Chapter 2
Fluid Properties
Problem 2.1
Calculate the density and speci c weight of nitrogen at an absolute pressure of
1 MPa and a temperature of 40 C.
Solution
Ideal gas law

=

From Table A.2, = 297 J/kg/K. The temperature in absolute units is =
273 + 40 = 313 K.
106 N/m2
=
297 J/kgK × 313 K
= 10 75 kg/m3
The speci c weight is
=
= 10 76 kg/m3 × 9 81 m/s2
= 105 4 N/m3
5
6 CHAPTER 2. FLUID PROPERTIES
Problem 2.2
Find the density, kinematic and dynamic viscosity of crude oil in traditional units
at 100oF.
Solution
From Fig. A.3, =6 5 × 10 5 ft2 s and = 0 86
The density of water at standard condition is 1.94 slugs/ft3 so the density of crude
oil is 0 86 × 1 94 =1 67 slugs/ft3 or 1 67 × 32 2 =53 8 lbm/ft3
The dynamic viscosity is = 1 67 × 6 5 × 10 5 =1 09 × 10 4 lbf·s/ft2
Problem 2.3
Two parallel glass plates separated by 0.5 mm are placed in water at 20oC. The
plates are clean, and the width/separation ratio is large so that end e ects are
negligible. How far will the water rise between the plates?
Solution
The surface tension at 20oC is 7 3 × 10 2 N/m. The weight of the water in the
column is balanced by the surface tension force.
= 2 cos
where is the width of the plates and is the separation distance. For water
against glass, cos ' 1 Solving for gives
2 × 7 3 × 10 2 N/m
= =

0 5 × 10 3 m × 998 kg/m3 × 9 81 m/s2
= 0 0149 m = 29 8 mm
7
Problem 2.4
The kinematic viscosity of helium at 15oC and standard atmospheric pressure (101
kPa) is 1.14×10 4 m2 s. Using Sutherland s equation, nd the kinematic viscosity
at 100oC and 200 kPa.
Solution
From Table A.2, Sutherland s constant for helium is 79.4 K and the gas constant is
2077 J/kgK. Sutherland s equation for absolute viscosity is
µ Å›3 2
+
=
+
The absolute viscosity is related to the kinematic viscosity by = . Substituting
into Sutherland s equation
µ Å›3 2
+
=
+
or
µ Å›3 2
+
=
+
From the ideal gas law

=

so
µ Å›5 2
+
=
+
The kinematic viscosity ratio is found to be
µ Å›5 2
101 373 288 + 79 4
=
200 288 373 + 79 4
= 0 783
The kinematic viscosity is
= 0 783 × 1 14 × 10 4 = 8 93 × 10 5 m2 s
8 CHAPTER 2. FLUID PROPERTIES
Problem 2.5
Air at 15oC forms a boundary layer near a solid wall. The velocity distribution
in the boundary layer is given by

= 1 exp( 2 )

where = 30 m/s and = 1 cm. Find the shear stress at the wall ( = 0)
Solution
The shear stress at the wall is related to the velocity gradient by

= | =0

Taking the derivative with respect to of the velocity distribution

= 2 exp( 2 )

Evaluating at = 0
30
| =0= 2 = 2 × = 6 × 103 s 1
0 01
From Table A.2, the density of air is 1.22 kg/m3 and the kinematic viscosity is
1.46×10 5 m2 s. The absolute viscosity is = = 1 22×1 46×10 5 = 1 78×10 5
N·s/m2 The shear stress at the wall is

= | =0= 1 78 × 10 5 × 6 × 103 = 0 107 N/m2