# N°3 – Flows to be transmitted training – NVQ level

The diameter of the piping depends on the flow to be transmitted.
If you have already studied the file “Practical calculation of water and air flows – Part 2”, go straight on to the last paragraph of this exercise.

The flow to be transmitted is:
• Proportional to the power to be transported (the greater the power of transfer, the greater the flow will be).
• Inversely proportional to the temperature loss or gain of the water (in heating, if the water cools only slightly), it means the flow is important).

With ΔT in [K]

#### Question

What is the water flow rate, in [m³/h] (US gpm), for an installation of 150 [kW] running at a regime of 80/60 [°C] (176/140°F)?
qv = 6.46 [m³/h] (28.4 US gpm)

Explanation:
qv = P / (1.16 x ∆T)
= 150 / (1.16 x ((80 – 60))
= 150 / (1.16 x 20)
= 6.46 [m³/h] (28.4 US gpm)

#### Question

What is the water flow rate, in [m³/h] (US gpm) for an installation of 150 [kW] running at a temperature regime of 75/65 [°C] (167/149°F)?
qv = 12.93 [m³/h] (56.9 USgpm)

Explanation:
qv = P / (1.16 x ∆T)
= 150 / (1.16 x ((75 – 65))
= 150 / (1.16 x 10)
= 12.93 [m³/h] (56.9 USgpm)

#### Question

What is the water flow rate, in [m³/h] (US gpm) for an installation of 150 [kW] running at a regime of 50/40 [°C] (122/104°F)?
qv = 12.93 [m³/h] (56.9 USgpm)

Explanation:
qv = P / (1.16 x ∆T)
= 150/ (1.16 x ((50 – 40))
= 150 / (1.16 x 10)
= 12.93 [m³/h] (56.9 USgpm)

#### Question

What is the water flow rate in [l/h] (US gpm) in a radiator of 1500 [W] running at a regime of 80/60 [°C] (176/140°F)?
qv = 64.6 [l/h] (0.28 USgpm)

Explanation:
qv = P / (1.16 x ∆T)
= 1500 / (1.16 x ((80 – 60))
= 1500 x (1.16 x 20)
= 64.6 [l/h] (0.28 USgpm)

Installations with the same power may have very different flow rates depending on the temperature regimes

#### Question

Three perfectly identical buildings in architectural terms (with the same heat loss factor), have heating networks as defined in the diagrams below.
The hot water flow rates which supply them (and the corresponding piping) are very different.
Calculate in [m³/h] the flow rates of the pumps in the three buildings: From left to right:
qv1 = 3.44 [m³/h]; (15.1 US gpm)qv2 = 6.89 [m³/h]; (30.3 USgpm)
qv3 = 3.44 [m³/h]; (15.1 US gpm)
qv4 = 6.89 [m³/h]; (30.3 USgpm)

Explanation:
qv1 = P / (1.16 x ∆T) = 80 / (1.16 x ((80-60))
= 80 / (1.16 x 20) = 3.44 [m³/h] (15.1 US gpm)

qv2 = P / (1.16 x ∆T) = 80 / (1.16 x ((50-40))
= 80 / (1.16 x 10) = 6.89 [m³/h] (30.3 US gpm)

qv3 = P / (1.16 x ∆T) = 80 / (1.16 x ((80-60))
= 80 / (1.16 x 20) = 3.44 [m³/h] (15.1 US gpm)

qv4 = P / (1.16 x ∆T) = 80 / (1.16 x ((50-40))
= 80 / (1.16 x 10) = 6.89 [m³/h] (30.3 US gpm)