Successfully reported this slideshow.

Materials for Passive Solar Heating


Published on

Brief presentation for Materials Design and Selection
MME Dept
IIT Kharagpur

Published in: Technology
  • Be the first to comment

  • Be the first to like this

Materials for Passive Solar Heating

  1. 1. Brief presentation onMaterials for Passive Solar Heating By 09MT3018 PIYUSH VERMA Department of Metallurgical and Materials Engineering, Indian Institute of Technology Kharagpur
  2. 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. 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. 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. 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. 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. 7.  Considering both Model 1 & Model 2 the best materials are Cement, Brick and concrete.