Drainage System Project

8/18/2019 Drainage System Project

1/17

Question 1

A)

Determine the total Loading Unit (L.U.) of the system

Given the design flow rate, demand time, frequency of use of each sanitary fitment, demandunit of each fitment can be found.

Usage ratio =

Since the demand of one basin is 1 Loading Unit (L.U.), one basin is our base appliance.

Proportion of base appliance ratio = ( .)

Proportion of base appliance flow rate = (.)

Demand figure = proportion of base appliance ratio x proportion of base appliance flow rate

Demand unit = round up number of the demand figure = loading unit of each fitment

Fitment QuantityDemand

unit

total Loading unit

(L.U.) of each type of

fitment(Quantity x Demand

unit

Basin 10 1 10

Sink 2 5 10

Shower 10 5 50

Clothwasher

2 5 10

Hence, the total Loading Unit (L.U.) of the system = 10 +10 +50+ 10 = 80

Given in the question Calculations

Fitment

Flow

rate(L/s)

Demand

time (s)

Frequency

Of use(s)

Usage

ratio

proportionof base

appliance

ratio

proportionof base

appliance

flow rate

Demand

figure

Demand

unit

Basin 0.12 30 1800 0.017 1.00 1.00 1 1

Sink 0.3 60 1800 0.033 2.00 2.50 5 5

Shower 0.18 300 6000 0.050 3.00 1.50 4.5 5

Cloth

washer 0.2 450 9000 0.050 3.00 1.67 5 5


2/17

Determine the simultaneous demand in L/s of the system

Fitment n

P= Usage

ratio m

Probability of

occurrence Accumulating

Basin 10 0.017 0 P0= 0.845

1 P1= 0.143 0.9892 P2= 0.011 0.999

Sink 2 0.033 0 P0= 0.934

1 P1= 0.064 0.999

Shower 10 0.05 0 P0= 0.599

1 P1= 0.315 0.914

2 P2= 0.075 0.988

3 P3= 0.010 0.999

clothwasher 2 0.05 0 P0= 0.903

1 P1= 0.095 0.998

Based on probability theory, probability of having a sanitary fitment in use at a certain time

can be calculated.

= ( !! ( )!) (1 )− Where n = total number of fittings having the same probability

m= number of fitting in use at any one time

Pm = probability of occurrence

Consider the first case where a group of 10 basins, with a certain design demand time and

frequency of use, the probability factor of a particular number of draw off points occurring at

any one time, P indicated above the chart on the third column from the left, can be computed.

P =

Since the design intent is to serve 99% of all conditions, in the “accumulating” column as

long as shows nearly close to 0.99 meaning with this number of fitments or less in use at one

single time can serve 99% of all times.

For the basin case, 2 basins become the simultaneous demand.

For the sink case, 1 sink becomes the simultaneous demand.

For the shower case, 3 showers become the simultaneous demand.

For the cloth washer case, 1 cloth washer becomes the simultaneous demand.


3/17

Hence, the simultaneous demand of the system, 0.24+0.3+0.54+0.2=1.28 L/s

B) Determine the pump rating and size of the pneumatic tank

The booster system at 20/F serves the toilets three floors above (21/F, 22/F and 23/F) and

three floors below (17/F, 18/F and 19/F). Assume vertical pipe loss=0.1mH/m run.

Technically, the cut- in pressure of the pump only occurs from the booster system at 20/F to

the water point at 23/F since that is the highest water point which has to be satisfied with the

minimum pump pressure need for the entire booster system. In addition, the cut-out pressure

of the pump only occurs at the 17/F since gravity will add up along the pipe going down.

Hence, we only need to consider two conditions: pump cut-in pressure from 20/F to 23/F and

pump cut-out pressure from 20/F to 17/F.

Consider three floors below (17/F, 18/F, and 19/F) and ignore the top three floors for a

moment.

Given maximum residual pressure 4.5bar = 45mH

Head gain from the booster system (20/F) to 17/F = 3.5mH x3=10.5mH

Pump cut-out pressure = 45m-10.5 + 0.1x3.5x3 = 35.55mH

Consider three floors above (21/F, 22/F and 23/F) and ignore the down three floors for a

moment.

Given minimum residual pressure 1.0 bar = 10mH

Head loss from the booster system (20/F) to 23/F = 3.5mH x3=10.5mH

Pump Cut-in pressure= 10mH+10.5+ 0.1x3.5mx3H= 21.55mH

To sum up, pump cut-in pressure is 21.55mH and pump cut-out pressure is 35.55mH in the

booster system which is acceptable since their pressure difference is within the range of 10 -

20 mH.

Fitment Flow rate (L/s) Simultaneous demand

Flow rate x

simultaneous demand(L/s)

Basin 0.12 2 0.24

Sink 0.3 1 0.3

Shower 0.18 3 0.54Cloth

washer0.2 1 0.2


4/17

Each floor has a male, female and disabled toilet: 4 basins for male, 5 basins for female and

1 basin for disabled toilet. Therefore, each floor has 10 basins.

Since the booster system serves the basins in toilets three floors above and three floors below,

The simultaneous demand for 6 floors = 1 LU x 6 floors x 10 basins = 60 LU

Flow rate based on 60LU (See Pipe sizing Chart on the next page, graph 3)

For a 10 minute cut-out cycle, this ‘effective volume’ = 0.9 L/s x 10min x 60sec =540 L

P1 V1 = pressure and volume at initial conditions

(P1 =100kPa (abs) for normal atm. Condition)

P2V2 = pressure volume at ‘Pump Cut-in’

P3V3 = pressure volume at ‘Pump Cut-out’

V2 – V3 = effective capacity of tank = 540L = 0.54m3

P2 = 21.55mH (gauge) = 100 + 215.5 = 315.5kPa (abs) (lower limit)P3 = 35.55mH (gauge) = 100 +355.5 = 455.5kPa (abs) (upper limit)

According to Ideal gas law = 315500 ( +0.54) =455500 = 1.22 = 100000 =455500∗1.22 =5.54 The actual size of the pneumatic tank = 5.54m

3

Pump rating: Pump Cut-in pressure= 21.55mH; Pump Cut-out pressure = 35.55mH


5/17

Pipe Sizing Chart

Source: Plumbing Engineering Services Design Guide (2002)

0.9 L/s


6/17

Question 2

A) Determine whether the existing hot water system is sufficient for the present demand

and give recommendations if the analysis shows dissatisfaction of the system

Hour

Consumption

(L )

0:00 16801:00 1260

2:00 590

3:00 170

4:00 160

5:00 250

6:00 170

7:00 840

8:00 1680

9:00 2015

10:00 67011:00 250

Hour

Consumption

(L )

12:00 8013:00 85

14:00 505

15:00 1090

16:00 1085

17:00 2100

18:00 2770

19:00 3360

20:00 4200

21:00 4620

22:00 378023:00 2100

Table A: Peak hourly loads (above)

Table B: Peak hourly loads

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 : 0 0

1 : 0 0

2 : 0 0

3 : 0 0

4 : 0 0

5 : 0 0

6 : 0 0

7 : 0 0

8 : 0 0

9 : 0 0

1 0 : 0 0

1 1 : 0 0

1 2 : 0 0

1 3 : 0 0

1 4 : 0 0

1 5 : 0 0

1 6 : 0 0

1 7 : 0 0

1 8 : 0 0

1 9 : 0 0

2 0 : 0 0

2 1 : 0 0

2 2 : 0 0

2 3 : 0 0

C o n s u m p t i o n ( L )

Hour (hr)

Demand Profile of a Hot Water


7/17

Peak

hours

litres

used

litres per

hour

1 4620 4620

2 8820 4410

3 12600 4200

4 15960 39905 18730 3746

6 20830 3472

7 22930 3276

8 24015 3002

9 25105 2789

10 25610 2561

11 25695 2336

12 25775 2148

13 26025 2002

Peak

hours

litres

used

litres per

hour

14 26695 1907

15 28710 1914

16 30390 1899

17 31230 183718 31400 1744

19 31650 1666

20 31810 1591

21 31980 1523

22 32570 1480

23 33830 1471

24 35510 1480

24hr average:

35510/24 = 1479.6 =1480 L/hrTable B: Peak hourly loads (above)

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 5000 10000 15000 20000 25000 30000 35000 40000

R e c o v e r y ( L / h r )

Storage-litres (L)

Storage make-up ratio curve peak hour 1peak hour 2

peak hour 3

peak hour 4

peak hour 5

peak hour 6

peak hour 7

peak hour 8

peak hour 9

peak hour 10

peak hour 11

peak hour 12

peak hour 13

peak hour 14

peak hour 15

peak hour 16

peak hour 17

peak hour 18

peak hour 19

peak hour 20

peak hour 21

peak hour 22

peak hour 23

peak hour 24

Min recovery rate =1480 L/hr

Recovery/storage curve


8/17

Given that

-Existing hot water tank = 8000 L

-Existing hot water heater Q= 120kW

-Design cold and hot water temperature = Tw=10°C & Th= 65°C

-Heat capacity of water, cp= 4.2 kJ/kgK

Assume water density as 1kg/L = ∆120= 4.2 (6 5 1 0) = 0.52 /

= 0.52 1 / 3600 /ℎ

= 1872ℎ In the existing system, the 8000L hot water tank is at a recovery rate of 1872L/hr. However, in

the storage make-up ratio curve, under the condition of a 8000L hot water demand, the required

recovery rate is at least roughly 2300L/hr. Hence, the existing system is not sufficient for the

present hot water demand.

Recommendations:

If the hot water tank storage to 8000L is kept,

= ∆ = 80001ℎ 1ℎ3600 11 4.2 (6 5 1 0) =147.6 The power output of the heater requires to be larger so as to satisfy the required hot water

demand instead of the existing smaller power output of 120kW. Moreover, finding a heater with

a recovery rate of exactly 2300 L/hr on the existing market is not very possible although

2300L/hr recovery rate is the theoretical value based on the plot. Perhaps, we can pick a

2500L/hr recovery rate heater that should be available on the market for our need.

Or if the power output of the heater is kept to 120kW, with 1872L/hr recovery rate, based on the

storage make-up ratio, the hot water tank should require around 12000L


9/17

B) Sizing the primary and secondary pipework

Consider a primary circuit of a hot water system,

Design flow rate = 2.0 L/s

Cold water = 20 °C (density 998 kg/ m3)

Hot water = 65° C (density 980 kg/m3)

The height difference between boiler and calorifier = 8m

Pressure difference of two columns of water=∆ℎ= (998 – 980) x 9.81x 8 = 1413 PaAssume the overall hydraulic length of the pipe = 24m (measured pipe length + length of fittings)

Allowable pressure drop = 1413Pa/ 24m = 59 Pa/m

From the pipe sizing chart for copper pipe, at design flow =2.0L/s,

Based on average 989kg/m3,

Head loss per meter run = 59 = 0.006 /,(see Pipe Sizing Chart on the next page) 76mm pipe suits the requirement.

Consider secondary pipework of a hot water system

The secondary return pipe has to be sized based on the minimum

flow rate to determine the possible heat loss from hot water pipe.

Given from the diagram that supply temperature of water=65°C

Return temperature of water = 55°C

Consider insulated supply pipe size =54mm OD (~21W/m run heat

loss), with the assumption of head loss = 0.1 mH/m at 3.5L/s (see

Pipe Sizing Chart on the next page)

Supply pipe length = 20m

Heat loss= 21W/m x 20m = 320W

Assume water density as 1kg/L

Temperature drop ∆ = 5°C (from 65°C - 60°C at the end of supply) = ∆ 21 20 = 4200 5°C =0.02/ This is the minimum flow rate in the secondary circuit to ensure the end of the supply still has

60°C

As for the secondary return, = ∆ ℎ 20 = 0.02 4200 (6055)°C ℎ = 21


10/17

Since the secondary return has very similar conditions, including same temperature drop, same

mass flow rate, same pipe length, the size of secondary return should be 54mm OD same as the

secondary supply pipe.

Pipe Sizing Chart

Source: Plumbing Engineering Services Design Guide (2002)

For primary

circuit (orange)

For secondary

supply circuit

(red)


11/17

Question 3

A) Wobbe index is equal to higher heating value (calorific value) in MJ/m3 divided by square

root of gas specific gravity. The Wobbe index is used to compare the combustion energy output

of different fuel gases, such as natural gas, liquefied petroleum gas (LPG), and town gas, in an

appliance, like a water heater. If two fuels have identical Wobbe indices, then for a given

pressure and valve setting, the energy output will also be the same. For example, under the same

pressure and valve setting condition, if gas A with a Wobbe index of 24 and gas B also with the

same Wobbe index of 24, although they both have different calorific values and specific gravity

values, both gases will give the same energy output. Hence, we can make use of any fuels that

have the identical Wobbe index and give out same combustion energy output to substitute one

another.

Town

gas LPG

calorific value (MJ/m3) 17.27 104.9

specific gravity 0.52 2.05

wobbe index= √ 24 73Consider 1 water heater and 1 table top cooker

Water heater rating = 102.2 MJ/hr

Table top cooker rating = 51.3 MJ/hr

Total = 102.2 +51.3 =153.5 MJ/hr =././ =8.89/ℎ

The Pole formula is used for determining the flow of gas in pipes.

=0.0071 ℎ Where Q = flow (m

3/hr),

d= diameter of pipe (mm),

h =pressure drop (mbar), assume 100 Pa = 1mbar,

l= length of pipe (m), in this case, measured length of pipe: 5+8+1 =14m, length of fitting around

4m

s =specific gravity of gas (town gas =0.52)

8.89=0.0071 1

0.5218 = 2 7 Gas supply pipes from the gas meter to the individual appliances = 32mm (available on the

market)


12/17

B) Determine pressure drop of steam pipe work and then size the steam pipe

Given:

A boiler plant producing 180°C superheated steam at 600kPa (absolute) = 6 bar

Minimum pressure at the heater at 150 °C = 475kPa =4.75 bar (absolute) from steam table

Pressure drop of steam pipe work = 6bar -4.75 bar = 1.25bar (absolute)

Steam loading of the heater =300kg/hr

Overall equivalent length of the steam pipe = 40m

Consider the heat loss from pipework = 3.5% of heater load per 100m

Hence, revised steam load = 300kg/hr + 300 x(3.5% x 40/100)=304kg/hr

Steam pressure drop factor table

6bar =43.54 (blue circle)

4.75bar= (29.27+30.27)/2=29.77

= where F= pressure factor, P1 = factor based on theinlet pressure, P2 = factor based on the pressure at a

distance of L metres, L= equivalent length of pipe (m) = .−. = 0 . 3 4 From pipeline capacity table (see the next page),

F=0.3, capacity =487.3kg/hr> 304kg/hr

Size =32mm

Steam = 304kg/hr at 6 bar

Specific volume = 0.316m3/kg at 6 bar

Volume flow = 96.064 m3/hr = 96.064/3600= 0.027m

3/s

32mm pipe area = 0.0008m2

Thus pipe velocity = 0.027m3/s / 0.0008 = 33.75m/s

The pipe size is within the normal design velocity range ( 24-36m/s)


13/17


14/17

Question 4

A) Given:

60% water depth for the drain pipe at peak hours

A measured velocity, v0.6= 1.2 m/s

450mm drain pipe is laid to fall at 1:225

Proportional flow chart:

. = ℎ = 0.6 . =1.05

= 1.2 / 1.05 = 1.14/ , = 4 = 0.454 = 0.159 , =

= 1 . 1 4 0.159

=0.181/

=181/ . =0.65 .181 =0.65 ,. = 118 / Lets assume full potential as ¾ bore flow.


15/17

. =0.9. = 0.181 0.9 . =0.1629/ . =162.9 /Amount of discharge to reach its full potential=162.9118=44.9L/s

Since design daily discharge per occupant =225 L per day, ½ of the discharge is recorded in a 6

hour period,

Originally, with the existing drain12 ( 225)6ℎ =118/12 ( 225)6ℎ 1ℎ3600 =118/ =22656

To reach its full potential (3/4 bore flow)12 ( 225)6ℎ =162.9/12 ( 225)6ℎ 1ℎ3600 =162.9/ =31277 Therefore, the drain can accommodate 31277 – 22656 = 8621 people on top of the original

population.

B)

= Where Q is rain water flow (L/s)C is the impermeability factor or run-off coefficient

A is drainage or catchment area (m2)

I is rainfall intensity (m/hr) = +0.5 Where Af is the catchment floor area (m

2),

Aw is the area of vertical wall (m2)

A is catchment area (m2)

= 1010

= 100 Determine: the time of concentration of the storm water for the roof = 0.14465 ( . .)

Where tc is time of concentration (min)

H is average fall (m per 100m) from the summit of catchment to the point of drainage

outlet


16/17

L is the largest distance from catchment boundary to the point of drainage outlet (m)

A is the catchment area (m2)

= 0.14465 ( 1 0 + 5(1).(100).)

=1.37

Determine the extreme mean intensity of the rainfall, i , for a 10 year rainstorm return period

= 603( +4.4). Where td duration of rainfall in min. In this case td is equal to tc at a peak discharge of the

rainstorm system

= 603(1.37+4.4

). = 279 /ℎ Determine stormwater flow rate to be handled by the system, Q = 0.278/

Where Q is rain water flow (L/s)

C is the impermeability factor or run-off coefficient, in this case, assume 0.95 for roof

A is drainage or catchment area (m2)

I is rainfall intensity (m/hr)

= 0.951002791000 0.278

= 7 . 3 6

Determine the expected bore flow of the storm water pipe

Based on empirical formula of vertical stacks

= Where Q is rain water flow (L/s), k is a coefficient to determine which bore flow is expected, D

is the diameter of pipe (mm)

7.36=(100) = 3.4 10− So 1 / 4 bore full is expected of the storm water pipe.


17/17

Drainage System Project

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