Configuration of direct heating equipment
"Direct heating" means the heat goes straight from a hot body (a radiator or a heater) to the room. In a hot-water system the parts are:
- Boiler: burns gas, oil or wood, or uses electricity or a heat pump, to heat the water (typically 55 to 80 °C).
- Circulating pump: pushes the water round the loop.
- Supply pipe (hot) and return pipe (cooler).
- Radiators: metal panels with a large surface that give heat by warm surface and moving air (convection).
- Expansion tank: water gets about 3 percent bigger when heated from cold to 80 °C. This tank has a cushion of air or a rubber bag to absorb the extra volume. Without it, pressure rises and a safety valve opens.
- Air vent: lets trapped air escape so radiators do not get cold spots.
- Valves: a thermostatic radiator valve (TRV) opens and closes with room temperature.
The system is closed: the same water goes round again and again. Only heat leaves it.
Design of direct heating and piping
Step 1: how much heat does each room lose?
Heat loss through a wall, window or roof is Q = U × A × ΔT, where U is how easily heat passes (W/m²K), A is the area and ΔT is the inside-outside temperature difference. Add all the walls, windows and the air leakage. The radiator must give at least this much.
Step 2: how much water flows?
The heat carried by water is Q = m × c × ΔT, with c = 4.186 kJ/kg·K. Choose ΔT (for example 20 K between supply and return). Then the flow follows. In litres per hour: Q (kW) = flow × 4.186 × ΔT ÷ 3600.
Step 3: pipe size
Pipe area A = flow ÷ speed. Keep water speed between about 0.3 and 1 m/s. Faster water makes noise and wears the pipe; slower water needs wide, costly pipes.
Step 4: layout
- Two-pipe system: every radiator connects to a hot supply and a return. All get nearly the same temperature. Most common.
- One-pipe system: one loop passes through all radiators in a row; the last radiator is cooler.
- Balancing valves share the flow so far rooms also get enough.
- Pipes slope slightly toward an air vent, and have bends or loops to allow for expansion.
- Insulate pipes that pass through cold spaces.
Key formulas and definitions
- Q = m × c × ΔT (heat carried by water, c = 4.186 kJ/kg·K)
- Q (kW) = flow (L/h) × 4.186 × ΔT ÷ 3600
- Heat loss: Q = U × A × ΔT
- Pipe area A = flow ÷ speed; d = √(4A ÷ π)
- Water expands about 3 percent from 10 °C to 80 °C
Worked examples
1. A room loses 2 kW. Supply and return differ by 20 K. Find the water flow in L/h.
Q = flow × 4.186 × ΔT ÷ 3600, so flow = 2 × 3600 ÷ (4.186 × 20) = 7200 ÷ 83.7 = 86 L/h.
2. A loop carries 500 L/h with ΔT = 15 K. What heat does it deliver?
Q = 500 × 4.186 × 15 ÷ 3600 = 8.72 kW.
3. A room has walls, windows and roof of total area 30 m² with average U = 0.6 W/m²K. Inside 20 °C, outside 0 °C. Find the heat loss.
Q = U × A × ΔT = 0.6 × 30 × 20 = 360 W.
4. A pipe must carry 0.3 L/s at 0.8 m/s. Find the pipe diameter.
Flow = 0.3 L/s = 0.0003 m³/s. A = 0.0003 ÷ 0.8 = 0.000375 m². d = √(4 × 0.000375 ÷ 3.14) = 0.0219 m, about 22 mm. Choose the next standard size, 22 mm.
5. A system has 1000 L of water. It is heated from 10 °C to 80 °C and grows by 3 percent. What volume must the expansion tank take?
3 percent of 1000 L = 30 L. The tank must absorb at least 30 L (plus a safety margin).
Common mistakes
- Thinking heat is used up in the water. The water only carries heat; the radiator gives it to the room.
- Forgetting the expansion tank. Hot water takes more space and pressure can burst a pipe.
- Mixing kW and W, or L/h and L/s. 1 kW = 1000 W; 1 L/s = 3600 L/h.
- Making the pipe too narrow to save money. Water speed rises, the pipe whistles and wears out.