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 6.53
For water at 20C, Table A.3,
= 998 kg/m3 and
= 0.001 kg/m-s. With no minor losses, the
gravity head matches the Moody friction loss in the pipe:
Problem 6.54*
A swimming pool W by Y by h deep is to be emptied by gravity through the long pipe shown in
Fig. P6.54. Assuming an average pipe friction factor fav and neglecting minor losses, derive a
formula for the time to empty the tank from an initial level ho.
Neglect minor losses.
Solution 6.54*
With no driving pressure and negligible tank surface velocity, the energy equation can be
combined with a control-volume mass conservation:
Problem 6.55
The reservoirs in Fig. P6.55 contain water at 20C. If the pipe is smooth with L = 4500 m and
d = 4 cm, what will the flow rate in m3/h be for z = 100 m?
Neglect minor losses.
Solution 6.55
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. The energy equation from surface 1
to surface 2 gives
Problem 6.56
The Alaska Pipeline in the chapter opener photo has a design flow rate of 4.4E7 gallons per day
of crude oil at 60C (see Fig. A.1). (a) Assuming a galvanized-iron wall, estimate the total
pressure drop required for the 800-mile trip. (b) If there are nine equally spaced pumps, estimate
the horsepower each pump must deliver.
Neglect minor losses.
Solution 6.56
From Fig. A.1 for crude oil at 60C,
= 860 kg/m3 and
= 0.004 kg/m-s. The pipe diameter is
4 ft. For galvanized iron,
= 0.0005 ft, hence
/D = 0.0005/4 = 0.000125. Convert the data to
Problem 6.57
Apply the analysis of Prob. 6.54 to the following data. Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m,
D = 5 cm, and
= 0. (a) By letting h = 1.5 m and 0.5 m as representative depths, estimate the
average friction factor. Then (b) estimate the time to drain the pool.
Neglect minor losses.
Solution 6.57
For water, take
= 998 kg/m3 and
= 0.001 kg/ms. The velocity in Prob. 6.54 is calculated from
the energy equation:
Problem 6.58
For the system in Prob. P6.53, a pump is used at night to drive water back to the upper reservoir.
If the pump delivers 15,000 W to the water, estimate the flow rate.
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 6.58
For water at 20C, Table A.3,
= 998 kg/m3 and
= 0.001 kg/m-s. Since the pressures and
velocities cancel, the energy equation becomes
Problem 6.59
The following data were obtained for flow of 20C water at 20 m3/hr through a badly corroded
5-cm–diameter pipe which slopes downward at an angle of 8: p1 = 420 kPa, z1 = 12 m,
p2 = 250 kPa, z2 = 3 m. Estimate (a) the roughness ratio of the pipe; and (b) the percentage
change in head loss if the pipe were smooth and the flow rate the same.
Neglect minor losses.
Solution 6.59
The pipe length is given indirectly as L = z/sin
= (9 m)/sin8 = 64.7 m. The steady flow
energy equation then gives the head loss:
Problem 6.60
In the spirit of Haaland’s explicit pipe friction factor approximation, Eq. (6.49), Jeppson [20]
proposed the following explicit formula:
(a) Is this identical to Haaland’s formula and just a simple rearrangement? Explain.
10 0.9
1 / 5.74
2.0log 3.7 Re
()
d
d
f
− +
(b) Compare Jeppson’s formula to Haaland’s for a few representative values of (turbulent) Red
and
/d and their errors compared to the Colebrook formula (6.48). Discuss briefly.
Neglect minor losses.
Solution 6.60
(a) No, it looks like a rearrangement of Haaland’s formula, but it is not. Haaland started with
Colebrook’s smooth-wall formula and added just enough
/d effect for accuracy. Jeppson started
Problem 6.61
What level h must be maintained in Fig. P6.61 to deliver a flow rate of 0.015 ft3/s through the
1
2-in
commercial-steel pipe?
Neglect minor losses.
Solution 6.61
For water at 20C, take
= 1.94 slug/ft3 and
= 2.09E−5 slug/fts. For commercial steel, take
Problem 6.62
Water at 20C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as
shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient,
what horsepower pump is needed?
Neglect minor losses.
Solution 6.62
Problem 6.63
A tank contains 1 m3 of water at 20C and has a drawn-capillary outlet tube at the bottom, as in
Fig. P6.63. Find the outlet volume flux Q in m3/h at this instant.
Neglect minor losses.
Solution 6.63
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. For drawn tubing, take
0.0015
mm, or
/d = 0.0015/40 0.0000375. The steady-flow energy equation, with p1 = p2 and V1 0,
gives
Problem 6.64
For the system in Fig. P6.63, solve for the flow rate if the fluid is SAE 10 oil. Is the flow laminar
or turbulent?
Neglect minor losses.
Solution 6.64
For SAE 10 oil at 20C, take
= 870 kg/m3 and
= 0.104 kg/ms. For drawn tubing, take
Problem 6.65
In Prob. 6.63 the initial flow is turbulent. As the water drains out of the tank, will the flow revert
to laminar motion as the tank becomes nearly empty? If so, at what tank depth? Estimate the
time, in h, to drain the tank completely.
Neglect minor losses.
Solution 6.65
Recall that
= 998 kg/m3,
= 0.001 kg/ms, and
/d 0.0000375. Let Z be the depth of water in
Problem 6.66
Ethyl alcohol at 20C flows through a 10-cm horizontal drawn tube 100 m long. The fully
developed wall shear stress is 14 Pa. Estimate (a) the pressure drop, (b) the volume flow rate, and
(c) the velocity u at r = 1 cm.
Neglect minor losses.
Solution 6.66
For ethyl alcohol at 20C,
= 789 kg/m3,
= 0.0012 kg/ms. For drawn tubing, take
Problem 6.67
A straight 10-cm commercial-steel pipe is 1 km long and is laid on a constant slope of 5. Water at
20C flows downward, due to gravity only. Estimate the flow rate in m3/h. What happens if the
pipe length is 2 km?
Neglect minor losses.
Solution 6.67
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. If the flow is due to gravity only,
then the head loss exactly balances the elevation change:
Problem 6.68*
The Moody chart cannot find V directly, since V appears in both ordinate and abscissa.
(a) Arrange the variables (hf , d, g, L, ν) into a single dimensionless group, with hf d 3 in the
numerator, denoted as ξ , which equals (f Red2/2). (b) Rearrange the Colebrook formula (6.48) to
solve for Red in terms of ξ. (c) For extra credit, solve Example 6.9 with this new formula.
Neglect minor losses.
Solution 6.68*
(a) The variables (hf , d, g, L, ν), which do not contain mass {M}, are easily arranged into a
unique dimensionless group with hf d 3 in the numerator:
Problem 6.69
For Prob. 6.62 suppose the only pump available can deliver only 80 hp to the fluid. What is the
proper pipe size in inches to maintain the 3 ft3/s flow rate?
Neglect minor losses.
Problem 6.62
Water at 20C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as
shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient,
what horsepower pump is needed?
Neglect minor losses.
Solution 6.69
For water at 20C, take
= 1.94 slug/ft3 and
= 2.09E−5 slug/fts. For cast iron, take
Problem 6.70
Ethylene glycol at 20C flows through 80 meters of cast iron pipe of diameter 6 cm. The
measured pressure drop is 250 kPa. Using a non-iterative formulation, estimate the flow rate in
m3/h.
Neglect minor losses.
Solution 6.70
For ethylene glycol at 20C, Table A.3,
= 1117 kg/m3 and
= 0.0214 kg/m-s. The head loss is
Problem 6.71*
It is desired to solve Prob. 6.62 for the most economical pump and cast-iron pipe system. If the
pump costs $125 per horsepower delivered to the fluid and the pipe costs $7000 per inch of
diameter, what are the minimum cost and the pipe and pump size to maintain the 3 ft3/s flow
rate? Make some simplifying assumptions.
Neglect minor losses.
Problem 6.62
Water at 20C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as
shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient,
what horsepower pump is needed?
Neglect minor losses.
Solution 6.71
For water at 20C, take
= 1.94 slug/ft3 and
= 2.09E−5 slug/fts. For cast iron, take
0.00085 ft. Write the energy equation (from Prob. 6.62) in terms of Q and d:
Problem 6.72
Modify Prob. P6.57 by letting the diameter be unknown. Find the proper pipe diameter for which
the pool will drain in about 2 hours flat.
Neglect minor losses.
Problem 6.57
Apply the analysis of Prob. 6.54 to the following data. Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m,
D = 5 cm, and
= 0. (a) By letting h = 1.5 m and 0.5 m as representative depths, estimate the
average friction factor. Then (b) estimate the time to drain the pool.
Neglect minor losses.
Solution 6.72
Recall the data: Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m, and
= 0, with water,
= 998 kg/m3
and
= 0.001 kg/ms. We apply the same theory as Prob. 6.57:
Problem 6.73
For 20ºC water flow in a smooth, horizontal 10-cm pipe, with Δp/L = 1000 Pa/m, the writer
computed a flow rate of 0.030 m3/s. (a) Verify, or disprove, the writer’s answer. (b) If verified,
use the power-law friction factor relation, Eq. (6.41), to estimate the pipe diameter that will triple
this flow rate. (c) For extra credit, use the more exact friction factor relation, Eq. (6.38), to solve
part (b).
Neglect minor losses.
Solution 6.73
(a) For water at 20ºC, ρ = 998 kg/m3, and μ = 0.0010 kg/ms. The pressure-drop relation is