Volumetric Flow Rate and Flow Velocity

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In this article I introduce the volumetric flow rate and the flow velocity as further physical quantities and show you two examples how you can calculate them.

Volumetric Flow Rate

The volumetric flow rate $Q$ or $\dot V$ indicates how much volume $V$ of a gas or liquid flows through a pipe or channel in a certain period of time $t$ . The question is therefore asked in relation to a heating system:

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

This question is asked in thermodynamics, because the volumetric flow rate is an important parameter for calculating the heat output of a radiator. The unit of volumetric flow rate is cubic meters per hour $\frac{m^3}{h}$ or liters per hour $\frac{l}{h}$ .

Volumetric Flow Rate Formula

Volume Flow Rate visualisation — Volumetric Flow Rate visualisation

The volumetric flow rate can be calculated with the help of the volume and the time, with the pipe cross-sectional area $A$ and the flow velocity $\omega$ (small Omega), but also with the heat flow $\dot Q$ , the specific heat capacity $c$ , the density $\rho$ and the temperature difference $\Delta\vartheta$ . The following formulas will give you an overview:

volumetric flow rate = volume / time -> $\dot V=\frac{V}{t}$

volumetric flow rate = pipe cross-sectional area * Flow velocity -> $\dot V=A \cdot \omega$

Volumetric flow rate= heat flow / density * specific heat capacity * temperature difference -> $\dot V=\frac{\dot Q} {\rho \cdot c \cdot \Delta\vartheta}$

Flow Velocity

The flow velocity is indicated differently in the literature. It is known as $\omega$ (small Omega), $v$ and $c$ . I chose the little Omega $\omega$ . The flow velocity indicates the velocity of a flow in a pipe and is used for pipe flow calculations. The unit is meters per second $\frac{m}{s}$ .

Flow Velocity Formula

Flow velocity = volumetric flow rate / pipe cross-sectional area -> $\omega=\frac{\dot V}{A}$

Example Calculations

In order to understand the topic a little better, I will show you in the following example calculations how you can use the formulas.

1. Example Calculation – Volumetric Flow Rate and Flow Velocity

Task

In a medium-heavy pipe with a nominal diameter of DN 15 (according to DIN EN 10255), 7 liters of water are to be conveyed in a period of one minute. IMPORTANT: “DN” (Diameter Nominal) is used in europe and internationally. In the US “NPS” (Nominal Pipe Size) is used. To keep it simple we calculate this example in DN.

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

Given

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

Wanted

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

Solution

volumetric flow rate $\dot V$

First we calculate the volumetric flow rate with the following formula and then use the given values.

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

Since the volumetric flow rate is required in the unit m³/h, the result still has to be converted.

Important: The following applies to the conversion into cubic meters/hour or into liters/minute or liter/second:

1 cubic metre = 1000 litres
1 hour = 60 minutes ⇒ 60 / 1000 = 0,06
1 hour = 3600 seconds ⇒ 3600 / 1000 = 3,6

1 cubic meter/hour [m³/h] = 16.67 liters/minute [l/min]
⇒ Divisor for conversion: 0.06 ⇒ x m³/h / Divisor = y l/min
1 liter/minute [l/min] = 0.06 cubic meters/hour [m³/h]
⇒ Divisor for conversion: 16.67 ⇒ x l/min / Divisor = y m³/h

1 cubic meter/hour [m³/h] = 0.278 litres/second [l/s]
⇒ Divisor for conversion: 3.6 ⇒ x m³/h / Divisor = y l/s
1 liter/second [l/s] = 3.6 cubic meters/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 $\dot V$ thus amounts to 0.42 m³/h.

Pipe Cross-Sectional Area $A$

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

To determine the inner diameter $d$ and the resulting radius $r$ of a medium-weight pipe of DN 15, we have to look into a conversion table (or you know the values by heart). A look at DIN EN 10255 or the nominal width tables of Wikipedia (GER) will help here.

The nominal diameter DN 15 therefore has an outside diameter of $d$ = 21,3 mm = 2,13 cm and an inside diameter of $d$ = 16 mm = 1,6 cm. The inside diameter is used for the calculation. The radius resulting from $r = \frac{d}{2}$ is then 0.8 cm.

We use the following formula and use the given values:

$\ A=\pi \cdot r^2 \Longrightarrow A=\pi \cdot (0,008 m)^2=\underline{\underline{0,0002 m^2}}$

The pipe cross-sectional area thus amounts to 0.0002 m².

Flow velocity $\omega$

First, we convert the volumetric flow rate from m³/h to m³/s, since the final result of the flow velocity is to be given in m/s. The flow velocity is to be calculated in m³/s. The flow velocity is to be calculated in m³/h. Here not the divisor 3,6 is used but 3.600, because cubic meters remain and are not converted into liters (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 in the following formula and obtain the flow velocity.

$\ \omega=\frac{\dot V}{A} \Longrightarrow \omega=\frac{0,0001166 m^3/s}{0,0002 m^2}=\underline{\underline{0,58 m/s}}$

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

2. Example Calculation – Calculate Volumetric Flow Rate for Radiators

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

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Task

One room has a room heating load of 850 W and an installed radiator with a heating capacity 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 volumetric flow rate?
How many litres of water per hour does the radiator need to keep the room at a standard outside temperature of -14 °C to a room temperature of 22 °C?
What is the pipe cross-sectional area in m²?
What is the resulting flow velocity in the unit m/s?

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

Given

Room heating load $\Phi_{HL,i}$ = $850 W$
Radiator output $\dot Q$ = $900 W$ at 75/55/22
Temperature spread at 75/55: $\Delta\vartheta = 20 K$
Specific heat capacity: $c = 1.16\frac{Wh}{kg\cdot K}$
Density: $\rho = 1\frac{kg}{l}$

Wanted

volumetric flow rate in l/h
Pipe cross-sectional area in m²?
Flow velocity in m/s

Solution

What performance specification is used?

It depends. Theoretically, it is possible to calculate the volumetric flow rate with both performance data. However, since we strive for an optimal and energy-efficient operation of the heating system, it is recommended to calculate with the room heating load $\Phi_{HL,i}$ .

This has the following reason: many radiators in old and renovated buildings are far oversized. If we calculated the volumetric flow rate with the oversized radiator output, we would supply more energy to the room than it requires according to the calculated room heating load.

However, if 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.

Volumetric Flow Rate $\dot V$

The questions about standard outside temperature and room temperature may be a little confusing, but they are easier to answer than expected. The room heating load of 850 W is calculated with the help of building physics, the local standard outside temperature and the desired room temperature. I recommend my contribution to the heating load calculation.

So we can calculate the volumetric flow rate without any problems with the given values. To calculate the volumetric flow rate for the radiator we use the following formula:

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

To understand how the unit l/h results, you can shorten the units of the given sizes below.

$\dot V=\frac{W \cdot kg \cdot K \cdot l}{Wh \cdot kg \cdot K} = \frac{l}{h}$

In individual formula collections, the above-mentioned formula for the volumetric flow rate is represented by a factor of 0.86. The factor 0.86 contains the constants of the specific heat capacity and the density of water and is calculated as follows:

$\frac{1 kg \cdot K \cdot l} {1,16 Wh \cdot 1kg} = 0,86 \frac{K \cdot l}{Wh}$

We can also simplify the formula and use a factor of 0.86 instead of density and specific heat capacity for water:

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

With this formula we can now calculate the volumetric flow rate and use the given values.

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

The volumetric flow rate $\dot V$ for the given radiator is thus 33.55 l/h.

Pipe cross-sectional area $A$

The pipe cross-sectional area can be determined as in the first example calculation:

The nominal diameter DN 10 has an inside 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 use the given values:

$\ A=\pi \cdot r^2 \Longrightarrow A=\pi \cdot (0,00625 m)^2=\underline{\underline{0,000123 m^2}}$

The pipe cross-sectional area thus amounts to 0.000123 m².

Flow velocity $\omega$

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

First we convert the volumetric flow rate from l/h to m³/h and then to m³/s, since the final result of the flow velocity is to be given in m/s. The flow velocity can be calculated in the same way as in the first example.

$\ \frac{33,55 l/h}{1000}=\underline{\underline{0,03355 m^3/h}}}$

$\ \frac{0,03355 m^3/h}{3600}=\underline{\underline{0,00000932 m^3/s}}$

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

$\\omega=\frac{\dot V}{A} \Longrightarrow \omega=\frac{0,00000932 m^3/s}{0,000123 m^2}=\underline{\underline{0,076 m/s}}}$

The flow velocity $\omega$ in the connection pipe to the radiator is thus 0.076 m/s.

Conclusion

I hope with these example calculations I could bring the matter a little closer to you. Now I recommend you to read the article about the continuity law because it explains the peculiarity of the behavior of the volumetric flow rate and the flow velocity when changing the pipe diameter.

Further information about the volumetric flow rate and the flow velocity can be found on the following pages. If you have questions, suggestions or criticism, I look forward to your comments.

Greetings, Martin

Further links and sources:
Wikipedia – Volumetic Flow Rate
Wikipedia – Flow Velocity

Continuity Equation with Example calculation

What is Density?

Heat Flow Rate - Performance in Heating Technology

What is Specific Heat Capacity?

Energy and Work in Heating Technology

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Volumetric Flow Rate and Flow Velocity