3. Domestic Demand: # J Elevatio n ( m ) Domestic Area DensityPopulation Q ( m² )( m3/d ) J134.033256.820.05162.8432.98 J229.066493.580.05324.6865.75 J325.305778.820.05288.9458.51 J425.206637.210.05331.8667.20 J526.704675.620.05233.7847.34

4. For the 1 st junction : P domestic = Area*Population Density = 3256.82* 0.05= 163 Q domestic = P domestic *150 *1.35 /1000 = 32.98 m³/d Average Demand for capita / day =150 L/c/d The value 1.35 = 25% loses in the network and the equivalent commercial demand which was taken 10% domestic demand. Domestic Demand:

5. For public demand: Average Demand for capita / day = 40 L/c/d Q public = # employee * 40 * 1.25 (25% UFW) Public Demand:

6. # J Public Area # Employees Q ( m² )( m3/d ) J10 00 J20 00 J30 00 J4548.38109.685.5 J51922.40384.4819

7. For junction 4 which has a public demand, Area = 548.38 m², Let # of employee = 0.2*Area =109.68 =110 person Q public =110*40*1.25/1000 = 5.5 m³ /day Public Demand:

8. Commercial Demand: For junction which has a commercial demand, Let # of = 1200 person/day Average Demand for capita / day = 30 L/c/d Q Commercial = 1200*30*1.25/1000 = 45 m³ /day (25% UFW)

9. Industrial Demand: For junction which has an industrial demand, Let # of workers = 850 person Average Demand for capita / day = 30 L/c/d Add 30% for cleaning works.. etc Q Industrial = 850*30*1.55/1000 = 40 m³ /day (25% UFW, 30% Others)

10. School Demand: For junction which has a school demand, # of pupils/school = 1200 pupil Average Demand for capita / day = 10 L/c/d Q school = 1200*10*1.25/1000 = 15 m³ /day (25% UFW)

11. Hospital Demand: For junction which has a hospital demand, Let # of beds = 160 Average Demand for capita / day = 300 L/c/d Q Hospital = 160*300*1.25/1000 = 60 m³ /day (25% UFW)

12. Mosque Demand: For junction which has a mosque demand, # of prayers/day =500 person Average Demand for capita / day = 25 L/c/d Q Mosque = 500*25*1.25/1000 = 15.7 m³ /day (25% UFW)

13. Agricultural Demand: For junction which has a agricultural demand, Agricultural demand varies according to the type of trees. Demand for Green areas = 5 m 3 /Donum/Week Average Demand / day = ??? m 3 /d Don’t Forget (25% UFW)

14. Tourist Demand: For junction which has a tourist demand, Demand for Hotels = 180 L/C/d Average Demand / day = ??? m 3 /d Don’t Forget (25% UFW)

15. Total Demand: Q total of network = ∑ Q nodes m 3 /day Average Demand: For each type of demand Q avg_i = ∑ Q nodes / 24 m 3 /hr Where (i) is the type of demand

16. Q total of network (m 3 /day) Example..

17. Questions