# January 29

```College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
January 29 Homework Solutions
1.6
IF V is a velocity, ℓ, a length, and  a fluid property having dimensions of L2T-1, which of the
following combinations are dimensionless: (a) Vℓ, (b) Vℓ/, (c) V2, (d) V/ℓ?
Velocity has dimensions of LT-1 and the length, ℓ., has dimensions of L. Check each combination
of variables to see if it is dimensionless:
(a) Vℓ has dimensions of (LT-1)(L)(L2T-1) = L4T-2 is not dimensionless.
(b) Vℓ/ has dimensions of (LT-1)(L)/(L2T-1) = L0T0 = 1 is dimensionless.
(c) V2 has dimensions of (LT-1)2(L2T-1) = L4T-3 is not dimensionless.
(a) V/ℓ has dimensions of (LT-1)(L)-1(L2T-1)-1 = L-3T is not dimensionless.
1.17
Clouds can weigh thousands of pounds due to their liquid water content. Often this
content is measured in grams per cubic meter (g/m3). Assume that a cumulus cloud
occupies a volume of one cubic kilometer, and its liquid water content is 0.2 g/m 3. (a) what
is the volume of this cloud in cubic miles? (b) How much does the water in the cloud
weigh in pounds?
(a)
3  1000
1 km  1 km 

3
m
km
3
3


ft
mi 
mi
  1 km3 
  0.240 mi 3
0.3048 m 5280 ft 
1
.
609344
km


Note that the conversion factor of 1.609344 km/mi is exact. Travelers often use the
approximation of 1.6 = 8/5 km/mi to get a good approximate of the relationship between miles
and kilometers.
(b) For this part we want to compute the weight as the product of mass times gravity, W = mg.
Here the mass the product of the water content and the cloud volume. We will assume
standard gravity of 9.80665 m/s2 to compute the weight.
W  mg 
0.2 g
m3
3
lb f
kg 9.80665 m 1 N  s 2
 1000 m 
1 km3 
 4.41x10 5 lb f

2
kg  m 4.4482 N
 km  1000 g
s
The first two terms are the known variables (water content and cloud volume; the remaining
terms are the acceleration of gravity and unit conversion factors.
Jacaranda (Engineering) 3333
E-mail: [email protected]
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
January 29 homework solutions
1.34
ME 390, L. S. Caretto, Spring 2008
Page 2
A closed tank having a volume of 2 ft3 is filled with 0.30 lbf of a gas. A pressure gage
attached to the tank reads 12 psi when the gas temperature is 80oF. There is some
question as to whether the gas in the tank is oxygen or helium. Which do you think it is?
Assume that we an apply the ideal gas equation to this problem: PV = mRT where we have all
the data to solve for R = PV/(mT). We can compute the value of R and compare it to tabulated
values of R in Table 1.7 for helium and oxygen. Assume that the atmospheric pressure is the
standard pressure of 14.696 psi and that g = 32.174 ft/s2.

PV g Pgage  Patm
R

mT
WT


32.174 ft  12 lb f 14.696 lb f 

 2 ft 3



2
2
2
V
1528 ft  lb f
s
in
in




slug  R
1 slug  ft ft 2
(0.30 lb f )(80  459.69) R 
lb f  s 2 144 in 2

The gas constants for helium and oxygen from Table 1.7 are 1.242x104 ft&middot;lbf/(slug&middot;R) and
1.554x103 ft&middot;lbf/(slug&middot;R), respectively. The value found above is much closer to the R value for
oxygen so we conclude that the gas in the tank is oxygen.
1.49
For air at standard atmospheric pressure the values of the constants that appear in the
Sutherland equation (Eq. 1.10) are C = 1.458x10-6 kg/(m&middot;s&middot;K1.5) and S = 110.4 K. Use these
values to predict the viscosity of air at 10oC and 90oC and compare with values given tin
Table B.4 in Appendix B.
The Sutherland equation for viscosity is m = CT 0.5/(T + S). Plugging in the constants given for the
two temperatures of 10oC (283.15 K) and 90oC (363.15 K) gives the following results:
1.458 x10 6 kg
10oC 
1.5
CT

T S
283.15 K 1.5
ms K
283.15 K  110.4 K
0.5
1.458 x10 6 kg
 90oC
1.5
CT


T S
363.15 K 1.5
ms K
363.15 K  110.4 K
0.5

1.765 x10 5 kg 1.765 x10 5 N  s
.

ms
m2
2.13x10 5 kg 2.13x10 5 N  s
.


ms
m2
These values agree well with the values from Table B-4 on page 763:  = 1.76x10-5 N&middot;s/m2 at
10oC and  = 2.14x10-5 N&middot;s/m2 at 90oC.
January 29 homework solutions
1.53
ME 390, L. S. Caretto, Spring 2008
Page 3
For a parallel plate
arrangement of the type
shown in Fig. 1.3 (copied at
right) it is found that when
the distance between plates
is 2 mm, a shearing stress of
150 Pa develops at the upper
plate when it is pulled at a
velocity of 1 m/s. Determine
the viscosity of the fluid
between the plates. Express
The basic relationship between shear stress, viscosity, and velocity gradient is  =  du/dy. For
the figure shown here, the velocity profile is linear so that du/dy = U/b. From the data given we
can find the velocity gradient and use that to compute the viscosity for the given shear stress of
150 Pa.

1.85




du U
dy
b
150 Pa
1N
Pa  m 2  0.300 N  s
1m
m2
s
0.001 m
2 mm
mm
To measure the water depth in a large open tank with opaque walls, an open vertical glass
tube is attached to the side of the tank. The height of the water column in the tube is then
used as a measure of the depth of water in the tank. (a) For a true water depts. in the tank
of 3 ft, make use of Eq. 1.22 (with  = 0) to determine the percent error due to capillarity as
the diameter of the glass tube is changed. Assume a water temperature of 80 oF. Show
your results on a graph of percent error versus tube diameter, D, in the range 0.1 in &lt; D &lt;
1.0 in. (b) If you want the error to be less than 1%, what is the smallest tube diameter
allowed?
Equation 1.22 gives the capillary rise, h = 2cos/R. For water at 80oF we find  = 0.00491 lbf/ft
and  = 62.22 lbf/ft3 from Table B-1 on page 761. Substituting these numbers,  = 0, and R = D/2
(because we are asked to plot the results versus D) into the formula for capillary rise gives
2 cos  4 cos 
h


R
D
4
0.00491 lb f
ft
62.22 lb f
ft 3
cos 0

D
0.00031565 ft 2
D
This gives the capillary rise in ft, when D is given in ft. The relative error in the depth is simply the
capillary rise divided by the true depth of 3 ft. Defining this error and using a unit conversion
factor so that we can use D in inches gives the following working equation.
January 29 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
0.00031565 ft 2 12 in
0.0012626 in
h
D
ft
error 


depth
3 ft
D
Calculating this equation and plotting results in a spreadsheet gives the results shown below.
D (in)
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
error
1.26%
0.84%
0.63%
0.51%
0.42%
0.36%
0.32%
0.28%
0.25%
0.23%
0.21%
0.19%
0.18%
0.17%
0.16%
0.15%
0.14%
0.13%
0.13%
(b) For the error to be less than 1% requires that
error 
0.0012626 in
0.0012626 in
 0.01  D 
 0.12626 in
D
0.01
So the minimum allowed diameter is 0.126 in.
```