What are the energy considerations for my off-grid house?

Who really knows? The best that can be done is to come up with some plausible figures then see how the system operates in practice. The main tactic to get a reasonably economic system will be to undersize somewhat initially then add components as experience dictates. However, the following over-precise calculations should serve as a basic sanity check on the whole enterprise.

# Energy Use

First, what energy is needed?

Any predominantly solar based system which can take you through the winter should give plenty of energy in the summer. Therefore, the following calculations are based on the assumption of just about squeaking through the winter with a bit of care: deferring some activities to the summer or, at least, bursts of good weather and storing some energy, particularly as heat, for the worst periods.

## Space Heating

### Conduction

Outline design for this initial calculation could be taken as a 12 m (E/W) x 6 m (N/S) x 2.4 m (height) box with a 2.4 m high asymmetric gabled roof with warm attic. The roof continues down below the 2.4 height for an addition 1.2 m to the north of the main 6 m floor area (for the thermal store).

Total area (including thermal store) =

- E, W and S main walls: (6 × 2 + 12) × 2.4 = 57.6 m².
- North wall: 1.6 × 12 = 19.2 m².
- Gables: (6 × 2.4 / 2) × 2 = 14.4 m².
- Floor: 12 x 7.2 = 86.4 m².
- South facing roof: √(2.4² × 2) × 12 = 40.73 m².
- North facing roof: √(3.2² + (6 - 2.4 + 1.2)²) × 12 = 69.23 m²
- Thermal store E and W walls: ((1.6 × 1.2) + (0.8 × 1.2 / 2)) × 2 = 4.8 m².

giving 292.36 m². Call it 300 m².

Assuming a U value of 0.1 W·m⁻²·K⁻¹ overall (actually better for the roof but that's exposed to the cold sky) conduction heat loss is 30 W for every °C difference between inside and outside.

Taking Scottish building regs heating system sizing numbers for the outside temperature (-4°C) and a reasonably comfy 22°C inside (allowing a bit of cooling from the heat storage) the total heat loss by conduction is 780 W.

### Ventilation

#### Sensible heat

Building regs require 0.45 air changes per hour (though you can probably get away with a lot less using humidity and, maybe, CO₂ sensors). Actually, I'm not sure that number is exactly what they require - I read somewhere else it was so many litres per second per occupant or something. Anyway, habitable volume (excluding roof and thermal store) is 6 × 12 × 2.4 = 172.8 m³ so that's about 78 m³ per hour, 0.022 m³ per second at 1.3 kg·m⁻³, 0.0286 kg per second. Specific heat capacity of dry air at sea level is 1.005 kJ·kg⁻¹·K⁻¹ so the ventilation heat loss in sensible heat would be 0.0286 × 1.005 × (22 - -4) = 747 W.

#### Latent heat

There's also a loss of heat through transport of evaporated water. Assuming slightly different conditions from above:-

Inside | Outside | |
---|---|---|

Temperature | 20°C | 0°C |

Relative humidity | 50% | 100% |

Equilibrium vapour pressure | 2.338 kPa | 0.6105 kPa |

Vapour pressure | 1.169 kPa | 0.6105 kPa |

Molar density | 0.48 mol·m⁻³ | 0.269 mol·m⁻³ |

Net loss | 0.211 mol·m⁻³ | |

Air flow rate | 0.022 m³·s⁻¹ | |

Water vapour flow rate | 0.004642 mol·s⁻¹ | |

Heat of vapourization (at 300K, 17°C) | 43 kJ·mol⁻¹ | |

Energy loss rate | 199.6 W |

#### Heat recovery ventilation

Call it 200 W. Total ventilation heat loss is therefore 747 + 200 = 947 W.

In practice, ventilation losses would be a bit smaller if it is assumed that it is made dependent on actual humidity levels (using a humidity, and perhaps CO₂ sensor) and assuming some care is taken to avoid producing excess humidity inside when the conditions outside make ventilation least effective (high humidity and low temperature).

Still, this rate would mean losing more heat through ventilation than through conduction: hence the decision to use mechanical heat-recovery ventilation (MHRV). At already reasonably high levels of insulation adding MHRV is more cost effective than adding more insulation. Assuming 80% efficient heat recovery this reduces ventilation losses to 190 W.

Total heat loss is now 780 (conduction) + 190 (ventilation) = 970 W.

This is somewhat conservative as, depending on the location, outside temperatures are likely to stay much below zero for only short periods and the thermal mass should help carry the house through those.

## Domestic Hot Water

Typical requirement is for 50 litres per day of hot water. Being slightly generous and assuming this is all at 55°C (actual usage is normally diluted with cold water to somewhat less) and assuming the water needs to be raised from 5°C this gives a heat requirement of 50 × (55 - 5) × 4.184 = 10 460 kJ per day which requires an average of 121 W - say 120 W for round numbers.

## Electricity

From October 2007 to March 2008 I used an average of 240 W of electricity for lighting, computing, refrigeration, cooking and running the central heating pump (plus possibly a little bit of domestic hot water via the immersion heater though most was heated by gas). This was for being at home most of the time.

This was with a very inefficient old fridge (consuming around 130 W pretty much continuously and 14 W even with the compressor turned off (I've no idea what for, and no, there's no light on) and using fairly inefficient cooking methods (over-large electric oven and pans on electric rings rather than induction hob, microwave and pressure cooker). Almost all my lighting was, of course, CFL.

However, the mechanical heat recovery ventilation discussed above also needs electricity: about 50 W for a good system. I think that keeping basic electricity consumption below 150 W on average over many weeks should not be difficult, giving a total consumption, including ventilation, of 200 W.

## Overall Energy Use

On the face of it, the above calculations might indicate a total energy requirement of 1290 W (970 W (space heating) + 120 W (DHW) + 200 W (electricity)). However, pretty much all of the electrical energy consumed will contribute to space heating (except perhaps some of the heat from the exhaust fan of the MHRV). Similarly, it should be possible to recover a little of the energy in the DHW for space heating.

Also, there is heat gain from building occupants. A single sedentary adult might contribute around 120 W.

Therefore, I assume 600 W (space heating), 120 W (DHW) and 200 W (electricity).

# Energy Sources

The energy collectors for the house will most likely be a mix of wind, solar thermal and photovoltaic.

It's really difficult to get good data on the likely availability of energy from these sources. There are a number of reasons:

- Results are so site and installation specific.
- Data is often for the energy entered into the store, not that available. If the energy store is full (thermal store is hot or battery is charged) then the dumped energy is not counted.
- Meteorological solar radiation measurements are usually made in the horizontal plane so effectively discount direct radiation when the sun is near to the horizon (the critical case in the winter, of course).
- Surprisingly often, the data is only available for the spring, summer and autumn but often not for the winter.
- Sometimes the installation is not well explained. Often the orientation of solar panels is not stated explicitly, for example.

## Electricity Generation

An ideal site would have the opportunity to produce some hydroelectricity. However, I'm not betting this whole project on the assumption of finding a suitable plot so, for feasibility calculations, I'm assuming a mix of PV and wind.

Having looked at quite a lot of data I'm going to make up some simple rules of thumb. With any luck they won't be out by much more than a factor of two:

- In the summer the sun shines one tenth of the time (2.4 hours per day) and the wind blows for an hour a day.
- In the winter the sun shines for an hour a day and the wind blows for one tenth of the time.

These would be relative to "nominal" outputs of a panel at 1000 W·m⁻² insolation and a turbine output rating at, say, 10 m·s⁻¹ wind speed.

PV costs about £4 to £5 per watt peak. Wind turbines cost a bit
less, say £2 to £3 per nominal watt including tower and stuff.
On that basis it might be thought that wind is the obvious answer.
However, there can be long periods of no wind whereas the sun
comes up every day and produces *some* output from the PV
even if it's very small. This reduces the need for energy
storage capacity (batteries). Also, PV is a lot less hassle;
you just stick it on the roof and maybe clean it once in a while.
Wind turbines seem to need a bit of work to keep going and the
general engineering to run them safely is a bit more hassle.

Taking these points into consideration I come up with the following:

- 1600 W of PV costing about £8000, generating about 67 W on average in winter and 160 W on average in summer.
- 1600 W of wind, probably two turbines slightly optimistically rated at 1 kW each expecting to actually get the equivalent of 800 W, costing (including poles, cables, etc) £2500 each giving a total of £5000 and generating between them 160 W on average in winter and 67 W in summer.

Total generation both summer and winter would then be 227 W at a cost of about £13 000.

Yes, practical generation is likely to cost nearly £60/W before including the costs of storage (batteries) and conversion (charge control and inversion). Saving £10 000, or even £5000, on the price of the plot for not having services available, though, would make this seem somewhat less insane.

## Solar Domestic Hot Water

For solar thermal energy collection I envisage a combination of commercial evacuated tube water heating collectors and site-built building-integrated warm air collectors.

Since these panels are intended to provide near-year-round heating they are optimized for winter operation. This means mounting them near vertical to make the best possible use of the low winter sun. The consequent reduction in summer output should not be a problem as there's always likely to be energy to spare then.

With PV a simple assumption of the equivalent of one hour's output per day in winter is sufficient. This is because the output of the PV panels is very roughly proportional to the insolation so one hour of full sunshine or two hours of half sunshine are approximately equivalent. This only works with solar thermal when the average temperature of the fluid in the collectors is about the same as the ambient temperatures. In the more normal, and desirable, case where the fluid is warmer the losses are significant and depend heavily on the temperatures involved.

Guessing wildly, I'm going to assume that there's the equivalent per day of one hour at 800 W·m⁻² (slightly less than the one hour at 1000 W·m⁻² assumed above for PV) split as a quarter of an hour at that power, forty minutes at 400 W·m⁻² and another hour and forty minutes at 200 W·m⁻².

Of course, realistically days will be hugely variable and the different levels of power available will blend together. The following calculations are only quoted to more than one digit of accuracy to make it easier to follow what's going on. The real idea is a sanity check that the proposed size of panels could plausibly provide roughly the right amount of hot water.

Taking the Navitron SFB20-47 panel as an example (and a quite possible candidate for actual use) with the SPF test report at scf782en.pdf the output of the panel is:

P = A·(G·η₀ - a1·ΔT - a2·(ΔT)²)

where, with figures for a single SFB20-47 panel:

- P
- Power output (W)
- A
- Aperture (1.747 m²)
- G
- Insolation (W·m⁻²)
- η₀
- Zero-temperature-difference efficiency (0.561)
- a1
- Linear heat loss coefficient (1.61 W·m⁻²·K⁻¹)
- a2
- Quadratic heat loss coefficient (0.004 W·m⁻²·K⁻²)
- ΔT
- Difference between mean collector fluid temperature and ambient (K or °C)

Sizing the panel to raise the required daily amount of DHW (50 l) from 45°C to 60°C (55°C output with 5°C drop in the heat exchanger) in the quarter hour of bright sunshine we get the requirement for a 120 tube (6 panel) array (or fewer larger panels, depending on the economics at the exact time of purchase).

The detailed calculations are in an OpenOffice Spreadsheet but, in summary, the following table assumes the use of the bright quarter hour to heat the hottest water (45°C to 60°C) as noted above, the medium brightness forty minutes to raise water from 27°C to 45°C and the dullest hour and bit to raise water from 5°C to 27°C. Though theoretically you should use the mid-temperature as the average in the panel actually, with the quadratic term there, it's more conservative to use something a bit closer to the warmest. I've used 5°C less than the final temperature. In each case the assumed external temperature is 0°C.

Insolation | Duration | Temp. in | Temp. out | P | Energy | Vol. Heated |
---|---|---|---|---|---|---|

800 W·m⁻² | 15 mins | 45°C | 60°C | 3 649 W | 3 284 kJ | 52.13 l |

400 W·m⁻² | 40 mins | 27°C | 45°C | 1 610 W | 3 864 kJ | 51.11 l |

200 W·m⁻² | 100 mins | 5°C | 27°C | 784 W | 4 707 kJ | 50.94 l |

The total energy for the day is 11 855 kJ giving an average power of 137 W which matches well with the assumed requirement of 120 W.