Stability -I : Chapter 1

1. 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/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 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 = π²h
= 3.1416 x 4 x 4 x 10
=502.656m³

We know that:
Density = (Mass/Volume)
0.9 = 400/ (Volume of the oil)

Hence, Volume of the oil = 400/0.9
=444.44m³

Depth of oil = volume/area
Area of cylinder   =( π²r)
 =(3.1416 x 4 x 4)
 =502656m³

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

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.

Solution :-

Given :
Volume of the tank = 2400m³
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)
= 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

Depth of oil = (volume of oil /area)
= 2352/(L x B)
= (2352/ 200)
=11.76 m