# Fundamentals of Engineering

## 11 (A): Fluid Properties

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

## Practice Problems

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

1. $$1000 \, kg/m^3$$
2. $$1048 \, kg/m^3$$
3. $$1125 \, kg/m^3$$
4. $$1215 \, kg/m^3$$

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

1. $$106 \, kN/m^3$$
2. $$120 \, kN/m^3$$
3. $$133 \, kN/m^3$$
4. $$165 \, kN/m^3$$

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

1. $$0.00070 \, m^3/kg$$
2. $$0.00092 \, m^3/kg$$
3. $$0.00103 \, m^3/kg$$
4. $$0.00138 \, m^3/kg$$

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. $$1.51 \times 10^{-5} \, m^2/s$$
2. $$1.82 \times 10^{-5} \, m^2/s$$
3. $$2.20 \times 10^{-5} \, m^2/s$$
4. $$2.47 \times 10^{-5} \, m^2/s$$

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 $$1.2 \, m/s$$

1. $$779 \, N$$
2. $$865 \, N$$
3. $$1018 \, N$$
4. $$1131 \, N$$

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. $$1.5 \, N \cdot s/m^2\,$$
2. $$1.7 \, N \cdot s/m^2\,$$
3. $$2.1 \, N \cdot s/m^2\,$$
4. $$2.6 \, N \cdot s/m^2\,$$

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

1. $$0.32 \, N/m$$
2. $$0.46 \, N/m$$
3. $$0.52 \, N/m$$
4. $$0.65 \, N/m$$

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.

1. $$34 \, mm$$
2. $$42 \, mm$$
3. $$47 \, mm$$
4. $$59 \, mm$$

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 $$20 \, mm$$. $$\sigma = 0.025 \, N/m$$.

1. $$8 \, Pa$$
2. $$10 \, Pa$$
3. $$14 \, Pa$$
4. $$25 \, Pa$$

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