Chapter 5, Problem 5.1P

A liquid mixture containing 40.0 wt% n-octane and the balance n-decane flows into a tank mounted on a balance. The mass in kg indicated by the scale is plotted against time. The data fall on a straight line that passes through the points (t = 3 min, m = 150 kg) and (t= 10 min, m = 250 kg).

(a) Estimate the volumetric flow rate of the liquid mixture.

(b) What does the empty tank weigh?

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Interpretation Introduction

(a)

Interpretation:

Volumetric flow rate of the mixture is to be calculated.

Concept introduction:

Volumetric flow rate of a fluid is the volume of a fluid that passes per unit time. It is mathematically represented as follows:

Volumetric flow rate = $\frac{m_{\text{flow rate}}}{{\rho }_{\text{avg}}}$

Here, $m_{\text{flow rate}}$ is mass flow rate and ${\rho }_{\text{avg}}$ is average density.

Answer to Problem 5.1P

19.85 m³/min

Explanation of Solution

Given Information: Density of n-Octane = $\rho$ ₁ = 703 kg/m³ Density of n-Decane = $\rho$ ₂ = 730 kg/m³ Mole fraction of n-Octane = x₁ = 0.4 Mole fraction of n-Decane = x₂ = 0.6

Average density of the mixture = _avg    ${\rho }_{\text{avg}}=x_1{\rho }_1+x_2{\rho }_2=$

$\left(0.4\times 703\right)$ + $\left(0.6\times 730\right)$ kg/m³ =719.2 kg/m³

Given graph is a straight line with slope as follows:

= $\frac{250-150}{10-3}$ = 14.28 kg/ min = Mass flow rate Volumetric flow rate = $\frac{\text{Mass flow rate}}{\rho avg}$

Putting the values, = $\frac{14.28}{719.2}$ m3/min = $\frac{14.28}{719.2}$

$\times 1000\text{ L/min}$

=19.85 m³/min

Interpretation Introduction

(b)

Interpretation:

Weight of empty tank is to be calculated.

Concept introduction:

A straight line equation is y = mx + c, where,

y = y-axis coordinate

x = x- axis coordinate

m = slope of line

c = intercept of straight line

Weight of a substance = Mass of substance $\times$ acceleration due to gravity

Answer to Problem 5.1P

1050.16 N

Explanation of Solution

The equation is a straight line with equation

m = at + c

where,

m = mass

a = slope of line

t = time

c = intercept

Mass of empty tank is mass at t = 0

i.e.

m = 0 + c

Mass of empty tank = intercept of the straight line

a = $\frac{250-150}{10-3}\text{ kg/min}$

a = 14.28 kg/min

At t = 3 min, m =150 kg

So,

150 = (14.28) (3) + c

c = 107.16

Therefore, mass of empty tank = 107.16 kg

Now, weight can be calculated using the following formula:

$W=m\times g$

Here, g is acceleration due to gravity.

Putting the values,

W = 107.16 $\times$ 9.8 kg m/s² = 1050.16 N