SOLAR ENERGY

Utilization

ENGS-44 Sustainable Design

Benoit Cushman-Roisin8 April 2014

Recapitulation

1. We know how much energy the sun provides as a function of

- latitude of location- orientation of surface (window, roof)- month of year- hour of day

- cloudiness

2. We know the energy need of the buildingas a function of

- R-values of walls, windows, roof, etc.- respective surfaces of walls, roof, etc.- air infiltration

- how cold it is outside

Solar Heat Gain Factors(SHGFs)

Heat Loss (HL)

Degree-Days (DD)

Cloudiness factor (%)

The question now is: How much of the need (part 2) can we meet with the sun (part 1)?

In building design, there are basically three passive solar techniques:

1. Direct gain (= let the sun enter through windows)

2. Trombe wall (= enhanced direct gain)

3. Greenhouse (= enhanced trombe wall)

Caution!These techniques, if used at all, need to be used extremely carefully, for it is very easy to focus on cold winter days and then have a building that is uncomfortably warm in summer.

Calculations Recipe for Direct Gain

1. Determine square-feet of glazing (windows) on East (Ae), South (As), West (Aw) and North (AN) sides of the building.

2. Adjust these areas downward for shading by overhangs, vegetation or neighboring structures

3. Select a month and pick the values SHGFe, SHGFs, SHGFw, and SHGFN.

4. Correct the SHGF’s for cloudiness (% sun).

5. Correct for partial reflection by window glass(87% or applicable solar heat gain coefficient (SHGC)).

6. Multiply and add for each side of the building:Solar heat gain per day of the month =

SHG = SGHFe x Ae + SHGFs x As + SHGFw x Aw + SHGFN x AN

6. Multiply by number of days in the month.

7. Repeat for other months of the heating season and add the numbers.

Example: Salt-box house in Lebanon, NH

Near 40oN → SHGFs, in BTUs/(ft2.day), and cloudiness factors, in %

House structure

Need to multiply by 0.87to account for reflectionat window surface

East South West North Total R-value U = 1/R

Window areas 64 162 35 10 271 ft2 1.92 0.5208

External walls 1,898 ft2 21.37 0.0468

Roof 1,520 ft2 31.97 0.0313

Heating month East South West North Cloudiness

September 906 1344 906 238 57%

October 712 1582 712 176 55%

November 508 1596 508 126 46%

December 427 1550 427 104 46%

January 514 1626 514 122 46%

February 733 1642 733 168 55%

March 946 1388 946 228 56%

April 1105 976 1105 308 54%

May 1200 716 1200 430 57%

Add infiltration: I = 4,220 BTUs/(day.oF) → HL = 6,660 + 4,220 = 10,880 BTUs/(day.oF)

September 207

October 558

November 858

December 1,269

January 1,450

February 1,223

March 1,038

April 653

May 308

Climatologicaldegree-daysfor Lebanon, NH

Compare energy demand to solar supply, month after month:

September Demand is HL x Degree-days= (10,880 BTUs/day.oF) x (207 oF.days) = 2.252 x 106 BTUs

Supply is (SHGFeastAeast + …)(0.87 window reflection)(57% cloudiness)= (906 x 64 + 1344 x 162 + 906 x 35 + 238 x 10)(0.87)(0.57)= 153,631 BTUs/day

There are 30 days in September → 153,631 x 30 = 4.609 x 106 BTUs

Good news: Supply is more than enough to cover the demand !

Energy

demand

Solar

supply Difference

September 2.252 4.609 + 2.357

October 6.071 4.873 -1.198

November 9.335 3.723 - 5.612

December 13.807 3.653 - 10.154

January 15.776 3.914 - 11.862

February 13.306 4.599 - 8.707

March 11.293 4.845 - 6.447

April 7.105 3.814 - 3.291

May 3.351 3.676 +0.325

Values in million BTUsfor each month

In winter, solar energy is rarely enough, but it does make a significant contribution. The danger is to provide too much heat the rest of the year.

Similar calculations for the remaining heating months of the year. Results are:

In the winter months, when the solar energy input fails to meet the building demand, additional heat must be supplied from a furnace or other source (solar panels on roof? geothermal heat?)

Alternatively, one can decrease the demand by increasing the insulation of the building, for example, by drawing curtains at night.

or …

one can be clever and get more free energy from the sun !

For example, what happens if one increases the window area by 20% on the southern side of the building?

This does two things, one negative and one positive:

1. It increases the heat loss because the R-value of a window is less thanthat of a wall (R value drops from 21.97 to 1.92):

→ HL increases from 10,822 to 11,192 BTUs/(day . oF)→ October demand increases from 6.039 to 6.245 million BTUs

2. It increases the capture of solar energy:

→ October solar gain increases from 4.873 to 5.633 million BTUs

The October gap is reduced from 1.198 to 0.538 million BTUsa reduction of 63%.

There is a better way to get more sun without more conductive heat loss…

Except for a small amount of reflection, most of the solar radiation goes through glass because glass is almost perfectly transparent to radiation in the visible spectrum. (We can see through windows!)

This radiation is not absorbed by the air in the room but rather by the opaque surfaces it falls upon, like the floor or walls.

The receiving surface heats up and, in steady state, emits back the same amount of heat, mostly through convection.

Heat is lost through conductive loss through the window (small R-value).

Absorber-storage wall (Trombe wall):

But since glazing creates a relatively large conductive heat loss, consider placing a thick piece of better insulating material just inside

Improved Trombe wall:

With vent holes through the storage wall to bring some of the heat from the greenhouse into the living space.

Absorber wall combined with greenhouse:

A variation…

The greenhouse may be stifling during the day and too cold at night for comfort, but it may be just fine to grow plants … and food, too!

Should interior space get too hot, a passive solution is the

Solar Chimney

A solar chimney — often referred to as a thermal chimney — is a way of improving the natural ventilation of buildings by using convection of air heated by passive solar energy. A simple description of a solar chimney is that of a vertical shaft utilizing solar energy to enhance the natural stack ventilation through a building.

The solar chimney has been in use for centuries, particularly in the Middle East and Near East by the Persians, as well as in Europe by the Romans. (Source: Wikipedia)

Then, one can think of saving the extra daytime heat for use at night.

J. Ka

chado

rian

The P

assive So

lar H

ouse, 1997

, page

39.

Heat storage:

Heat content = c x M x T↑ ↑ ↑

heat mass temperaturecapacity

(BTUs/lb oF) (lb) (oF)

In buildings, we deal with volumes more than masses:

M = x V  ↑ ↑

density volume(lb/ft3) (ft3)

Heat content = H x V x T

where H = c x = specific heat per volume, in BTUs/(ft3 x oF)

Thermal mass inside a building is adequate for smoothing day-night temperature variations.

For smoothing seasonal temperature fluctuations (i.e., storing summer heat for use in the following winter), one needs to resort to a geothermal system.

Specific heat H of various substances and materialsOn a volume basis:

Air 0.0182 ← extremely lowWater 62.44 ← very high

Concrete 30.1 ← quite highConcrete block 28.8Sheetrock 13.0

Plywood 9.86Particle board 15.5

Asphalt roofing shingle 21.0

values in BTUs/(ft3 x oF)

When the sun shines on a wall or floor:

AQdt

dTVH

AQTVHdt

d

where

dAV

IQ

IQ

sin

cos for vertical wall

for horizontal floor

)()(

)(sin

outsideroomroomfloorfloorroom

roomair

roomfloorfloorfloorfloor

floorfloor

TTHLTTUAdt

dTVH

TTUAIAdt

dTVH

)(

)(

sin

outsideroom

roomfloorfloor

floor

TTHL

TTUA

IA

Heat received from sun:

Heat flowing from floor to room:

Heat through walls, etc.:

Heat budgets for floor and room air:

Heat exchange between floor and room:

Warm air created next to floor rises and convects through the room:

)( roomfloorfloor TTUA

TAUQ

↑

U due to convection 33.020.0 TU

(Newton’s Law of convection)

If the heated surface is a vertical wall:

)( roomwallwall TTUA

TAUQ

33.031.0 TU

In these expressions, T is in oF and U in BTUs/(ft2 x hour x oF).

Specific heat of air is almost nil, and we can assume steady state for the room budget:

)()(0 outsideroomroomfloorfloor TTHLTTUA

of which the solution is:

HLUA

THLTHLTT

HLUA

THLUTAT

floor

outsidefloorroomfloor

floor

outsidefloorfloorroom

The heat budget for the floor then becomes:

)(sin outsidefloorfloor

floorfloor

floorfloorfloor TT

HLUA

HLUAIA

dt

dTVH

= instant adaptation of air temperature to a weighted average between floor and outside temperatures

thermal inertia gain from sun loss to the outside

Examples of calculations:

1. Average indoor temperature is adequate but swings too much from day to night.→ Not enough thermal mass

Roomtemperature

2. Indoor temperature well smoothed between day and night but not high enough in average→ Not enough solar intake; need to increase glazing

Room temperature

3. About correct balance of solar intake and thermal mass:

Roomtemperature

A final remark

It is important to keep in mind that in a passive-solar design, the building must accomplish the following three functions simultaneously:

1. Collection of solar energynot too little and not too muchwith appropriate glazing, overhangs, etc.

2. Storage of energy collectedwith appropriate amount and placement of thermal mass

3. Distribution of heatwith facilitation of natural ventilation into the desired areas.