This calculation assumes:

• A base heating requirement of 100 W/m²
• An extra 20 W for the “Flur + WC” area
• Different target indoor temperatures:

1. 1.OG: 20 °C (assumed ΔT = 15 K)
2. EG: 22 °C (assumed ΔT = 18 K)

• The total pipe length in the house is 741 m, which is used to allocate the 100 m² area proportionally to each circuit.

Total pipe lengths:

• 1.OG circuits: 59 + 51 + 72 + 70 + 82 = 334 m
• EG circuits: 30 + 78 + 70 + 69 + 75 + 85 = 407 m
• Total: 741 m

For a circuit with pipe length X, the allocated area is:

$$ Area_{circuit} = \\frac{X}{741} \\times 100 \\; m² $$

^ Circuit       ^ Pipe Length (m) ^ Allocated Area (m²)          ^
| Raum 1 (a)    | 59              | 59 / 741 × 100 ≈ **7.96**    |
| Raum 1 (b)    | 51              | 51 / 741 × 100 ≈ **6.88**    |
| Raum 2 (a)    | 72              | 72 / 741 × 100 ≈ **9.71**    |
| Raum 2 (b)    | 70              | 70 / 741 × 100 ≈ **9.45**    |
| Raum 3        | 82              | 82 / 741 × 100 ≈ **11.07**   |

**Total 1.OG Area ≈ 45.07 m²**

This calculation assumes:

• A base heating requirement of 100 W/m²
• An extra 20 W for the “Flur + WC” area
• Different target indoor temperatures:

1. 1.OG: 20 °C (assumed ΔT = 15 K)
2. EG: 22 °C (assumed ΔT = 18 K)

• The total pipe length in the house is 741 m, which is used to allocate the 100 m² area proportionally to each circuit.

Total pipe lengths:

• 1.OG circuits: 59 + 51 + 72 + 70 + 82 = 334 m
• EG circuits: 30 + 78 + 70 + 69 + 75 + 85 = 407 m
• Total: 741 m

For a circuit with pipe length X, the allocated area is:

$$ Area_{circuit} = \\frac{X}{741} \\times 100 \\; m² $$

^ Circuit       ^ Pipe Length (m) ^ Allocated Area (m²)          ^
| Raum 1 (a)    | 59              | 59 / 741 × 100 ≈ **7.96**    |
| Raum 1 (b)    | 51              | 51 / 741 × 100 ≈ **6.88**    |
| Raum 2 (a)    | 72              | 72 / 741 × 100 ≈ **9.71**    |
| Raum 2 (b)    | 70              | 70 / 741 × 100 ≈ **9.45**    |
| Raum 3        | 82              | 82 / 741 × 100 ≈ **11.07**   |

**Total 1.OG Area ≈ 45.07 m²**

The EG area is the remainder: 100 – 45.07 ≈ 54.93 m². For each EG circuit, the allocated area is:

$$ Area_{circuit} = \\frac{Pipe\\ Length}{407} \\times 54.93 \\; m² $$

^ Circuit       ^ Pipe Length (m) ^ Allocated Area (m²)                ^
| Flur + WC     | 30              | 30/407 × 54.93 ≈ **4.05**          |
| Wohnen (1)    | 78              | 78/407 × 54.93 ≈ **10.52**         |
| Wohnen (2)    | 70              | 70/407 × 54.93 ≈ **9.44**          |
| Wohnen (3)    | 69              | 69/407 × 54.93 ≈ **9.31**          |
| Wohnen (4)    | 75              | 75/407 × 54.93 ≈ **10.13**         |
| Küche         | 85              | 85/407 × 54.93 ≈ **11.45**         |

**Total EG Area ≈ 54.93 m²**

For each circuit, the heating load is calculated as:

• For most circuits:

$$ Q = \\text{Area} \\times 100 \\; W $$

• For Flur + WC:

$$ Q = \\text{Area} \\times 100 + 20 \\; W $$

The heat delivered by water is:

$$ Q = \\dot{m}\\, c_p\\, \\Delta T, $$  

where:

• \\( c_p \\) ≈ 4180 J/(kg·K) (water’s specific heat)
• \\( \\Delta T \\) is the temperature drop
• \\( \\dot{m} \\) is the mass flow rate in kg/s (≈ L/s)

To compute the flow in L/min, use:

* For 1.OG (ΔT = 15 K):

$$ Flow = \\frac{Q \\times 60}{4180 \\times 15} \\quad (L/min) $$

* For EG (ΔT = 18 K):

$$ Flow = \\frac{Q \\times 60}{4180 \\times 18} \\quad (L/min) $$

^ Circuit       ^ Allocated Area (m²) ^ Q (W)              ^ Flow (L/min) Calculation             ^ Flow (L/min) ^
| Raum 1 (a)    | ~7.96               | 7.96 × 100 = 796     | (796×60)/(4180×15) ≈ 0.76            | ~0.76        |
| Raum 1 (b)    | ~6.88               | 6.88 × 100 = 688     | (688×60)/(4180×15) ≈ 0.66            | ~0.66        |
| Raum 2 (a)    | ~9.71               | 9.71 × 100 = 971     | (971×60)/(4180×15) ≈ 0.93            | ~0.93        |
| Raum 2 (b)    | ~9.45               | 9.45 × 100 = 945     | (945×60)/(4180×15) ≈ 0.90            | ~0.90        |
| Raum 3        | ~11.07              | 11.07× 100 = 1107    | (1107×60)/(4180×15) ≈ 1.06           | ~1.06        |

**Total 1.OG Flow ≈ 4.31 L/min**

^ Circuit       ^ Allocated Area (m²) ^ Q (W)                                 ^ Flow (L/min) Calculation                  ^ Flow (L/min) ^
| Flur + WC     | ~4.05               | 4.05×100 + 20 = 405 + 20 = 425             | (425×60)/(4180×18) ≈ 0.34                  | ~0.34        |
| Wohnen (1)    | ~10.52              | 10.52×100 = 1052                           | (1052×60)/(4180×18) ≈ 0.84                 | ~0.84        |
| Wohnen (2)    | ~9.44               | 9.44×100 = 944                             | (944×60)/(4180×18) ≈ 0.75                  | ~0.75        |
| Wohnen (3)    | ~9.31               | 9.31×100 = 931                             | (931×60)/(4180×18) ≈ 0.74                  | ~0.74        |
| Wohnen (4)    | ~10.13              | 10.13×100 = 1013                           | (1013×60)/(4180×18) ≈ 0.81                 | ~0.81        |
| Küche         | ~11.45              | 11.45×100 = 1145                           | (1145×60)/(4180×18) ≈ 0.91                 | ~0.91        |

**Total EG Flow ≈ 4.39 L/min**

• These figures are first-order estimates.
• The allocation assumes the pipe length is proportional to the area served. In a real system, room boundaries and specific heating demands may adjust these values.
• Actual balancing in a floor heating system may also require manifold adjustments and fine-tuning based on in-situ measurements.

This completes the advanced calculation in Dokuwiki format.

For each circuit, the heating load is calculated as:

• For most circuits:

$$ Q = \\text{Area} \\times 100 \\; W $$

• For Flur + WC:

$$ Q = \\text{Area} \\times 100 + 20 \\; W $$

The heat delivered by water is:

$$ Q = \\dot{m}\\, c_p\\, \\Delta T, $$  

where:

• \\( c_p \\) ≈ 4180 J/(kg·K) (water’s specific heat)
• \\( \\Delta T \\) is the temperature drop
• \\( \\dot{m} \\) is the mass flow rate in kg/s (≈ L/s)

To compute the flow in L/min, use:

* For 1.OG (ΔT = 15 K):

$$ Flow = \\frac{Q \\times 60}{4180 \\times 15} \\quad (L/min) $$

* For EG (ΔT = 18 K):

$$ Flow = \\frac{Q \\times 60}{4180 \\times 18} \\quad (L/min) $$

^ Circuit       ^ Allocated Area (m²) ^ Q (W)              ^ Flow (L/min) Calculation             ^ Flow (L/min) ^
| Raum 1 (a)    | ~7.96               | 7.96 × 100 = 796     | (796×60)/(4180×15) ≈ 0.76            | ~0.76        |
| Raum 1 (b)    | ~6.88               | 6.88 × 100 = 688     | (688×60)/(4180×15) ≈ 0.66            | ~0.66        |
| Raum 2 (a)    | ~9.71               | 9.71 × 100 = 971     | (971×60)/(4180×15) ≈ 0.93            | ~0.93        |
| Raum 2 (b)    | ~9.45               | 9.45 × 100 = 945     | (945×60)/(4180×15) ≈ 0.90            | ~0.90        |
| Raum 3        | ~11.07              | 11.07× 100 = 1107    | (1107×60)/(4180×15) ≈ 1.06           | ~1.06        |

**Total 1.OG Flow ≈ 4.31 L/min**

^ Circuit       ^ Allocated Area (m²) ^ Q (W)                                 ^ Flow (L/min) Calculation                  ^ Flow (L/min) ^
| Flur + WC     | ~4.05               | 4.05×100 + 20 = 405 + 20 = 425             | (425×60)/(4180×18) ≈ 0.34                  | ~0.34        |
| Wohnen (1)    | ~10.52              | 10.52×100 = 1052                           | (1052×60)/(4180×18) ≈ 0.84                 | ~0.84        |
| Wohnen (2)    | ~9.44               | 9.44×100 = 944                             | (944×60)/(4180×18) ≈ 0.75                  | ~0.75        |
| Wohnen (3)    | ~9.31               | 9.31×100 = 931                             | (931×60)/(4180×18) ≈ 0.74                  | ~0.74        |
| Wohnen (4)    | ~10.13              | 10.13×100 = 1013                           | (1013×60)/(4180×18) ≈ 0.81                 | ~0.81        |
| Küche         | ~11.45              | 11.45×100 = 1145                           | (1145×60)/(4180×18) ≈ 0.91                 | ~0.91        |

**Total EG Flow ≈ 4.39 L/min**

• These figures are first-order estimates.
• The allocation assumes the pipe length is proportional to the area served. In a real system, room boundaries and specific heating demands may adjust these values.
• Actual balancing in a floor heating system may also require manifold adjustments and fine-tuning based on in-situ measurements.

This completes the advanced calculation in Dokuwiki format.