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10.5: Steady-Flow Devices

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    A steady-flow device is any device that will have a continuous flow of material through it. Some examples of steady-flow devices include pipes, nozzles, diffusers, and pumps. Generally, the material flowing through the device is a gas or liquid, and if the device in any way changes the velocity of the fluid then that fluid will exert a force on the steady flow device in return.

    A team of firefighters operating a fire hose in a training exercise.
    Figure \(\PageIndex{1}\): The nozzle on the fire hose is an example of a steady-flow device. Because the nozzle changes the velocity of the water as it exits the hose, it will take a force to hold the nozzle in place.

    In order to determine the forces at play on a steady-flow device, we will start with our impulse-momentum equation.

    \[ \vec{J} = m \vec{v}_f - m \vec{v}_i \]

    Because this is a continuous process, it doesn't really make sense to have initial and final velocities. Instead, we will have inlet and outlet velocities. Also, the mass will need to be changed to the mass flow rate (the mass entering or leaving the device per unit time) to deal with the continuous-flow nature of the system.

    A symmetrical trapezoid with the longer base on the left side. This wider side represents the entering mass flow rate and inlet velocity, which is changed by the operation of the device to a different exiting mass flow rate and outlet velocity (which exits the device through the narrower base of the right side of the trapezoid).
    Figure \(\PageIndex{2}\): The mass flow rate is a more appropriate measure for steady-flow devices than standard mass.

    Dividing our initial impulse-momentum equation by time on both sides will give us the desired mass flow rate on the right, while the time on the left will cancel out the time component of the impulse.

    \[ \frac{\vec{F} * t}{t} = \frac{m_{out}}{t} \vec{v}_{out} - \frac{m_{in}}{t} \vec{v}_{in} \]

    Simplifying this equation, we will arrive at our final equation, which relates the force our steady-flow device exerts on the fluid to the mass flow rates and velocities at the inlet and outlet. The force the fluid exerts on the device would simply be equal and opposite to the force below.

    \[ \vec{F} = \dot{m}_{out} \vec{v}_{out} - \dot{m}_{in} \vec{v}_{in} \]

    One final note is that these equations are vector equations. If the device is changing the direction of the flow of a fluid you will need to break the force and velocities into \(x\) and \(y\) components and split the above equation into \(x\) and \(y\) components.

    Finding Mass Flow Rate:

    If the mass flow rate in or out of your device is not given directly, you may need to find those values. First we can use a simple identity: we know the mass flow rate will be equal to the density of the fluid \((\rho)\) times the volumetric flow rate. Furthermore, the volumetric flow rate can be related to the geometry of the device, in that it will be equal to the average velocity of the fluid at the inlet or outlet times the cross-sectional area at the inlet or outlet. Putting this all together, we arrive at the following formula.

    \[ \dot{m} = \rho \dot{V} = \rho \vec{v} A \]

    Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/R8A35UeoNyE.

    Example \(\PageIndex{1}\)

    A firefighter supports a hose as shown below. The hose has a volumetric flow rate of 60 gal/min and the nozzle reduces in diameter from 4 cm to 2 cm. What force will the firefighter have to exert, in Newtons, to keep the hose in place?

    Two firefighters hold the body of a fire hose as it sprays water, while a third firefighter controls the direction of the nozzle of the hose.
    Figure \(\PageIndex{3}\): A firefighter supporting the nozzle of a hose. Photo by Macomb Paynes, CC BY-NC-SA 2.0.
    Solution
    Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/rl4A1N_BNGA.

    Example \(\PageIndex{2}\)

    A 90-degree elbow joint redirects the flow along a pipe of diameter 3 cm. If water (density=1000 kg/m3) is traveling through the pipe with an average speed of 5 m/s, what is the magnitude and direction of the force the water exerts on the elbow joint?

    A 90° elbow joint connects one length of pipe that stretches from the lower left of the image to the upper left to a second length of pipe stretching from the upper left of the image to the upper right. Water flows into this assembly at the opening in the lower left, and exits at the opening in the upper right.
    Figure \(\PageIndex{4}\): problem diagram for Example \(\PageIndex{2}\). An inverted-L-shaped assembly of two pipe segments connected by an elbow joint has water entering through the bottom left opening and leaving through the upper right opening.
    Solution
    Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/znKq2quYMWk.

    This page titled 10.5: Steady-Flow Devices is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jacob Moore & Contributors (Mechanics Map) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.