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. 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

Answer 2

Answer:

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

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:

$m_g=\rho\times V$

$m_g=700\times 50\times 10^{-3}$

$m_g=35\,kg$

(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)

$\dot{V_g}=\frac{1150\,kg\times m^3}{(1\times60)\,s\times 700\,kg}$

$\dot{V_g}=0.02738095\,m^3.s^{-1}$

$\dot{V_g}=0.02738095\times 1000\,L.s^{-1}$

$\dot{V_g}=27.38095\,L.s^{-1}$

(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:

$mass= density\times volume$

and by the law of conservation of mass & volume:

$m_g+m_k=\rho_n\times V_n$

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.

$V_g\times 700+V_k\times 820= (V_k+V_g)\times 780$

$70V_g+82V_k=78V_k+78V_g$

$4V_k=8V_g$

$\frac{V_k}{V_g} =\frac{8}{4}$

V_k : V_g = 2 : 1