Question

A pump steadily delivers water through a hose terminated by a nozzle. The exit of the nozzle has a diameter of \(2.5 \mathrm{~cm}\) and is located \(4 \mathrm{~m}\) above the pump inlet pipe, which has a diameter of \(5.0 \mathrm{~cm}\). The pressure is equal to 1 bar at both the inlet and the exit, and the temperature is constant at \(20^{\circ} \mathrm{C}\). The magnitude of the power input required by the pump is \(8.6 \mathrm{~kW}\), and the acceleration of gravity is \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the mass flow rate delivered by the pump, in \(\mathrm{kg} / \mathrm{s}\).

Short answer

The mass flow rate delivered by the pump is approximately 4.48 kg/s.

Step-by-step solution

Define the known variables

List all the known values: - Inlet diameter, \(D_1 = 5.0 \text{ cm} = 0.05 \text{ m}\) - Nozzle exit diameter, \(D_2 = 2.5 \text{ cm} = 0.025 \text{ m}\) - Height difference, \(z = 4 \text{ m}\) - Pressure at inlet and exit, \(P = 1 \text{ bar} = 100000 \text{ Pa}\) - Power input, \(P_{\text{pump}} = 8.6 \text{ kW} = 8600 \text{ W}\) - Gravity, \(g = 9.81 \text{ m/s}^2\) - Temperature, \(T = 20^\text{°}C\) - Density of water at \(20^\text{°}C\), \(\rho = 998 \text{ kg/m}^3\)

Calculate the cross-sectional areas

Compute the cross-sectional areas at the inlet and the exit: Given the diameter at each point, use the formula for the area of a circle, \(A = \pi \frac{D^2}{4}\): For the inlet, \(D_1 = 0.05 \text{ m}\): \(A_1 = \pi \frac{(0.05)^2}{4} = 1.9635 \times 10^{-3} \text{ m}^2\) For the nozzle exit, \(D_2 = 0.025 \text{ m}\): \(A_2 = \pi \frac{(0.025)^2}{4} = 4.9087 \times 10^{-4} \text{ m}^2\)

Apply Bernoulli’s Equation

Utilize Bernoulli’s equation to relate the fluid dynamic properties between two points along the streamline: \[ \frac{{P_1}}{{\rho}} + \frac{{v_1^2}}{{2}} + gz_1 = \frac{{P_2}}{{\rho}} + \frac{{v_2^2}}{{2}} + gz_2 \] Since \(P_1 = P_2 = 100000 \text{ Pa}\) and \(z_1 = 0 \text{ m}\) while \(z_2 = 4 \text{ m}\): \[ 0 + \frac{{v_1^2}}{{2}} = \frac{{v_2^2}}{{2}} + (4 \times 9.81) \] Hence, after simplification: \[ \frac{{v_1^2}}{{2}} = \frac{{v_2^2}}{{2}} + 39.24 \] Solve for \(v_1\) in terms of \(v_2\): \[ v_1^2 = v_2^2 + 78.48 \]

Apply Continuity Equation

By using the continuity equation, the relationship can be established between the velocities and cross-sectional areas: \[ A_1 v_1 = A_2 v_2 \] Substitute \(A_1\) and \(A_2\) into the equation: \[ 1.9635 \times 10^{-3} v_1 = 4.9087 \times 10^{-4} v_2 \] Solve for \(v_1\): \[ v_1 = \frac{ 4.9087 \times 10^{-4} v_2 }{ 1.9635 \times 10^{-3}} = 0.25 v_2 \] Also, substitute \(v_1 = 0.25v_2\) into the Bernoulli’s equation from Step 3: \[ (0.25v_2)^2 = v_2^2 + 78.48 \rightarrow 0.0625v_2^2 = v_2^2 + 78.48 \rightarrow 0.9375v_2^2 = 78.48 \rightarrow v_2^2 = 83.66 \rightarrow v_2 \approx 9.14 \text{ m/s} \]

Solve for \(v_1\)

Using the continuity equation value substituted for \(v_1\): \[ v_1 = 0.25 \times v_2 = 0.25 \times 9.14 \approx 2.285 \text{ m/s} \]

Calculate the mass flow rate

Use the relation \(\dot{m} = \rho A_1 v_1\): \[ \dot{m} = 998 \times 1.9635 \times 10^{-3} \times 2.285 \] Compute the product: \[ \dot{m} \approx 4.48 \text{ kg/s} \]

Key concepts

Bernoulli’s equation

Bernoulli’s equation is a fundamental principle in fluid dynamics that helps us understand how fluid properties like pressure, velocity, and height change along a streamline. It’s particularly useful when analyzing fluid flow through pumps and hoses. The equation is expressed as:
\[ \frac{{P_1}}{{\rho}} + \frac{{v_1^2}}{{2}} + gz_1 = \frac{{P_2}}{{\rho}} + \frac{{v_2^2}}{{2}} + gz_2 \]
Where:

• \( P_1 \) and \( P_2 \) are the fluid pressures at two points
• \( \rho \) is the fluid density
• \( v_1 \) and \( v_2 \) are the fluid velocities
• \( z_1 \) and \( z_2 \) are the heights above a reference level
• \( g \) is the acceleration due to gravity

By simplifying the equation appropriately, we can find unknowns such as the velocity of the fluid at different points which is crucial for determining the mass flow rate.

Continuity equation

The continuity equation ensures that the mass flow rate is conserved in a fluid flow. For incompressible fluids like water, it tells us that the product of cross-sectional area and velocity remains constant. Mathematically, it is represented as:
\[ A_1 v_1 = A_2 v_2 \]
Where:

• \( A_1 \) and \( A_2 \) are the cross-sectional areas at two points
• \( v_1 \) and \( v_2 \) are the velocities at these points

This is particularly useful for calculating the velocity at the nozzle exit when the diameter and initial velocity are known.

Fluid dynamics

Fluid dynamics is the study of how fluids (liquids and gases) move. Understanding fluid dynamics allows us to analyze the behavior of fluids in various situations, such as the flow through pipes and nozzles. When utilizing the principles of fluid dynamics, concepts like Bernoulli’s equation and the continuity equation become invaluable. These principles help explain how variables like pressure, velocity, and height influence fluid flow within different regions of a system.

Cross-sectional area calculation

To determine the velocity and mass flow rate of fluid, it's essential to calculate the cross-sectional area of the points where measurements are made. The area for a circular pipe or nozzle can be found using the formula:
\[ A = \frac{{\text{π} D^2}}{4} \]
Where:

• \( A \) is the cross-sectional area
• \( D \) is the diameter
• π (pi) is approximately 3.14159

Accurate calculation of areas is crucial in applying the continuity equation. For instance, in the exercise, the areas were needed to find fluid velocities at different sections of the hose.

Density of water

The density of water is a significant property when performing fluid dynamic calculations. At 20°C, water has a density of approximately 998 kg/m³. This value is essential when using Bernoulli’s equation and the mass flow rate formula:
\[ \text{Mass flow rate} = \rho A v \]
Where:

• \( \rho \) is the fluid density
• \( A \) is the cross-sectional area
• \( v \) is the fluid velocity

Understanding the density helps in determining the amount of fluid mass flowing through the system per unit time. It’s a key factor when transitioning from volume flow rate to mass flow rate.