Electric water heaters lose heat through:

• Tank side wall

• Top cover 

• Bottom cover

The loss occurs by conduction through insulation + convection to air.

1️ - Basic Heat Loss Formula

Heat loss through insulation is calculated using formula:   Q = k × A × ΔT / d​

Where:

Q   : Heat loss (W)

k    : Thermal conductivity of insulation (W/m·K)

A   : Surface area (m²)

ΔT : Temperature difference (°C)

d   : Insulation thickness (m)

2️ - Temperature Difference

ΔT=T water − T ambient

Typical values:

Water temperature      :   65°C

Ambient temperature  :   20°C

Then ΔT=45°C

3️ - Tank Surface Area

For a cylindrical tank:

Side area : Aside=π * D * H 

Top & bottom head - Atop= π * D2 /4

Total area : A=Aside+2 Atop​

4️ - Thermal Conductivity of Insulation

Thermal conductivity ( k ) For typical materials used in water heaters  :

Polyurethane foam (PU)  :  0.022 – 0.028 W/m.K

Rock wool                        :  0.035 – 0.045 W/m.K

Glass wool                       :  0.038 – 0.050 W/m.K

Most heaters use:  Rigid PU foam k=0.025 W/m.K

5️ - Example Calculation

Cylindrical water heater data :-

• Volume = 50 L

• Diameter = 0.46  m

• Height = 0.60 m

Surface area

Side area = π × 0.450 × 0.60 = 0.87  m2

Top area = π × 0.450 × 0.460 /4 = 0.17 m2

Bottom area =π × 0.450 × 0.460 /4 =  0.17 m2

Total area = 0.16+0.16+0.99 = 1.199 m2

Insulation thickness , d= 35 mm = 0.035 m

Heat loss

Q = k × A × ΔT / d​ = 0.025 * 1.31* 45 / 0.04 = 38.55 W 

6️ - Daily Energy Loss

Energy loss / day = Q ×24 = 38.55 *24 ≈ 0.93 kWh/day

7️ - Suggested Insulation Thickness

30 L   :   25 – 30 mm

50 L   :   30 – 35 mm

80 L   :   35 – 40 mm

100 L :   40 – 45 mm

8️ -  Target standby heat loss:

50 L    :  30 – 40 W

80 L    :  35 – 50 W

100 L  :  45 – 60 W

This ensures compliance with European energy efficiency regulations.

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