40 ft long, with a = 1 in and b =
1
2
in, connects two reservoirs which
differ in surface height by 20 ft. Compute the flow rate in ft3/s through the annulus if the fluid is
water at 20C.
Neglect minor losses
Solution 6.87
For water, take
= 1.94 slug/ft3 and
= 2.09E−5 slug/fts. For commercial steel, take
Problem 6.88
An oil cooler consists of multiple parallel-plate passages, as shown in Fig. P6.88. The available
pressure drop is 6 kPa, and the fluid is SAE 10W oil at 20C. If the desired total flow rate is
900 m3/h, estimate the appropriate number of passages. The plate walls are hydraulically
smooth.
Neglect minor losses
Solution 6.88
For SAE 10W oil,
= 870 kg/m3 and
= 0.104 kg/ms. The pressure drop remains 6 kPa no
matter how many passages there are (ducts in parallel). Guess laminar flow, Eq. (6.63),
h
D
N
Problem 6.89
An annulus of narrow clearance causes a very large pressure drop and is useful as an accurate
measurement of viscosity. If a smooth annulus 1 m long with a = 50 mm and b = 49 mm carries
an oil flow at 0.001 m3/s, what is the oil viscosity if the pressure drop is 250 kPa?
Neglect minor losses
Solution 6.89
Assuming laminar flow, use Eq. (6.73) for the pressure drop and flow rate:
Problem 6.90
A rectangular sheet-metal duct is 200 ft long and has a fixed height H = 6 in. The width B,
however, may vary from 6 to 36 inches. A blower provides a pressure drop of 80 Pa of air at
20C and 1 atm. What is the optimum width B that will provide the most airflow in ft3/s?
Neglect minor losses
Solution 6.90
For air at 20C and 1 atm, take
= 1.20 kg/m3 and
= 1.8E-5 kg/m-s. The pressure drop is
Problem 6.91
Heat exchangers often consist of many triangular passages. Typical is Fig. P6.91, with L = 60 cm
and an isosceles-triangle cross section of side length a = 2 cm and included angle
= 80. If the
average velocity is V = 2 m/s and the fluid is SAE 10 oil at 20C, estimate the pressure drop.
Neglect minor losses
Solution 6.91
For SAE 10 oil, take
= 870 kg/m3 and
= 0.104 kg/ms. The Reynolds number based on side
Problem 6.92
A large room uses a fan to draw in atmospheric air at 20C through a 30 cm by 30 cm commercial-
steel duct 12 m long, as in Fig. P6.92. Estimate (a) the air flow rate in m3/hr if the room
pressure is 10 Pa vacuum; and (b) the room pressure if the flow rate is 1200 m3/hr.
Neglect minor losses
Solution 6.92
For air, take
= 1.2 kg/m3 and
= 1.8E−5 kg/ms. For commercial steel,
= 0.046 mm. For a
square duct, Dh = side-length = 30 cm, hence
/d = 0.046/300 = 0.000153. The (b) part is easier,
Problem 6.93
In Moody’s Example 6.6, the 6-inch diameter, 200-ft-long asphalted cast iron pipe has a pressure
drop of about 280 lbf/ft2 when the average water velocity is 6 ft/s. Compare this to an annular
cast iron pipe with an inner diameter of 6 in and the same annular average velocity of 6 ft/s.
(a) What outer diameter would cause the flow to have the same pressure drop of 280 lbf/ft2?
(b) How do the cross-section areas compare, and why? Use the hydraulic diameter
approximation.
Neglect minor losses
Solution 6.93
Recall the Ex. 6.6 data,
= 0.0004 ft. For water at 68F, take
= 1.94 slug/ft3 and
= 2.09E-5 slug/ft-sec. The hydraulic diameter of an annulus is Dh = 2(Ro – Ri),
Problem 6.94
Air at 20C flows through a smooth duct of diameter 20 cm at an average velocity of 5 m/s. It
then flows into a smooth square duct of side length a. Find the square duct size a for which the
pressure drop per meter will be exactly the same as the circular duct?
Neglect minor losses.
Solution 6.94
For air at 20C and 1 atm, take
= 1.20 kg/m3 and
= 1.8E-5 kg/m-s. Compute the pressure
drop in the circular duct:
Problem 6.95
Although analytical solutions are available for laminar flow in many duct shapes [34], what do we
do about ducts of arbitrary shape? Bahrami et al. [57] propose that a better approach to the pipe
result, f Re = 64, is achieved by replacing the hydraulic diameter Dh by A, where A is the area of
the cross section. Test this idea for the isosceles triangles of Table 6.4. If time is short, at least
try 10, 50, and 80. What do you conclude about this idea?
Neglect minor losses.
Solution 6.95
We can see for the triangles in Table 6.4 that the values of f ReDh are all less than 64, by as much
as 25%. If we denote Bahrami’s idea as DB = A, the new Reynolds number is based on
Problem 6.96
A fuel cell [59] consists of air (or oxygen) and hydrogen micro ducts, separated by a membrane
that promotes proton exchange for an electric current, as in Fig. P6.96. Suppose that the air side,
at 20C and approximately 1 atm, has five 1 mm by 1 mm ducts, each 1 m long. The total flow
rate is 1.5E-4 kg/s. (a) Determine if the flow is laminar or turbulent. (b) Estimate the pressure
drop.
Neglect minor losses.