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AP Physics 2 : Flow Rate

Study concepts, example questions & explanations for ap physics 2, all ap physics 2 resources, example questions, example question #81 : fluids.

solving flow rate problems

The velocity of the water can be determined from the following formula:

solving flow rate problems

We need to calculate the volumetric flow rate and the cross-sectional area. For the flow rate:

solving flow rate problems

Rearrange to solve for volumetric flow rate:

solving flow rate problems

Next, calculate cross-sectional area:

solving flow rate problems

Now we can solve for the velocity:

solving flow rate problems

Example Question #2 : Flow Rate

solving flow rate problems

To find the answer to this question, we'll need to use the continuity equation to determine the flow rate, which will be the same in both pipes.

solving flow rate problems

We'll also need to calculate the area of the pipe using the equation:

solving flow rate problems

Use the continuity equation for incompressible fluids. 

solving flow rate problems

The cross sectional area of the garden hose at both ends are circular regions. Rewrite the equation replacing areas with the formula for an area of a circle and solve for the velocity at the second point.

solving flow rate problems

Example Question #4 : Flow Rate

solving flow rate problems

This is a volume flow rate problem. Because the water is an incompressible fluid, we can apply the flow rate equation:

solving flow rate problems

Find the surface area of the pond:

solving flow rate problems

Substitute into the flow rate equation:

solving flow rate problems

Example Question #5 : Flow Rate

solving flow rate problems

When the area halves, the velocity of the fluid will double. However, the volume flow rate (the product of these two quantities) will remain the same. In other words, the volume of water flowing through location 1 per second is the same as the volume of water flowing through location 2 per second. 

Example Question #6 : Flow Rate

solving flow rate problems

The volumetric flow rate of fluid is found using the equation:

solving flow rate problems

In this problem the cross-section of the pipe is a circle. The area of the cross-section is:

solving flow rate problems

The volumetric flow rate is:

solving flow rate problems

Example Question #7 : Flow Rate

solving flow rate problems

In this problem the cross-section of the pipe is a square. The area of the cross-section is:

solving flow rate problems

Example Question #8 : Flow Rate

solving flow rate problems

In this problem, the cross-section of the pipe is a circle, which is

solving flow rate problems

The area of the exit cross-section is:

solving flow rate problems

Plug in these variables into the continuity equation and solve:

solving flow rate problems

Example Question #9 : Flow Rate

solving flow rate problems

In this problem,

The cross-section of the pipe is a circle, which is:

solving flow rate problems

Plugging our variables into the continuity equation gives us

solving flow rate problems

Example Question #1 : Flow Rate

solving flow rate problems

The fluid in both sections of the pipe have the same cross-sectional area

The fluid in both sections of the pipe are not related

The fluid in both sections of the pipe have the same velocity

The fluid in both sections of the pipe have the same volumetric flow rate

Based on the continuity equation, both parts of the flow must have the same volumetric flow rate.

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Physics LibreTexts

14.7: Fluid Dynamics

  • Last updated
  • Save as PDF
  • Page ID 4059

Learning Objectives

  • Describe the characteristics of flow
  • Calculate flow rate
  • Describe the relationship between flow rate and velocity
  • Explain the consequences of the equation of continuity to the conservation of mass

The first part of this chapter dealt with fluid statics, the study of fluids at rest. The rest of this chapter deals with fluid dynamics, the study of fluids in motion. Even the most basic forms of fluid motion can be quite complex. For this reason, we limit our investigation to ideal fluids in many of the examples. An ideal fluid is a fluid with negligible viscosity . Viscosity is a measure of the internal friction in a fluid; we examine it in more detail in Viscosity and Turbulence . In a few examples, we examine an incompressible fluid—one for which an extremely large force is required to change the volume—since the density in an incompressible fluid is constant throughout.

Characteristics of Flow

Velocity vectors are often used to illustrate fluid motion in applications like meteorology. For example, wind—the fluid motion of air in the atmosphere—can be represented by vectors indicating the speed and direction of the wind at any given point on a map. Figure \(\PageIndex{1}\) shows velocity vectors describing the winds during Hurricane Arthur in 2014.

Figure is a pressure map of Hurricane Arthur traveling up the East Coast. The low pressure center is indicated as the blue dot. Wind speed is highest near the low pressure center with the winds moving in a counterclockwise direction around it.

Another method for representing fluid motion is a streamline . A streamline represents the path of a small volume of fluid as it flows. The velocity is always tangential to the streamline. The diagrams in Figure \(\PageIndex{2}\) use streamlines to illustrate two examples of fluids moving through a pipe. The first fluid exhibits a laminar flow (sometimes described as a steady flow), represented by smooth, parallel streamlines. Note that in the example shown in part (a), the velocity of the fluid is greatest in the center and decreases near the walls of the pipe due to the viscosity of the fluid and friction between the pipe walls and the fluid. This is a special case of laminar flow, where the friction between the pipe and the fluid is high, known as no slip boundary conditions. The second diagram represents turbulent flow , in which streamlines are irregular and change over time. In turbulent flow, the paths of the fluid flow are irregular as different parts of the fluid mix together or form small circular regions that resemble whirlpools. This can occur when the speed of the fluid reaches a certain critical speed.

Figure A is the schematic of the laminar flow shown as layers of fluid moving in parallel lines. Figure B is the schematics of the turbulent flow shown as layers of fluid moving in irregular, colliding paths.

Flow Rate and its Relation to Velocity

The volume of fluid passing by a given location through an area during a period of time is called flow rate \(Q\), or more precisely, volume flow rate. In symbols, this is written as

\[Q = \frac{dV}{dt} \label{14.13}\]

where \(V\) is the volume and \(t\) is the elapsed time. In Figure \(\PageIndex{3}\), the volume of the cylinder is \(Ax\), so the flow rate is

\[Q = \frac{dV}{dt} = \frac{d}{dt} (Ax) = A \frac{dx}{dt} = Av \ldotp\]

The SI unit for flow rate is m 3 /s, but several other units for \(Q\) are in common use, such as liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters (10 −3 m 3 or 10 3 cm 3 ).

Figure is a schematic of a uniform pipeline with the cross-section area A. Fluid flows through the pipeline. Volume of fluid V passes a point P in time t.

Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, consider the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size and shape of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. Figure \(\PageIndex{3}\) illustrates the volume flow rate. The volume flow rate is \(Q = \frac{dV}{dt} = Av\), where A is the cross-sectional area of the pipe and v is the magnitude of the velocity.

The precise relationship between flow rate \(Q\) and average speed \(v\) is

\[Q = Av,\]

where \(A\) is the cross-sectional area and \(v\) is the average speed. The relationship tells us that flow rate is directly proportional to both the average speed of the fluid and the cross-sectional area of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. Figure \(\PageIndex{3}\) illustrates how this relationship is obtained. The shaded cylinder has a volume \(V = Ad\), which flows past the point \(P\) in a time \(t\). Dividing both sides of this relationship by \(t\) gives

\[\frac{V}{t} = \frac{Ad}{t} \ldotp \label{eq14.14A} \]

We note that \(Q = \frac{V}{t}\) and the average speed is \(v = \frac{d}{t}\). Thus Equation \ref{eq14.14A} becomes

\[Q = Av.\]

Figure \(\PageIndex{4}\) shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. The flow is continuous because they are no sources or sinks that add or remove mass, so the mass flowing into the pipe must be equal the mass flowing out of the pipe. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for arbitrary points 1 and 2,

\[\begin{split} Q_{1} & = Q_{2}, \\ A_{1} v_{1} & = A_{2} v_{2} \ldotp \end{split} \label{14.14}\]

This is called the equation of continuity and is valid for any incompressible fluid (with constant density). The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: It emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.

Figure is the schematic of a pipeline that narrows from the cross section area A1 to the cross section area A2. Fluid flows through the pipeline. Volume of fluid V1 passes a point 1, located at the wide cross-section part, in time t. Volume of fluid V2 passes a point 2, located in the narrow cross-section part, in time t.

Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, so the equation must be applied with caution to gases if they are subjected to compression or expansion.

Example 14.5: Calculating Fluid Speed through a nozzle

A nozzle with a diameter of 0.500 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water:

  • in the hose and
  • in the nozzle.

We can use the relationship between flow rate and speed to find both speeds. We use the subscript 1 for the hose and 2 for the nozzle.

  • We solve the flow rate equation for speed and use \(\pi r_{1}^{2}\) for the cross-sectional area of the hose, obtaining $$v = \frac{Q}{A} = \frac{Q}{\pi r_{1}^{2}} \ldotp$$ Substituting values and using appropriate unit conversions yields $$v = \frac{(0.500\; L/s)(10^{-3}\; m^{3}/L)}{(3.14)(9.00 \times 10^{-3}\; m)^{2}} = 1.96\; m/s \ldotp$$
  • We could repeat this calculation to find the speed in the nozzle v 2 , but we use the equation of continuity to give a somewhat different insight. The equation states $$A_{1} v_{1} = A_{2} v_{2} \ldotp$$Solving for v 2 and substituting \(\pi r^{2}\) for the cross-sectional area yields $$v_{2} = \frac{A_{1}}{A_{2}} v_{1} = \frac{\pi r_{1}^{2}}{\pi r_{2}^{2}} v_{1} = \frac{r_{1}^{2}}{r_{2}^{2}} v_{1} \ldotp$$ Substituting known values, $$v_{2} = \frac{(0.900\; cm)^{2}}{(0.250\; cm)^{2}} (1.96\; m/s) = 25.5\; m/s \ldotp$$

Significance

A speed of 1.96 m/s is about right for water emerging from a hose with no nozzle. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.

The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.

Mass Conservation

The rate of flow of a fluid can also be described by the mass flow rate or mass rate of flow. This is the rate at which a mass of the fluid moves past a point. Refer once again to Figure \(\PageIndex{3}\), but this time consider the mass in the shaded volume. The mass can be determined from the density and the volume:

\[m = \rho V = \rho Ax \ldotp\]

The mass flow rate is then

\[\frac{dm}{dt} = \frac{d}{dt} (\rho Ax) = \rho A \frac{dx}{dt} = \rho Av,\]

where \(\rho\) is the density, A is the cross-sectional area, and v is the magnitude of the velocity. The mass flow rate is an important quantity in fluid dynamics and can be used to solve many problems. Consider Figure \(\PageIndex{5}\). The pipe in the figure starts at the inlet with a cross sectional area of A 1 and constricts to an outlet with a smaller cross sectional area of A 2 . The mass of fluid entering the pipe has to be equal to the mass of fluid leaving the pipe. For this reason the velocity at the outlet (v 2 ) is greater than the velocity of the inlet (v 1 ). Using the fact that the mass of fluid entering the pipe must be equal to the mass of fluid exiting the pipe, we can find a relationship between the velocity and the cross-sectional area by taking the rate of change of the mass in and the mass out:

\[\begin{split} \left(\dfrac{dm}{dt}\right)_{1} & = \left(\dfrac{dm}{dt}\right)_{2} \\ \rho_{1} A_{1} v_{1} & = \rho_{2} A_{2} v_{2} \ldotp \end{split} \label{14.15}\]

Equation \ref{14.15} is also known as the continuity equation in general form. If the density of the fluid remains constant through the constriction—that is, the fluid is incompressible—then the density cancels from the continuity equation,

\[A_{1} v_{1} = A_{2} v_{2} \ldotp\]

The equation reduces to show that the volume flow rate into the pipe equals the volume flow rate out of the pipe.

Figure is a schematic of fluid flowing in a uniform pipeline with the cross-section area A. Volume of fluid V delta t passes the pipeline during the time delta t.

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Physics library

Course: physics library   >   unit 9, volume flow rate and equation of continuity.

  • What is volume flow rate?
  • Bernoulli's equation derivation part 1
  • Bernoulli's equation derivation part 2
  • Finding fluid speed exiting hole
  • More on finding fluid speed from hole
  • Finding flow rate from Bernoulli's equation
  • What is Bernoulli's equation?
  • Viscosity and Poiseuille flow
  • Turbulence at high velocities and Reynold's number
  • Venturi effect and Pitot tubes
  • Surface Tension and Adhesion

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Great Answer

Video transcript

One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

This topic is relevant for:

GCSE Maths

Here we will learn about flow rate, including how to calculate it and how to use it to solve problems involving volume and capacity.

There are also flow rate worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is flow rate?

Flow rate is the term used to describe the rate at which a fluid or substance flows into or out of an object.

To calculate flow rate we can use information about the change in capacity (or volume) of the substance and the amount of time taken for that change to occur.

It is important in GCSE mathematics questions to look at the unit of time being used and whether the question is referring to a volume of fluid. The flow rate could be given as a compound measure such as depth per second or volume per minute.

Typical GCSE mathematics questions involve having to find the time taken for a shape to be filled or for the depth of a container to reach a specific height.

For example,

A tap is used to fill a container in the shape of a cuboid measuring 1.5 \ m by 2 \ m by 0.4 \ m. The tap releases water at a flow rate of 5 litres per minute. 

Find the time taken for the container to be filled.

flow rate image 1

Rate of output

We measure flow rate in metres cubed per second (m^3/s) , or in litres per second (l/s) . Litres are the more common measures used for liquid volume. We also need to remember the conversion factor 1m ^{3} = 1000 litres. The rate of output is the amount of time it takes to discharge an amount of liquid.

The diagram shows a swimming pool.

solving flow rate problems

The swimming pool is empty and is then filled with water at a constant rate of 25 litres per minute. 

Work out how long it will take for the swimming pool to be completely full of water. Give your answer in hours. 

We need to find the volume of the swimming pool, so we can find out the amount of water it can hold. 

Volume of the swimming pool = 1.8m ^{3}

Using the conversion factor 1m ^{3} = 1000 Litres, the swimming pool holds 1800 litres of water. 

If the pool is filled with water at a constant rate of 25 litres per minute, this is our rate of output .

1800 ÷ 25 = 72 minutes 

It takes 72 minutes to fill the pool. 

We are asked to give our answer in hours, 72 minutes = 1 hour 12 minutes or 1.2 hours. 

How to solve flow rate problems

In order to solve flow rate problems:

  • Use information provided to calculate the flow rate. \text{Flow rate}=\frac{\text{change in volume}}{\text{time taken}}

Use information provided to calculate volume that will be changing.

Calculate the required value.

Explain how to solve flow rate problems

Explain how to solve flow rate problems

Flow rate worksheet

Get your free flow rate worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on   compound measures

Flow rate  is part of our series of lessons to support revision on  compound measures . You may find it helpful to start with the main compound measures lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

  • Compound measures
  • Mass density volume
  • Population density
  • Pressure force area
  • Speed distance time

Flow rate examples

Example 1: finding the time taken to fill a container.

Water flowing from a hose pipe fills a 10 litre bucket in 2 minutes. Find the time it would take to fill a container in the shape of a cuboid measuring 2 metres by 1 metre by 50 \ cm.

Use information provided to calculate the flow rate.

The flow rate of the water is 10 \div 2 = 5 litres per minute.

2 Use information provided to calculate volume that will be changing.

Volume of the cuboid = 2 \times 1 \times 0.5 = 1 m^{3}.

3 Calculate the required value.

We can work out the time by dividing the volume by the flow rate.

We need to convert the volume so that the units fit in with the volume.

So the time will be 200 minutes or 3 hours 20 minutes.

Example 2: finding the volume after an allotted time

A tap has been left running. Water flows from the tap at a rate of 250 \ cm^3 \ per \ minute.

How many litres of water will come out of the tap in half an hour?

We have been given the flow rate, 250 \ cm^3 \ per \ minute.

Here we need to find the volume. But we have been given a time – half an hour is 30 minutes.

We need to adapt the flow rate formula,

\text{Volume}=\text{flow rate}\times \text{time taken}

\begin{aligned} \text{Volume}&=\text{flow rate}\times \text{time taken}\\\\ \text{Volume}&= 250\times 30 \\\\ \text{Volume}&=7500 cm^3 \end{aligned}

Example 3: finding the height after an allotted time

Water flowing from a tap fills a 500 \ ml jug in 80 seconds. 

The same tap will be used to fill this cylindrical container, with a radius of 12 \ cm.

flow rate example 3 image 1

Find the height of the water, h , after 3 minutes.

We will be finding the volume of the cylinder in cm^{3}, so we will convert to flow rate so that the units match.

500 \ ml = 500 \ cm^3

The flow rate of the water is 500 \div 80 =6.25 \ cm^3 per second.

The cross-sectional area of the cylinder is a circle.

The volume of the cylinder of water will be \pi \times 12^2 \times h=144\pi h \ cm^3 .

We need to calculate the volume.

The flow rate is in cm^3 per second, so we need to convert the time to seconds to fit.

3 minutes = 180 seconds

\begin{aligned} \text{Volume}&=\text{flow rate}\times \text{time taken}\\\\ 144 \pi h &= 6.25\times 180 \\\\ h &= \frac{6.25\times 180}{144 \pi}\\\\ h&=2.4867...\\\\ h&=2.49 \ \text {cm} \ \text{(2 dp)} \end{aligned}

Example 4: finding the time taken for a container to empty

A container full of water is found to have a leak. 

The container is in the shape of a trapezium prism shown below and water is leaking from point P.

solving flow rate problems

4 minutes after being completely full, the water level in the container has dropped by 2 \ cm.

Assuming the flow rate of the leak is constant, find the total time for all the water to leak out of the container. 

As the shape of the container is a trapezium, use the volume of the water lost to find the flow rate.

The volume lost = 10 \times 12 \times 2 = 240 \ cm^{3}.

The flow rate of the water is 240 \div 4 = 60 \ cm^3 per minute.

The cross-sectional area of the container is a trapezium.

The volume of the trapezium = \frac{20 + 16}{2} \times 10 \times 12=2160 \ cm^{3}.

This is the volume of the full tank.

2160 \div 60 = 36 minutes

The tank would take a total of 36 minutes to empty.

Common misconceptions

  • Forgetting to convert units of volume or time

It is important to ensure the correct units are being used throughout a problem. If the flow rate is given in cm^3 per minute it is important to make sure all measurements are converted to centimetres and that time is given in minutes.

Practice flow rate questions

1. A tap can fill a 500 \ ml jug in 8 seconds. Calculate the flow rate, giving the units.

0.0625 \ ml per second

GCSE Quiz False

62.5 \ ml per second

GCSE Quiz True

0.016 \ ml per second

62.5 seconds per ml

Use the formula,

2. A tap produces a flow rate of 30 \ cm^3 per second. How long will it take to fill a container with a volume of 195 \ cm^{3}.

6.5 seconds

5850 seconds

0.154 seconds (to 3 dp)

195 seconds

and rearrange to find the time taken.

3. A hose pipe is used to fill a water butt. Water flows at a rate of 3.5 litres per minute and fills the water butt in 12 minutes and 52 seconds. Calculate the volume of the water butt to the nearest litre.

Convert the time to decimalised form, then use the formula

and rearrange to find the volume.

The volume will be 45 litres, to the nearest litre.

4. A hose can produce a flow rate of 6 cubic metres per hour. How long would it take the same hose to fill the swimming pool below? The cross-section of the swimming pool is a trapezium. Give your answer in hours and minutes.

flow rate practice question 4

8 hours 45 minutes

8 hours 15 minutes

8 hours 25 minutes

The volume of the swimming pool is

\text{Volume}=\frac{1}{2}\times (4+1.5)\times 6\times 3=49.5 .

Using the formula

we rearrange to find the time taken.

So the time is 8.25 hours or 8 hours 15 minutes.

5. A swimming pool has dimensions shown in the diagram below. It took 19 hours 12 minutes to fill the swimming pool from a hose. Find the flow rate of the water from the hose in m^3 per hour.

flow rate practice question 5

5 \ m^3 per hour

5.02 \ m^3 per hour ( 2 d.p.)

5.83 \ m^3 per hour ( 2 d.p.)

5.86 \ m^3 per hour ( 2 d.p.)

Find the volume of the swimming pool.

Then use the formula,

6. A container in the shape of a cone with radius 6 \ cm and height 27 \ cm , is leaking water from its apex.

flow rate practice question 6

The container is originally full and after 20 minutes the depth of water has reduced by 9 \ cm. Calculate the flow rate of the leak in cm^3 per minute. Give your answer to 3 significant figures.

39.7 \ cm^3 per minute (to 3 sf)

35.8 \ cm^3 per minute (to 3 sf)

29.3 \ cm^3 per minute (to 3 sf)

24.7 \ cm^3 per minute (to 3 sf)

Use similar triangles to find the radius of the reduced cone.

flow rate practice question 6 explanation image 1

Find the volume of the small cone and the large cone and then subtract the volumes to find the change in volume.

flow rate practice question 6 explanation image 2

Then we can find the flow rate.

flow rate practice question 6 explanation image 3

Flow rate GCSE questions

1. The diagram shows a container in the shape of the cuboid. The base of the container is a rectangle measuring 32 \ cm by 24 \ cm.

flow rate gcse question 1

A tap produces a flow rate of 4 litres per minute.

The tap can fill the container in 4 minutes and 48 seconds.

Using the fact that 1 \ litre = 1000 \ cm^3 , find the height of the container in centimetres.

2. The organiser of a fun run needs to provide 120 cups of water for competitors to drink from during a race. Each cup is a cylinder of radius 4 \ cm and height 10 \ cm as shown in the diagram.

flow rate gcse question 2

The organiser needs to fill all 120 cups with water from a tap which has a flow rate of 3.5 litres per minute. There is only 16 minutes available to get all of the cups filled.

Will the organiser get all of the cups filled in the available time? You must show your workings. Use the fact that 1 \ litre = 1000 \ cm^{3}.

Volume of 1 cup = \pi \times {{4}^{2}}\times 10=502.65… cm^3

Total volume = 502.65… \times 120= 60318.5 \ cm^3

Finding time by dividing by 3500 or equivalent = 17.23… minutes

Answer of ‘no’ with correct workings

3. The image shows a full swimming pool that needs cleaning, so will be emptied by a pump. After 30 minutes the water level has decreased by 40 \ cm.

flow rate gcse question 3

How much extra time will be required for the swimming pool to be completely empty? Give your answer in minutes.

Volume of decrease = 40 \times 500 \times 700=14 \ 000 \ 000 \ cm^3

Flow rate =14 \ 000 \ 000 \ cm^3 \div 30=466 \ 666.666 … \ cm^3 \ \text{per minute}

Remaining volume

Area of cross-section =(120\times 700)+(\frac{1}{2}\times 500 \times 140)=119 \ 000 \ cm^2

Volume =119 \ 000 \times 500 = 59 \ 500 \ 000 \ cm^3

Remaining time =59 \ 500 \ 000\div 466 \ 666.6…=125.5 \ \text{minutes}

Alternatively

Volume of decrease = 0.4 \times 5 \times 7=14 \ cm^3

Flow rate =14 \ m^3 \div 30=0.466666 … \ m^3 \ \text{per minute}

Area of cross-section =(1.2\times 7)+(\frac{1}{2}\times 5 \times 1.4)=11.9 \ m^2

Volume =11.9 \times 5 = 59.5 \ m^3

Remaining time =59.5\div 0.466666…=125.5 \ \text{minutes}

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  • Physics Formulas

Flow Rate Formula

As we all know, liquids flow. One can say how slow or fast it flows. But can someone tell at what rate Fluid flows? To compute that let’s study the flow rate. The flow rate of a liquid is how much fluid passes through an area in a particular time. Flow rate can be articulated in either in terms of velocity and cross-sectional area, or time and volume. As liquids are incompressible, the rate of flow into an area must be equivalent to the rate of flow out of an area. This is identified as the equation of continuity.

Flow rate is the quantity of fluid flowing in the specified time. It is expressed in litres per meter  (lpm) or gallons per metre  (gpm)  . It is articulated as

Flow rate formula

Where, the flow area is A and

the flow velocity is v.

Flow rate can also be articulated as in a given time  (t) the capacity of fluid stored  (C) . It is also articulated as

Flow rate in terms of capacity

Where, the capacity of fluid stored is C and the time taken to flow is t

Flow rate formula  has extensive applications in fluid dynamics to compute the velocity, area or flow rate.

Flow Rate Solved Examples

Provided underneath are the questions grounded on flow rate which may be useful for you.

Problem 1:  Compute the flow rate of fluid if it is moving with the velocity of 20 m/s through a tube of diameter 0.03 m. Answer:

Velocity of fluid flow v =20m/s

Diameter of pipe d=0.03m

Area of cross-section formula

A={(3.14)/4}(0.03)(0.03)

A =(0.785)(0.0009)

solving flow rate problems

Flow rate is given by Q = vA=(20)(0.000706)

Flow Rate Formula-1

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  • 14.6 Bernoulli’s Equation
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Explain the terms in Bernoulli’s equation
  • Explain how Bernoulli’s equation is related to the conservation of energy
  • Describe how to derive Bernoulli’s principle from Bernoulli’s equation
  • Perform calculations using Bernoulli’s principle
  • Describe some applications of Bernoulli’s principle

As we showed in Figure 14.27 , when a fluid flows into a narrower channel, its speed increases. That means its kinetic energy also increases. The increased kinetic energy comes from the net work done on the fluid to push it into the channel. Also, if the fluid changes vertical position, work is done on the fluid by the gravitational force.

A pressure difference occurs when the channel narrows. This pressure difference results in a net force on the fluid because the pressure times the area equals the force, and this net force does work. Recall the work-energy theorem,

The net work done increases the fluid’s kinetic energy. As a result, the pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube.

There are many common examples of pressure dropping in rapidly moving fluids. For instance, shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The reason is that the high-velocity stream of water and air creates a region of lower pressure inside the shower, whereas the pressure on the other side remains at the standard atmospheric pressure. This pressure difference results in a net force, pushing the curtain inward. Similarly, when a car passes a truck on the highway, the two vehicles seem to pull toward each other. The reason is the same: The high velocity of the air between the car and the truck creates a region of lower pressure between the vehicles, and they are pushed together by greater pressure on the outside ( Figure 14.29 ). This effect was observed as far back as the mid-1800s, when it was found that trains passing in opposite directions tipped precariously toward one another.

Energy Conservation and Bernoulli’s Equation

The application of the principle of conservation of energy to frictionless laminar flow leads to a very useful relation between pressure and flow speed in a fluid. This relation is called Bernoulli’s equation , named after Daniel Bernoulli (1700–1782), who published his studies on fluid motion in his book Hydrodynamica (1738).

Consider an incompressible fluid flowing through a pipe that has a varying diameter and height, as shown in Figure 14.30 . Subscripts 1 and 2 in the figure denote two locations along the pipe and illustrate the relationships between the areas of the cross sections A , the speed of flow v , the height from ground y , and the pressure p at each point. We assume here that the density at the two points is the same—therefore, density is denoted by ρ ρ without any subscripts—and since the fluid is incompressible, the shaded volumes must be equal.

We also assume that there are no viscous forces in the fluid, so the energy of any part of the fluid will be conserved. To derive Bernoulli’s equation, we first calculate the work that was done on the fluid:

The work done was due to the conservative force of gravity and the change in the kinetic energy of the fluid. The change in the kinetic energy of the fluid is equal to

The change in potential energy is

The energy equation then becomes

Rearranging the equation gives Bernoulli’s equation:

This relation states that the mechanical energy of any part of the fluid changes as a result of the work done by the fluid external to that part, due to varying pressure along the way. Since the two points were chosen arbitrarily, we can write Bernoulli’s equation more generally as a conservation principle along the flow.

Bernoulli’s Equation

For an incompressible, frictionless fluid, the combination of pressure and the sum of kinetic and potential energy densities is constant not only over time, but also along a streamline:

A special note must be made here of the fact that in a dynamic situation, the pressures at the same height in different parts of the fluid may be different if they have different speeds of flow.

Analyzing Bernoulli’s Equation

According to Bernoulli’s equation, if we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, let us consider some specific situations that simplify and illustrate its use and meaning.

Bernoulli’s equation for static fluids

First consider the very simple situation where the fluid is static—that is, v 1 = v 2 = 0 . v 1 = v 2 = 0 . Bernoulli’s equation in that case is

We can further simplify the equation by setting h 2 = 0 . h 2 = 0 . (Any height can be chosen for a reference height of zero, as is often done for other situations involving gravitational force, making all other heights relative.) In this case, we get

This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h 1 h 1 , and consequently, p 2 p 2 is greater than p 1 p 1 by an amount ρ g h 1 ρ g h 1 . In the very simplest case, p 1 p 1 is zero at the top of the fluid, and we get the familiar relationship p = ρ g h p = ρ g h . ( Recall that p = ρ g h ( Recall that p = ρ g h and Δ U g = − m g h . ) Δ U g = − m g h . ) Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid is ρ g h ρ g h . Although we introduce Bernoulli’s equation for fluid motion, it includes much of what we studied for static fluids earlier.

Bernoulli’s principle

Suppose a fluid is moving but its depth is constant—that is, h 1 = h 2 h 1 = h 2 . Under this condition, Bernoulli’s equation becomes

Situations in which fluid flows at a constant depth are so common that this equation is often also called Bernoulli’s principle , which is simply Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) Bernoulli’s principle reinforces the fact that pressure drops as speed increases in a moving fluid: If v 2 v 2 is greater than v 1 v 1 in the equation, then p 2 p 2 must be less than p 1 p 1 for the equality to hold.

Example 14.6

Calculating pressure.

Substituting known values,

Significance

Applications of bernoulli’s principle.

Many devices and situations occur in which fluid flows at a constant height and thus can be analyzed with Bernoulli’s principle.

Entrainment

People have long put the Bernoulli principle to work by using reduced pressure in high-velocity fluids to move things about. With a higher pressure on the outside, the high-velocity fluid forces other fluids into the stream. This process is called entrainment . Entrainment devices have been in use since ancient times as pumps to raise water to small heights, as is necessary for draining swamps, fields, or other low-lying areas. Some other devices that use the concept of entrainment are shown in Figure 14.31 .

Velocity measurement

Figure 14.32 shows two devices that apply Bernoulli’s principle to measure fluid velocity. The manometer in part (a) is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity ( v 1 = 0 v 1 = 0 ) in front of it, while fluid passing the other tube has velocity v 2 v 2 . This means that Bernoulli’s principle as stated in

Thus pressure p 2 p 2 over the second opening is reduced by 1 2 ρ v 2 2 1 2 ρ v 2 2 , so the fluid in the manometer rises by h on the side connected to the second opening, where

(Recall that the symbol ∝ ∝ means “proportional to.”) Solving for v 2 v 2 , we see that

Part (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air-speed indicators in aircraft.

A fire hose

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.

Example 14.7

Calculating pressure: a fire hose nozzle.

where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds v 1 v 1 and v 2 v 2 . Since Q = A 1 v 1 Q = A 1 v 1 , we get

Similarly, we find

This rather large speed is helpful in reaching the fire. Now, taking h 1 h 1 to be zero, we solve Bernoulli’s equation for p 2 p 2 :

Substituting known values yields

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IV Flow Rate Calculation NCLEX Reviewer & Practice Questions (60 Items)

IV Flow Rate Calculations Nursing Test Banks for NCLEX RN

Welcome to your free NCLEX reviewer and practice questions quiz for IV flow rate calculations and formulas. This quiz aims to help student nurses review and test their competence in the intravenous flow rate calculation.

IV Flow Rate Calculation Nursing Test Bank

The nursing test bank for IV flow rate calculations below is separated into two sets of quizzes. Included topics are IV flow rate calculation, calculating for drops per minute, calculating for milliliters per hour, and total infusion time. If you need a quick review, please read the IV flow rate reviewer below.

Quiz guidelines:

  • Take Your Time: No rush! Read each question thoroughly and pick the best answer at your own pace.
  • Complete the Quiz: Ensure you answer ALL questions. Only after you’ve answered every item will the score and rationales be available.
  • Review Before Submitting: Once you’re done, you’ll be directed to the Quiz Summary . Double-check your answers or see if you’ve missed any before clicking the Finish Quiz button.
  • Learn from Rationales: After submitting, click View Questions to understand the explanation for each answer.
  • Share Your Thoughts: We’d love your feedback, scores, and questions! Please share them in the comments below.
  • Free access. Guess what? Our test banks are 100% FREE. Skip the hassle – no sign-ups or registrations here. A sincere promise from Nurseslabs: we have not and won’t ever request your credit card details or personal info for our practice questions. We’re dedicated to keeping this service accessible and cost-free, especially for our amazing students and nurses. So, take the leap and elevate your career hassle-free!

Quizzes included in this guide are:

IV Flow Rate Calculations Reviewer & Formulas

The IV flow rate study guide below will help refresh your memory on calculating IV flow rates, including a refresher on the formulas and nursing considerations when maintaining IV therapy.

For more information about IV fluids , visit our IV Fluids and Solutions Guide and Cheat Sheet

How to Calculate IV Flow Rate?

  • To calculate IV flow rates, the nurse must know the total volume of fluid to be infused and the specific time for the infusion. 
  • Intravenously administered fluids are prescribed most frequently based on milliliters per hour to be administered. The volume per hour prescribed is administered by setting the flow rate, which is counted in drops per minute.
  • Milliliters per hour (mL/h) . Calculated by dividing the total infusion volume by the total infusion time in hours
  • Number of drops per one (1) minute (gtts/min). Calculated by multiplying the total infusion volume to the drop factor and then dividing by the total infusion time in minutes. 
  • Infusion time. Total volume to infuse divided by milliliters per hour being infused. 
  • Generally, macrodrops have a drop factor of 10, 12, 15, or 20 drops/mL .
  • Microdrip sets, on the other hand, have a drop factor of 60 drops/mL . 

How to Regulate IV Fluids?

  • Size of the catheter. A catheter with a larger bore allows solution to flow faster. 
  • Height of the IV bag. The higher the IV bag, the faster the infusion will flow. 
  • Position of the insertion site. A change in the position of the client’s arm may decrease the flo, while elevation on a pillow may increase flow rate. If the IV is inserted into the antecubital area, the solution can flow freely if the client extends the arm and can be obstructed if the client bends the arm at the elbow. 
  • Monitoring and regulating the rate of the infusion is a responsibility of the nurse. 
  • A slower rate is usually necessary for older adults or those who are at risk of fluid overload (e.g., heart disease or client with head injury ). 
  • A faster IV flow rate is therapeutic for patients who have lost large amounts of body fluids and those who are severely dehydrated. 
  • Never increase the rate of infusion if it is running behind schedule. Check for obstructions and collaborate with primary care providers to determine the patient’s ability to tolerate an increased flow rate. 
  • Flow rate is regulated by tightening or releasing the IV tubing clamp and counting the drops for 15 seconds then multiplying the number 4 to get drops per minute.
  • Sometimes, the IV rate order will say “to keep open” (TKO) or “keep vein open” (KVO). This order does not specify the Milliliters per hour. Generally, KVO is infused at 50 mL/h.

Flow-Control Devices

  • Flow-control devices are any manual, mechanical, or electronic infusion device used to regulate the IV flow rate. These devices may include manual flow regulators, elastomeric balloon pumps, and electronic infusion devices. 
  • EIDs are often used in acute care settings and use positive pressure to deliver a preset fluid volume at preset limits. 
  • They are programmed to regulate the IV flow rate in either drops per minute or milliliters per hour. 
  • EIDs use gravity to maintain the flow of the IV fluid. They sense the rate and amount of IV fluid. 
  • An alarm is set off if there is air in the tubing, the bag is empty, or the flow is obstructed. However, the nurse should still conduct regular evaluations of the IV site. 
  • Another type of flow-control device that can deliver several medications and fluids (from either bags, bottles, or syringes) at the same time, at multiple rates. 
  • Multichannel pumps usually have two to four channels with each channel that can be programmed independently. 
  • Are nonelectric devices used to regular IV flow rate. These are in-line devices with a manual regulator that controls the amount of fluid to be administered. 
  • Rotating a dial sets the flow rate. 
  • Are disposable, portable, and nonelectric pumps that are prefilled with medication and connect to the client’s needleless connector to deliver a controlled rate of medication.

General Nursing Considerations

  • Monitor for infiltration or irritation. Inspect the insertion site for fluid infiltration. If present, stop the infusion and remove the catheter. Restart the infusion at another site and start supportive treatment by elevating or applying heat to the site. 
  • Look for signs of infiltration. Infiltration occurs when the IV fluid is not flowing into the client’s vein but into surrounding tissues. Signs of infiltration include swelling or puffiness, coolness, pain at the insertion site, and tenderness in the area. 
  • Monitor for signs of phlebitis. Phlebitis is the inflammation of the vein. Signs include pain and tenderness, swelling, and warmth in the area. If phlebitis occurs, stop infusion and restart at another site. Do not use the injured vein again. 
  • Regularly monitor IV flow rate. Monitor IV flow rate regularly (every hour) even if the solution is administered through an IV pump. 
  • Assess for fluid overload. Regularly assess the patient for signs of fluid overload: increased heart rate , increased respirations, and increased lung congestion. 
  • Risk for fluid overload. IV flow-control devices should be used for older and pediatric patients when administering IV fluids. These age groups are at risk for complications of fluid overload.
  • Proper documentation. Document all findings on the IV flow sheet or in the computer. Including the total amount of fluid administered, and any adverse responses of the client. 

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solving flow rate problems

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15 thoughts on “IV Flow Rate Calculation NCLEX Reviewer & Practice Questions (60 Items)”

Just wanted to let you know that question 25 is incorrect. It should be 137 gtts/min not 138 gtts/min. In your explanation, you’re not following the rules of BEDMAS where you should divide 275/120 first before multiplying by 60. If you follow BEDMAS you get 137.4 which you would round down to 137 gtts/min

Their answer is correct.

A health care provider orders cefoxitin (Mefoxin) 0.5 g in 275 mL of D5W to infuse IVPB over 2 hours; Drop Factor: 60 gtt/mL. How many gtts/min will you regulate the IV? Fill in the blank and round to the nearest whole number for your final answer.

What we want? gtts/min

What we have: 275mL/2hr -> 137.5mL/hr 60gtt/mL

Convert mL/hr to mL/min: 137.5mL/hr * 1hr/60min -> 2.2916666mL/min

Dimensional Analysis: 60gtt/1mL * 2.2916666mL/1min -> 137.5gtt/min or 138gtt/min rounded

The NTG question is incorrect. The question states 5 mcg/min, but the problem is solved using 15 mcg/min.

No matter how you do the math, 275 / 2 x 60 / 60 = 137.5 = 138. The answer is not 137.4. :)

FACTS! THIS IS CORRECT. EVEN AS STATED ABOVE 275/120 X 60 =137.5 = 138

Great practice! Thank you for setting this up! Question 23 reads ” df of 12 gtts per min, how many gtts per min?” It should read Gtts per mL.

I want to clarify ques no 3?

Hi Venilla,

What is your clarification regarding question number 3?

Great practice before my ATI dosage calculations exam!

Why nowadays rationale and whether correct or wrong not showing, kindly respond.

Due to performance issues, in the meantime, we have disabled the rationales from showing after each question. Rationales are still available but only at the end of the quiz. Please bear with us while we try to resolve this problem.

unit of measure drip rate measure is gtt/min gtt/min = 60gtt/ml x 275ml/2hr x 1hr/60min = 16500gtt/120min = 137.5gtt/min 0.5gtt cannot be given hence gtt/min = 138gtt/min

Great opportunity!

Hey good morning!

It seems as if the nursing test bank for IV Flow rate puts you in a continuous two page loop back and forth with no questions present. Maybe the link is broken or missing?

Hi Jeffrey, I checked and it’s working fine on my end. I think your browser is blocking opening new tabs, but I have updated the links and they should open in the same tab/window. Thank you for letting us know!

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State standards, tennessee - math.

Students must be able to solve one- and two-step equations with variables (pre-algebra/algebra 1 level work).

(In advance, prepare to show students a short video clip [attached; also available on the internet]. Make copies of the attached Flow Rate Worksheet , one per student.)

You have already identified which form of a medication works the fastest. Now we want to determine if anything else can be done to make the medicine work even faster once it has entered the body. I wonder what you'll predict will be the solution!

(Remind students of the Challenge Question that was introduced in lesson 1 of this unit.) One morning you don't feel very well, so your parents take you to see a doctor. The doctor suggests a generic antibiotic, and asks whether you would like to have the prescription filled in one of three ways: as a pill, in liquid form, or as a shot. Which delivery method should you choose? And, is there anything else you can do to feel better more quickly?

(Continue with a class discussion of microfluidic devices and how they are used in the development of medicine.) What is a microfluidic device? (Use lecture information provided in the Lesson Background section.)

(For the Test Your Mettle stage, lead in to the associated activity Making Model Microfluidic Devices Using JELL-O , in which students make model microfluidic devices at a larger than real-life scale.) While we do not have the materials necessary to create a true lab-on-a-chip device in the classroom, we can replicate the process using different materials on a much larger scale.

(After creating the devices, as described in the associated activity, continue on with the video and math worksheet.)

Now that we have created and tested our devices, let's watch a video of particles flowing through a microfluidic device. (For the Generating Ideas stage, show the attached 31-second video clip, Microfluidic Device Particle Flow Video . It shows particles moving through a lab-on-a-chip microfluidic device at various speeds, depending on the size of the channels.) Pay attention to what is occurring in each of the channels, and how that affects where the particles drop. (Then, ask the students:)

  • What do you notice about the movement of the particles? (Possible answers: Slower particles fall to the bottom of the channel quickly, the faster particles all end up in the "pit" at the end.)
  • How would these different flow rates affect the delivery of medicine? (Answer: Flow rate determines the amount of fluid that flows, and determines how particles are distributed.)
  • What is the most effective flow rate (slow, medium, fast)? (Answer: A medium flow rate leaves the particles well dispersed on the bottom of the channel.)
  • How is flow rate increased/decreased? (Answer: Flow rate is altered by the size of the channel and/or pressure of the fluid.)
  • How can we increase/decrease the "flow rate" in a human body? (Answer: Increasing heart rate increases how fast blood moves through the veins.)

(Introduce students to the flow rate equation and how to determine flow rate. Solve several example problems, as provided in the Lesson Background section. Conclude by handing out the Flow Rate Worksheet for students to complete as homework).

Lesson Background and Concepts for Teachers

Legacy Cycle Information: This lesson continues the Generate Ideas , Multiple Perspectives , Research and Revise stages of the legacy cycle. Students are reminded of the Challenge Question and learn about microfluidic devices and how they are used in the development of medicine, before beginning the associated activity. The next day, they complete the activity, watch the video and learn about flow rates, concluding with a worksheet of flow rate problems to solve.

Lecture Information: Microfluidic devices (see examples in Figure 1) are used in experiments called lab-on-a-chip experiments. In these experiments, the entire procedure is performed and viewed on a microscopic level. Lab-on-a-chip experiments usually take place on glass slides used with standard microscopes. The microfluidic device is irreversibly sealed to the glass slide.

Using this method enables us to understand what happens at the molecular level in the body and to any type of cell. For example, we can view how a disease affects cells in the body. In one instance, two groups of molecules were separated by a microscopic barrier. Then a scientist was able to collapse the barrier and observe how one group of molecules moved and was attracted to the other group. Then the scientist re-closed the barrier and observed the disease begin to take over (whether it killed the cells, attached to the cells, etc.). Because the molecules are in an environment similar to conditions in the body, we are able to view changes, often in real-time, as fluids, chemicals and/or cells pass through specially designed channels. These experiments are important because they enable testing of new medicines before using on animals and humans.

To design microfluidic devices, engineers must research how particles move in the body. Biologists describe what natural occurrences they wish to re-create for their experiments and the engineers develop channels, barriers and other passages that produce effects similar to those that occur in the body (a simulation). For instance, to see how molecules "mix" in the blood, we could design multiple channels that merge and then zigzag many times; the zigzagging route would cause the liquids to combine. If engineers have limited background in biology, they research the specific topic so they fully understand it. Microfluidic devices are often made using a soft-polymeric system that allows scientists and biomedical engineers to create a single master from which they can develop an infinite number of copies. This method is cost effective and allows for the simple and quick production of many devices. When exposed to oxidized plasma, the material used, polydimethylsiloxane (called PDMS), produces irreversible bonding to glass surfaces, such as microscope slides.

Example Problems: Students should be able to solve one- and two-step equations with variables. Go through a few examples with them:

Figure shows an equation which would read "Q is equal to V times pi times D squared all divided by 4".

where D is the inside diameter, and V is the velocity.

The figure would read "1. Solve for h, A = one-half times b times h (Answer is h = 24 divided by b) and 2. Solve for l, V = l times w times h (Answer: l = A divided by (w times h).

Also have students solve for variables when given information.

  • The area of a triangle is 36 inches, with a base of 4 inches. Find the height, using the information from #1, above. (Answer: h = 18 inches)
  • The volume of a rectangular box is 268 cm, with a length of 16 cm. The width is the same as the length. Find the height of the box. (Answer: h = 1.047 cm)
  • Making Model Microfluidic Devices Using JELL-O - Students make large-scale models of microfluidic devices using a bonding process similar to that used in the creation of lab-on-a-chip devices. They use disposable foam plates, bendable plastic straws, tape and JELL-O mix. From what they learn, students are able to answer the challenge question presented in lesson 1 of this unit.

flow rate: The amount of fluid that flows in a given time.

lab-on-a-chip (LOC): A device that combines one or more laboratory functions on a single chip that is less than a few square cm in size and handles extremely small fluid volumes.

microfluidic device: A device that deals with the behavior, precise control and manipulation of fluids that are geometrically constrained to a small, often sub-millimeter, scale.

Pre-Lesson Assessment

Discussion Questions: Ask the students the following questions and discuss as a class.

  • What do you do after you take medicine?
  • What types of advice have your parents offered to help you feel better?

Post-Introduction Assessment

Discussion Questions: After conducting the first half of the associated activity, and then watching the video, ask the students the following questions and discuss as a class.

  • How might flow rate be seen in the body? (Answer: In our blood streams, through veins and arteries.)
  • How might flow rate be used in the body? (Answer: To measure blood flow.)
  • How might the particles shown in the lab-on-a-chip video relate to medicine?

Lesson Summary Assessment or Homework

Worksheet: Have students complete the problems on the attached Flow Rate Worksheet . Review their answers to gauge their comprehension of the subject

The attached 31-second video clip, called "0.5microL_fast.avi," is also available on YouTube at http://youtu.be/YDD0WAAYw4s.

solving flow rate problems

Students are introduced to a challenge question. Towards answering the question, they generate ideas for what they need to know about medicines and how they move through our bodies, watch a few short videos to gain multiple perspectives, and then learn lecture material to obtain a basic understandin...

preview of 'How Antibiotics Work ' Lesson

Contributors

Supporting program, acknowledgements.

The contents of this digital library curriculum were developed under National Science Foundation RET grant nos. 0338092 and 0742871. However, these contents do not necessarily represent the policies of the NSF, and you should not assume endorsement by the federal government.

Last modified: June 18, 2019

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  • Calculator Technique for Solving Volume Flow Rate Problems in Calculus

Jhun Vert's picture

Flow Problems in Calculus Solved by a Formula from Physics | Differential Calculus

The following models of CASIO calculator may work with this method: fx-570ES, fx-570ES Plus, fx-115ES, fx-115ES Plus, fx-991ES, and fx-991ES Plus.  

The following calculator keys will be used for the solution

This is one of the series of post in calculator techniques in solving problems. You may also be interested in my previous posts: Calculator technique for progression problems and Calculator technique for clock problems ; both in Algebra.  

Flow Rate Problem Water is poured into a conical tank at the rate of 2.15 cubic meters per minute. The tank is 8 meters in diameter across the top and 10 meters high. How fast the water level rising when the water stands 3.5 meters deep.  

$r = \frac{2}{5}h$  

Volume of water inside the tank $V = \frac{1}{3}\pi r^2 h$

$V = \frac{1}{3}\pi (\frac{2}{5}h)^2 h$

$V = \frac{4}{75}\pi h^3$  

Differentiate both sides with respect to time $\dfrac{dV}{dt} = \frac{4}{25}\pi h^2 \dfrac{dh}{dt}$

$2.15 = \frac{4}{25}\pi h^2 \dfrac{dh}{dt}$  

When h = 3.5 m $2.15 = \frac{4}{25}\pi (3.5^2) \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = 0.3492 \, \text{m/min}$           answer  

Solution by Calculator

Where $Q$ = the discharge or volume flow rate $v$ = velocity of flow $A$ = cross-sectional area of flow  

The equivalent of the above elements in Calculus are: $Q = \dfrac{dV}{dt}$ where V is the volume and dV/dt is the volume flow (time) rates and

$v = \dfrac{dh}{dt} = \dfrac{dx}{dt} = \dfrac{dy}{dt} = \dfrac{ds}{dt}$ where h, x, y, s are distances and v is velocity.  

Thus, the formula Q = vA can be written as  

which is the formula we are going to use in our calculator.  

For the area A The general prismatoid is a solid in which the area of any section, say A y , parallel to and at a distance y from a fixed plane can be expressed as a polynomial in y not higher than third degree, or  

Common solids like cone, prism, cylinder, frustums, pyramid, and sphere are actually prismatoids in which any area parallel to a base is at most a quadratic function in height y. Thus,  

We can therefore use the Quadratic Regression in STAT mode of the calculator to find the area in relation with its distance from a predefined plane.

AC → 2.15 ÷ 3.5y-caret = 0.3492           answer  

To input the 3.5y-caret above, do 3.5 → SHIFT → 1[STAT] → 7:Reg → 6:y-caret  

What we just did was actually v = Q / A which is the equivalent of $\dfrac{dh}{dt} = \dfrac{dV/dt}{A}$ for this problem.  

Problem Water is being poured into a hemispherical bowl of radius 6 inches at the rate of x cubic inches per second. Find x if the water level is rising at 0.1273 inch per second when it is 2 inches deep?  

$V = \frac{1}{3}\pi h^2 [ \, 3(6) - h \, ]$

$V = \frac{1}{3}\pi (18h^2 - h^3)$  

Differentiate both sides with respect to time $\dfrac{dV}{dt} = \frac{1}{3}\pi (36h - 3h^2) \dfrac{dh}{dt}$  

When h = 2 inches, dh/dt = 0.1273 inch/sec $\dfrac{dV}{dt} = \frac{1}{3}\pi [ \, 36(2) - 3(2^2) \, ] (0.1273)$

$x = 7.9985 \, \text{in}^3\text{/sec}$           answer  

AC → 0.1273 × 2y-caret = 7.9985           answer  

I hope you enjoy this post. Next time you solve problems involving flow rate, try to use this calculator technique to save time.  

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6.2.3: Mixture Problems

  • Last updated
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  • Page ID 107200

  • Russell Herman
  • University of North Carolina Wilmington

There are many types of mixture problems. Such problems are standard in a first course on differential equations as examples of first order differential equations. Typically these examples consist of a tank of brine, water containing a specific amount of salt with pure water entering and the mixture leaving, or the flow of a pollutant into, or out of, a lake. We first saw such problems in Chapter \(1 .\)

In general one has a rate of flow of some concentration of mixture entering a region and a mixture leaving the region. The goal is to determine how much stuff is in the region at a given time. This is governed by the equation

\[\text{Rate of change of substance} = \text{Rate In} - \text{Rate Out}. \nonumber \]

This can be generalized to the case of two interconnected tanks. We will provide an example, but first we review the single tank problem from Chapter \(1 .\)

Example \(\PageIndex{1}\): Single Tank Problem

A 50 gallon tank of pure water has a brine mixture with concentration of 2 pounds per gallon entering at the rate of 5 gallons per minute. [See Figure \(\PageIndex{1}\).] At the same time the well-mixed contents drain out at the rate of 5 gallons per minute. Find the amount of salt in the tank at time \(t .\) In all such problems one assumes that the solution is well mixed at each instant of time.

clipboard_e38e8cbd09c0ef3d9bdafd668d65686b3.png

Let \(x(t)\) be the amount of salt at time \(t\) . Then the rate at which the salt in the tank increases is due to the amount of salt entering the tank less that leaving the tank. To figure out these rates, one notes that \(d x / d t\) has units of pounds per minute. The amount of salt entering per minute is given by the product of the entering concentration times the rate at which the brine enters. This gives the correct units:

\[\left(2 \dfrac{\text { pounds }}{\text { gal }}\right)\left(5 \dfrac{\text { gal }}{\text { min }}\right)=10 \dfrac{\text { pounds }}{\text { min }}. \nonumber \]

Similarly, one can determine the rate out as

\[\left(\dfrac{x \text { pounds }}{50 \text { gal }}\right)\left(5 \dfrac{\text { gal }}{\text { min }}\right)=\dfrac{x}{10} \dfrac{\text { pounds }}{\text { min }} \nonumber \]

Thus, we have

\[\dfrac{d x}{d t}=10-\dfrac{x}{10} \nonumber \]

This equation is easily solved using the methods for first order equations.

Example \(\PageIndex{2}\): Double Tank Problem

One has two tanks connected together, labeled \(\operatorname{tank} X\) and \(\operatorname{tank} Y\) , as shown in Figure \(\PageIndex{2}\).

clipboard_e6af625415be6ada43380a0bd95348a7d.png

Let tank X initially have 100 gallons of brine made with 100 pounds of salt. Tank Y initially has 100 gallons of pure water. Pure water is pumped into tank \(X\) at a rate of \(2.0\) gallons per minute. Some of the mixture of brine and pure water flows into tank \(Y\) at 3 gallons per minute. To keep the tank levels the same, one gallon of the \(Y\) mixture flows back into \(\tan \mathrm{k} X\) at a rate of one gallon per minute and 2.0 gallons per minute drains out. Find the amount of salt at any given time in the tanks. What happens over a long period of time?

In this problem we set up two equations. Let \(x(t)\) be the amount of salt in \(\operatorname{tank} X\) and \(y(t)\) the amount of salt in tank \(Y\) . Again, we carefully look at the rates into and out of each tank in order to set up the system of differential equations. We obtain the system

\[ \begin{aligned} \dfrac{d x}{d t} &=\dfrac{y}{100}-\dfrac{3 x}{100} \\ \dfrac{d y}{d t} &=\dfrac{3 x}{100}-\dfrac{3 y}{100} \end{aligned} \label{6.50} \]

This is a linear, homogenous constant coefficient system of two first order equations, which we know how to solve. The matrix form of the system is given by

\[\dot{\mathbf{x}}=\left(\begin{array}{cc} -\dfrac{3}{100} & \dfrac{1}{100} \\ \dfrac{3}{100} & -\dfrac{3}{100} \end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{c} 100 \\ 0 \end{array}\right)\nonumber \]

The eigenvalues for the problem are given by \(\lambda=-3 \pm \sqrt{3}\) and the eigenvectors are

\[\left(\begin{array}{c} 1 \\ \pm \sqrt{3} \end{array}\right) \nonumber \]

Since the eigenvalues are real and distinct, the general solution is easily written down:

\[\mathbf{x}(t)=c_{1}\left(\begin{array}{c} 1 \\ \sqrt{3} \end{array}\right) e^{(-3+\sqrt{3}) t}+c_{2}\left(\begin{array}{c} 1 \\ -\sqrt{3} \end{array}\right) e^{(-3-\sqrt{3}) t} \nonumber \]

Finally, we need to satisfy the initial conditions. So,

\[\mathbf{x}(0)=c_{1}\left(\begin{array}{c} 1 \\ \sqrt{3} \end{array}\right)+c_{2}\left(\begin{array}{c} 1 \\ -\sqrt{3} \end{array}\right)=\left(\begin{array}{c} 100 \\ 0 \end{array}\right),\nonumber \]

\[c_{1}+c_{2}=100, \quad\left(c_{1}-c_{2}\right) \sqrt{3}=0.\nonumber \]

So, \(c_{2}=c_{1}=50\) . The final solution is

\[\mathbf{x}(t)=50\left(\left(\begin{array}{c} 1 \\ \sqrt{3} \end{array}\right) e^{(-3+\sqrt{3}) t}+\left(\begin{array}{c} 1 \\ -\sqrt{3} \end{array}\right) e^{(-3-\sqrt{3}) t}\right),\nonumber \]

\[ \begin{aligned} &x(t)=50\left(e^{(-3+\sqrt{3}) t}+e^{(-3-\sqrt{3}) t}\right) \\ &y(t)=50 \sqrt{3}\left(e^{(-3+\sqrt{3}) t}-e^{(-3-\sqrt{3}) t}\right) . \end{aligned} \label{6.51} \]

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  • Mass Flow Rate Formula

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Mass Flow Rate Equation

The conservation of mass is a part of thermodynamics and it is a fundamental concept of physics. According to physics the mass of the substance should always remain constant. Hence, the mass is neither created nor destroyed, but it can modify from one form to another. The mass of any object can be determined by its volume, density, and area it occupies. 

The mass of a liquid substance passing per unit time is termed the mass flow rate. The mass flow rate can also be expressed as the rate of movement of liquid pass through a unit area. Here, the mass flow rate will vary as the density, velocity of the liquid varies. Also, the mass flow rate will increase as the cross-sectional area increases. So, the mass flow is directly proportional to the velocity of the liquid, density of the liquid, and area of the cross-section. The mass flow rate is the movement of mass per unit time. The mass flow rate can be expressed as m and the mass flow can be calculated in terms of  kg/s. 

Mass flow rate equation m = ρVA

Here, 

m denoted the mass flow rate

ρ denotes the density of the liquid.

V denotes the velocity of the liquid

A denotes the cross-sectional area. 

The cross-sectional area of a tube can be calculated as given below. 

Then, 

Here, r denotes the radius of the tube. 

By using the above mass flow rate of the water formula, we can also calculate velocity from mass flow rate. 

Problems Related to Mass Flow Rate 

Problem 1: Calculate the mass flow rate of a given fluid whose density is 700 kg/m 3 , velocity, and area of cross-section is 40 m/s and 30 cm 2 respectively.

As per given data

ρ = 700 kg/m 3

V = 40 m/s and

A = 30 cm 2

    = 0.30 m 2

The formula for calculating mass flow rate (m)

m = 700 × 40 × 0.30

m = 8400 kg/s

The mass flow rate m  for the above-given data is 8400 kg/s

Problem 2: Determine the density of the liquid? The mass flow rate of the liquid is 4800 kg/s. And the velocity and area of the cross-section are 30 m/s and 20 cm 2 respectively.

From the given values

m = 4800 kg/s

V = 30 m/s and

A = 20 cm 2

To calculate the density of the liquid from the above-given values. 

ρ = 4800 / ( 30 * 0.20 )

ρ =  4800 / 6

ρ =  800 kg/m 3

The density of the liquid ρ for the above-given data is 800 kg/m 3

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FAQs on Mass Flow Rate Formula

1. How Do You Calculate Mass Flow Rate?

Ans: Mass flow rate is calculated by producing the volume flow rate by the mass density of the fluid ( ρ). Further, the volume flow rate can be calculated by multiplying the cross-sectional area vector area (A) and the flow velocity of the mass elements (v).

2. What is the Mass Flow Rate in Fluid Mechanics?

Ans: The mass flow rate defines the mass of a liquid substance that passes per unit time. In other words, the mass flow rate formula shows the rate of movement of liquid passing through a unit area. The fuel mass flow rate formula is directly proportional to the density and velocity of the liquid with the cross-sectional area. 

3. What is the Relationship Between Mass Flow Rate And Volume Flow Rate?

Ans: The mass flow formula measures the number of molecules that pass through a flowing gas. Volumetric flow measures the space occupied by those molecules. As the gas is compressible, the volumetric flow rates can change significantly, when the pressure or temperature of the gas changes. 

Veteran economists say a carbon levy would cut emissions, cut inflation and raise billions, but see little prospect of adoption

silhouetted coal mining digger next to a dump truck at a coal mine

Two of the nation's most respected economists have put forward a bold plan they say can lower global carbon emissions by at least 6 per cent, super-charge a new green export industry for Australia, deliver much cheaper power bills and even dramatically cut the rate of inflation.

They also frankly admit their ambitious blueprint, due to be unveiled at the National Press Club today, is likely to be immediately shot down by the major political parties.

Professor Rod Sims and Professor Ross Garnaut have spent the past six months working on what they're calling a pathway to prosperity for Australia in a net-zero world. Their work is for The Superpower Institute, a think tank focused on the economic opportunities of climate change, founded by Professor Garnaut and now chaired by Professor Sims.

Under their plan, a "Carbon Solutions Levy" would be imposed on all fossil fuel extraction sites (around 105 coal and gas projects) and all fossil fuel imports (oil and diesel). The levy would be in place by 2030-31, set at the level of the European Union's carbon price (currently around $90 per tonne of emissions).

The pair suggest the levy would raise well over $100 billion a year, declining slowly as fossil fuels exit the market. That enormous pool of funds would then be used to over-compensate households, slashing their electricity bills by $440 a year and removing all petrol and diesel excise. They estimate this would significantly lower the Consumer Price Index by more than 1.5 percentage points.

"This is the mother of all cost-of-living events", Professor Sims told the ABC, highlighting the benefit to households. "Everyone is a winner except those fossil fuel companies", which he frankly admits "will hate it".

A headshot of Rod Sims. He wears a suit.

The sizeable funds raised by the levy could also be used to repair the budget and fund further tax cuts.

The other key pillar of the plan is to set aside at least $6 billion a year from the funds raised for a new "Superpower Innovation Incentive Scheme".

This scheme would help the first few early movers in green exports get off the ground, with grants covering up to 50 per cent of their capital costs.

The vision is for Australia to become a major manufacturer and exporter of metals and fuels made with renewable energy, including green iron, green aluminium, green polysilicon, green transport fuels and green urea.

Zero and low-emissions industry would be powered by large-scale solar and wind farms in remote areas including the Pilbara.

"We are consumed by how to reduce carbon emissions here", says Professor Sims, arguing the greater focus should be on reducing global emissions, which could be 6-9 per cent lower if Australia were to adopt this plan.

The Superpower Institute is less focused on the ongoing battles over lowering domestic emissions. It's more concerned Australia may miss out on the enormous opportunities in the great global transition to a net-zero future.

Europe has already legislated a Carbon Border Adjustment Mechanism (CBAM), which from 2026 will impose a levy on imports of carbon intensive products including aluminium, iron, steel, cement, and fertilisers.

Professor Sims says this means "green iron will be more competitive than black iron" in the European market from 2026.

He argues China will also need to import "green iron", due to its limited capacity to transition its own industry.

Ross Garnaut leans on a dead tree.

Professor Sims insists the plan being unveiled today is "not a Carbon Tax or an Emissions Trading Scheme". The Carbon Solutions Levy being proposed is confined to fossil fuel companies.

Nonetheless, the plan's authors acknowledge the levy will be branded a tax and will be initially rejected by the major parties, given the bruising history of climate and energy politics in Australia. Even if the scale of ambition in the plan was halved, they doubt either side of politics would embrace the idea.

"It doesn't matter", says Professor Sims. "This is a long-term game. The debate starts now. Every policy idea that involves complex economics takes years to permeate through consciousness".

But the longer Australia waits, the two economists argue, the more opportunities will slip away.

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  1. How to solve flow rate calculations problems

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  2. 12.1: Flow Rate and Its Relation to Velocity

    How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min? Strategy. Time and flow rate \(Q\) are given, and so the volume \(V\) can be calculated from the definition of flow rate. Solution. Solving \(Q = V/t\) for volume gives \[V = Qt. \nonumber \] Substituting known values yields

  3. 12.1 Flow Rate and Its Relation to Velocity

    Q = V t, 12.1. where V is the volume and t is the elapsed time. The SI unit for flow rate is m3/s, but a number of other units for Q are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters ( 10−3 m3 or 103 ...

  4. Finding flow rate from Bernoulli's equation

    About Transcript Sal solves a Bernoulli's equation example problem where fluid is moving through a pipe of varying diameter. Understand how pressure differentials and pipe dimensions influence flow rate and velocity. Created by Sal Khan. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted Max 11 years ago

  5. 14.5 Fluid Dynamics

    The mass flow rate is an important quantity in fluid dynamics and can be used to solve many problems. Consider Figure 14.28 . The pipe in the figure starts at the inlet with a cross sectional area of A 1 A 1 and constricts to an outlet with a smaller cross sectional area of A 2 A 2 .

  6. Flow Rate

    The volumetric flow rate of fluid is found using the equation: is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing. In this problem the cross-section of the pipe is a square. The area of the cross-section is: The volumetric flow rate is: A pipe narrows from a diameter.

  7. What is volume flow rate? (article)

    Q = A v A is the cross sectional area of a section of the pipe, and v is the speed of the fluid in that section. So, we get a new formula for the volume flow rate Q = A v that is often more useful than the original definition of volume flow rate because the area A is easy to determine.

  8. Volume Flow Rate & Mass Flow Rate

    The Organic Chemistry Tutor 7.33M subscribers Join Subscribe Subscribed 220K views 6 years ago New Physics Video Playlist This physics video tutorial provides a basic introduction into mass flow...

  9. 14.7: Fluid Dynamics

    The mass flow rate is an important quantity in fluid dynamics and can be used to solve many problems. Consider Figure \(\PageIndex{5}\). The pipe in the figure starts at the inlet with a cross sectional area of A 1 and constricts to an outlet with a smaller cross sectional area of A 2 .

  10. Volume flow rate and equation of continuity

    Conservation of energy tells you that the pressure in the reduced area will be lower because the velocity is increased (speeding a fluid up lowers it pressure, some what counter intuitive because we think of pressure in terms of force not potential energy) Flow rate (Q) = velocity * Area. Q1 = Q2 v1 * A1 = v2 * A2.

  11. Volume Flow Rates in Physics Problems

    0.25 m Use the equation of continuity: In terms of the stream width and depth d, the cross-sectional area of the stream is Use this in the continuity equation: You know that so you can solve the continuity equation for the width 5.0 You know that are the volume flow rate through pipes 1 and 2, respectively.

  12. Flow Rate

    In order to solve flow rate problems: Use information provided to calculate the flow rate. \text {Flow rate}=\frac {\text {change in volume}} {\text {time taken}} Flow rate = time takenchange in volume Use information provided to calculate volume that will be changing.

  13. Flow Rate Formula With Solved Examples

    Problem 1: Compute the flow rate of fluid if it is moving with the velocity of 20 m/s through a tube of diameter 0.03 m. Answer: Velocity of fluid flow v =20m/s Diameter of pipe d=0.03m Area of cross-section of the pipe, A= { (3.14)/4} (0.03) (0.03) A = (0.785) (0.0009) A Flow rate is given by Q = vA= (20) (0.000706) Q=

  14. 14.6 Bernoulli's Equation

    Bernoulli's equation in that case is. p 1 + ρ g h 1 = p 2 + ρ g h 2. We can further simplify the equation by setting h 2 = 0. (Any height can be chosen for a reference height of zero, as is often done for other situations involving gravitational force, making all other heights relative.) In this case, we get.

  15. IV Flow Rate Calculation Reviewer & Quiz (60 Questions)

    There are three commonly used ways on how to indicate flow rates: Milliliters per hour (mL/h). Calculated by dividing the total infusion volume by the total infusion time in hours Number of drops per one (1) minute (gtts/min).

  16. Microfluidic Devices and Flow Rate

    The next day, they complete the activity, watch the video and learn about flow rates, concluding with a worksheet of flow rate problems to solve. Lecture Information: Microfluidic devices (see examples in Figure 1) are used in experiments called lab-on-a-chip experiments. In these experiments, the entire procedure is performed and viewed on a ...

  17. Continuity Equation, Volume Flow Rate & Mass Flow Rate ...

    Solving Density Problems 13m. Dimensional Analysis 10m. Counting Significant Figures 5m. Operations with Significant Figures 10m. 2. 1D Motion / Kinematics 3h 56m. ... Volume Flow Rate & Mass Flow Rate Physics Problems. The Organic Chemistry Tutor. 556. views. Showing 1 of 10 videos. Load more videos. Click to get Pearson+ app Download the ...

  18. Calculator Technique for Solving Volume Flow Rate Problems ...

    Flow Rate Problem Water is poured into a conical tank at the rate of 2.15 cubic meters per minute. The tank is 8 meters in diameter across the top and 10 meters high. How fast the water level rising when the water stands 3.5 meters deep. Traditional Solution r h = 4 10 r = 2 5 h Volume of water inside the tank V = 1 3 π r 2 h V = 1 3 π ( 2 5 h) 2 h

  19. Quiz & Worksheet

    Print Worksheet. 1. Flow rate Q is equal to the _____ of fluid that passes through a slice of a pipe (or other container carrying moving fluid) each second. volume. area. mass. density. 2. The ...

  20. 6.2.3: Mixture Problems

    We first saw such problems in Chapter 1. In general one has a rate of flow of some concentration of mixture entering a region and a mixture leaving the region. The goal is to determine how much stuff is in the region at a given time. This is governed by the equation. Rate of change of substance = Rate In − Rate Out.

  21. Solving IV Flow Rate Problems Using Dimensional Analysis

    Solving IV Flow Rate Problems Using Dimensional Analysis • Dimensional Analysis (DA): A powerful method of solving problems in pharmacy, chemistry, physics, and engineering in which a given is multiplied by one or more ratios • IV: Abbreviation for intravenous, meaning administered into a vein. • drop factor: The number of drops (gtts) per mL.

  22. How to solve flow rate calculation problems 3

    Live TV from 100+ channels. No cable box or long-term contract required. Cancel anytime. This video illustrates how to solve flow rate calculation problems. Share this...

  23. Mass Flow Rate Formula

    By using the above mass flow rate of the water formula, we can also calculate velocity from mass flow rate. Problems Related to Mass Flow Rate . Problem 1: Calculate the mass flow rate of a given fluid whose density is 700 kg/m 3, velocity, and area of cross-section is 40 m/s and 30 cm 2 respectively. Solution: As per given data. ρ = 700 kg/m ...

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    Two of the nation's most respected economists have put forward a bold plan they say can lower global carbon emissions by at least 6 per cent, super-charge a new green export industry for Australia ...

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    Few companies have suffered as much from the Federal Reserve's interest rate campaign as Upstart (UPST-3.39%).Shares of the AI-based consumer lending company have fallen as much as 97% from peak ...