Question

Water at room temperature flows with a constant speed of \(8.00 \mathrm{~m} / \mathrm{s}\) through a nozzle with a square cross section, as shown in the figure. Water enters the nozzle at point \(A\) and exits the nozzle at point B. The lengths of the sides of the square cross section at \(A\) and \(B\) are \(50.0 \mathrm{~cm}\) and \(20.0 \mathrm{~cm}\), respectively. a) What is the volume flow rate at the exit? b) What is the acceleration at the exit? The length of the nozzle is \(2.00 \mathrm{~m}\). c) If the volume flow rate through the nozzle is increased to \(6.00 \mathrm{~m}^{3} / \mathrm{s}\). what is the acceleration of the fluid at the exit?

Short answer

Based on the given information about the water nozzle, calculate: a) the volume flow rate at the exit b) the acceleration of the fluid at the exit c) the acceleration of the fluid at the exit when the volume flow rate is increased to 6.00 m^3/s. a) The volume flow rate at the exit is 2.00 m^3/s. b) The acceleration at the exit is 604 m/s^2. c) The acceleration of the fluid at the exit with the increased volume flow rate is 5514 m/s^2.

Step-by-step solution

Convert all units to SI units (meters and seconds)

We must convert the side lengths of the square cross-sections to meters. Given measurements are A_side = 50.0 cm and B_side = 20.0 cm. Converting to meters, A_side = 50.0 / 100 = 0.50 m B_side = 20.0 / 100 = 0.20 m

Calculate the area of the cross sections at A and B

The cross-sectional areas are squares, so we can calculate their area as A_area = A_side^2 and B_area = B_side^2. A_area = (0.50 m)^2 = 0.25 m^2 B_area = (0.20 m)^2 = 0.04 m^2

Calculate the volume flow rate at the exit (point B)

Since the water is incompressible, and the flow rate is constant, we have entering_flow_rate = exiting_flow_rate. Thus, Q_A = A_area * v_A = B_area * v_B. We know A_area, B_area and v_A(8.00 m/s) from Step 2 and the given information. Solving for v_B: v_B = (A_area * v_A) / B_area = (0.25 m^2 * 8.00 m/s) / 0.04 m^2 = 50 m/s Now calculate the volume flow rate at the exit, Q_B: Q_B = B_area * v_B = 0.04 m^2 * 50 m/s = 2.00 m^3/s So, the volume flow rate at the exit is 2.00 m^3/s.

Calculate the acceleration at the exit

We'll use the following formula to find the acceleration: a = (v_B^2 - v_A^2) / (2 * L), where L is the length of the nozzle. a = (50 m/s)^2 - (8.00 m/s)^2) / (2 * 2.00 m) = 604 m/s^2 So, the acceleration at the exit is 604 m/s^2.

Calculate the acceleration of the fluid at the exit, with an increased volume flow rate

We are given the new volume flow rate, Q_new = 6.00 m^3/s. From this, we can find the new exit velocity, v_new, using: v_new = Q_new / B_area = (6.00 m^3/s) / (0.04 m^2) = 150 m/s Now, we can calculate the new acceleration at the exit, a_new, using the same formula as in Step 4: a_new = ((150 m/s)^2 - (8.00 m/s)^2) / (2 * 2.00 m) = 5514 m/s^2 So, the acceleration of the fluid at the exit with the increased volume flow rate is 5514 m/s^2. In summary: a) The volume flow rate at the exit is 2.00 m^3/s. b) The acceleration at the exit is 604 m/s^2. c) The acceleration of the fluid at the exit with the increased volume flow rate is 5514 m/s^2.

Key concepts

Understanding Volume Flow Rate

In fluid mechanics, volume flow rate is crucial to assess how much fluid passes through a given area in a certain timeframe. For instance, visualize a garden hose; the volume flow rate would be the quantity of water that can fill a bucket per second as it flows out of the hose. Mathematically, it's calculated using the formula:
\( Q = A \times v \),
where \( Q \) is the volume flow rate, \( A \) the cross-sectional area and \( v \) the velocity of the fluid. As shown in the exercise solution, when water exits the nozzle, it has the same volume flow rate as when it enters, following the principle of continuity for incompressible fluids.
Take the given problem: With a consistent initial speed and knowing the cross-sectional area at both points A and B, the exiting flow rate can be computed based on the input conditions. In simpler terms, given that water doesn't compress or expand significantly, whatever volume enters the nozzle per second must exit per second. This results in a calculable, unchanged flow rate, essential for many applications such as water supply systems or in finding the efficiency of fluid machinery.

Fluid Acceleration at the Nozzle Exit

The concept of fluid acceleration aids in understanding how swiftly the velocity of the fluid changes as it travels through a system like a nozzle. It's calculated in units of \( m/s^2 \) and represents the change in velocity per time. Think of the nozzle as a funnel: water speeds up significantly as it moves from a wide to a narrow end due to the decrease in cross-sectional area while maintaining the same flow rate.
In the presented exercise, knowing the initial and final velocities and the length of the nozzle enables us to calculate the acceleration using the formula:
\( a = \frac{v_B^2 - v_A^2}{2 \cdot L} \),
where \( v_B \) and \( v_A \) are the velocities at points B and A, respectively, and \( L \) is the length of the nozzle. This calculation helps to understand the dynamics of the fluid inside a nozzle, often used for controlling fluid speed in various industrial applications including engines and sprayers. As water velocity increases from A to B, significant acceleration results, which must be considered for pressure considerations and material strength of the nozzle.

Applying Bernoulli's Equation

Finally, Bernoulli's equation, a principle cornerstone of fluid dynamics, relates the speed of the fluid to its potential energy and pressure. In its form, it establishes that the total mechanical energy of the fluid remains constant along a streamline if it's incompressible and loses no energy to friction. The equation commonly looks like this:
\( P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \),
where \( P \) is the fluid pressure, \( \rho \) the fluid density, \( v \) the fluid velocity, \( g \) the acceleration due to gravity, and \( h \) the height above a reference point. In the context of this problem, if we assume negligible height change and disregard friction, Bernoulli's equation helps to relate the pressure and velocity of water at different points inside the nozzle. High-speed fluid at the exit of the nozzle will have low pressure compared to the entrance, a principle exploited in numerous devices such as the Venturi meter or atomizers.
While the provided steps do not use Bernoulli's equation directly, understanding its implications can be crucial for students getting a deeper insight into fluid behavior and relating pressure with flow velocity and height.