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IP Water flows at the rate of \(3.11 \mathrm{kg} / \mathrm{s}\) through a hose with a diameter of \(3.22 \mathrm{cm}\) (a) What is the speed of water in this hose? (b) If the hose is attached to a nozzle with a diameter of \(0.732 \mathrm{cm},\) what is the speed of water in the nozzle? \((\mathrm{c})\) Is the number of kilograms per second flowing through the nozzle greater than, less than, or equal to \(3.11 \mathrm{kg} / \mathrm{s}\) ? Explain.

Short Answer

Expert verified
(a) Speed in hose: Use Q = A*v to find it. (b) Speed in nozzle: Use continuity, v_nozzle = (A_hose/A_nozzle) * v_hose. (c) Mass flow rate remains 3.11 kg/s due to mass conservation.

Step by step solution

01

Convert Diameter to Radius and Area (Hose)

The diameter of the hose is given as \(3.22\, \text{cm}\). First, convert this to meters: \(3.22\, \text{cm} = 0.0322\, \text{m}\). Calculate the radius: \(r = \frac{0.0322}{2} = 0.0161\, \text{m}\). Next, calculate the cross-sectional area: \(A = \pi r^2 = \pi (0.0161)^2\, \text{m}^2\).
02

Calculate Speed of Water in Hose

Using the formula for volumetric flow rate \(Q = A \cdot v\), where \(Q\) is the volumetric flow rate, \(A\) is the cross-sectional area, and \(v\) is the speed, we can express this as \(v = \frac{Q}{A}\). The mass flow rate \(m = 3.11\, \text{kg/s}\), and using the density of water \(\rho = 1000\, \text{kg/m}^3\), we find \(Q = \frac{m}{\rho} = \frac{3.11}{1000}\, \text{m}^3/\text{s}\). Substitute \(Q\) and \(A\) to find \(v\).
03

Calculate Radius and Area of Nozzle

The diameter of the nozzle is \(0.732\, \text{cm} = 0.00732\, \text{m}\). The radius is \(r = \frac{0.00732}{2} = 0.00366\, \text{m}\). Calculate the nozzle area: \(A_{nozzle} = \pi r^2 = \pi (0.00366)^2\, \text{m}^2\).
04

Calculate Speed of Water in Nozzle

Using the principle of continuity, the volumetric flow rate through the nozzle must equal the volumetric flow rate through the hose: \(Q = A_{hose} \cdot v_{hose} = A_{nozzle} \cdot v_{nozzle}\). Rearrange to find: \(v_{nozzle} = \frac{A_{hose} \cdot v_{hose}}{A_{nozzle}}\). Substitute known values to solve for \(v_{nozzle}\).
05

Evaluate Water Flow Rate in Nozzle

According to the principle of conservation of mass, the mass flow rate is the same throughout the system. Therefore, the number of kilograms per second of water flowing through the nozzle is equal to \(3.11\, \text{kg/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volumetric Flow Rate
Volumetric flow rate is a fundamental concept in fluid dynamics, essentially representing how much fluid flows through a cross-section of a pipe or channel per unit of time. It is often denoted by the symbol \( Q \) and is typically measured in cubic meters per second (\( \, ext{m}^3/ ext{s}\)).
\[ Q = A \cdot v \]
Here, \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the fluid flowing through it.
  • The volumetric flow rate tells us how fast a volume of fluid is moving, which can vary based on things like pipe size or fluid velocity.
  • This measurement is crucial for determining the design and efficiency of systems like water distribution networks, chemical processing, and HVAC systems.
In this exercise, calculating the volumetric flow rate helped determine the water speed in both the hose and the nozzle. While the actual speed may change due to the size of the pipe, the volume rate remains constant between connected sections if no fluid is added or removed.
Continuity Equation
The continuity equation is a principle derived from the conservation of mass, essential for understanding fluid flow in tubular systems. It states that, for an incompressible and non-leaking system, the mass flow rate must be constant throughout. Mathematically, it's expressed as:
\[ A_1v_1 = A_2v_2 \]
\( A_1 \) and \( A_2 \) are the cross-sectional areas at two different points, while \( v_1 \) and \( v_2 \) are the fluid velocities at these points.
  • This equation helps in deducing speed changes, explaining why water flows faster in narrower sections.
  • It assumes a steady-state flow where all velocities and areas are fixed over time.
Applying the continuity equation in the exercise, we found that while the area decreases as water moves from the hose to a smaller nozzle, velocity increases to maintain a constant flow rate.
Remember, both the nozzle and the hose transmit fluid at the same flow rate, due to the conservation of mass.
Conservation of Mass
The conservation of mass is one of the fundamental principles of physics, stating that mass in a system must remain constant over time as long as it is isolated from external influences. In fluid dynamics, this principle applies to continuous and steady flows.
  • In any closed system where no mass is added or taken away, the mass flow rate entering is equal to the mass flow rate exiting.
  • This ensures that in connected pipes or conduits, the same amount of mass passes through without accumulating at any point.
This concept was crucial in solving the exercise, affirming that the speed and flow rate stay consistent from the hose to the nozzle. Even if the velocity of the water changes, the mass flow per second remains constant at \( 3.11 \, ext{kg/s} \), exemplifying mass conservation in fluid dynamics.

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