# Stability Chapter 1 – Density and relative density Solutions.

**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=0.7**

=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:**

**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}

**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**

**Hence, Ullage = (12-11.76)**** = 0.24m.**

**A tank 10m deep has vertical sides .its bottom consist of triangle 12mx12mx10m. find the mass of oil (of RD0.800) to be loaded , allowing 3 % of the volume of oil loaded for expansion. State the ullage on completion of loading.**

*Solution:*

**Depth of the tank = 10m****L, B, H, of triangle = (12 x 12 x 12****RD = 0.8**

**Allowing, 3% of the volume of the oil loaded for expansion.****Area of the triangle = 1/2 (base x height)**

**In triangle PQR, RQ in the height which divide the base at same length.****So In triangle PSR**

**SR ^{2} =PR^{2} –PS^{2}**

**= (12**^{2}– 5^{2})

**= (144 -25)**

**=119****So, SR = 10.9m****So area of the triangle = 1/2 (10×10.9)****=54.5m ^{2}**

**Volume of the tank = (10 x54.5)****= 545m ^{3}**

**Now, volume of the oil = (Total volume -free space)**

**let ‘V’ be the volume of the oil****545 = (V + 3/100 x V)****545 = (V + 3V/ 100)****545 = (103V /100)****V = (545 x 100) /103****V = 529.126m ^{3}**

**Mass of the oil = (density x volume)****= 0.8 x 529.126****=423.3 t**

**Depth of the oil = (volume of oil)/(area)****= (529.126 /54.5)****= 9.7087m**

**Ullage = ( 10 – 9.708)**

** = 0.292m**

**Ullage = ( 10 – 9.708)**

**= 0.292m**

**A rectangular tank measuring 25mx 12m x8m has an ullage pipe projecting 0.3m above the tank top. find the mass of SW in the tank when the ullage is 3.3m.**

*Solution :*

*Solution :*

**Volume of the rectangular tank = ( L x B x H)****= (25 x 12 x 8 )**

**Pipe above the tank top is 0.3m**

**Ullage inside the tank = (3.3 – 0.3)****= 3m**

**Depth of the SW = (8 – 3)****= 5m**

**Volume of the SW = (L x B x H )****= ( 25 x12 x5)****= 1500 m ^{3}**

**Mass of the SW = (volume x density)**** = (1500 x1.025)**** = 1537.5 t**