ELEM.PRINCIPLES OF CHEMICAL PROCESSES
ELEM.PRINCIPLES OF CHEMICAL PROCESSES
4th Edition
ISBN: 9781119571070
Author: FELDER
Publisher: WILEY
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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?

.

Expert Solution
Check Mark
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 = mflow rateρavg

Here, mflow rate is mass flow rate and ρavg is average density.

Answer to Problem 5.1P

19.85 m3/min

Explanation of Solution

Given Information: Density of n-Octane = ρ 1 = 703 kg/m3 Density of n-Decane = ρ 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     ρavg=x1ρ1 + x2ρ2=

(0.4×703) + (0.6×730) kg/m3 =719.2 kg/m3

Given graph is a straight line with slope as follows:

= 250150103 = 14.28 kg/ min = Mass flow rate Volumetric flow rate = Mass flow rateρavg

Putting the values, = 14.28719.2 m3/min = 14.28719.2

×1000 L/min

=19.85 m3/min

Expert Solution
Check Mark
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 × 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 = 250150103 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×g

Here, g is acceleration due to gravity.

Putting the values,

W = 107.16 × 9.8 kg m/s2 = 1050.16 N

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Chapter 5 Solutions

ELEM.PRINCIPLES OF CHEMICAL PROCESSES

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