GATE Numericals.post graduation ,march ,architecture

gate architecture numericals

1. GATE Numerical

2. NUMERICAL ESTIMATION & VALUATION HOUSING (FAR – DENSITY CALC) ACOUSTICS ILLUMINATION RUN OFF CALCULATIONS SCALE STURCTURES CPM & PERT IMPORATANT TOPICS

3. Rainfall The total quantity of runoff for an area of 18 Hec in a lateritic region (run off coeff= 0.5 and rainfall= 10mm/hr) is Run off coeff= 0.5 Rainfall= 10mm/hr = 0.01m/hr Q = Runoff coeff x Rainfall x Surface area Area = 18 Hec Q = 0.5 x 0.01 x 18000 Q = 90 m 3 /hr

4. Rainfall Rainfall = 100 mm = 0.1 m Storm water of central play area = 100% Storm water of rest of the complex = 70% Volume of water after runoff AB CPA OPA OIA 100 9 14 36 Retention period 1250 X 0.8 X 0.1 150 X 0.6 X 0.1 200 X 0.7 X 0.1 400 X 0.9 X 0.1 70 9 9.8 25.2

5. Rainfall Total Volume = 70 + 9 + 9.8 + 25.2 Total Volume = 114 m 3 Total Volume = l x b x h Height = ? 114 = 150 x h Central play area = 150sq m h = 0.76 m h = 760 mm

6. Q.42 A building with 100 sqm roof area is connected to a 72 cum rainwater collection tank. If the rainfall is 60 mm per hour and the loss during water storage is 20%, then the time taken to fill the tank completely is_______ hours. Roof area = 100 sqm 20% waste T ank = 72 cum Suppose X amount of water is coming 80 % of X is 72cum 72 = 80 X --------- 100 X = 90 cu m VOL = AREA X HEIGHT 90 = 100 X H H = 9 m = 900 mm 60 mm = 1hr 900 mm = T = 900 / 60 T = 15 hrs Rainfall

7. Brick paving area = 20% of 35000 Concrete paving area = 15% of 35000 Grass cover area = 65% of 35000 = 7000 m 2 = 5250 m 2 = 22750 m 2 Total surface area = 35000 sqm Water runoff by brick paving = 0.8 × 7000 × 0.07 Water runoff by concrete paving = 0.9 × 5250 x 0.07 Water runoff by grass cover = 0.5 × 22750 × 0.07 = 392m 3 = 330.75 m 3 = 796.25 m 3 Ra infall = 70 mm = 0.07 m Total run off = 392 + 330.75 + 796.25 Total run off = 1519 m 3 Rainfall

8. Q.34 A house located in Delhi has 111 m 2 of flat terrace area (runoff coefficient = 0.85) and 55 m 2 of ground area covered with grass (runoff coefficient = 0.15). If annual average rainfall is 611.8 mm, then rain water harvesting potential (L/year) from runoff will be ----------------- Flat terrace = 111 sqm Run off coeff terrace = 0.85 Ground area with grass = 55 sqm Annual rainfall = 611.8 mm = 0.6118 m Rain water harvesting potential = ? RWH = SURFACE AREA X RAINFALL X RW COEFF = (111 x 0.85 + 55 x 0.15 ) x 0.6118 Run off coeff grass = 0.15 = 102.6 x 0.6118 = 62.77068 cum 1 m 3 = 1000 L = 627706 L Rainfall

9. Construction A rectangular room (internal dimension 5m x 3m) is made if 250 mm walls. Calculate volume of concrete needed for 25mm Dampproof course Room size = 5 x 3 m Wall thickness = 250 mm = 0.25 m Area under wall = 19.25 – 15 Outer area = 5.50 x 3.50 m = 4.25 m 2 Vol of conc req for DPC = 4.25 x 0025 Inner area = 5 x 3 m = 19.25 m 2 = 15 m 2 = 0.10625 m 3

10. Q.51 A room of internal dimension 4m x 5m x 3.5m (LxBxH) has 20 cm thick walls and two doors of size 1m x 2m. The required area of Damp Proof Course (sq.m) is _____________ Room = 4 x 5 x 3.5 m Thickness of wall = 20 cm = 0.2 m 2 doors =1 x 2 m = 2m Area covered by walls = 4.4 x 5.4 - 4 x 5 = 3.76 m 2 Area covered by 2 doors = 1 x 2 x 0.2 = 0.4 m 2 Area for DPC = 3.76 – 0.4 = 3.36 m 2

11. Construction A rectangular room (internal dimension 5m x 3m) is made if 250 mm walls. Calculate volume of concrete needed for 25mm Dampproof course Room = 1.8 x 2.4 m = 4.32 sqm Tile = 300 x 300 mm = 90000 sqmm = 0.09 sqm Area of skirting = 1.8 x 0.6 x 2 + 2.4 x 0.6 + 1.5 x 0.6 Skirting ht = 600 mm Door opening = 900 mm = 0.6 m = 0.9 m = 4.5 sqm Area of tiling = 4.5 + 4.32 = 8.82 sqm No of tiles = 8.82 / 0.09 = 98 tiles

12. Q.47 One litre of acrylic paint can cover 16 sqm of wall area for the first coat and 24 sqm for the second coat. The walls of a lecture hall measuring 12m × 8m × 4m (L × B × H) need to be painted with two coats of this paint. The hall has total glazed fenestration area of 12 sqm. The number of 4 litre paint containers required will be __________ 1 lit covers FIRST COAT = 16 sqm SECOND COAT = 24 sqm Room size = 12 x 8 x 4 FENESTRATION = 12 sqm Surface area = 2 ( 12 x 4 + 8 x 4 ) = 2 ( 48+ 32 ) = 160 sq m = 160 - 12 = 148 First coat = 148 / 16 Second coat = 148 / 24 = 9.25 lit = 6.17 lit Total = 9.25+ 6.17 = 15.42 4 containers Construction

13. Number of tiles = 45 / 0.09 Driveway = 15m x 3m Tiles = 300mm x 300 mm = 45 sqm = 900000 sq mm = 0.09 sqm = 500 One packet = 30 tiles No of packets for 500 tiles = 500 / 30 = 16.6 = 17 packets Construction

14. Initial cost = 400000 Scrap value = 10 % of initial cost Time = 30 years Rate of interest = 5 % = 40000 Sinking fund = initial cost – scrap cost = 400000 – 40000 = 360000 = 360000 (0.05 / (1+0.05) 30 -1) = 5421 Rs ESTIMATION VALUATION

15. Area of building = 5000 sqm Site = 1 hectare = 10000 sqm Present value of land = Rs 100/sqm Present value of site land = 100 x 10000 = 10 lakhs Cost of construction = 5000 x 2500 Present construction cost = Rs 2500/sqm = 125 lakhs ESTIMATION VALUATION

16. Depreciation rate = 6% Value of the building after depreciation in 5 years = 125 (1 – r) n   = 91.7 lakhs = 125 × (1 – 0.06) 5   Present Value of Property = Land Value + Building Value = 10+ 91.7 = 101.7 lakhs ESTIMATION VALUATION

17. Annual rent = 10000 Present Value Of Land = Annual Rent ------------------------ Rate = 10000 x 100 ---------------- 5 = 20000 Rs Rate = 5 % ESTIMATION VALUATION

18. Annual net income = 22000 Rs Interest rate = 8% Capitalize Value x Rate of Interest = Net Income CV x 8 % = 22000 CV = 275000 Rs ESTIMATION VALUATION

19. FAR A residential plot of 20m frontage and 25 m depth is governed by the development regulations of maximum FAR of 2 and maximum ground coverage 50%. Up to what maximum height can the plot be built ? Plot = 20 x 25 m = 500 sqm FAR = 2 Ground coverage = 50% of plot area Built up area = 500 x 2 = 1000 sqm = 250 sqm No of floors = 1000 / 250 = 4 floors

20. FAR A group housing plot of 400 sqm abutting a 20 m road is permitted to have 35% ground coverage and maximum FAR 1.25 assuming 70 sqm of super built up area, calculate the maximum number of units that can be accommodated Plot = 400 sqm Ground coverage = 35% x 400 = 140 sqm Built up area = 1.25 x 400 FAR = 1.25 = 500 sqm Super built up area = 70 sqm = 7 units Max no of units to be built = 500 / 70

21. FAR A small commercial plot measuring 20 m x 30 m located in a community center is subjected to maximum ground coverage of 60% and FAR 3. With fullest land utilization and uniform floor areas how many floors can be built on the plot. Plot = 20 x 30 Ground coverage = 60% x 600 = 360 sqm Built up area = 3 x 600 FAR = 3 = 1800 sqm = 6 floors Total no of floors = 1800 / 360

22. FAR MIG tower = 4 towers = 8 storey = 400m 2 LIG tower = 3 towers = 7 storey 75% of FAR = 12800 Remaining FAR which can be utilize in LIG towers = 17066.67 - 12800 Total floor area = 4 x 8 x 400 = 12800 = 17066.67 = 4266.67 sqm Floor area for LIG = 4266.67 / 7x3 = 203.17

23. Plot = 3000 sqm Total b/up area 45 00 6000 6000 9000 9000 + 15 % of EWS (9000) 10350 FAR

24. 39132 50872 60133 39132 / 130440 x 100 Percentage inc = 30% 50872 / 169572 x 100 30% 60133 / 220444 x 100 30% Population 2021 Population 2031 (286577 x 0.3) + 286577 372550 ( 372550 x 0.3) + 372550 484315 FAR

25. Housing The relative distribution of different types of housing within total residential area of 150 hec is such that Builders housing group is 1/4 th of total area, cooperative housing society if half of the builders housing group and rest is the plotted hosing. If the net density of the plotted housing area is 350ppha, how many people will be accommodated there? Total residential area = 150 hec Builders housing group = 1/4 x 150 = 37.5 hec Plotted housing consumers = 150 – 37.5 – 18.75 Cooperative society housing = 1/2 x 37.5 = 18.75 = 93.75 Density of plotted housing = 350 ppha = 350 x 93.75 = 32812.5

26. Housing A town has basic employment of 25000 workers. If the basic : non basic ration is 1:2.5 and the workers dependency ration is 1:4, what is the population size of the town? Basic employment = 25000 Non Basic employment = 2.5 x 25000 = 62500 Total dependent population = 87500 x 4 Total workers = 25000 + 62500 = 87500 = 350000 Total population = 350000 + 87500 = 437500

27. Housing A community comprising 120 hec is having 60% of the land put to residential plots an a population of 30600. Calculate the net residential density? Total area of community sector = 120 hec Area of residential dev = 60% of 120 = 72 hec Net residential density = 30600 / 72 Population of community = 30600 = 425 pph

28. Housing A town of 225000 population in 1981 has exhibited decadal growth rates of 25% and 30% during 1991 and 2001 respectively estimated the population in 2011 having 40% decided growth rate. 1981 1991 2001 2011 - 25% 30% 40% 225000 281250 365625 511875 Years Growth Population

29. Housing In a residential community of 10000 population 20% are higher secondary school going children the expected enrollment is 80% and per capita gross floor space required is 3 sqm. The ground coverage permissibly is 40%. Indicate the land area required for the higher secondary building in hectares. Expected enrolment = 80% x 2000 = 1600 Population = 10000 Population of school going children= 20% x 10000 = 2000 Per capita gross floor pace = 3 sq m = 1.2 Ha 0.40a = 4800 = 12000 sqm Floor space req for 1600 children = 1600 x 3= 4800 sq m Ground coverage = 40% = 0.40a

30. Housing Residential area of 80 hectares has the following residential plots sub division. Each plot has won dwelling unit and the average household sizes 5 persons. The rest area is devoted to roads, schools, parks and shops. What are the gross and net density respectively of the area in persons per hectare (ppha) Total area = 80 hec = 800000 sqm = 8000 / 80 Gross density = total population / total area Each plot 1 DU Avg household size = 5 Plot size Numbers 500 sqm 100 300 sqm 500 200 sqm 1000 Total area Total DU Total population 50000 100 500 150000 500 2500 200000 1000 5000 400000 1600 8000 Net density = total population / net residential area = 8000/ 40 40 hec = 100 ppha = 200 ppha

31. Area of plot 1 = 500sqm No of units = 100 Total area = 500 x 100 = 50000 sqm = 5 Ha Area of plot 2 = 300sqm No of units = 120 Total area = 300 x 120 = 36000 sqm = 3.6 Ha Area of plot 3 = 200sqm No of units = 150 Total area = 200 x 150 = 30000 sqm = 3 Ha Total area = 5 + 3.6 + 3 Total area = 11.6 Ha Housing

32. Population = 100 x 5 Avg household size = 5 = 500 Population = 120 x 5 Avg household size = 5 = 600 Population = 150 x 5 Avg household size = 5 = 750 Total Population = 500 + 600 + 750 Total Population = 1850 Total Population Net Density = ------------------------- Area of Residential Plots 1860 Net Density = ----------- 11.6 Net Density = 160.3 persons per hectare Housing

33. Acoustics In a seminar room of area 200 sqm an height of 4m an total absorption power of 120m 2 Sabines, what is the reverberation time ? Vol of room = 200 x 4 = 800 cum Total absorption = 120 RT = 0.16 Vol / Absp = 0.16 x 800 / 120 = 1.06 sec

34. Acoustics An auditorium of size 30 m x 20 m is 8 m high. Assuming optimum reverberation time of 1.2 seconds existing absorption power of the hall is 300 m 2 sabine calculate the extra absorption required Vol = 4800 cum RT = 1.2 sec RT = 0.16 Vol / Absp 1.2 = 0.16 x 4800 / abs Abs = 640 m 2 sabines Total absorption = ? Extra Absorption reqd = 640 – 300 = 340 m 2 sabine

35. Room = 8 x 14 x 4 m 4 windows = 1.5 x 1 m 2 doors = 1 x 2 m - Closed - Open RT = ? Total surface area of cuboid = 2 ( lb +bh + hl ) Total surface room= 2 ( 8x14 + 14x4 + 2x2 ) = 400 sqm Minus openings = 400 – (1.5 x 4 + 2x2) = 390 sqm Acoustics

36. Vol of Room = 8 x 14 x 4 = 448 sqm Total absorption = 390 x 0.2 + 4 x 0.4 + 6 x 1 = 85.6 Sabine For an open window absorption coeff will be 1 – total incident sound is equal to total absorbed sound RT = 0.16 x 448 / 85.6 RT = 0.83 sec Acoustics

37. Room size = 10 x 10 x 4 Vol of room = 400 cum Reverberation time = 1.2 sec 1.2 = 0.16 (400/360a) Total surface area = 10 x 4 x 4 + 10 x 10 x 2 = 360 sqm (V) Volume = L X B X H (S) Total Absorption = Surface Area X Absorption Coeff a = 0.148 Openings = 16 sqm If the room is closed RT = 0.16 (400/344a) If the room is open openings= 360- 16 = 344 sqm RT = 0.16 (400/344 X 0.148) RT = 1.26 sec Acoustics

38. Q.44 For a room with dimensions 4m × 3m × 3m (L × B × H), the details of indoor acoustical treatment are as follows. The reverberation time in seconds at 1000 Hz is _______ Dim = 4m × 3m × 3m Area of wall = 4 x 3 x 2 + 3 x 3 x 2 = 42 42 ((0.3 x 0.4) + (0.7 x 0.1)) 42 x 0.19 7.98 Sabine Area of ceiling = 4 x 3 = 12 12 ((0.4 x 0.6) + (0.6 x 0.1)) 12 x 0.3 3.6 Sabine Total absorption wall Total absorption ceiling Acoustics

39. Area of floor = 4x 3 = 12 12 (1 x 0.1) 12 x 0.1 1.2 Sabine Total absorption floor RT = 0.16 x 36 ------------ 12.78 Total absorption = 7.98 + 3.6 + 1.2 RT = 0.45 sec = 12.78 Volume = 4 x 3 x 3 = 36 Acoustics

40. Scale If the scale of the map is 1:30000, then 1 sq cm of the map would represent__________Ha 1 sq c m On map = 1 cm x 1 cm On ground = 30000 cm x 30000 cm = 900000000 sq cm = 90000 sq m = 9 Ha

41. A building site measures 96 sq.cm on a scale of 1:12500. The actual area it represents (in hectare, in integer) is ___________. Scale

42. Scale A 4 cmx 4 cm area on a mao represented a land area of 16 hectares on ground. If this map is transformed to a scale of 1:5000 the same ground area will be represented by On map = 4 cm x 4 cm On ground = 16 hec = 16000 sqm = 400 m x 400 m Current Scale = 1 : 1000 = 8 cm On ground = 40000 cm x 40000 cm = 40000 cm x 40000 cm If Scale = 1 : 5000 On map = 40000 / 5000 On map = 8 cm x 8 cm = 64 sq cm

43. MAP GROUND Plot = 150 x 40 Road = 16mm Road = 4m = 4000 mm 16 / 4000 Equating to find scale 1 / 250 1 : 250 Actual Length = 150 x 250 Actual width = 40 x 250 = 37500 sq mm = 3.75 sq m = 10000 sq mm = 10 sq m Plot = ? Scale

44. Total area = 37.5 x 10 = 375 sq m Total Built Up = FAR X Site Area = 375 x 1.2 FAR = 1.2 Max built up area = 450 sq m Scale

45. 360000 m 2 = 600 x 600 25 cm 2 = 5 x 5 MAP GROUND 5 cm 600 m 1 cm 600 / 5 120 m 1 : 12000 1 cm 12000 cm Scale

46. Lighting Calculate the no of light fixtures required in an office room of 8m x 7m, requiring an illumination level of 400lux on the work plane. Each fixture has a rate output of 7350 lumens. Assume a utilization factor of 0.5 and maintenance factor of 0.8. Room area = 8 x 7 m No of lamps = Area x illumination / output x Uf x Mf = 56 x 400 / 7350 x 0.5 x 0.8 = 7.61 = 8 Output = 7350 Uf = 0.5 Mf = 0.8

47. Classroom size = 10 x 8 x 2.7 m Output = 5000 lumens But including utilization factor = 0.5 Maintenance factor = 0.8 Lumen output = 5000 x 0.5 x 0.8 Illumination = 500 lux = 500 lumen /m 2 Total Lumen Req No Of Lamps = ----------------------- Lumen Of Lamp Total lumen req = illumination x area of room =500 x 10 x 8 500 x 10 x 8 = ----------------------- 5000 x 0.5 x 0.8 =20 lamps Lighting

48. Surface conductance = 20 w/m 2 o C Q = U X A X T Absorbance = 0.66 U value = 1.2 w/m 2 o C Solar gain factor = ? SHFC = U X A / C Ts = To + (I x A) / C SHGC = Q / I = 1.2 X 0.66 / 20 = 0.03 Lighting

49. Thermal conductance (k value) Surface conductance (C - value) Brick Wall Plastering Internal surface External surface 1.2 0.5 8.0 9.5 Thickness = 200 mm = 0.2 m = 20 mm = 0.02 m R value = thickness / k value 0.16 0.04 0.125 0.105 R value = 1 / C value Lighting

50. Total R value = 0.16 + 0.04 + 0.125 + 0.105 + 0.17 U = 1/R U = 1/ 0.6 U = 1.67 W/m 2 O C Thermal resistance = Total R value = 0.6 Resistance (R value) 50mm wall cavity = 0.17 Lighting

51. Flux Luminous Efficacy = -------- Power Flux Power = -------- Efficacy Illumination= 300 Efficacy = 60 300 Power = -------- 60 Power = 5 W/m 2 Lighting

52. Room size = 12 x 10 x 3.5 Temp diff = 12 deg C Specific heat = 1250 J/m 3 o C 1 ton = 3.5 kW Vol of room x no of air exchanges Rate = ---------------------------------------------- 3600 12 x 10 x 3.5 x 3 = ------------------------- 3600 = 0.35 cum/sec Cooling load = 1250 × 0.35 × 12 = 5.25 kW = 1.5 TR Lighting

53. Q.47 A 5m × 5m × 3m room has four 230 mm thick external brick walls. Total wall fenestration is 10 sqm. The temperature difference between indoor and outdoor is 2 degC. The air to air transmittance values for 230 mm thick brick wall and 200 mm thick aerated concrete block wall are 2.4 and 1.7 W/sqm degC respectively. If the brick walls are replaced with the aerated concrete block walls, then the change in conductive heat flow through the walls is _________W. Room = 5 x 5 x 3 m Wall thickness = 230 mm Transmittance (U) 2.4 1.7 Q = U X A X T Temp (T) = 2 degC = 2.4 x ((4 x5 x3) -10) x 2 = 1.7 x ((4 x5 x3) -10) x 2 = 2.4 x 50 x 2 = 1.7 x 50 x 2 = 240 = 170 Q = 240 – 170 = 70 W Lighting

54. Q.42 A room is mechanically ventilated through four air-conditioning ducts. The opening area of each duct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between the ambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m 3 °C. Assuming one Ton of refrigeration (TR) equals 3.5 kW, the cooling load of the room in Tons will be _____________ Area = 0.35 sqm Velocity = 0.5 m/s Temp = 10 °C Specific heat = 1250 J/m 3 °C Q = A x V x S x T = 0.35 x 0.5 x 1250 x 10 x 4 = 8750 W No of ducts = 4 = 8.75 kW 1 ton = 3.5kW = 8.75 / 3.5 = 2.5 Ton Lighting

55. Q.31 A brick wall 19 cm thick has a thermal conductivity 0.811 W/m °C. The outside and inside surface conductance of the wall are 16 W/m 2 °C and 8 W/m 2 °C respectively, then the U-value of the wall in W/m 2 °C is ____________ R = 0.19 / 0.811 1 / U = 1/ Co + 1/ Ci + R Co = 16 W/m 2 °C Ci = 8 W/m 2 °C 1 / U = 1/ 16 + 1/ 8 + 0.19 / 0.811 U = 2.30 to 2.45 W/ m 2 °C

56. Q.45 In a dance hall the indoor and outdoor temperatures are 28°C and 18°C respectively. There is an internal heat gain of 5 kW and the specific heat of air (on volume basis) is 1300 J/m 3 °C, then the necessary cross sectional area (m 2 ) of a duct with an air velocity of 2 m/s required for cooling by ventilation is ____________ Indoor temp = 28 °C Outdoor temp = 18°C Internal heat gain = 5 kW Heat of air = 1300 J/m 3 °C Air velocity = 2 m/s V olume of air passed in 1 sec = 2 x area Volume = 2A Heat in Vol of Air = Vol of Air x Specific Heat x Temp Diff Cross sectional area = A = 2A x 1300 x 10 = 2600A Internal heat = 5 kW = 5000 W 1 Watt = 1 Joule per second Internal Heat = Heat Vol 5000 = 2600A A = 0.19 m 2 1 kW = 1000 J/s

57. Q.48 For an activity, ‘optimistic time duration’ is 4 days, ‘pessimistic time duration’ is 11 days and ‘most-likely time duration’ is 8 days. The PERT value of time duration is______ days ( up to one decimal place ). O = 4 P = 11 M = 8 = 47 / 6 T = 7.8 days CPM PERT

58. O = 12 P = 17 M = 15 = 89 / 6 T = 14.8 MONTHS CPM PERT

59. CPM PERT For a PERT network analysis and activities likely to have optimistic duration of 2 day, pessimistic duration of 7 days and most likely duration of 3 days. The expected time duration that may be allocated would be ______ ET = OT + 4 MLT + PT / 6 = 2 x 4 x 3 x 7 / 6 = 3.5 days

60. Q.41 The optimistic time, the pessimistic time and the most likely time of a job are 6, 13 and 8 days respectively. The variance for this job is _______________ OT = 6 PT = 13 MLT = 8 = ((13 – 6 ) / 6) 2 = 1.36 CPM PERT