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.

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

Mass of oil loaded = 3860.03 MT

Ullage at load port = 1.75 M

1) Total Vol of tank: LxBxH = 21m x 15m x 16m = 5040 m3

2) 3% left for expansion. so max we can load is 5040m3 - 3% of 5040m3 = 4888.8m3. 4888.8m3 is the GOV at 42 deg C. ( i.e the max vol cargo should reach @ 42degC)

3) GOV x VCF = Std vol @ 15degC so 4888.8 x 0.9761 (vcf for 42degC) = 4771.9577       m3 @ 15degC

4) Now with this std vol we have to get the GOV for 23degC. This is because we have to load the cargo at 23deg C so we have to consider GOV for same temperature.           Std vol / VCF = GOV which will be 4771.9577 / 0.9929 = 4806.0809 m3.                        (VCF considered here is for 23degC as we need GOV for same temp which the             cargo has to be loaded).

5) So cargo to be loaded @ 23degC = 4806.0809 m3, which in metric ton will be            GOV x VCF x WCF = 4806.0809 x 0.9929 x 0.8099 = 3860.0366 MT.

6)  For final ullage, final vol i.e GOV @ 23degC (the loadport temp) is 4806.0809m3.           VOL = LxBxH so 4806.0809 = 21x15x(ht of cargo). so ht of cargo = 15.2574m.                 FINAL ULLAGE = Height of tank + height of ullage point - height of cargo                                           = 16m + 1m - 15.2574m                                                                                               =  1.7426m 

Note: Ullages mentioned in ullage tables are against the gross observed volume of                 tank.   

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

• Standard Volume ( @ 15 Deg Cel ) = 4888.8 x 0.9929 = 4854.0894 m3
• Volume at 42 deg cel =( Standard vol /0.9761) = 4854.0894 /0.9761 =4972.9427 m3
• Load Port and Dis Port Volume Difference = + 84.14273 m3 ( If temp rise, Vol Increase)
• Tank Innage at Discharge Port = 4972.9427 / (21x15 ) = 15.79 m
• Discharge Port Ullage = 1 + (16-15.79) = 1.21 m