Consider for 1CUM of concrete is to be prepare,For that

M15 is a nominal mix, its proportion is 1:2:4

for 1cum of final mix(Dence) is required, 1.50 to 1.55 Cum
of loose ingredients with the required proportion we want
to add.

Based on that take avg of 1.50 & 1.55 is 1.52

Total proportion = 1+2+4 = 7

For Cement Qty = 1.52/7=0.217 Cum of cement required to
prepare 1Cum of M15 concrete.But cement is always taken in
Weight.For that the unit weight of cement is 1440 KG/Cum
Total cement required = 0.217x1440= 312.68 Kgs.

Note: For 1m3 wet concrete = 1.52m3 dry concrete approximately
SP.Wt of concrete= 1440 kg/m3 (or) 1.44 t/m3
1 bag of cement = 50 Kg
Example:- for 20 m3 volume of concrete.

a) Quantity of cement required = (1:2:4)
=1/(1+2+4) x 1.52 × 20 = 4.14m3 x 1440/50 =119.26 bags

b)Quantity of Sand required = (1:2:4)
=2/(1+2+4)×1.52x20=8.28m3 OR =4.14 x 2=8.28 m3.

c)Quantity of course aggregate = (1:2:4)
=4/(1:2;4) x 1.52 x20 = 16.56m3 OR =4.14 x 4= 16.56m3.

If qty. calculate in Kg just multiply by its Sp. wt of sand
or aggregate.

ratio is 1:2:4 sum is equle 7
1 bag cement is equle to 1.25
total quantity resum is 200 cft
dry Volume of total quantity which is 200 cft to multiply
1.54 will be the value of total quantity is 308 cft.
which is the total quantity 308 multiply 1(which is cement
ratio)and divided 7 which is the sum of ratio it will
become 44 and now we convert cft to cement bag which is the
value of 1.25 for bag.

44 divided by 1.25 is 35.2 the Sum is 35.2 Cement of Bags.
same for sand. which value was came after dry
308 multiply ratio of sand which is 2 then divided sum of
the ratio which is 7 will become the value of Sand is=88 cft

for crash 308 multiply 4 which is the value of crush and
devided by sum of value which is 7 total sum will become
176 cft.

Total sum of quantity of 1:2:4 mean.
Cement=35.2 Bags
Sand= 88 cft
Crush= 176 cft

we have a texmo make submersible water pump 5hp 10 stage
what will be the total head & is it any thumb rule for
calculating the head please reply urgently