hydraulicheabgleich
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Advanced Floor Heating Calculations for a 100 m² House
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.OG: 20 °C (assumed ΔT = 15 K)
- 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.
Step 1: Area Allocation
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:
$$ Area_{circuit} = \\frac{X}{741} \\times 100 \\; m² $$
1.OG Area Allocation
^ 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**
Advanced Floor Heating Calculations for a 100 m² House
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.OG: 20 °C (assumed ΔT = 15 K)
- 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.
Step 1: Area Allocation
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:
$$ Area_{circuit} = \\frac{X}{741} \\times 100 \\; m² $$
1.OG Area Allocation
^ 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**
EG Area Allocation
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**
Step 2: Heating Load (Q) Calculation
For each circuit, the heating load is calculated as:
- For most circuits:
$$ Q = \\text{Area} \\times 100 \\; W $$
- For Flur + WC:
$$ Q = \\text{Area} \\times 100 + 20 \\; W $$
Step 3: Required Flow Calculation
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):
$$ Flow = \\frac{Q \\times 60}{4180 \\times 18} \\quad (L/min) $$
1.OG Calculations (ΔT = 15 K)
^ 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**
EG Calculations (ΔT = 18 K)
^ 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**
Final Notes
- 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.
Step 2: Heating Load (Q) Calculation
For each circuit, the heating load is calculated as:
- For most circuits:
$$ Q = \\text{Area} \\times 100 \\; W $$
- For Flur + WC:
$$ Q = \\text{Area} \\times 100 + 20 \\; W $$
Step 3: Required Flow Calculation
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):
$$ Flow = \\frac{Q \\times 60}{4180 \\times 18} \\quad (L/min) $$
1.OG Calculations (ΔT = 15 K)
^ 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**
EG Calculations (ΔT = 18 K)
^ 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**
Final Notes
- 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.
hydraulicheabgleich.txt · Last modified: 2025/03/08 10:29 by burim