Heat Transfer Triangle
kW
l/s
(DTxSHC)
With any mathematical triangle, you need to have two pieces of information to work out the third.
You cover the part of the triangle you want to calculate and this shows you what to do for working it out.
The component in the top of the triangle will always be divided by either of the components below.
The components in the bottom of the triangle will either be multiplied or divided by one another, depending on what you're trying to calculate.
Calculating Flow Rate
kW
l/s
(DTxSHC)
Let's start by working out the required litres per minute (l/m) to deliver 8.5kW with a 5K DT and SHC of 3.8 (75:25 water to glycol mix)
Cover the bottom left part of the triangle (the bit you want to find), the equation you're left with is kW/(DTxSHC)
Always start with the part of the equation that is in brackets, so: 5x3.8=19. Next: 8.5/19=0.447 l/s (3dp). Finally, multiply this answer by 60 to convert from l/s to l/m: 0.447x60=26.82 l/m
Calculating Output
kW
l/s
(DTxSHC)
This time, let's work out how many kW can be achieved with a given flow rate of 40 l/m, this time using a 5K DT and SHC of 4.2 (pure water).
Cover the top part of the triangle (the bit you want to find), the equation you're left with is l/sx(DTxSHC).
As always, start with the brackets, so: 5x4.2=21. Next, convert l/m to l/s: 40/60=0.667 l/s (3dp). Finally: 0.667x21=14.01kW (2dp).
Calculating Delta T
kW
l/s
(DTxSHC)
The third and final function of the heat transfer triangle is to claculate Delta T (DT). For this example we will use 11.2kW with a flow rate of 0.533 l/s and a SHC of 4.2.
Cover only the DT part of the triangle (the bit you want to find), the equation you're left with is kW divided by l/s divided by SHC.
This time there are no brackets, so: 11.2/0.533=21.013 (3dp), 21.013/4.2=5.00 k (2dp)
Delta T x S.H.C
kW
l/s
(DTxSHC)
Multiplying the Delta T by the Specific Heat Capacity calculates the amount of energy required to change the temperature of a substance by 1 degree kelvin.
When the SHC is expressed as 4.2 this is kiloJoules (kJ) and therefore the units are kJ/Kg/K. If the SHC is expressed as 4200 this is Joules and the units would be J/g/K.
Therefore, this part of the equation triangle calculates the amount of energy per unit of mass. Which is why it's also known as the Mass Flow Rate Triangle.
Heat Transfer Triangle
DEP | MITSUBISHI ELECTRIC EUROPE | UK
Created on March 10, 2025
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Transcript
Heat Transfer Triangle
kW
l/s
(DTxSHC)
With any mathematical triangle, you need to have two pieces of information to work out the third.
You cover the part of the triangle you want to calculate and this shows you what to do for working it out.
The component in the top of the triangle will always be divided by either of the components below.
The components in the bottom of the triangle will either be multiplied or divided by one another, depending on what you're trying to calculate.
Calculating Flow Rate
kW
l/s
(DTxSHC)
Let's start by working out the required litres per minute (l/m) to deliver 8.5kW with a 5K DT and SHC of 3.8 (75:25 water to glycol mix)
Cover the bottom left part of the triangle (the bit you want to find), the equation you're left with is kW/(DTxSHC)
Always start with the part of the equation that is in brackets, so: 5x3.8=19. Next: 8.5/19=0.447 l/s (3dp). Finally, multiply this answer by 60 to convert from l/s to l/m: 0.447x60=26.82 l/m
Calculating Output
kW
l/s
(DTxSHC)
This time, let's work out how many kW can be achieved with a given flow rate of 40 l/m, this time using a 5K DT and SHC of 4.2 (pure water).
Cover the top part of the triangle (the bit you want to find), the equation you're left with is l/sx(DTxSHC).
As always, start with the brackets, so: 5x4.2=21. Next, convert l/m to l/s: 40/60=0.667 l/s (3dp). Finally: 0.667x21=14.01kW (2dp).
Calculating Delta T
kW
l/s
(DTxSHC)
The third and final function of the heat transfer triangle is to claculate Delta T (DT). For this example we will use 11.2kW with a flow rate of 0.533 l/s and a SHC of 4.2.
Cover only the DT part of the triangle (the bit you want to find), the equation you're left with is kW divided by l/s divided by SHC.
This time there are no brackets, so: 11.2/0.533=21.013 (3dp), 21.013/4.2=5.00 k (2dp)
Delta T x S.H.C
kW
l/s
(DTxSHC)
Multiplying the Delta T by the Specific Heat Capacity calculates the amount of energy required to change the temperature of a substance by 1 degree kelvin.
When the SHC is expressed as 4.2 this is kiloJoules (kJ) and therefore the units are kJ/Kg/K. If the SHC is expressed as 4200 this is Joules and the units would be J/g/K.
Therefore, this part of the equation triangle calculates the amount of energy per unit of mass. Which is why it's also known as the Mass Flow Rate Triangle.