Question

Water is flowing in fully developed conditions through a 3 -cm-diameter smooth tube with a mass flow rate of \(0.02 \mathrm{~kg} / \mathrm{s}\) at \(15^{\circ} \mathrm{C}\). Determine \((a)\) the maximum velocity of the flow in the tube and \((b)\) the pressure gradient for the flow.

Short answer

Answer: The maximum velocity of the flow in the tube is 5.62 m/s, and the pressure gradient for the flow is -30952 Pa/m.

Step-by-step solution

Convert given quantities to SI units

First, let's convert the given quantities into SI units, mainly the diameter of the tube, which is given in centimeters. Diameter: \(3 cm = 0.03 m\)

Write down given data

Now, let's write down all the given data in one place: - Diameter (D): \( \ensuremath{0.03 m}\) - Mass flow rate (m˙): \(\ensuremath{0.02 kg/s}\) - Temperature (T): \(\ensuremath{15^{\circ}C} = \ensuremath{288.15 K}\)

Determine the water properties

To obtain the water properties (density - \(\rho\) and dynamic viscosity - \(\mu\)) at the given temperature (\(15^{\circ}C\)), we can refer to a standard water properties table or use an online calculator. At \(15^{\circ}C\): - Density (\(\rho\)): \(\ensuremath{999 kg/m^3}\) - Dynamic viscosity (\(\mu\)): \(\ensuremath{1.13 * 10^{-3} \,Pa\cdot s}\)

Find the maximum velocity in the tube

In a fully developed flow, the velocity profile follows a parabolic shape where the maximum velocity (V_max) is at the center of the tube. To find the maximum velocity, we can use the mass flow rate and the parabolic velocity profile equation: $$V_{max} = \frac{2m˙}{\rho\,\pi\left(\frac{D}{2}\right)^2}$$ Plugging in the given values: $$V_{max} = \frac{2(0.02)}{999\pi \left(\frac{0.03}{2}\right)^2} = \ensuremath{5.62 m/s} $$ The maximum velocity of the flow in the tube is \(\ensuremath{5.62 m/s}\).

Find the pressure gradient for the flow

To determine the pressure gradient, we will use the Hagen-Poiseuille equation for fully developed flow in a smooth tube: $$ \frac{dP}{dx} = \frac{-32\mu Q}{\pi D^4}$$ We also know that the volume flow rate (Q) can be calculated from the mass flow rate (m˙) and density (\(\rho\)) as follows: $$ Q = \frac{m˙}{\rho}$$ By plugging in the given values, we can find the volume flow rate: $$ Q = \frac{0.02}{999} = \ensuremath{2*10^{-5} m^3/s} $$ Now, we can find the pressure gradient: $$ \frac{dP}{dx} = \frac{-32(1.13 * 10^{-3})(2*10^{-5})}{\pi (0.03)^4} = \ensuremath{-30952 Pa/m}$$ The pressure gradient for the flow is \(\ensuremath{-30952 Pa/m}\). To summarize: \((a)\) The maximum velocity of the flow in the tube is \(\ensuremath{5.62 m/s}\). \((b)\) The pressure gradient for the flow is \(\ensuremath{-30952 Pa/m}\).

Key concepts

Mass Flow Rate

Mass flow rate is a crucial concept in fluid dynamics. It tells us how much mass of a fluid passes through a specific point in a given time.
In our scenario with water, the mass flow rate is given as

• \(0.02 \text{ kg/s}\).
• This means that every second, 0.02 kilograms of water flows through the tube.

To understand mass flow rate better, imagine it as a conveyor belt where water molecules move past a point. The rate at which these molecules pass by determines the mass flow rate.
It's a product of the fluid density \( (\rho)\), the cross-sectional area of the tube \( (A)\), and the velocity \((v)\) of the water:\[m˙ = \rho A v\]This relationship shows how mass flow rate connects to other variables, like density and velocity. By knowing any two, we can find the third.

Pressure Gradient

The pressure gradient is like a driving force behind fluid flow. It describes how pressure changes along the direction of flow.
In simpler terms, it’s the difference in fluid pressure between two points over a distance.
This concept helps explain why fluids flow from high-pressure areas to low-pressure areas. It’s similar to how water flows downhill due to gravity.

• The pressure gradient is typically denoted as \( \frac{dP}{dx} \), where \( P \) is pressure and \( x \) is distance.
• In our problem, the pressure gradient was found to be \(-30952 \text{ Pa/m} \).
• This negative value shows that pressure decreases along the flow direction, pushing the water forward.

Knowing the pressure gradient helps to determine the rate of fluid flow, making it an essential concept in fluid dynamics.

Velocity Profile

The velocity profile describes how fluid speed varies across the tube’s diameter. For laminar flow, this profile is parabolic.
The fluid's center moves fastest, while the velocity decreases towards the tube walls.
This happens because fluid layers experience differing frictional forces.

• In the given exercise, the maximum velocity at the tube's center is
• \(5.62 \text{ m/s}\).

A fully developed flow implies that the velocity profile has settled into its most stable form.
The velocity at any point can be found using the maximum velocity and a mathematical function to describe the parabola:

• Each molecule has its own path and speed, combined to shape the parabolic profile.
• This relationship is vital for calculating how quickly a fluid travels through a pipe or tube.

Hagen-Poiseuille Equation

This equation is fundamental in explaining the flow of viscous fluids through a smooth tube.
Named after Hagen and Poiseuille, it connects the flow to various factors, such as:

• Tube radius,
• fluid viscosity,
• and pressure drop.

The Hagen-Poiseuille Equation is:\[\frac{dP}{dx} = \frac{-32\mu Q}{\pi D^4}\]which was used to find the pressure gradient in our example.
This equation is valid for laminar flows (not turbulent) and is a powerful tool for predicting how fluids behave under set conditions.

• It describes how changes in tube diameter or fluid viscosity can significantly impact the pressure and thus the flow rate.
• Understanding this equation helps engineers design efficient piping systems crucial for many industrial applications.