4. Let us consider a process flow diagram describing the water conditioning setup for a salt-water fish tank. This process removes all of the dirt from the inlet stream and adds oxygenand salt to a final salt-to-water ratio of 0.0375, and an oxygen-to-water ratio of 0.0015.The inlet stream, which contains water, salt, dissolved oxygen, and dirt enters asedimentation unit. This unit concentrates all of the dirt and removes it at a concentrationof 20.0 wt% dirt. This dirt is then subsequently filtered out of the system. While theconcentrating of dirt is the only separation achieved in the sedimentation unit, it also createstwo additional outlet streams, which feed the oxygenator and the salinator. In theoxygenator, oxygen is bubbled into the water to achieve an oxygen-to-water ratio of0.0020. In the salinator, salt tablets are dissolved into the stream as needed.Sa20OOgatorc. What is the flow rate of stream 6?Ser*11W = 0.9625,a. Calculate the flow rate and concentration of the final product stream 11?kgt₁ = 52.0*115 = 0.0361,*110 = 0.0014b. What is the flow rate of stream 3?d. What is the flow rate of stream 9?Ⓒ100%th=0.050-kghkghth=0.625-kg9the 2.00300e. What is the flow rate and composition of stream 8?kgthe = 10.0Xaw = 0.7801, Xas = 0.0195,*ao = 0.00047

Answered: 4. Let us consider a process flow…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Related questions

Question

4. Let us consider a process flow diagram describing the water conditioning setup for a salt-
water fish tank. This process removes all of the dirt from the inlet stream and adds oxygen
and salt to a final salt-to-water ratio of 0.0375, and an oxygen-to-water ratio of 0.0015.
The inlet stream, which contains water, salt, dissolved oxygen, and dirt enters a
sedimentation unit. This unit concentrates all of the dirt and removes it at a concentration
of 20.0 wt% dirt. This dirt is then subsequently filtered out of the system. While the
concentrating of dirt is the only separation achieved in the sedimentation unit, it also creates
two additional outlet streams, which feed the oxygenator and the salinator. In the
oxygenator, oxygen is bubbled into the water to achieve an oxygen-to-water ratio of
0.0020. In the salinator, salt tablets are dissolved into the stream as needed.
Sa
20
O
Ogator
c. What is the flow rate of stream 6?
Ser
*11W = 0.9625,
a. Calculate the flow rate and concentration of the final product stream 11?
kg
t₁ = 52.0
*115 = 0.0361,
*110 = 0.0014
b. What is the flow rate of stream 3?
d. What is the flow rate of stream 9?
Ⓒ100%
th=0.050-
kg
h
kg
h
th=0.625-
kg
9
the 2.003
00
e. What is the flow rate and composition of stream 8?
kg
the = 10.0
Xaw = 0.7801, Xas = 0.0195,
*ao = 0.0004
7

Expert Solution

Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

Given that-

Water feed, W=50 kg/hr

At product stream 11,

Final salt to water ratio=0.0375

Final oxygen to water ratio= 0.0015

At oxygenator stream 4,

Oxygen to water ratio=0.002

(a)

Now, first of all the final balance at product stream (11) is calculated as follows-

Considering initial water= 1, fractions of oxygen, water and salt in product stream are-

$x_{O_{2}} = \frac{0.0015}{1 + 0.0015} = \frac{0.0015}{1.0015} = 0.00149 x_{s a l t} = \frac{0.0375}{1 + 0.0375} = \frac{0.0375}{1.0375} = 0.03614 x_{w a t e r} = 1 - (0.03614 + 0.00149) = 0.96237$