Flow rate, power and temperature difference of water are linked through the formula:

P = q_{v} × 1.16 × ΔT

With:

– P in [kW]

– q_{v} in [m³/h]

– 1.16: Volumetric heat of water in [kWh/m³ K]

– ∆T: Temperature difference gained or lost by water in [°C] (or [K])

This formula is logical:

The greater the water flow qv, the more the transmitted power (of heating up or cooling down) is increased.

But taken alone the flow amount is not enough to judge the distributed power (of heating up or cooling down). Hence we can send as much cold water as we like through a radiator, but it won’t get any warmer.

For there to be heating or cooling, the temperature of water must vary by ΔT.

When working on boilers or chillers, their power is generally calculated in [kW] from flow rates expressed in [m³/h]. When working on radiators or convector ventilators, the Watt and the [l/h] are more suitable. We therefore use them as units:

**P
**=**
q**_{v}×**
1.16 **×**
ΔT***With:**–
P in [W]**–
q*_{v}*
in [l/h]**–
1.16: Volumetric heat of water in [kWh/m³ K]**–
∆T: Range of water temperature loss or gain in [°C] (or [K])*

#### Question

Using the formula P = q_{v} × 1.16 × ΔT, calculate the power in [kW] of a boiler which heats by 15 [°C] (15 [K]) a water flow of 7.5 [m³/h] (33 US gmp)?

P = q

_{v}× 1.16 × ΔT

P = 7.5 × 1.16 × 15 =131 [kW]

#### Question

What is the power in [kW] of a boiler which heats 12 [m³/h] (52.8 US gmp) of water from: 70 [°C] to 80 [°C] (from 158 °F to 176 °F)?

ΔT = 80 [°C] – 70 [°C] = 10 [K]

P = q_{v}× 1.16 × ΔT

P = 12 × 1.16 × 10 = 139 [kW]

#### Question

What is the refrigeration capacity in [kW] of a chiller capable of cooling by 6 [°C] (6 [K]) the temperature of a water flow of 4.8 [m³/h] (21.13 US gmp)?

P = 4.8 × 1.16 × 6 = 33 [kW]

#### Question

What is the power of a radiator in which 80 [l/h] (0.35 US gmp) of water cools by 20 [°C] (20 [K])?

P = 80 × 1.16 × 20 = 1,856 [W]

#### Question

What is the refrigeration capacity in [W] of a cooling coil in a convector ventilator in which the water flow rate of 370 [l/h] (1.63 US gmp) heats up from 7 [°C] à 12 [°C] (from 44.6 °F to 53.6 °F)?

ΔT = 12 [°C] – 7 [°C] = 5 [K])

P = 370 × 1.16 × 5 = 2,146 [W]

#### Question

What is the power of a radiator fed with a flow of 0.045 [m³/h] (0.198 US gmp) at a regime of 75 / 60 [°C] (167 / 140 °F)?

q

_{v}= 0.045 [m³/h] = 45 [l/h]

ΔT = 75 [°C] – 60 [°C] = 15 [°C] (15 [K])

P = 45 × 1.16 × 15 = 783 [W]