Answer:
The 1st plant with a storage reservoir is a better option as compared to that of the 2nd plant.
Explanation:
Suppose the factor for variation in hourly demand is 2 So the average hourly demand is given as
Average Hourly Demand=Factor x Average Daily Demand
AHD=2 x 18000 m3
AHD=36000 m3
For the first pump
The Quantity in storage tank is 3975 m3
So the amount of pumping required is
[tex]Q_{pump1}=AHD-Q_{reservoir}\\Q_{pump1}=36000-3975 \\Q_{pump1}=32025 m^3[/tex]
For this value the pump will work for following hours
[tex]t_{pump1}=\frac{Q_{pump}}{pumping rate_1}\\t_{pump1}=\frac{32025}{1750}\\t_{pump1}=18.3 \, hours[/tex]
So the pump 1 can complete the demand of the town by working for 18.3 hours.
Now in order to complete the demand, the second pump is given as
[tex]Q_{pump2}=AHD\\Q_{pump2}=36000 m^3[/tex]
For this the pump will work for as
[tex]t_{pump2}=\frac{Q_{pump2}}{pumping rate_2}\\t_{pump2}=\frac{36000}{2250}\\t_{pump2}=16 \, hours[/tex]
So the pump 2 requires 16 hours to complete the demand of the town.
Here it is important to note that the realistic demand of the water can vary from the average value and thus when there is a drastic requirement of water in certain cases, the pump 2 will fail. Also pump 2 has to be run continuously and will produce excessive water which will be wasted if the hourly demand is less than that of the production value.
In context of this, the 1st plant with a storage reservoir is a better option as compared to that of the 2nd plant.
