How to calculate the quantity of water for a given concrete mix.?

Obviously, water quantity cannot be decided based on the quantity mix design proportions only. There are other data which need to be known in order to be able to specify water content quantity for example, types of structural element, maximum aggregate size, and required workability. If these data are available, then the quantity of water can be easily computed from Table 6.3.3 of ACI 211.1-91.

Required slump can be determined based on the structural element type. Maximum aggregate size and slump value can be used to determine quantity of water for 1 meter cubic of concrete.

As per Is code 456-200 clauses 6.1.2

maximum free water cement ratio for M15(1:2:4) is 0.50

for 1 cubic metre concrete

1+2+4=7

cement content =1/7 *1 =0.1428 m3

density =mass/volume

density of cement =1440kg/m3 (note :- density of cement varies with grade and brand)

mass of cement =0.1428*1440=205.632 kg (note:- please consider minimum cement content also i.e. also given in clauses 6.1.2 of Is code)

w/c as per IS code is 0.50 for M15

w/c=0.50

water content=0.50*205.632=102.8 m3 =103 litre

Note:-w/c depends on various factors like moisture content of fine/coarse aggregate,workability .....

it has a simple way to solve that.

The taken water ratio of your mix design (multiply)Weight of cement.

W/C*Quantity of cement= Quantity of water

Yes, thank you Ricardo for your good answer.

If we need to calculate Water quantity for concrete, first find the cement content for the volume. Therefore, Required amount of water = 0.5 X 50 kg = 25 litres / 50 kg cement bag. For Design mix, the W/C Ratio will depend upon the workability, strength requirements.

The quantity of water calculates the base one depend on the water-cement ratio.

W/C=AS PER IS CODE RATIO(DEPEND ON GRADE OF CONCRETE)

Calculate the existing moisture content in the aggregates (fine & coarse) and it will be deducted from the water content, which is calculated from water cement ratio

Moisture content in Coarse aggregate - x1

Moisture content in fine aggregate - x2

Moisture content based on water cement ration (example -0.355) = 0.355*450kg/cum = 159.75

Net water content is = 159.75 - x1 - x2

The mix 1:2:4 represents the M15 grade of concrete, which means the concrete formed with this ratio has a compressive strength of 15 kN/m^2. 1:2:4 represents the amount of cement, fine aggregate (sand) and coarse aggregate to be added during batching of concrete.

 Calculation of Water Quantity for Concrete If we need to calculate Water quantity for concrete, first find the cement content for the volume. Therefore, Required amount of water = 0.5 X 50 kg = 25 litres / 50 kg cement bag. For the Design mix, the W/C ratio will depend upon the workability, strength requirements.

Dear Sir

First of all you need to know the quantity of cement.

1:2:4 Means M15 Grade

Cement require for 01 Cu.m Concrete is:

Wet volume of concrete is 1.54 Cu.m = (1.54/1+2+4)*1 = 0.22 Cu.m

Now 0.22 Cu.m Cement = 0.22*1440 Kg/m3 (Density of cement) =316.80 kg cement = 316.80 kg

Now W/c Normally require for M15 Grade is 0.50

So water require: W/C = 0.50 =

C*0.50 = Water

316.80*0.50=Water

Water = 158.40 Litre/Cu.m

for Mix 1:2:4 water requirement would be-

Plain concrete(moderate)

The Minimum cement content for M15=240 kg/ m3

and min water cement ratio =water/cement =0.60 =water/240=0.60 Min water requirement for mix=240*.60=120 liters

It is also depend upon climate condition,method of mixing and the purpose of use.

Can you please tell me where you got the wet of concrete is 1.54 cu/m

The W/C ratio varies from 0.4 to 0.6 as per IS 10262(2009) for a normal mix

To calculate the amount of water required for a concrete mix, we need to calculate the quantity of cement required.

Assume the required amount of cement is 50kg

Amount of water required = W/C ratio x cement required.

= 0.5 x 50

= 25 litres/50kg

For the design mix, W/C ratio depends upon the workability and strength requirements.

The accurate quantity of water for a mix depends your selection of water cement ratio, compressive strength, workability, durability, maximum nominal size of aggregate, grading,type of aggregates , absorption and the degree of control. The question of 1:2:4 could be on mix by volume or mass. W/C ratio is usually determined by mass.

IS:456 is s good code to follow for design a mix.

Actually water content is determined from the W/c ratio of different mixes as per IS 456:2000. but roughly for normal mix (1:2:4), we can calculate as

vol. of cement = 0.0354 cu.m

vol. of C.A+ F.A=2*0.0354+4*0.0354=0.211 cu.m

Vol. of water = 11.67 litre for 1 bag + 5% of total aggregate

= 11.67+(0.05*0.211)*1000=11.67+10.55= 22.22 litre

Here we have 1:2:4 mix is given, which is the standard ratio for M15 grade.

I do the calculation for 1 m3 concrete.

for cement content, c= 1/(1+2+4)= 1/7 m3

weight of cement = {1/7/(.035)}*50 = 204kg

total weight of cement for mix = 1.54*204 = 310.20kg

(total weight of cement for mix =310.20kg) >( minimum cement content - 300kg for moderate exposure condition)

for exact water content, we should know about exposure conditions, for different exposure conditions water content will be different. here I mention different exposure conditions with their maximum w/c ratio.

mild-.55

moderate - .50

sever - .45

very sever - .45

extreme - .40

I do the calculation for moderate exposure condition (you can do it for others exposure condition also, but the method will be the same. for other calculation you should check cement content for minimum and maximum cement content with different exposure condition)

w/c = .50

w = .5*310.20

w = 155.10 kg = 155.10 liter

so, water content for m15 grade with moderate exposure is 155.10 liter.