Thermal Behavior Study for Traditional and Modern Residential Buildings of the Kathmandu Valley by Subik Shrestha

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Thermal Behavior Study for Traditional and Modern Residential Buildings of the Kathmandu Valley

An academic work for Passive Solar Architecture (Year 4 part 2), Bachelor of Architecture course at Institute of Engineering, Tribhuvan University By: Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha Original work- 2010; Edited- 2016

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Introduction to the research work This paper analyzes the thermal behaviors of two residential building types of the Kathmandu Valley: 1. Traditional Newari building 2. Modern Nepali building The main intention of this paper is to study the thermal characteristics of each building type. As such, one building case has been selected for each of the two building types and thermal behavior analysis has been conducted for its walls, floors, and roofs. The method adopted for understanding the thermal behavior is through the calculation of U-values for each building element. The major reference for the calculations has been taken from Diamant (1986) and Pohl (2011) and theoretical understanding from DeKay and Brown (2001). From the results of the analysis, it has been identified that traditional buildings perform better than the modern counterparts in terms of the thermal performance. In terms of the design suggestions, ways of improving the thermal performance of walls, floors, and roofs have been discussed for the traditional building. Similarly, ways of improving the thermal performance of walls have been discussed for the modern building. In addition, some additional designs are also recommended to improve the thermal behavior through use of green roofs and solarium.

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Table of Contents 01 Traditional Newari building

02 Modern Nepali building

Introduction U-value calculation: existing and proposed designs of walls, floors, roofs Calculation of shadow angles

Introduction U-value calculation: existing and proposed designs of walls, floors, roofs Solarium design Green roof design

Acknowledgements Professor Dr. Sushil Bajracharya, Institute of Engineering, Tribhuvan University for the supervision. Civil Homes Private Limited for the drawings.

01 Traditional Newari building Location: Mangabazaar, Lalitpur Latitude: 27o 40’ 26.56” Longitude: 85o 19’ 32” Altitude: 4346 feet Height: 3 and half floors Ground floor: shop, store room 1st-3rd floors: rooms Top floor: kitchen, dining, and room This is a typical Newari building constructed nearly 100 years ago. During the process of renovation, a few modern materials were used for construction but the house still retains its function as a traditional residential building. Initially, the house was only three stories high but later during renovation, a small room was added on the top floor. By selecting this building, we can learn about the passive solar characters of the original technology of a traditional Newari building—the historic building type of the Kathmandu Valley. Along with the study of “environmental” concerns, the impacts of removing certain elements from the original technology has also be studied. The building lies in the traditional settlement of Mangal Bazaar area. The major types of settlement patterns are linear cluster along with courtyard systems. The nature of the settlement pattern, i.e. compact-row of neighborhoods create a micro climate of their own. This type of traditional planning system protects the building from cold breeze in winters and also decreases heat loss. The building is oriented toward eastwest direction and the south face has openings which help to harvest a good amount of solar

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Building materials Walls There are three 540 mm parallel load bearing brick walls running from north to south and two similar walls running from east to west. The burnt bricks are used in the exterior face whereas sundried bricks are used for the interior faces of the same wall. Mud mortar is used in between the brick courses in both exterior and interior walls. Roofs At present, corrugated zinc sheets are used for covering the roof of the attic. The wooden dhalins (wooden beams) are used for supporting the CGI sheets. But, we can see the remaining portions of roof tiles (jhingati) as well, which gives a clear idea of the original roofing material used in the past.

energy inside. The building is closed only on some portions of the north and east walls whereas all other sides are exposed to the outdoors. There are two courtyards on the eastern and western faces. The ground floor is used for a shop and as a store. The top floor consists of kitchen and dining along with an extra room, so the two floors on the top and bottom floors that are not being used as living spaces act as buffer zones to reduce the transmission of heat. The maximum height of each floor is very low, i.e. they do not exceed six feet probably to reduce the volume of air insideâ&#x20AC;&#x201D; lesser the volume of air, lesser the energy required to heat it.

Floors The floors are made up of wooden dhalin along with bamboo sheets. Mud mortar is used for covering the floor.

The Building Envelope Existing wall U-value calculation for the following composition of wall Brick outer wall (burnt) 230mm Brick inner wall (sun dried) 260mm Mud plaster 30mm + 20mm Internal Surface Resistance (1/fi) = 0.123 m2 oC/W External Surface Resistance (south facing) (1/fe) = 0.1 m2 oC /W

Resistivity of burnt brick (1/kb) = 0.83 m oC/W Resistivity of sun dried brick (1/ksb) = 2.68 m oC/W Resistivity of mud (1/km) = 2.17 m oC/W Resistance of wall (R) = (1/kb x t) + (1/ksb x t) + (1/ km x t) = (0.83 x .230) + (2.68 x 0.260) + (2.17 x 0.050) m2 o C/W = (0.1909 + 0.6968 + 0.1085) m2 oC/W =0.996 m2 oC /W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =0.996 + 0.123 + 0.1 m2 oC/W =1.219 Hence, U-value= 1/ Ra-a = 0.82 W/ m2 oC Proposed wall The burnt brick wall thickness has been reduced to 110mm (according to old brick dimension) and 380mm sun dried bricks added in the interior with 30mm mud plaster finishing and 20mm mud mortar. Brick wall (burnt) 110mm Sundried brick wall 440mm Mud plaster 30mm Internal Surface Resistance (1/fi) = 0.123 m2 oC /W External Surface Resistance (south facing) (1/fe) = 0.1 m2 oC /W Resistivity of burnt brick (1/kb) = 0.83 m oC/W Resistivity of sun dried brick (1/ksb) = 2.68 m oC/W Resistivity of mud (1/km) = 2.17 m oC/W Resistance of wall (R) = (1/kb x t) + (1/ksb x t) + (1/ km x t) = (0.83 x 0.11) + (2.68 x 0.44) + (2.17 x 0.03) m2 o C/W =0.0913 + 1.179 + 0.065 =1.2182m2 oC/W

Â© Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Ground floor plan

Section

Â© Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Therefore, air to air resistance (Ra-a) = R + 1/fi + 1/fe =1.218 + 0.123 + 0.1 m2 oC/W =1.441 m2 oC/W Hence, U-value= 1/ Ra-a = .69 W/m2 oC Existing floor U-value calculation for the following composition of floor: Mud mortar 50mm Wooden dhalin 75mm +75mm Ply wood 25mm Internal Surface Resistance (up) (1/fi) = 0.105 m2 C/W Internal Surface Resistance (down) (1/fi) = 0.149 m2 oC/W Resistivity of mud mortar (1/km) = 2.17 m oC/W Resistivity of plywood (1/ kp) = 7.25 m oC/W Resistivity of wooden dhalin (1/kw) = 6.25 m oC/W Resistance of floor (R) = (1 km / x t) + (1/ kw x t) + (1/kw x t) = (2.17 x 0.05) + (6.25 x 0.15) + (7.25 x 0.025) m2 o C/W =(0.1085 + .937 + .1812) m2 oC/W =1.2055 m2 oC/W o

Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =1.205 + 0.105 + 0.149 m2 oC/W =1.491 m2 oC/W Hence, U-value= 1/ Ra-a = 0.67 W/m2 oC Proposed floor Mud mortar used instead of cement mortar and 2mm bamboo mat added in the underside of the

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

floor with 75mm air cavity Mud mortar 50mm Wooden dhalin 75mm +75mm Ply wood 25mm Air cavity 75mm Bamboo mat 2mm Internal Surface Resistance (up) (1/fi) = 0.105 m2 o C/W Internal Surface Resistance (down) (1/fe) = 0.149 m2 oC/W Resistivity of mud mortar (1/km) = 2.17 m oC/W Resistivity of wooden dhalin (1/kw) = 6.25 m oC/W Resistivity of plywood (1/ kp) = 7.25 m oC/W Resistivity of air cavity (1/ka) = 38.45 m oC/W Resistivity of bamboo mat (1/kbm) = 6.25 m oC/W Resistance of floor (R) = (1/km x t) + (1/kw x t) + (1/ kp x t) + (1/ka x t) + (1/kbm x t) = (2.17 x 0.05) + (6.25 x 0.150) + (7.25 x 0.025) + (38.45 x0.075) + (6.25 x 0.002) m2 oC/W =0.1085 + .937 + .181 + 2.883 + .0125 =4.234 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =4.234 + 0.105 + 0.149 m2 oC/W =4.488 m2 oC/W Hence, U-value= 1/ Ra-a = 0.22 W/m2 oC Existing roof U-value calculation for the following composition of roof Zinc sheet 2mm

External Surface Resistance (south facing) (1/fe) = 0.044 m2 oC/W Resistivity of (1/kzn) =0.00086 m oC/W Resistance of floor (R) = (1/kzn x t) =(0.00086x 0.002) m2 oC/W =.00000172 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =.00000172 + 0.105 + 0.044 m2 oC/W =0.149 m2 oC/W Hence, U-value= 1/ Ra-a = 6.7 W/m2 oC Proposed roof 2mm bamboo mat added in the underside of the floor with 75mm air cavity. Jhingati tiles 50mm Mud mortar 50mm Bitumen Layer 2mm Ply Wood 12mm Air cavity 75mm Internal Surface Resistance (1/fi) = 0.105 m2 oC/W External Surface Resistance (south facing) (1/fi) = 0.044 m2 oC/W Resistivity of Jhingati tiles (1/kj) = 0.83 m oC/W Resistivity of Mud mortar (1/km) = 2.17 m oC/W Resistivity of Bitumen (1/kbi) = 1.74 m oC/W Resistivity of Air Cavity (1/ka) = 38.45 m oC/W Resistivity of Ply Wood (1/kp) = 7.25 m oC/W Resistance of floor (R) = (1/kj x t) + (1/km x t) + (1/ kbi x t) + (1/ka x t) + (1/kp x t) =(0.83 x 0.05) + (2.17 x 0.05) + (1.74 x 0.002) + (38.45 x 0.075) + (7.25 x 0.012) m2 oC/W =.0415 + .1085 + .000348 + 2.883 + .087 m2 oC/W

Internal Surface Resistance (1/fi) = 0.105 m2 oC/W

=3.12 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =3.12 + 0.105 + 0.044 m2 oC/W =3.269 m2 oC/W Hence, U-value= 1/ Ra-a = 0.305 W/m2 oC

Calculation of shadow angles For Summer Sun May 1, 10:00 am Solar Altitude Angle (a) = 59.94o Solar Azimuth Angle (α) = 109o Wall Azimuth Angle (south facing) (ω) = 215o So, δ = ω – α =215 o - 109 o = 106 o This shows that the sun’s rays do not strike the front façade (SW) at 10:00am of May 01. So the calculation of Vertical Shadow Angle is not possible. However, we can consider the conditions as on 02:00pm of May 1. Thus, May 1, 02:00pm Solar Altitude Angle (a) = 59.94o Solar Azimuth Angle (α) = 251o Wall Azimuth Angle (south facing) (ω) = 215o So, δ = ω – α =215o - 251o = 36o (taking +ve value) We have, Tan ε = Tan a x Sec δ (where ε is the Vertical Shadow Angle) = Tan 59.94 o x Sec 36 o Therefore, ε = 64.91o For Winter Sun Nov 3, 12:00 noon Solar Altitude Angle (a) = 46.59o

Solar Azimuth Angle (α) = 180o Wall Azimuth Angle (south facing) (ω) = 215o So, δ = ω – α =215o - 180o = 35o We have, Tan ε = Tan a x Sec δ (where ε is the Vertical Shadow Angle) = Tan 46.59 o x Sec 35 o Therefore, ε = 52.23 o Analysis of Overhangs From the case study, Height of the opening (h) =1000 mm Overhang required (a) =? ε= 65o For blocking the summer sun, Tan ε= h/a Tan 65o =1000mm/a

02 Modern Nepali residence Location: Sunakothi, Lalitpur Latitude: 27o 37’ 20” Longitude: 85o 19’ 21” Altitude: 4700 feet Plot Area: 2750 SF Plinth Area: 875 SF Length of Building: 36 feet Breadth of Building: 29 feet Height: 2 1/2 stories Ground floor occupancy: Living room, dining room, kitchen, and common toilet First Floor occupancy: Master bedroom, two bedrooms Top floor occupancy: store

Materials Walls: brick, concrete shear wall Structure: concrete Finishes: external texture, paint Floorings: marble, parquets, tiles, granite Window frame: wooden Since the case taken is situated in civil homes housing, the microclimate of the house is influenced by it. Land Form- Constructed on land with fairly flat land Water bodies- No water bodies found in or around the site Vegetation- Greenery present on the northern

a=1000mm/2.14 So, a=460 mm Projection of 460 mm is proposed Similarly, it allows the entry of the winter sun as well.

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Site plan (copyright- civilhomes)

Section (copyright- civilhomes)

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

side and a patch of lawn on the south eastern region External wall texture and color- The exposed external wall is cement plaster and weather coat paints of orange colored texture. The building is oriented on east west long axis. Living room lies on the northern side. Bed rooms are on the northern and south eastern side. First floor terrace is on the east side.

Building envelope Existing wall Cement Plaster 12mm Brick wall (burnt) 230mm Cement Plaster 12mm Internal Surface Resistance (1/fi) = 0.123 m2 oC/W External Surface Resistance (south facing) (1/fe) = 0.1 m2 oC/W Resistivity of burnt brick (1/kb) = 0.83 m oC/W Resistivity of cement plaster (1/kc) = 0.714 m oC/W Resistance of wall (R) = (1/kb x t) + 2(1/kc x t) = (0.83 x 0.23) + 2(0.714 x 0.012) m2 oC/W = 0.208 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe = 0.208 + 0.123 + 0.1 m2 oC/W = 0.431 m2 oC/W Hence, U-value= 1/ Ra-a = 2.319 W/m2 oC

better option for walls is including a cavity. Since air has high conductivity and can also work well as cavity with walls, suitably thick cavity needs to be designed. The main importance however is to increase the resistivity of the wall so that thermal comfort could be maintained in the living areas. Cement Plaster 12mm Brick wall (burnt) 230mm Cement Plaster 12mm Air Cavity 50mm Ply Board 12mm Internal Surface Resistance (1/fi) = 0.123 m2 oC/W External Surface Resistance (south facing) (1/fe) = 0.1 m2 oC/W Resistivity of burnt brick (1/kb) = 0.83 m oC/W Resistivity of cement plaster (1/kc) = 0.714 m oC/W Resistivity of air cavity (1/ka) = 38.45 m oC/W Resistivity of ply board (1/kp) = 7.25 m oC/W Resistance of wall = (1/kb x t) + 2(1/kc x t) + (1/ka x t) + (1/kp x t) = (0.83 x 0.23) + 2(0.714 x 0.012) + (38.45 x 0.05) + (7.25 x 0.012) m2 oC/W = 2.217 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe = 2.217 + 0.123 + 0.1 m2 oC/W = 2.44 m2 oC/W Hence, U-value= 1/ Ra-a = 0.409 W/m2 oC

Proposed wall Since the standard value of conductivity for the external wall is approximately 0.4-0.7 W/m2 oC, it is required to significantly decrease the U value. A

Existing parquet floor Parquet 12mm Parquet joint 25mm Screed 25mm RCC Slab 100mm Cement Plaster 12mm Internal Surface Resistance (up) (1/fi) = 0.105 m2 o C/W Internal Surface Resistance (down) (1/fi) = 0.149 m2 oC/W Resistivity of parquet (1/kpq) = 6.25 m oC/W Resistivity of IPS or parquet joint (1/kips) = 0.714 m o C/W Resistivity of screed (1/ksc) = 0.714 m oC/W Resistivity of RCC Slab (1/krcc) = 0.69 m oC/W Resistivity of cement plaster (1/kc) = 0.714 m oC/W Resistance of floor= (1/kpq x t) + (1/kips x t) + (1/ksc x t) + (1/krcc x t) + (1/kc x t) = (6.25 x 0.012) + (0.714 x 0.025) + (0.714 x 0.025) + (0.69 x0.1) + (0.714 x 0.012) m2 oC/W =0.188 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =0.188 + 0.105 + 0.149 m2 oC/W =0.44 m2 oC/W Hence, U-value= 1/ Ra-a = 2.27 W/m2 oC Existing marble floor Marble 25mm Cement Mortar 25mm Screed 25mm RCC Slab 100mmCement Plaster 12mm Internal Surface Resistance (up) (1/fi) = 0.105 m2 C/W

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

Internal Surface Resistance (down) (1/fi) = 0.149 m2 oC/W Resistivity of marble (1/kmr) = 0.34 m oC/W Resistivity of cement mortar (1/kcm) = 0.714 m o C/W Resistivity of screed (1/ksc) = 0.714 m oC/W Resistivity of RCC Slab (1/krcc) = 0.69 m oC/W Resistivity of cement plaster (1/kc) = 0.714 m oC/W Resistance of floor (R) = (1/kmr x t) + (1/kcm x t) + (1/ ksc x t) + (1/krcc x t) + (1/kc x t) = (0.34 x 0.012) + (0.714 x 0.025) + (0.714 x 0.025) + (0.69 x0.1) + 0.714 x 0.012) m2 oC/W =0.126 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =0.126 + 0.105 + 0.149 m2 oC/W =0.38 m2 oC/W Hence, U-value= 1/ Ra-a = 2.63 W/m2 oC Existing roof Chinese tiles 25mm Cement mortar 25mm RCC slab 100mm Cement Plaster 12mm Internal Surface Resistance (1/fi) = 0.105 m2 oC/W External Surface Resistance (south facing) (1/fi) = 0.044 m2 oC/W Resistivity of Chinese tiles (1/kmt) = 0.69 m oC/W Resistivity of Cement mortar (1/kcm) = 0.714 m o C/W Resistivity of RCC slab (1/krcc) = 0.69 m oC/W Resistivity of Cement plaster (1/kc) = 0.714 m oC/W Resistance of floor (R) = (1/kmt x t) + (1/kcm x t) + (1/ krcc x t) + (1/kc x t) = (0.69 x 0.025) + (0.714 x 0.025) + (0.69 x 0.1)+ (0.714 x 0.012) m2 oC/W

=0.11 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =0.11 + 0.105 + 0.044 m2 oC/W =0.26 m2 oC/W Hence, U-value= 1/ Ra-a = 3.82 W/m2 oC It is not a practically feasible task to bring changes in roof so as to increase the resistivity. In the design the rooms below slopped roof are not living spaces. So it may not require insulation. However, a roof garden is proposed to increase the thermal efficiency of the bedrooms that lie below.

Solarium Design • • • • • •

On the backyard of the kitchen which lies on the western side 5 feet by 12 feet in dimension Entirely of glazing on walls and roof Roof is inclined to trap the southern sun Vegetations grown will add to the aesthetics and acts as back yard and kitchen garden Functional to store day light and transfer during the night time

Design of roof garden The terrace on the top floor is utilized as garden so that the thermal conductivity of the slab is decreased. The existing R value of the terrace floor is very less than required. So it is recommended to

ÂŠ Shreedhara Bajracharya, Shreejika Shrestha, Subik Kumar Shrestha

bring some changes and meet the standards. Along with the green grass, some plantations are also done.

Design of the green roof Green bed Soil 150 mm Filter membrane stone boulders Drainage layer Bitumen damp proofing 50 mm RCC Slab 100mm Cement Plaster 12mm Internal Surface Resistance (up) (1/fi) = 0.105 m2 o C/W Internal Surface Resistance (down) (1/fi) = 0.149 m2 oC/W Resistivity of soil with organic matter (1/ks) = 5 m o C/W Resistivity of bitumen (1/kb) = 5.88 m oC/W Resistivity of RCC Slab (1/krcc) = 0.69 m oC/W Resistivity of cement plaster (1/kc) = 0.714 m oC/W Resistance of floor (R) = (1/ks x t) +(1/kb x t) + (1/ krcc x t) + (1/kc x t) = (5x 0.15) + (5.88 x 0.5) + (0.69 x0.1) + (0.714 x 0.012) m2 oC/W =3.76 m2 oC/W Therefore, Air to air resistance (Ra-a) = R + 1/fi + 1/fe =3.76 + 0.105 + 0.149 m2 oC/W =4.0 m2 oC/W Hence, U-value= 1/ Ra-a = 0.25 W/m2 oC

References Pohl, J. 2011. Building Science: Concepts and Application. John Wiley and Sons. DeKay, M. and Brown G.Z. 2001. Sun, Wind and Light. John Wiley and Sons. Diamant, R.M.E. 1986. Thermal and Acoustic Insulation. Butterworths.

The skylight is also provided so as to trap the sun rays, yet the problem of glaring needs to be considered.