Brief presentation for Materials Design and Selection MME Dept IIT Kharagpur

1. Brief presentation on
Materials for Passive Solar Heating
By
09MT3018
PIYUSH VERMA
Department of Metallurgical and Materials Engineering,
Indian Institute of Technology Kharagpur

2. 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

3.  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

4.  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

5.  By solving equation 1 and 2 , and eliminating free variable w,
we get
Q=(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

6.  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)

7.  Considering both Model 1 & Model 2
the best materials are Cement, Brick and concrete.