Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Problem 6.31
A laminar flow element or LFE (Meriam Instrument Co.) measures low gas-flow rates with a
bundle of capillary tubes packed inside a large outer tube. Consider oxygen at 20C and 1 atm
flowing at 84 ft3/min in a 4-in-diameter pipe. (a) Is the flow turbulent when approaching the
element?(b) If there are 1000 capillary tubes, L = 4 in., select a tube diameter to keep Red below
1500 and also to keep the tube pressure drop no greater than 0.5 lbf/in2. (c) Do the tubes selected
in part (b) fit nicely within the approach pipe?
Neglect minor losses.
Solution 6.31
For oxygen at 20C and 1 atm (Table A.4), take R = 260 m2/(s2K), hence
= p/RT = (101350Pa)/[260(293K)] = 1.33 kg/m3 = 0.00258 slug/ft3.
Problem 6.32
SAE 30 oil at 20C flows in the 3-cm-diameter pipe in Fig. P6.32, which slopes at 37. For the
pressure measurements shown, determine (a) whether the flow is up or down and (b) the flow
rate in m3/h.
Neglect minor losses.
Solution 6.32
For SAE 30 oil, take
= 891 kg/m3 and
= 0.29 kg/ms. Evaluate the hydraulic grade lines:
Problem 6.33
Water at 20C is pumped from a reservoir through a vertical tube 10 ft long and 1/16th inch in
diameter. The pump provides a pressure rise of 11 lbf/in2 to the flow. Neglect entrance losses.
(a) Calculate the exit velocity. (b) Approximately how high will the exit water jet rise?
(c) Verify that the flow is laminar.
Neglect minor losses.
Solution 6.33
For water at 20C, Table A.3,
= 998 kg/m3 = 1.94 slug/ft3, and
= 0.001 kg/m-s = 2.09E-5 slug/ft-s.
Problem 6.34
Derive the time-averaged x-momentum equation (6.21) by direct substitution of Eqs. (6.19) into
the momentum equation (6.14). It is convenient to write the convective acceleration as
2
u(u ) (uv) (uw)
t x y z
d
d
= + +
which is valid because of the continuity relation, Eq. (6.14).
Neglect minor losses.
Solution 6.34
Into the x-momentum eqn. substitute u = u + u’, v = v + v’, etc., to obtain
Problem 6.35
In the overlap layer of Fig. 6.9a, turbulent shear is large. If we neglect viscosity, we can replace
Eq. (6.24) with the approximate velocity-gradient function
( , , )
w
du fcn y
dy
=
Show that, by dimensional analysis, this leads to the logarithmic overlap relation (6.28).
Neglect minor losses.
Solution 6.35
There are four variables, and we may list their dimensions in the (MLT) system:
Problem 6.36
The following turbulent-flow velocity data u(y), for air at 75F and 1 atm near a smooth flat wall,
were taken in the University of Rhode Island wind tunnel:
y, in:
0.025
0.035
0.047
0.055
0.065
u, ft/s:
51.2
54.2
56.8
57.6
59.1
Estimate (a) the wall shear stress and (b) the velocity u at y = 0.22 in.
Neglect minor losses.
Solution 6.36
For air at 75F and 1 atm, take
= 0.00230 slug/ft3 and
= 3.80E−7 slug/fts. We fit each data
point to the logarithmic-overlap law, Eq. (6.28):
Problem 6.37
Two infinite plates a distance h apart are parallel to the xz plane with the upper plate moving at
speed V, as in Fig. P6.37. There is a fluid of viscosity
and constant pressure between the plates.
Neglecting gravity and assuming incompressible turbulent flow u(y) between the plates, use the
logarithmic law and appropriate boundary conditions to derive a formula for dimensionless
wall shear stress versus dimensionless plate velocity. Sketch a typical shape of the profile u(y).
Neglect minor losses.
Solution 6.37
The shear stress between parallel plates is constant, so the centerline velocity must be exactly
u = V/2 at y = h/2. Anti-symmetric log-laws form, one with increasing velocity for 0 y h/2,
Problem 6.38
Suppose in Fig. P6.37 that h = 3 cm, the fluid is water at 20°C (
= 998 kg/m3,
= 0.001 kg/ms),
and the flow is turbulent, so that the logarithmic law is valid. If the shear stress in the fluid is
15 Pa, estimate V in m/s.
Neglect minor losses.
Solution 6.38
Just as in Prob. 6.37, apply the log-law at the center between the wall, that is, y = h/2, u = V/2.
Problem 6.39
By analogy with laminar shear,
=
du/dy. T. V. Boussinesq in 1877 postulated that turbulent
shear could also be related to the mean-velocity gradient
turb =
du/dy, where
is called the
eddy viscosity and is much larger than
. If the logarithmic-overlap law, Eq. (6.28), is valid with
turb
w, show that
u*y.
Neglect minor losses.
Solution 6.39
Differentiate the log-law, Eq. (6.28), to find dudy, then introduce the eddy viscosity into the
turbulent stress relation:
Problem 6.40
Theodore von Kármán in 1930 theorized that turbulent shear could be represented by
turb =
dudy where
=
2y2du/dy is called the mixing-length eddy viscosity and
0.41 is
Kármán’s dimensionless mixing-length constant [2,3]. Assuming that
turb
w near the wall,
show that this expression can be integrated to yield the logarithmic-overlap law, Eq. (6.28).
Neglect minor losses.
Solution 6.40
This is accomplished by straight substitution:
Problem 6.41
Two reservoirs, which differ in surface elevation by 40 m, are connected by 350 m of new pipe
of diameter 8 cm. If the desired flow rate is at least 130 N/s of water at 20C, can the pipe
material be (a) galvanized iron, (b) commercial steel, or (c) cast iron?
Neglect minor losses.
Solution 6.41
Applying the extended Bernoulli equation between reservoir surfaces yields
Problem 6.42
Fluid flows steadily, at volume rate Q, through a large horizontal pipe and then divides into two
small pipes, the larger of which has an inside diameter of 25 mm and carries three times the flow
of the smaller pipe. Both small pipes have the same length and pressure drop. If all flows are
turbulent, at Red near 104, estimate the diameter of the smaller pipe.
Neglect minor losses.
Solution 6.42
For turbulent flow, the formulas are algebraically complicated, such as Eq. (6.38). However, in
the low Reynolds number region, the Blasius power-law approximation, Eq. (6.39), applies,
Problem 6.43
A reservoir supplies water through 100 m of 30-cm-diameter cast iron pipe to a turbine that
extracts 80 hp from the flow. The water then exhausts to the atmosphere. Neglect minor losses.
(a) Assuming that f 0.019, find the flow rate (which results in a cubic polynomial). Explain
why there are two legitimate solutions. (b) For extra credit, solve for the flow rates using the
actual friction factors.
Neglect minor losses.
Solution 6.43
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/m-s. The energy equation yields a
relation between elevation, friction, and turbine power:
Problem 6.44
Mercury at 20C flows through 4 m of 7-mm-diameter glass tubing at an average velocity of
5 m/s. Estimate the head loss in m and the pressure drop in kPa.
Neglect minor losses.
Solution 6.44
For mercury at 20C, take
= 13550 kg/m3 and
= 0.00156 kg/ms. Glass tubing is considered
hydraulically “smooth,”
/d = 0. Compute the Reynolds number:
Problem 6.45
Oil, SG = 0.88 and
= 4E−5 m2/s, flows at 400 gal/min through a 6-in asphalted cast-iron pipe.
The pipe is 0.5 miles long and slopes upward at 8 in the flow direction. Compute the head loss
in ft and the pressure change.
Neglect minor losses.
Solution 6.45
First convert 400 gal/min = 0.891 ft3/s and
= 0.000431 ft2/s. For asphalted cast-iron,
Problem 6.46
The Keystone Pipeline in the chapter-opener photo has a diameter of 36 inches and a design flow
rate of 590,000 barrels per day of crude oil at 40ºC. If the pipe material is new steel, estimate the
pump horsepower required per mile of pipe.
Neglect minor losses.
Solution 6.46
From Fig. A.1, for crude oil at 40ºC, read μ ≈ 0.0053 kg/m-s ÷ 47.88 = 0.0000111 slug/ft-s. The
Problem 6.47
The gutter and smooth drainpipe in Fig. P6.47 remove rainwater from the roof of a building. The
smooth drainpipe is 7 cm in diameter. (a) When the gutter is full, estimate the rate of draining.
(b) The gutter is designed for a sudden rainstorm of up to 5 inches per hour. For this condition,
what is the maximum roof area that can be drained successfully?
Neglect minor losses.
Solution 6.47
If the velocity at the gutter surface is neglected, the energy equation reduces to
Problem 6.48
Follow up Prob. P6.46 with the following question. If the total Keystone pipeline length, from
Alberta to Texas, is 2147 miles, how much flow, in barrels per minute, will result if the total
available pumping power is 8,000 hp? Assume a galvanized iron pipe.
Neglect minor losses.
Problem 6.46
The Keystone Pipeline in the chapter-opener photo has a diameter of 36 inches and a design flow
rate of 590,000 barrels per day of crude oil at 40ºC. If the pipe material is new steel, estimate the
pump horsepower required per mile of pipe.
Neglect minor losses.
Solution 6.48
For crude oil at 40ºC, read μ ≈ 0.0053 kg/m-s ÷ 47.88 = 0.0000111 slug/ft-s. The density, with
SG = 0.86, is 0.86(1.94) = 1.67 slug/ft3. For galvanized iron, ε ≈ 0.0005 ft, hence
Problem 6.49
The tank-pipe system of Fig. P6.49 is to deliver at least 11 m3/h of water at 20C to the reservoir.
What is the maximum roughness height
allowable for the pipe?
Neglect minor losses.
Solution 6.49
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. Evaluate V and Re for the expected
flow rate:
Problem 6.50
Ethanol at 20C flows at 125 U.S. gal/min through a horizontal cast-iron pipe with L = 12 m and
d = 5 cm. Neglecting entrance effects, estimate (a) the pressure gradient, dp/dx; (b) the wall shear
stress,
w; and (c) the percent reduction in friction factor if the pipe walls are polished to a
smooth surface.
Neglect minor losses.
Solution 6.50
For ethanol (Table A-3) take
= 789 kg/m3 and
= 0.0012 kg/ms. Convert 125 gal/min to
0.00789 m3/s. Evaluate V = Q/A = 0.00789/[
(0.05)2/4] = 4.02 m/s.
Problem 6.51
The viscous sublayer (Fig. 6.9) is normally less than 1 percent of the pipe diameter and therefore
very difficult to probe with a finite-sized instrument. In an effort to generate a thick sublayer for
probing, Pennsylvania State University in 1964 built a pipe with a flow of glycerin. Assume a
smooth 12-in-diameter pipe with V = 60 ft/s and glycerin at 20C. Compute the sublayer
thickness in inches and the pumping horsepower required at 75 percent efficiency if L = 40 ft.
Neglect minor losses.
Solution 6.51
For glycerin at 20C, take
= 2.44 slug/ft3 and
= 0.0311 slug/fts. Then
Problem 6.52
The pipe flow in Fig. P6.52 is driven by pressurized air in the tank. What gage pressure p1 is
needed to provide a 20C water flow rate Q = 60 m3/h?
Neglect minor losses.
Solution 6.52
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. Get V, Re, f:
