Q.A tank measuring 21m*15m*16m has an ullage port extending 1 m above the top of the tank. It is to be loaded with crude oil at 23°C. 3% of the tank vol. is to be left for expansion when maximum temperature expected during the outage is 42°C. ( Density of oil =0.81@ 15°C in vacuum). Find the mass of oil loaded and ullage at load port. Given From ASTM tables- VRF for 23°C = 0.9929, VRF for 42°C = 0.9761, WRF = 0.81 - 0.0011 = 0.8089 Sir pls solve this question step by step as it would be great help for me. Awaiting for your replies.

Answer 1 Answers

Volume of tank = 5040 m3

3 % of Volume = 151.2 m3

1)Observed Volume of cargo( @ 23 deg Cel ) = (5040 - (5040/100)*3) =4888.8 m3

2)Standard Volume ( @ 15 Deg Cel ) = 4888.8 x 0.9929 = 4854.0894 m3

3)Standard weight ( @ 15 Deg Cel in Vaccum) = 4854.0894 x0.81 =3931.8125 Tonnes ( in Vaccum)

4)Standard weight ( @ 15 Deg Cel in Air) =4854.0894 x0.8089= 3926.47291 Tonnes( in Air)

REVERSE CALCULATION FOR FINDING VOLUME AT DISCHARGE PORT

1)Standard Volume ( @ 15 Deg Cel ) = 4888.8 x 0.9929 = 4854.0894 m3

2)Volume at 42 deg cel =( Standard vol /0.9761) = 4854.0894 /0.9761 =4972.9427 m3

3) Load Port and Dis Port Volume Difference = + 84.14273 m3 ( If temp rise, Vol Increase)

4) Tank Innage at Discharge Port = 4972.9427 / (21x15 ) = 15.79 m

5) Discharge Port Ullage = 1 + (16-15.79) = 1.21 m