Calculation of Fine Aggregate Quantity for Circular Water Tank Construction

Given Parameters:
- Diameter of Tank = 3.5 m
- Height of Tank = 5 m
- Grade of Concrete = M25

Assumptions:
- Thickness of Tank Wall = 150 mm
- Ratio of Cement : Sand : Aggregate = 1 : 1 : 2

Formulae:
- Volume of Tank = πr^2h, where r is radius = d/2
- Volume of Concrete = Volume of Tank + Volume of Tank Wall
- Quantity of Fine Aggregate = (Volume of Concrete x (1/4) x 2) / (1+2), where 1:2 is the ratio of sand and aggregate in concrete mix.

Calculation Steps:
1. Calculation of Tank Radius
- Given Diameter of Tank = 3.5 m
- Tank Radius (r) = Diameter/2 = 3.5/2 = 1.75 m

2. Calculation of Volume of Tank
- Given Height of Tank = 5 m
- Volume of Tank = πr^2h = 3.14 x 1.75^2 x 5 = 48.125 m^3

3. Calculation of Volume of Tank Wall
- Given Thickness of Tank Wall = 150 mm = 0.15 m
- Volume of Tank Wall = (2πrh) x Thickness
- Inner radius of tank = r - Thickness = 1.75 - 0.15 = 1.60 m
- Volume of Tank Wall = (2 x 3.14 x 1.60 x 5) x 0.15 = 7.56 m^3

4. Calculation of Volume of Concrete
- Volume of Concrete = Volume of Tank + Volume of Tank Wall
- Volume of Concrete = 48.125 + 7.56 = 55.685 m^3

5. Calculation of Quantity of Fine Aggregate
- Given Ratio of Cement : Sand : Aggregate = 1 : 1 : 2
- Quantity of Fine Aggregate = (Volume of Concrete x (1/4) x 2) / (1+2)
- Quantity of Fine Aggregate = (55.685 x 0.5 x 2) / 3 = 18.56 m^3

Therefore, the quantity of fine aggregate required for construction of a circular water tank of 3.5 m diameter and 5 m height, if M25 cement concrete is used is 18.56 m^3.