The Tank Volume: To get the tank volume, we have to get the total hourly demand and the hourly pumping.

The Pump: The pump will operate 18 hours/ day The pump flow: Let Q total in our network = m 3 /day, then Q pump = Q total / #of pumping hours = / 18 = m 3 /hr Where : Q total = ∑ Q for all types of demand at all nodes in the network (m 3 /day)

Choose (6) consecutive hours to shut off the pump in the day. You have to choose hours which has minimum demand to get minimum tank volume. The hourly Pumping

We have to find the demands per hour in the network using the pattern factors for every type of demand. Q domestic at a given hour = domestic hour factor*Q avg_domestic Q public at a given hour = public hour factor*Q avg_public ….. etc Find the total demand required at every hour Q total at a given hour = Q domestic + Q Public + Q mosque …etc, at the given hour The total hourly demand:

Time ( hr ) Domestic Pattern Qdomestic School Pattern Qschool Public Pattern Qpublic mosque Pattern QmosqueQtotal m3/h

If (Q total – Q pump ) > 0, then water from tank is need. If (Q total – Q pump ) < 0, then water goes to tank. ∑ Q From tank = ∑ Q To tank Tank Volume

Time ( hr ) QtotalQpumpQtotal - Qpump From tankTo tank m3/h Tank Volume: 24

Tank Volume: Many trials should be made. Choose the most economical tank which has the minimum volume. An additional volume has to be added to the total volume as a provision for fire or repair. This volume has its own pipes; it means that it is not available for common use. Vi=∑ Q From tank = ∑ Q To tank So, Tank Volume = 1.05 * Vi +240

Tank Dimensions: Assume a tank with 6-15 m base diameter, then find its height. Height of tank 5-10 m. Max. Height of tank= 10m

Questions