Volumetric Flow Rate and Flow Velocity

by Martin Schlobach | Last updated: Oct 18, 2024 | 0 comments

Some of the links on my blog are affiliate links. This means that if you click on the link and buy the item, I will receive a commission at no extra cost to you. My opinion remains unaffected.

In this article, I will introduce you to volume flow and flow velocity as further physical variables and show you how to calculate them using two examples.

Table of Contents

1 Volumetric flow rate
2 The flow velocity
3 Example calculations
- 3.1 Example calculation – volume flow and flow velocity
- 3.2 Example calculation – Calculating the volume flow for radiators
4 Conclusion

Volumetric flow rate

The volumetric flow rate ( $\dot V$ ) (also known as volume flow) indicates how much volume (V) of a substance (e.g. a gas or liquid) flows through a pipe or channel in a certain period of time (t). In building services engineering, the following questions could therefore be asked:

“How much heating water flows through a heating system in a given time?”

“How much air flows through a ventilation network in a certain time?”

As water and air are the heat transfer media in building technology and transport heat, the volume flow is an important variable for calculating the heat flow. The volume flow rate can therefore be used to calculate how much heat needs to be transported to a room in order for it to maintain the desired room temperature. The unit of volume flow is cubic metres per hour (m³/h) or litres per hour (l/h).

You can calculate the volumetric flow rate using the volume (V) and the time (t) or with the pipe cross-sectional area (A) and the flow velocity $\omega$ (small omega). The following formulas provide an overview:

Volume flow = volume / time

$\boxed{\dot V=\frac{V}{t}$}$

Volume flow rate = pipe cross-sectional area * flow velocity

$\boxed{\dot V=A \cdot \omega$}$

Volumetric flow rate in building services engineering – an important variable

The volume flow rate is one of the most important variables in building services engineering and can also be calculated using the heat flow $\dot Q$ , the specific heat capacity (c), the density ( $\rho$ – small raw) and the temperature difference ( $\Delta\vartheta$ delta Theta).

Note: The following formula is one of the most important formulas in building services engineering and is also important for hydronic balancing, as you can see in the fifth step of the series on“Hydronic Balancing DIY“.

Volume flow = heat flow / density * specific heat capacity * temperature difference

$\boxed{\dot V=\frac{\dot Q} {\rho \cdot c \cdot \Delta\vartheta}}$

Tip: The specific heat capacity c and the density $\rho$ of air and water are not always easy to remember. To simplify things, you can derive two factors from the specific heat capacity and density of air and water to make your calculations easier. You can use these factors, as these values for water and air only change slightly in certain temperature ranges. The factors for water and air are derived below.

Simplification: Factor for the medium water

The density of water at 20 °C is around 1,000 kg/m³. You can simplify this value and use it for the temperature range from 0 to 100 °C. The specific heat capacity of water is c = 4,182 J/(kg*K) = 0.00116 kWh/(kg*k).

The factor for the medium water results from the product of the specific heat capacity and the density of water, as can be seen in the following calculation

$\boxed{0.00116 \frac{kWh} {kg \cdot K} \cdot 1,000 \frac{kg} {m³} = 1.16 \frac{kWh} {m³ \cdot K} = 1.16 \frac{Wh} {l \cdot K}}$

As the units in the formula for the volume flow are shortened, you can use the factor 1.16 without units for the density and the specific heat capacity. The formulas for calculating the volumetric flow rate with the factor for the medium water are as follows. The factor is 1.16 or the reciprocal 0.86.

$\boxed{\dot V=\frac{\dot Q} {1.16 \cdot \Delta\vartheta} = 0.86 \cdot \frac{\dot Q} {\Delta\vartheta} [l/h]}$

Simplification: Factor for the medium air

The density of air at 20 °C is around 1.2 kg/m³. You can simplify this value for the temperature range from -20 to 30 °C. The specific heat capacity of air is c = 1,005 J/(kg*K) = 0.000273 kWh/(kg*k).

The factor for the medium water results from the product of the specific heat capacity and the density of water, as can be seen in the following calculation

$\boxed{0.000273 \frac{kWh} {kg \cdot K} \cdot 1.2 \frac{kg} {m³} = 0.00033 \frac{kWh} {m³ \cdot K} = 0.33 \frac{Wh} {l \cdot K}}$

As the units in the formula for the volumetric flow rate are shortened, you can use the factor 0.33 without units for the density and the specific heat capacity. The formulas for calculating the volumetric flow rate with the factor for the medium water are as follows. The factor is 0.33 or the reciprocal 3.03.

$\boxed{\dot V=\frac{\dot Q} {0.33 \cdot \Delta\vartheta} = 3.03 \cdot \frac{\dot Q} {\Delta\vartheta} [m³/h]}$

The flow velocity

The flow velocity is specified differently in the literature. It is known as $\omega$ (small omega), $v$ and $c$ . I have opted for the small omega $\omega$ . The flow velocity, or flow speed, indicates the speed of a flow. The unit is metres per second $\frac{m}{s}$ .

Flow velocity = volume flow / pipe cross-sectional area

$\boxed{\omega=\frac{\dot V}{A}}$

Example calculations

How do you calculate volumetric flow rate? To help you understand the topic a little better, I will show you how to apply the formulas in the following example calculations.

Example calculation – volume flow and flow velocity

Task:

In a medium-duty pipe with a nominal diameter of DN 15 (according to DIN EN 10255), 7 litres of water are to be transported in a period of one minute.

What volumetric flow rate results from the given values in the unit m³/h?
What is the cross-sectional area of the pipe in m²?
What is the resulting flow velocity in m/s?

Given:

medium-weight pipe, nominal diameter: DN 15 according to DIN EN 10255
Time: t = 1 min
Volume: V = 7 litres

Wanted:

Volume flow $\dot V$ in m³/h
Pipe cross-sectional area $A$ in m²
Flow velocity $\omega$ (small omega) in m/s

Solution:

Volumetric flow rate $\dot V$

Firstly, we calculate the volume flow using the following formula and then insert the given values

$\ \dot V=\frac{V}{t} =\frac{7l}{1min}=\underline{\underline{7 l/min}}$

As the volume flow is required in the unit m³/h, the result must still be converted.

Important: The following applies to the conversion to cubic metres/hour or litres/minute or litres/second

1 cubic metre = 1000 litres
1 hour = 60 minutes ⇨ 60 / 1000 = 0.06
1 hour = 3600 seconds ⇨ 3600 / 1000 = 3.6

1 cubic metre/hour [m³/h] = 16.67 litres/minute [l/min]
⇨ Divisor for conversion: 0.06 ⇨ x m³/h / divisor = y l/min
1 litre/minute [l/min] = 0.06 cubic metres/hour [m³/h]
⇨ Divisor for conversion: 16.67 ⇨ x l/min / divisor = y m³/h

1 cubic metre/hour [m³/h] = 0.278 litres/second [l/s]
⇨ Divisor for conversion: 3.6 ⇨ x m³/h / divisor = y l/s
1 litre/second [l/s] = 3.6 cubic metres/hour [m³/h]
⇨ Divisor for conversion: 0.278 ⇨ x l/s / divisor = y m³/h

$\ \dot V=\frac{7l/min}{16.67} =\underline{\underline{0.42 m^3/h}}$

The volumetric flow rate $\dot V$ is therefore 0.42 m³/h.

Pipe cross-sectional area $A$

In the next step, we calculate the pipe cross-sectional area $A$ , which is determined using Pi ( $\pi$ ) and the radius $r$ of the pipe.

To determine the internal diameter $d$ and the resulting radius $r$ of a medium-diameter pipe of DN 15, we need to look at a conversion table (or you know the values by heart). A look at DIN EN 10255, the nominal diameter tables from Wikipedia or the post in the glossary will help here.

The nominal diameter DN 15 therefore has an inner diameter of $d$ = 16 mm = 1.6 cm. The radius resulting from $r = \frac{d}{2}$ is then 0.8 cm.

We use the following formula and insert the given values:

$\ A=\pi \cdot r^2 =\pi \cdot (0.008 m)^2=\underline{\underline{0.0002 m^2}}$

The pipe cross-sectional area is therefore 0.0002 m².

Flow velocity $\omega$

First, we convert the volume flow from m³/h to m³/s, as the final result of the flow velocity should be given in m/s. The divisor 3.6 is not used here, but 3,600, as cubic metres are retained and not converted into litres (1 hour = 3600 seconds).

$\ \frac{0.42 m^3/h}{3600}=\underline{\underline{0.0001166 m^3/s}}$

We can now enter the calculated results into the following formula and obtain the flow velocity

$\ \omega=\frac{\dot V}{A} =\frac{0.0001166 m^3/s}{0.0002 m^2}=\underline{\underline{0.58 m/s}}$

The flow velocity $\omega$ is therefore 0.58 m/s.

Example calculation – Calculating the volume flow for radiators

In the second example calculation, we calculate the volume flow for a radiator. It is important to know this in order to be able to make a presetting for hydronic balancing on the radiator.

Task:

A room has a room heating load of 850 W and an installed radiator with a heating output of 900 W at system temperatures of 75/55/22. The connection pipe to the radiator has a nominal diameter of DN 10.

Is the room heating load or the maximum radiator output used to calculate the volume flow?
How many litres of water per hour does the radiator need to keep the room at a room temperature of 22 °C at a design outside temperature of -14 °C?
What is the pipe cross-sectional area in square metres?
What is the resulting flow velocity in m/s?

Important: To calculate the radiator volume flow rate, we need the heat flow $\dot Q$ , the temperature spread $\Delta\vartheta$ and the factor from the density of water $\rho$ and the specific heat capacity of water $c$

Given:

Space heating load $\Phi_{HL,i}$ = 850 W
Radiator output $\dot Q$ = 900 W at 75/55/22
Temperature spread at 75/55: $\Delta\vartheta$ = 20 K
Factor from density and specific heat capacity: 0.86

Wanted:

Power used
Volume flow in l/h
Pipe cross-sectional area in m²?
Flow velocity in m/s

Solution:

Which performance specification is used?

It depends. Theoretically, you can calculate the volume flow with both performance specifications. However, as we are aiming for optimum and energy-efficient operation of the heating system, it is advisable to calculate with the space heating load $\Phi_{HL,i}$ .

This is for the following reason: Many radiators in old and renovated buildings are far oversized. If we were to calculate the volume flow using the oversized radiator output, we would supply the room with more energy than it requires according to the calculated space heating load.

If, on the other hand, the radiator is too small and has a lower output than the calculated room heating load, it would make little sense to calculate with the radiator output, as the room would be undersupplied. The room heating load is therefore decisive for optimum and efficient operation.

Volume flow $\dot V$

The questions about the standard outdoor temperature and the room temperature are perhaps a little confusing, but easier to explain than you might think. The room heating load of 850 W is calculated with the help of the building physics, the standard outside temperature prevailing on site and the desired room temperature. I recommend you read my article on calculating the heating load.

We can therefore calculate the volume flow without any problems using the given values. To calculate the volume flow for the radiator, we use the following formula

$\dot V= 0.86 \cdot \frac{\dot Q} {\Delta\vartheta}$

We can now use this formula to calculate the volume flow and insert the given values.

$\dot V= 0.86 \cdot \frac{850 W} {20K}= \underline{\underline{36.55\frac{l}{h}}}$

The volume flow $\dot V$ for the given radiator is therefore 36.55 l/h.

Pipe cross-sectional area $A$

We can determine the pipe cross-sectional area as in the first example calculation:

The nominal diameter DN 10 has an inner diameter of $d$ = 12.5 mm = 1.25 cm. The radius resulting from $r = \frac{d}{2}$ is then 0.625 cm.

We use the following formula and insert the given values:

$\ A=\pi \cdot r^2 =\pi \cdot (0.00625 m)^2=\underline{\underline{0.000123 m^2}}$

The pipe cross-sectional area is therefore 0.000123 m².

Flow velocity $\omega$

We can determine the flow velocity as in the first example calculation:

First we convert the volume flow from l/h to m³/h and then to m³/s, as the final result of the flow velocity should be given in m/s.

$\ \frac{36.55 l/h}{1000}=\underline{\underline{0.03655 m^3/h}}$

$\ \frac{0.03655 m^3/h}{3600}=\underline{\underline{0.00001015 m^3/s}}$

We can now enter the calculated results into the following formula and obtain the flow velocity

$\ \omega=\frac{\dot V}{A} =\frac{0.00001015 m^3/s}{0.000123 m^2}=\underline{\underline{0.0825m/s}}$

The flow velocity $\omega$ in the connecting pipe to the radiator is therefore 0.0825 m/s.

Conclusion

I hope that these example calculations have given you a better understanding of the subject. I recommend that you now read the article on the continuity law, as this explains the special features of the behaviour of the volume flow and the flow velocity when the pipe diameter changes.

If you have any questions, suggestions or criticism, I look forward to your comments.

Best regards, Martin

Further links and sources:
Wikipedia – Volumetric flow rate
Wikipedia – Flow velocity

About Me

Hi, my name is Martin and I’m a passionate engineer in the field of buildings technology. Here you can read who I am and why I write this blog.

Posts you might also like

Continuity Equation with Example calculation

[latexpage]In this article, I introduce the Continuity Equation, explain why it is important to know it and show the context of the continuity equation with an example calculation. Since this article refers to the article on volumetric flow rate and flow velocity, I...

What is Density?

[latexpage]In this article I show you the physical quantity "density" and an example how you can calculate it with the help of volume and mass. Basics of Density The density $\rho$ (rho) indicates the ratio of mass to volume of a substance. The density of a substance...

Heat Flow Rate – Performance in Heating Technology

[latexpage]In this small article I will introduce you to the heat flow rate, also called heat performance, as another basic and important quantity of heating technology. The heat flow rate is given as $\dot{Q}$ (Q point). Performance P In order to understand the heat...