Page 43 - GRIHA Manual Volume IV - Introduction to National Rating System
P. 43
34 Griha Manual: Volume 4
plan, which would clearly demonstrate the use/construction/specifications of low-
energy technologies.
16.3 Appraisal (maximum-4 points)
16.3.1 Structural application
Use of low-energy materials/efficient technologies in structural application clearly
demonstrating a minimum 5% reduction in the embodied energy, when compared with
equivalent products for the same application, for 100% structural system used in a building,
meeting the equivalent strength requirements, as per all compliance clauses (2 points).
16.3.2 non-structural application
Use of low-energy materials/efficient technologies (not based on the utilization of
industrial waste), which are used for non-structural applications, such as infill wall
system and cause a minimum 5% reduction in the embodied energy, when compared
with equivalent products for the same application, for 100% infill wall system used in
a building, meeting the equivalent strength requirements, as per all the compliance
clauses (2 points).
Methodology
The steps to be followed for calculating the reduction in embodied energy have been explained as
below with the help of an example.
Structural application
Consider the following example of a building. As per the conventional design practice, the structural
system for the building was designed and the following quantities were calculated.
Total quantity of concrete required= 3000 m 3
Total quantity of steel required=100 m 3
Now by using certain technologies, the volume of materials gets reduced for the same structural
requirements as follows.
The total quantity of concrete required reduces by 28% and the total quantity of steel increases
by 0.5%. Therefore,
Total quantity of concrete required for design case=2160 m 3
Total quantity of steel required for design case=100.5 m 3
The embodied energy values considered are
4
Concrete = 2530 MJ/cu m
5
Steel = 471 600 MJ/cu m
Therefore, the total embodied energy of structure in base case
= (2530×3000) + (471 600×100) = 54 750 000 MJ
The total embodied energy of structure in design case
= (2530×2160) + (471 600×100.5) = 47 395 800 MJ
Therefore, reduction in the embodied energy of the structure
= {(54 750 000–47 395 800)/54 750 000}×100 = 13.43%
4 TERI Knowledge Bank for Sustainable Materials & Reddy B. V. Venkatarama & Jagadish K. S. - Embodied energy of common and
alternative building materials and technologies. Energy and Environment (Elsevier), 2001.
5 Ibid.
plan, which would clearly demonstrate the use/construction/specifications of low-
energy technologies.
16.3 Appraisal (maximum-4 points)
16.3.1 Structural application
Use of low-energy materials/efficient technologies in structural application clearly
demonstrating a minimum 5% reduction in the embodied energy, when compared with
equivalent products for the same application, for 100% structural system used in a building,
meeting the equivalent strength requirements, as per all compliance clauses (2 points).
16.3.2 non-structural application
Use of low-energy materials/efficient technologies (not based on the utilization of
industrial waste), which are used for non-structural applications, such as infill wall
system and cause a minimum 5% reduction in the embodied energy, when compared
with equivalent products for the same application, for 100% infill wall system used in
a building, meeting the equivalent strength requirements, as per all the compliance
clauses (2 points).
Methodology
The steps to be followed for calculating the reduction in embodied energy have been explained as
below with the help of an example.
Structural application
Consider the following example of a building. As per the conventional design practice, the structural
system for the building was designed and the following quantities were calculated.
Total quantity of concrete required= 3000 m 3
Total quantity of steel required=100 m 3
Now by using certain technologies, the volume of materials gets reduced for the same structural
requirements as follows.
The total quantity of concrete required reduces by 28% and the total quantity of steel increases
by 0.5%. Therefore,
Total quantity of concrete required for design case=2160 m 3
Total quantity of steel required for design case=100.5 m 3
The embodied energy values considered are
4
Concrete = 2530 MJ/cu m
5
Steel = 471 600 MJ/cu m
Therefore, the total embodied energy of structure in base case
= (2530×3000) + (471 600×100) = 54 750 000 MJ
The total embodied energy of structure in design case
= (2530×2160) + (471 600×100.5) = 47 395 800 MJ
Therefore, reduction in the embodied energy of the structure
= {(54 750 000–47 395 800)/54 750 000}×100 = 13.43%
4 TERI Knowledge Bank for Sustainable Materials & Reddy B. V. Venkatarama & Jagadish K. S. - Embodied energy of common and
alternative building materials and technologies. Energy and Environment (Elsevier), 2001.
5 Ibid.
