3.3. The specific gravity of gasoline is approximately 0.70. (a) Estimate the ma

Question

3.3. The specific gravity of gasoline is approximately 0.70. (a) Estimate the mass (kg) of 50.0 liters of gasoline. (b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s. (e) Estimate the average mass flow rate (Ib/min) delivered by a gasoline pump. (d) Gasoline and kerosene (specific gravity 0.82) are blended to obtain a mixture with a specific gravity of 0.78. Calculate the volumetric ratio (volume of gasoline/volume of kerosene) of the two compounds in the mixture, assuming Vblend Vgasoline + Vkersene

Explanation / Answer

Part a

Specific gravity of gasoline = 0.70

Density of gasoline / density of water = 0.70

Density of gasoline = 0.70 x density of water

Density of gasoline = 0.70 x 1000 kg/m3 = 700 kg/m3

Part a

Volume of gasoline = 50 L x 1m3/1000 L = 0.050 m3

Mass of gasoline = volume x density

= 0.050 m3 x 700 kg/m3 = 35 kg

Part b

Mass flow rate = density x volumetric flow rate

1150 kg/min = 700 kg/m3 x volumetric flow rate

volumetric flow rate = (1150 kg/min) / (700 kg/m3)

= 1.643 m3/min x 1min/60s x 1000 L/m3

= 27.38 L/s

Part C

Average mass flow rate = 1150 kg/min x 2.205lbm/kg

= 2535.75 lbm/min

Part d

Density of mixture = specific gravity x density of water = (mass of mixture / volume of mixture)

specific gravity x density of water = (mass of gasoline + mass of kerosene) / (volume of gasoline + volume of kerosene)

= (Volume x density of gasoline + Volume x density of kerosene) / (volume of gasoline + volume of kerosene)

=density of water [ (Volume x specific gravity of gasoline + Volume x density of kerosene)] / (volume of gasoline + volume of kerosene)

0.78 = (volume of gasoline x 0.70 + volume of kerosene x 0.82) / (volume of gasoline + volume of kerosene)

(0.78-0.70) x volume of gasoline = (0.82-0.78) x volume of kerosene

Volume of gasoline / volume of kerosene = 0.50

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