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. (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,

Answer 1

a)
`m_g=35\,kg`

b)
`\dot{V_g}=27.38095\,L.s^(-1)`

c)
`\dot{m_g}=2535.313\,lbm.min^(-1)`

d) V_k : V_g = 2 : 1

Step-by-step explanation:

Given:

specific gravity of gasoline,
`s_g=0.7`

∴density of gasoline,
`\rho_g=700\,kg.m^(-3)`

(a)

volume of gasoline,
`V_g=50\,L`

We know ,

`1L=10^(-3)\,m^3`

& mass is related as:

`m_g=\rho* V`

`m_g=700* 50* 10^(-3)`

`m_g=35\,kg`

(b)

We have,

mass flow rate of gasoline,
`\dot{m_g}=1150\,kg.min^(-1)`

So, volume flow rate:(assuming the flow to be in-compressible)

`\dot{V_g}=(1150\,kg* m^3)/((1*60)\,s* 700\,kg)`

`\dot{V_g}=0.02738095\,m^3.s^(-1)`

`\dot{V_g}=0.02738095* 1000\,L.s^(-1)`

`\dot{V_g}=27.38095\,L.s^(-1)`

(c)

∵1 kg = 2.20462 lbm

∴1150 kg = 1150×2.20462 lbm

So,

`\dot{m_g}=2535.313\,lbm.min^(-1)`

(d)

specific gravity of kerosene,
`s_k=0.82`

∴density of kerosene,
`\rho_k=820\,kg.m^(-3)`

new density after mixing kerosene with gasoline,
`\rho_n= 780\, kg.m^(-3)`

We know that:

`mass= density* volume`

and by the law of conservation of mass & volume:

`m_g+m_k=\rho_n* V_n`

where
`V_n`= new volume

So,

`V_g.\rho_g+V_k.\rho_k=(V_k+V_g).\rho_n`

where
`V_k`=volume of kerosene.

`V_g* 700+V_k* 820= (V_k+V_g)* 780`

`70V_g+82V_k=78V_k+78V_g`

`4V_k=8V_g`

`(V_k)/(V_g) =(8)/(4)`

V_k : V_g = 2 : 1

Answer 2

a. Answer;

35.0 kg

Solution;

Mass of gasoline = ρGVg

= (sp grG)ρrefVG

Mass of G =(0.7)(1000 kg/m³) × (50.0 L)(1 m³/1000 L)

= 35.0 kg

b. Answer;

= 27.38 L/s

Explanation;

Volumetric flow of gasoline (Vg) = Mg/ρg = Mg/ (sp grG)ρ ref

= 1150 kg/min (m³/ (0.70 × 1000 kg))

= 1.643 m³/min

In liters/sec; 1.643 m³/min× (1 min/60 s) ×(1000L/M³)

= 27.38 L/s

Volumetric flow of gasoline is 27.38 L/s

c. Answer;

0.50 L gasoline/ L kerosene

Explanation;

ρmix = (SP grmix)(ρ ref) = Mmix/Vmix

= 0.78 = (Vg (0.70) + Vk (0.82))/( Vg +Vk)

Rearranging the equation

0.78 Vg + 0.78 Vk = 0.70Vg + 0.82 Vk

(0.78-0.70) Vg = (0.82 -0.78) Vk

Solving for the ratio;

Vg/Vk = (0.82 -0.78)/(0.78-0.70)

= 0.04/0.08

= 0.5

= 0.50 L gasoline/ L kerosene