Updated: Jun 2, 2020
1. A rectangular tank measures 16mx15mx6m. How many tonnes of oil of RD 0.78 can it hold?
Given : L =16m , B = 15m, H = 6m, RD=0.78
Volume of rectangle =(LxBxH)=16x15x6=1440m3
Hence, Mass =1123.2 tonnes
2. A cylindrical tank of diameter 8m is 10m high.400t of oil of RD 0.9 is poured into it. Find the ullage, assuming π to be 3.1416.
Given: Diameter = 8m,
Radius = (d/2)m=(8/2)m=4m Height =10m Mass = 400t RD =0.9
Volume of the cylindrical tank = π2h
= 3.1416 x 4 x 4 x 10=502.656m3
Density = (Mass/Volume) 0.9 = 400/ (Volume of the oil)
Hence, Volume of the oil = 400/0.9
Depth of oil = volume/area Area of cylinder=( π2r)
=(3.1416 x 4 x 4)
We can calculate Depth of oil =(volume of oil)/( area of cylinder) = (444. 44)/(50.2656 ) =8.8418 m
Hence,ullage =(10 – 8.8418)
3.A tank of 2400m3 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.
Given : Volume of the tank = 2400m3 Depth of tank =12m RD =0.7
According to the question, 2% of the volume is allowed for expansion.
Density = Mass/ Volume Since volume =(L x B x H) = 2400m3( given )
So, Area = (volume / depth) = (2400/12) =200m2
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 = 48m3
Volume of the oil = (volume of the tank – free space ) = ( 2400 -48) = 2352 m3
Mass of the oil = (volume x density) =2352 x 0.7