Building Services Applications

Design Criteria:

Following is design criteria as mentioned in below table for room G-14

As the room G-14 belongs to the educational building as especially lecture class room so recommended specifications for comfort environment are listed below:

Table No. 1

Diversity:

We have to consider, what is the lowest temperature in that area according to previous records then corresponding factor will be used by referring BS standards. To analyse the past climate change in that area and figure out the critical points of temperature, pressure and humidity etc. then accordingly standard factor will be applied.You must consider seasonal variations in that area then accordingly factor will be applied.Consider the design and orientation of room, like either the room is exactly face to sun or hidden one then accordingly factor will be applied.

Heating System Design:

Heat Loads:

Given data and assumptions are

The room is naturally ventilated directly with outdoor air. Assume minimum ventilation rates are met of 8l/s per person.

 Occupancy in each classroom is 30 pupils and 1 teacher

 U Values:

• Wall = 0.13 W/sqm.K
• Ground Floor = 0.1 W/sqm.K
• Roof = 0.1 W/sqm.K
• Glazing/doors = 1.4 W/sqm.K
• Infiltration rate = 0.2 ACH
• Internal Temperature = 20°C
• External Temperature = -3°C

 Base your heating loads on the combination of ventilation, infiltration and conduction. Neglecting internal gains.

Solution:

Here I have assumed all areas as below:

Total area of walls per room is 25sqm.

Plan Area of ground floor is 12sqm.

Plan Area of roof is 12sqm.

Area of doors/glazing is 3sqm

1. Fabric Heat Loads.

Qf = (ΣUA) x ΔT (ΔT= Interior T -Exterior T = 23K)

Qf = (0.13×30+0.1×12+0.1×12+1.4×3) x 23 =  151.6W (Ans).

Now total Fabric loads for 14 class rooms will be 14×151.6 = 2122.84 W. (Ans)

And as individually I have considered same area for all types of room likewise, Room Type-1, Room Type-2 etc.

So;

Fabric heat loads for room type-1 is = 151.6 x6= 909.6 W.

Fabric heat loads for room type-2 is = 151.6 x1= 151.6 W.

Fabric heat loads for room type-3 is = 151.6 x6= 909.6 W.

Fabric heat loads for room type-2 is = 151.6 x1= 151.6 W.

1. Infiltration Heat Loads

Qi = (NV/3) x ΔT

N = air changes/hour

V = Volume

 Vent rate = 10 litres/second/person= 10×3600 =36000 litres/hour/person=36000×31= 1116000 L/hr

Occupancy = 31 people

Volume (in L) = 64 cum x1000= 64000litre.

N Air changes per hour = 1116000 litres per hour ÷ V (litres in the room)

                                       = 1116000 ÷ 64000 = 17.34 per hour.

Qi = (17.34×64000/3) x 23 = 8556000 L.K per hr = 8556.0 cum.K/hr

So;

Infiltration Heat Loads for room type-1 is = 8556.0 x6=  51336cum.K/hr.

Infiltration Heat Loads for room type-2 is = 8556.0 x1=  8556cum.K/hr.

 Infiltration Heat Loads for room type-3 is = 8556.0 x6=  51336cum.K/hr.

Infiltration Heat Loads for room type-2 is = 8556.0 x1= = 8556 cum.K/hr.

1. Ventilation Heat Loads

Qv = (v x ρ x Cp) x ΔT

Qv = (10 x 1.2 x 1.02) x 23= 331.2KJ.

V = Ventilation Rate in l/s

ρ = density of air = 1.2kg/m^3

Cp = Specific heat capacity of air = 1.02kJ/kg.K

So;

 Fabric heat loads for room type-1 is = 331.2×6= 1987.2 KJ.

Fabric heat loads for room type-2 is = 331.2×1= 331.2 KJ.

 Fabric heat loads for room type-3 is = 331.2×6= 1987.2 KJ.

Fabric heat loads for room type-2 is = 331.2×1= 331.2 KJ.

Select Radiators:

1. Operating Factor:

Radiator output values are given as the wattage that the radiator is capable of giving up when the ΔT is 50C between the average system T and the room T. When the ΔT is not 50C, an operating factor has to be applied.

1. a) Average system temperature.

Flow = 70C

Return = 50C

Average = 60 C

1. b) Room Temperature = 20C
2. c) ΔT = Average system T – Room T

= 60C – 20C

= 40K

1. d) Use Operating factor chart to read O.F. value

For ΔT=40, O.F. = 0.748

This means that when there is only a 40C temperature difference between the radiator and the room, it only gives out 0.748 times as much power as the value on the manufac­turer sheet.

Manufacturer value (at ΔT=50) = 1000 W

Actual output (at ΔT=40) = 1000W x 0.748 = 748W

Here I have selected 2 radiators in each room of same size and also that are used in each type of room, as for uniformity.

Layout and pressure drops:

Figure 1 Drainage layout

Figure 2 INLET WATER LINE GROUND FLOOR

Figure 3 INLET WATER LINE FIRST FLOOR

In this layout we have below mention of following equipment’s.

14 Nos. of Washbasin and each required 6 l/sec of water.

8 Nos. of Toilet and each required 6 l/sec of water.

1 Nos. of kitchen and it required 10 l/sec of water.

Here we had considered Pump having Head 23= mtr.

Here we have considered Below mention pipe for conveying of water from pump to various portion from where can use the water for our daily routine purpose.

Main Header will be of 40 mm pipe having length 12 mtr.

Distribution from main header will be from 25 mm pipe having length 36 mtr.

And from 25 mm pipe we will have tapping of 15 mm pipe which will connect to our each

Equipment and length of 15mm pipe is 54 mtr.

Here we have considered all the minor losses as Zero and Friction loss is calculated below.

Where,

friction factor is f
the length of pipe is L
the inner diameter of the pipe is D
the velocity of the liquid is v
the gravitational constant  is g

1. Calculation for 40 mm pipe.

F = 0.4, L = 12 mtr. D = 0.04 Mtr., v = 2 m/s ( Considered for domestic Use.) and g = 9.8 m/s2.

H1= 0.4 x 12/0.04 x 2*2/2*9.81

H1= 2354 mtr.

1. Calculation for 25 mm pipe.

F = 0.4, L = 36mtr. D = 0.025Mtr., v = 2 m/s ( Considered for domestic Use.) and g = 9.8 m/s2.

H2= 0.4 x 36/0.025 x 2*2/2*9.81

H2= 11301mtr.

1. Calculation for 25 mm pipe.

F = 0.4, L = 52mtr. D = 0.015Mtr., v = 2 m/s ( Considered for domestic Use.) and g = 9.8 m/s2.

H2= 0.4 x 52/0.015 x 2*2/2*9.81

H2= 27206mtr.

Now,

P[end] = P[start] – Friction Loss – Fittings Loss – Component Loss + Elevation[start-end] + Pump Head

where

P[end] = Pressure at end of pipe
P[start] = Pressure at start of pipe
Elevation[start-end] = (Elevation at start of pipe) – (Elevation at end of pipe)
Pump Head = 0 if no pump present

P[end] = P[start] – Friction Loss – Fittings Loss – Component Loss + Elevation[start-end] + Pump Head

     = 5 Bar – (2354+11301+27206) + 8 + 25

    = 5 Bar – 40861 mtr.

    = 50000 mtr( of water column) – 40861 mtr.

   = 9139 mtr.

The above calculation shows that 4 bar Pr. Will be dropped and we will have output pr. Of 1 bar approx..

6. Design variation

Yes It is possible to use air for heat pumps but it may cause fluctuation in cost and project time. Some of the advantages and disadvantages due to of design variation are enlisted as below along with brief reason:

Advantages of design variation:

1. Less cost than the use of carbon.
2. Less dangerous then the use of carbon.
3. Efficient operating capacity.
4. May cause high quality operation.
5. Reduce environmental impacts.

Disadvantages of design variations:

1. May cause delay in project.
2. High risk as new innovation in market.
3. May cause huge fluctuations in cost.
4. There may be adverse reactions which cause environmental effects.
5. There may be risk of corrosion.
6.  

Lighting Design

User Requirement Specification

The user specification requirements are those that are to be considered before installation of any project, because they effect on the cost of project. The quality of work majority effected by user requirements. User requirement are likewise, do the project within defined deadlines at very low cost by compensating the each and every thing of high quality. User always try to get work very efficiently. In the case of electric work he will want at end as like, fully comfortable, less power consumption, efficient resources, efficient capacity of work, less power losses, less capacitance losses, less maintenance cost, high durability and serviceability. These are the few user requirement specifications that are to be considered.

Lighting Design

Assume room G-14 within the building used in your examples above will now we redeveloped into a permanent gallery space. Design for the electric lighting of the gallary selecting appropriate fittings and proving function with appropriate calculations. Prove your design meets lighting standards with an appropriate software calculation- for general and emergency. From the software generate a standard output room layout with lux contours on the working plane. Provide one page with illustrations of your room including an isometric view