Fluid Mechanics Midterm Exam Reviewer

Density is defined as mass per unit volume. To calculate the density of the fluid, divide its mass (500 kg) by its volume (0.5 m³). Performing the calculation: 500 kg ÷ 0.5 m³ = 1000 kg/m³. This indicates that for every cubic meter of the fluid, the mass is 1000 kg, which aligns with the properties of water, a common reference fluid.

Specific gravity is the ratio of a substance's density to the density of water. Since the specific gravity of the liquid is 0.85, it means the liquid is 85% as dense as water. The density of water is approximately 1000 kg/m³. Therefore, to find the density of the liquid, multiply 1000 kg/m³ by 0.85, resulting in 850 kg/m³.

To find the average flow velocity, we use the formula \( v = \frac{Q}{A} \), where \( Q \) is the volumetric flow rate and \( A \) is the cross-sectional area of the pipe. The area \( A \) can be calculated using \( A = \pi r^2 \), where \( r \) is the radius of the pipe (0.05 m). Thus, \( A \) is approximately 0.00785 m². Dividing the flow rate (0.05 m³/s) by the area gives an average velocity of about 6.37 m/s.

In a hydraulic press, the principle of Pascal's law states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid. The pressure exerted on the input piston is equal to the pressure on the output piston. Using the formula \( P = \frac{F}{A} \), we can find the output force. Given the areas of the pistons, the output force can be calculated as follows:

\[
\text{Output Force} = \text{Input Force} \times \frac{\text{Area of Output Piston}}{\text{Area of Input Piston}} = 200 \, \text{N} \times \frac{0.10 \, \text{m}²}{0.02 \, \text{m}²} = 1000 \, \text{N}.
\]

Thus, the output force is 1000 N.

According to the principle of conservation of mass, the flow rate must remain constant in a closed system. This means that the product of the cross-sectional area and the velocity at both sections must be equal. At Section 1, the area is larger, leading to a lower velocity. When the pipe narrows at Section 2, the area decreases, resulting in an increased velocity. By applying the equation of continuity (A1V1 = A2V2), we can calculate that the velocity at Section 2 is 8 m/s, given the initial conditions.

Hydrostatic pressure is calculated using the formula \( P = \gamma \times h \), where \( P \) is the pressure, \( \gamma \) is the specific weight of the fluid, and \( h \) is the depth. Given \( \gamma = 9810 \, \text{N/m}^3 \) and \( h = 5 \, \text{m} \), the calculation is \( P = 9810 \times 5 = 49,050 \, \text{Pa} \). This pressure results from the weight of the water above the specified depth, reflecting the force exerted by the water column.

In steady, incompressible flow, the principle of conservation of mass applies, often referred to as the continuity equation. When the cross-sectional area of a pipe decreases, the same volume of fluid must pass through a smaller space over the same time interval. To maintain a constant mass flow rate, the fluid must move faster in the narrower section of the pipe. This relationship shows that a decrease in area results in an increase in fluid velocity, confirming that the statement is true.