- Density, Specific Volume, Specific Weight, and Specific Gravity
- Shear Stress and Viscosity
- Surface Tension and Capillarity

**1)**
A can of beer has a mass \(343.7 \, g\) when full and \(13.7 \, g\)
when empty. The fluid volume is \(330 \, mL\). Calculate the density
of beer

- \(1000 \, kg/m^3\)
- \(1048 \, kg/m^3\)
- \(1125 \, kg/m^3\)
- \(1215 \, kg/m^3\)

**The correct answer is (A) **

**Solution:** Density is mass divided by volume and both are given.
$$ \begin{align}
\rho &= {m \over V} \\\\
&= {343.7 \, g - 13.7 \, g \over 330 \, mL} \\\\
&= {0.33 \, kg \over 0.330 \times 10^{-3} \, m^3} \\\\
&= 1000 \, kg/m^3
\end{align} $$
**The correct answer is (A)**

**2)**
The specific gravity of mercury is \(13.6\). Determine the specific
weight

- \(106 \, kN/m^3\)
- \(120 \, kN/m^3\)
- \(133 \, kN/m^3\)
- \(165 \, kN/m^3\)

**The correct answer is (C) **

**Solution:**
Note that \(SG = \rho / \rho_w\)
$$ \begin{align}
\gamma &= \rho \cdot g \\\\
&= SG \cdot \rho_w \cdot g \\\\
&= 13{,}600 \, kg/m^3 \cdot 9.81 \, m/s^2 \\\\
&= 133 \, kN/m^3
\end{align} $$
**The correct answer is (C)**

**3)**
\(8 \, kg\) of fluid \(A\) of density \(750 \, kg/m^3\) is mixed with
\(12 \, kg\) of water. If the resulting mixture is homogeneous what is
the resulting specific volume

- \(0.00070 \, m^3/kg\)
- \(0.00092 \, m^3/kg\)
- \(0.00103 \, m^3/kg\)
- \(0.00138 \, m^3/kg\)

**The correct answer is (C) **

**Solution:** First compute the volumes of the fluid A and water.
$$ \begin{align}
V_{A} &= {m \over \rho} = {8 \, kg \over 750 \, kg/m^3} = 0.01067 m^3 \\\\
V_{w} &= {m \over \rho} = {12 \, kg \over 1000 \, kg/m^3} = 0.01200 m^3 \\\\
\end{align} $$
Note that Specific Volume is defined in the Thermo Section of the
Reference Handbook under "State Functions".
$$ \begin{align}
v_{mixture} &= { \text{total volume} \over \text{total mass}} = { V_{A} + V_{w} \over m_A + m_w} \\\\
&= { 0.01067 m^3 + 0.01200 m^3 \over 10 \, kg + 12 \, kg} \\\\
&= 0.00103 \, m^3/kg
\end{align} $$
**The correct answer is (C)**

**4)**
At \(20^{\circ}C\) and \(1 \, atm\), the dynamic viscosity of air is
\(1.82 \times 10^{-5} \, N \cdot s / m^2\), and density is
\(1.205 \, kg/m^3\). Calculate the kinematic viscosity

- \(1.51 \times 10^{-5} \, m^2/s \)
- \(1.82 \times 10^{-5} \, m^2/s \)
- \(2.20 \times 10^{-5} \, m^2/s \)
- \(2.47 \times 10^{-5} \, m^2/s \)

**The correct answer is (A) **

**Solution:** The formula for kinematic viscosity is given in
the handbook. So it is a matter of simple substitution
$$ \begin{align}
\nu &= {\mu \over \rho} \\\\
&= {1.82 \times 10^{-5} \, N \cdot s / m^2 \over 1.205 \, kg/m^3} \\\\
&= 1.51 \times 10^{-5} \, m^2/s
\end{align} $$
**The correct answer is (A)**

**5)**
Two cylindrical bearings of length \(250 \, mm\) each have a
\(50 \, mm\) diameter rod being pulled through. The lubricant that fills
the \(0.25 \, mm\) gap between the rod and bearings is an oil of
viscosity \(3.0 \, N \cdot s/m^2\). Calculate the force required to pull
the rod through both bearings at a velocity of \(1.2 \, m/s\)

- \(779 \, N\)
- \(865 \, N\)
- \(1018 \, N\)
- \(1131 \, N\)

**The correct answer is (D) **

**Solution:** Since shear stress \( \tau = F/A\)
$$ F = A \cdot \mu \cdot {dv \over dy} $$
The area, \(A\), is the cylindrical surface of the rod that is contact
with the bearing. Since there are two bearings the total area is
\( 2 \pi D L \)
$$ \begin{align}
F &= (2 \pi D L) \cdot \mu \cdot {dv \over dy} \\\\
&= 2 \pi \times .050 \, m \times 0.25 \, m \times
3.0 \, N s/m^2 \times {1.2 \, m/s \over 0.00025 \, m} \\\\
&= 1131 \, N
\end{align} $$
**The correct answer is (D) **

**6)**
A rotating cylinder viscometer is constructed with concentric cylinders.
The inner cylinder has a diameter \(12.0 \, cm\) and length \(20 \, cm\).
The gap between the cylinders is \(1.0 \, mm\). It is found that the
torque required to rotate the inner cylinder at \(250 \, rpm\) is
\(15.0 \, Nm\). Determine the viscosity of the fluid

- \(1.5 \, N \cdot s/m^2\, \)
- \(1.7 \, N \cdot s/m^2\, \)
- \(2.1 \, N \cdot s/m^2\, \)
- \(2.6 \, N \cdot s/m^2\, \)

**The correct answer is (C) **

**Solution:**
The velocity of the outer surface of the inside cylinder is first
calculated by converting \(rpm\) to \(m/s\)
$$ \begin{align}
v &= rpm \times \left( \frac{\pi D}{60} \right) \\
&= 250 \times \left( \frac{\pi \cdot 0.12}{60} \right) m/s \\
&= 1.5708 \, m/s \\\\
\end{align} $$
Assuming a linear velocity profile we can write the following
relationship
$$ \begin{align}
T &= F \times R \\
&= \left( \tau \cdot A \right) \times \left( \frac{D}{2} \right) \\
&= \left( \mu {dv \over dy} \cdot A \right) \times \left( \frac{D}{2} \right) \\
&= \mu \times {\Delta v \over \Delta y} \times(\pi D L) \times
\left( \frac{D}{2} \right)
\end{align} $$
Substitute and solve for \( \mu \)
$$15 Nm = \mu \times {1.5708 \, m/s \over 0.001 m} \times
(\pi \cdot 0.12m \cdot 0.2m) \times
\left( \frac{0.12m}{2} \right)
$$
$$\mu = 2.11 \, N \cdot s/m^2$$
**The correct answer is (C)**

**7)**
The surface tension of a liquid is to be measured using a liquid film
suspended on a U-shaped wire frame with a \(10 \, cm\) long movable side.
If the force needed to move the wire is \(0.104 \, N\), determine the
surface tension of this liquid

- \(0.32 \, N/m \)
- \(0.46 \, N/m \)
- \(0.52 \, N/m \)
- \(0.65 \, N/m \)

**The correct answer is (C) **

**Solution:** Surface tension is force per unit length. Here since
a thin film has two surfaces (one on each side) so remember to
double the length
$$ \begin{align}
\sigma &= {F \over 2L} \\\\
&= {0.104 \, N \over 2 \times 0.1 \, m} \\\\
&= 0.52 \, N/m
\end{align} $$
**The correct answer is (C)**

**8)**
A glass tube of inside diameter \(0.5 \, mm\) is inserted into water.
The surface tension of water is \(0.0728 \, N/m\). Calculate the
capillary rise.

- \(34 \, mm\)
- \(42 \, mm\)
- \(47 \, mm\)
- \(59 \, mm\)

**The correct answer is (D) **

**Solution:** Using the capillary rise formula

$$ h = {4 \sigma \cos{\beta} \over \gamma d}$$
For water the contact angle \(\beta\) is assumed to be zero.
\(\gamma\) for water is given in the handbook (also easy to calculate).
$$
\begin{align}
h &= {4 \times 0.0728 \, N/m \times 1 \over
9810 \, N/m^3 \times 0.0005 \, m } \\
&= 0.059 \, m \\
&= 59 \, mm
\end{align} $$
**The correct answer is (D)**

**9)**
Find the gage pressure inside a soap bubble of diameter \(20 \, mm\).
\( \sigma = 0.025 \, N/m\).

- \( 8 \, Pa\)
- \( 10 \, Pa\)
- \( 14 \, Pa\)
- \( 25 \, Pa\)

**The correct answer is (B) **

**Solution:**
The formula for this problem is not given in the FE Handbook, but it is
easy to derive. If you consider one half of the sphere, the force balance
between the surface tension force(s) and internal pressure can be written
as
$$ 2 \times \text{circumference} \times \text{surface tension}
= \text{area of crossection} \times \text{gage pressure}$$
$$ 2 \times (\pi D) \times \sigma
= \left( \frac{\pi \cdot D^2}{4} \right) \times p$$
Solve for \(p\)
$$ \begin{align}
p &= \left( \frac{8 \sigma}{D} \right) \\
&= \left( \frac{8 \times 0.025 \, N/m}{0.020 m} \right) \\
&= 10 \, Pa
\end{align} $$
Note that the factor \(2\) is required for bubbles because you have to
account for both inside and outside surfaces. Note that if you were
calculating the pressure inside a drop then there is only one surface

**The correct answer is (B)**