Problem 1.C1
Sometimes we can develop equations and solve practical problems by knowing nothing more
than the dimensions of the key parameters in the problem. For example, consider the heat loss
through a window in a building. Window efficiency is rated in terms of “R value,” which has units
of (ft2·hr·F)/Btu. A certain manufacturer offers a double-pane window with R = 2.5. The same
company produces a triple–pane window with R = 3.4. In either case the window dimensions
are 3 ft by 5 ft. On a given winter day, the temperature difference between the inside and
outside of the building is 45F.
( a ) Develop an equation for the amount of heat lost in a given time period Δt , through a
window of area A , with a given R value, and temperature difference ΔT . How much heat (in
Btu) is lost through the double-pane window in one 24-h period?
( b ) How much heat (in Btu) is lost through the triple-pane window in one 24-h period?
( c ) Suppose the building is heated with propane gas, which costs $3.25 per gallon. The propane
burner is 80 percent efficient. Propane has approximately 90,000 Btu of available energy per
gallon. In that same 24-h period, how much money would a homeowner save per window by
installing triple-pane rather than double pane windows?
( d ) Finally, suppose the homeowner buys 20 such triple pane windows for the house. A typical
winter has the equivalent of about 120 heating days at a temperature difference of 45 8 F. Each
triple-pane window costs $85 more than the double-pane window. Ignoring interest and inflation,
how many years will it take the homeowner to make up the additional cost of the triple-pane
windows from heating bill savings?
Solution 1.C1
(a) The function Q = fcn(t, R, A, T) must have units of Btu. The only combination of units which
accomplishes this is:
Not a very good investment. We are using ‘$’ and ‘windows’ as “units” in our equations!
Problem 1.C2
When a person ice-skates, the ice surface actually melts beneath the blades, so that he or she
skates on a thin film of water between the blade and the ice.
(a) Find an expression for total friction force F on the bottom of the blade as a function of skater
velocity V, blade length L, water film thickness h, water viscosity
, and blade width W.
( b ) Suppose an ice skater of total mass m is skating along at a constant speed of V 0 when she
suddenly stands stiff with her skates pointed directly forward, allowing herself to coast to a stop.
Neglecting friction due to air resistance, how far will she travel before she comes to a stop?
(Remember, she is coasting on two skate blades.) Give your answer for the total distance
traveled, x , as a function of (Vo, m, L, h,
, W).
(c) Compute x for the case Vo = 4 m/s, m = 100 kg, L = 30 cm, W = 5.0 mm, and h = 0.10 mm.
Do you think our assumption of negligible air resistance is a good one?
Solution 1.C2
(a) The skate bottom and the melted ice are like two parallel plates:
, (a)
V VLW
F A Ans.
hh
= = =
Problem 1.C3
Two thin flat plates, tilted at an angle α , are placed in a tank of liquid of known surface tension
ϒ and contact angle θ , as shown in Fig. C1.3. At the free surface of the liquid in the tank, the
two plates are a distance L apart and have width b into the page. The liquid rises a distance h
between the plates, as shown.
( a ) What is the total upward ( z -directed) force, due to surface tension, acting on the liquid
column between the plates?
( b ) If the liquid density is ρ , find an expression for surface tension ϒ in terms of the other
variables.
Solution 1.C3
(a) Considering the right side of the liquid column, the surface tension acts tangent to the local
Problem 1.C4
Oil of viscosity
and density
drains steadily down the side of a tall, wide vertical plate,
as shown in Fig. C1.4. In the region shown, fully developed conditions exist; that is, the velocity
profile shape and the film thickness d are independent of distance z along the plate. The vertical
velocity w becomes a function only of x , and the shear resistance from the atmosphere is
negligible.
( a ) Sketch the approximate shape of the velocity profile w(x), considering the boundary
conditions at the wall and at the film surface.
( b ) Suppose film thickness δ , and the slope of the velocity profile at the wall, (dw / dx)wall , are
measured by a laser Doppler anemometer (to be discussed in Chap. 6). Find an expression for the
viscosity of the oil as a function of ρ , δ , (dw / dx)wall , and the gravitational acceleration g . Note
that, for the coordinate system given, both w and (dw / dx)wall are negative.
Solution 1.C4
(a) The velocity profile must be such that there is no slip (w = 0) at the wall and no shear
(dw/dx = 0) at the film surface. This is shown at right. Ans. (a
Problem 1.C5
Viscosity can be measured by flow through a thin–bore or capillary tube if the flow rate is low.
For length L, (small) diameter
,DL
pressure drop p, and (low) volume flow rate Q, the
formula for viscosity is
= D4p/(CLQ), where C is a constant.
(a) Verify that C is dimensionless. The following data are for water flowing through a 2–mm–
diameter tube which is 1 meter long. The pressure drop is held constant at p = 5 kPa.
T, °C:
10.0
40.0
70.0
Q, L/min:
0.091
0.179
0.292
(b) Using proper SI units, determine an average value of C by accounting for the variation with
temperature of the viscosity of water.
Solution 1.C5
(a) Check the dimensions of the formula and solve for {C}:
Problem 1.C6
The rotating–cylinder viscometer in Fig. C1.6 shears the fluid in a narrow clearance, r, as
shown. Assume a linear velocity distribution in the gaps. If the driving torque M is measured,
find an expression for
by (a) neglecting, and (b) including the bottom friction.
Solution 1.C6
(a) The fluid in the annular region has the same shear stress analysis as Prob. 1.49:
Problem 1.C7
Make an analytical study of the transient behavior of the sliding block in Prob. 1.45. (a) Solve
for V(t) if the block starts from rest, V = 0 at t = 0. (b) Calculate the time t1 when the block has
reached 98 percent of its terminal velocity.
Problem 1.45
A block of weight W slides down an inclined plane while lubricated by a thin film of oil, as in
Fig. P1.45. The film contact area is A and its thickness is h. Assuming a linear velocity
distribution in the film, derive a “terminal” (zero-acceleration) velocity V of the block.
Find the terminal velocity of the block if the block mass is 6 kg, A = 35 cm2 , θ = 15° , and the
film is 1-mm-thick SAE 30 oil at 20° C.
Figure P1.45
Solution 1.C7
Let x go down the slope, and write Newton’s law
W
h
V, x
μ
W
h
Ag
Problem 1.C8
A mechanical device that uses the rotating cylinder of Fig. C1.6 is the Stormer viscometer [29].
Instead of being driven at constant Ω , a cord is wrapped around the shaft and attached to a
falling weight W . The time t to turn the shaft a given number of revolutions (usually fi ve) is
measured and correlated with viscosity. The formula is
()
A
tWB
−
where A and B are constants which are determined by calibrating the device with a known fluid.
Here are calibration data for a Stormer viscometer tested in glycerol, using a weight of 50 N:
, kg/m·s:
0.23
0.34
0.57
0.84
1.15
t, sec:
15
23
38
56
77
(a) Find reasonable values of A and B to fit this calibration data. [Hint: The data are not very
sensitive to the value of B.]
(b) A more viscous fluid is tested with a 100 N weight and the measured time is 44 s. Estimate
the viscosity of this fluid.
Solution 1.C8
(a) The data fit well, with a standard deviation of about 0.17 s in the value of t, to the values
Problem 1.C9
The lever in Fig. C1.9 has a weight W at one end and is tied to a cylinder at the left end.
The cylinder has negligible weight and buoyancy and slides upward through a film of heavy oil
of viscosity
. (a) If there is no acceleration (uniform lever rotation) derive a formula for the
rate of fall V2 of the weight. Neglect the lever weight. Assume a linear velocity profile in the
oil film. (b) Estimate the fall velocity of the weight if W = 20 N, L1 = 75 cm, L2 = 50 cm,
D = 10 cm, L = 22 cm,
R = 1 mm, and the oil is glycerin at 20C.
Solution 1.C9
(a) If the motion is uniform, no acceleration, then the moments balance about the pivot:
Problem 1.C10
A popular gravity-driven instrument is the Cannon-Ubbelohde viscometer, shown in Fig. C1.10.
The test liquid is drawn up above the bulb on the right side and allowed to drain by gravity
through the capillary tube below the bulb. The time t for the meniscus to pass from upper to
lower timing marks is recorded. The kinematic viscosity is computed by the simple formula
= Ct, where C is a calibration constant. For
in the range of 100-500 mm2/s, the
recommended constant is C = 0.50 mm2/s2, with an accuracy less than 0.5%.
(a) What liquids from Table A.3 are in this viscosity range? (b) Is the calibration formula dimensionally
consistent? (c) What system properties might the constant C depend upon?
(d) What problem in this chapter hints at a formula for estimating the viscosity?
Solution 1.C10
(a) Very hard to tell, because values of
are not listed – sorry, I’ll add these values if I
remember. It turns out that only three of these 17 liquids are in the
Problem 1.C11
Mott [Ref. 49, p. 38] discusses a simple falling-ball viscometer, which we can analyze later in
Chapter 7. A small ball of diameter D and density
b falls through a tube of test liquid (
)
.
The fall velocity V is calculated by the time to fall a measured distance. The formula for
calculating the viscosity of the fluid is
This result is limited by the requirement that the Reynolds number (
VD/
) be less than 1.0.
Suppose a steel ball (SG = 7.87) of diameter 2.2 mm falls in SAE 25W oil (SG = 0.88) at 20C.
The measured fall velocity is 8.4 cm/s. (a) What is the viscosity of the oil, in kg/m-s?
(b) Is the Reynolds number small enough for a valid estimate?
Solution 1.C11
Relating SG to water, Eq. (1.7), the steel density is 7.87(1000) = 7870 kg/m3 and the oil density
is 0.88(1000) = 880 kg/m3. Using SI units, the formula predicts
Problem 1.C12
A solid aluminum disk (SG = 2.7) is 2 inches in diameter and 3/16 inch thick. It slides steadily
down a 14 incline that is coated with a castor oil (SG = 0.96) film one hundredth of an inch
thick. The steady slide velocity is 2 cm/s. Using Figure A.1 and a linear oil velocity profile
assumption, estimate the temperature of the castor oil.
2
()
18
bgD
V
−
=
Solution 1.C12
This problem reviews complicated units, volume and weight, shear stress, and viscosity. It fits
Problem 1.1
A gas at 20C may be considered rarefied, deviating from the continuum concept, when it
contains less than 1012 molecules per cubic millimeter. If Avogadro’s number is 6.023 E23
molecules per mole, what absolute pressure (in Pa) for air does this represent?
Solution 1.1
The mass of one molecule of air may be computed as
Problem 1.2
Table A.6 lists the density of the standard atmosphere as a function of altitude. Use these values
to estimate, crudely, say, within a factor of 2, the number of molecules of air in the entire
atmosphere of the earth.
Solution 1.2
Make a plot of density
versus altitude z in the atmosphere, from Table A.6:
1.2255 kg/m3
Problem 1.3
For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact with an
atmosphere at pressure pa, must undergo shear stress and hence begin to flow.
Hint: Account for the weight of the fluid and show that a no-shear condition will cause
horizontal forces to be out of balance.
Density in the Atmosphere
Solution 1.3
Assume zero shear. Due to element weight, the pressure along the lower and right sides must vary
linearly as shown, to a higher value at point C. Vertical forces are presumably in balance with
Problem 1.4
Sand, and other granular materials, definitely flow, that is, you can pour them from a container or
a hopper. There are whole textbooks on the “transport” of granular material. Therefore, is sand
a fluid? Explain.
Solution 1.4
Granular materials do indeed flow, at a rate that can be measured by “flowmeters”. But they are
not true fluids, because they can support a small shear stress without flowing. They may rest at a
Problem 1.5
The mean free path of a gas, l , is defined as the average distance traveled by molecules between
collisions. A proposed formula for estimating l of an ideal gas is
1.26 RT
=
What are the dimensions of the constant 1.26? Use the formula to estimate the mean free path of
air at 20C and 7 kPa. Would you consider air rarefied at this condition?
Solution 1.5
We know the dimensions of every term except “1.26”:
2
32
M M L
{ } {L} { } { } {R} {T} { }
LT LT
= = = = =
the flow are greater than about
100 ,
that is, greater than about 94
m.
Problem 1.6
Henri Darcy, a French engineer, proposed that the pressure drop 𝛥p for flow at
velocity V through a tube of length L could be correlated in the form
∆𝑝
𝜌 = 𝛼 𝐿 𝑉2
If Darcy’s formulation is consistent, what are the dimensions of the coefficient α?
Solution 1.6
From Table 1.2, introduce the dimensions of each variable:
Problem 1.7
Convert the following inappropriate quantities into SI units: (a) 2.283E7 U.S. gallons per
day; (b) 4.5 furlongs per minute (racehorse speed); and (c) 72,800 avoirdupois ounces per
acre.
Solution 1.7
(a) (2.283E7 gal/day) x (0.0037854 m3/gal) ÷ (86,400 s/day) = 1.0 m3/s Ans.(a)
Problem 1.8
Suppose we know little about the strength of materials but are told that the bending stress σ in a
beam is proportional to the beam half-thickness y and also depends on the bending moment M
and the beam area moment of inertia I . We also learn that, for the particular case M = 2900 inlbf,
y = 1.5 in, and I = 0.4 in4, the predicted stress is 75 MPa. Using this information and dimensional
reasoning only, find, to three significant figures, the only possible dimensionally homogeneous
formula σ
=
y f(M,I).
Solution 1.8
We are given that
= y fcn(M,I) and we are not to study up on strength of materials but only to
use dimensional reasoning. For homogeneity, the right hand side must have dimensions of
stress, that is,