Question

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.
(c) Estimate the average mass flow rate lbm/min delivered by a gasoline pump when the gas pump has a pump rate of 10 gallons per 2 minutes.

Answer 1

a) mass=35 kg

b) volumetric flow rate= 27.37
`(liters)/(s)`

c) average mass flow rate=29.21
`(lbm)/(min)`

Step-by-step explanation:

Specific gravity is defined as the relation between the density of one substance and the density of another reference substance (it is usual that water is used in this case). In this case the specific gravity is the relationship between the density of gasoline and the density of water. So:

`0.7=(densitygasoline)/(densitywater)`

Now you can know the density of gasoline with a simple mathematical operation and knowing that densitywater≅ 1000
`(kg)/(m^(3) )`:

densitygasoline=0.7*densitywater

densitygasoline=0.7*1
`(kg)/(liters)`

densitygasoline=0.7
`(kg)/(liters)`

a)

Now that the density is known, you can calculate the mass in 50 liters of gasoline.

By definition of density, you know that in 1 liter there are 700 kg of gasoline. So using The Rule of Three( tool that allows you to solve problems based on proportions), you can calculate the mass in 50 liters:

`(700 kg)/(1 liter) =(mass)/(50 liters)`

`mass=(0.7 kg* 50 liters)/(1 liter)`

mass=35 kg

b)

First, it is convenient to change units: from kg / min to kg / seconds. For that you know that 1 min= 60 seconds. So:

`(1150 kg)/(min) =(1150 kg)/(60 s)`

`(1150 kg)/(min) =(19.16 kg)/(s)`

By definition, density is the relationship between the mass and volume of a substance: This is:
`density=(mass)/(volume)`

Applied to this case, it would be
`density=(mass flow rate)/(volumetric flow rate)`

The mass flow rate and the density of the gasoline are known, so you can calculate the volumetric flow rate:

`volumetric flow rate= (mass flow rate)/(density gasoline)`

volumetric flow rate= 27.37
`(liters)/(s)`

c)

You know that the pump rate is 10 gallons per 2 minute. This is:

`(10 gallons)/(2 minutes) = 5 (gallons)/(minute)`

To calculate the average mass flow, you must relate this data to density and do the necessary unit conversions. Conversions are done similarly to what was previously applied in the previous exercises.

You have to know that:

• 1 gallon = 3.785 liters

• 1 kg= 2.205 kg

So:

`(5 gallons*3.785 liters)/(1gallons*min) =(18.925 liters)/(min)`

You know that the density is
`density=(mass flow rate)/(volumetric flow rate)`

The volumetric flow rate (pump rate) and the density of the gasoline are known, so you can calculate the average mass flow rate:

average mass flow rate=density*volumetric flow rate

`average mass flow rate=(0.7 kg)/(liters) *(18.925 liters)/(min)`

average mass flow rate=13.2475
`(kg)/(min)`

Converting the units:

`average mass flow rate=(13.2475 kg*2.205 lbm)/(1 kg*min)`

average mass flow rate=29.21
`(lbm)/(min)`