
A rectangular tank measures 16mx15mx6m. How many tonnes of oil of RD 0.78 can it hold?
SOLUTION:
Given : L =16m , B = 15m, H = 6m, RD=0.78
Volume of rectangle =(LxBxH)
=16x15x6
=1440m3
We know that :
Density = Mass /Volume
0.78 =Mass/1440m^{3}
Hence, Mass =1123.2 tonnes

A cylindrical tank of diameter 8m is 10m high.400t of oil of RD 0.9 is poured in to it. Find the ullage, assuming π to be 3.1416.
SOLUTION:
Given: Diameter = 8m,
Radius = (d/2)m
=(8/2)m
=4m
Height =10m
Mass = 400t
RD =0.9
We know that:
Volume of the cylindrical tank = π^{2}h
= 3.1416 x 4 x 4 x 10
=502.656m^{3}
We know that:
Density = (Mass/Volume)
0.9 = 400/ (Volume of the oil)
Hence, Volume of the oil = 400/0.9
=444.44m^{3}
Depth of oil = volume/area
Area of cylinder =( π^{2}r)
=(3.1416 x 4 x 4)
=502656m^{3}
We can calculate Depth of oil = (volume of oil)/ ( area of cylinder)
= (444. 44)/(50.2656 )
=8.8418 m
Hence, ullage = (10 – 8.8418) =1.1582m
 A tank of 2400m^{3} volume and 12 depth, has vertical side and horizontal bottom. Find how many tonnes of oil of RD 0.7 it can hold, allowing 2% of the tank for the expansion .state the ullage of loading.
Solution :
Given :
Volume of the tank = 2400m^{3}
Depth of tank =12m
RD =0.7
According to question 2% of the volume is allowed for expansion.
As we know that
Density = Mass/ Volume
Since volume =(L x B x H)
= 2400m^{3}( given )
So, Area = (volume / depth)
= (2400/12)
=200m^{2}
Mass = ( volume x density)
Mass = (2400×0.7)
=1680 t
Since 2% of the volume of the tank allowed for expansion
= (2/100) x 2400
= 48m^{3}
Volume of the oil = (volume of the tank – free space )
= ( 2400 48)
= 2352 m^{3}
Mass of the oil = (volume x density)
=2352 x 0.7
Depth of oil = (volume of oil /area)
= 2352/(L x B)
= (2352/ 200)
=11.76 m
In Q9 I found ullage by subtracting
Total depth – Depth of oil
I got 0.21m but by using your method answer is coming correct..can you please explain me🙏
Got its!!
It’s very helpful for me ..thank you sir
Fantastic innovation
In Q8 area of tank should be 1000m2 not 10000m2