By

09MT3018

PIYUSH VERMA

Materials for Passive Solar Heating

Department of Metallurgical and Materials Engineering, Indian Institute of Technology Kharagpur

Brief presentation on

What’s Passive heating??

Using solar energy incident on windows, skylights, greenhouses, clerestories, and mass walls, collectors in order to provide heating for a house.

Without the extensive use of pumps or fans typically used in active solar collector systems.

Because heating is needed only over the colder part of the year (Sept. to May), passive solar design must also eliminate unwanted solar heat gains during the summer. The use of techniques to eliminate solar gains and to cool a house with the use of active systems is often referred to as passive cooling

Objective:

To capture solar energy for home heating.

To extract heat at night , stored during the day.

Constraints:

Time-constant (time for heat to flow from the outer exposed surface to inner wall of the room): 12 hours

Wall-Thickness : 0.5m (architectural reasons )

Summation:

What materials maximize the thermal energy captured by the wall while retaining a heat-diffusion time of up to 12 hours

Model 1 :

objective function, heat content , Q, per kg of wall heated through a temperature range of ∆T,

Q = wρ(Cp) ∆T equation-1

w=wall thickness

ρ=density of wall material

Cp=specific heat capacity

∆T=Temp(outside subtracted inside)

By solving the Temperature profile:

(∂T/∂t)=(∂²T/∂²x)

we get, w=(2at)^(1/2) equation-2

By solving equation 1 and 2 , and eliminating free variable w, we getQ=(2t)^(1/2) ∆T a^(1/2) ρ(Cp)

a=λ/ ρ(Cp) (by the fact)

Q=(2t)^(1/2) * ∆T * *λ /a^(1/2) ]

So we need to maximize function

M(1) = [λ /a(1/2) ]

From equation-1

Another function say M(2) = a<= [(w)^(2)/2t]

So putting w=0.5m and t=12 * 60 * 60 seconds

M(2) = a <= 3 * 10^(-6) m^2 / sec

Model-2:

If cost is added , we need to minimize the cost,

C=w ρ Cm

(where w=wall thickness, ρ = density , Cm = relative cost)

eliminating w (as constraints t=12 hours and w=0.5m remains same) we get,

C=[ {t^(1/2)} * {a^(1/2)} * ρ Cm] ,

So we need to maximize,

M(3) = [{a^(1/2)} * ρ Cm] ^ (-1)

Considering both Model 1 & Model 2

the best materials are Cement, Brick and concrete.