(a) Interpretation: The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same. Concept introduction: Assume tank is perfectly mixed Mass balance over tank =ṁ i n − ṁ o u t = (volumetric inlet feed rate x Density of solution) - (volumetric outlet feed rate x Density of solution)

Question

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Answer 1

Question

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

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Answer 2

Question

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 6

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Answer 7

Chapter 10, Problem 10.14P

Interpretation Introduction

(a)

Interpretation:

The total mass balance on the tank needs to be determined and it is to be proved that the total mass of the liquid in the tank remains the same.

Concept introduction:

Assume tank is perfectly mixed

Mass balance over tank =ṁ in − ṁ out

= (volumetric inlet feed rate x Density of solution) -

(volumetric outlet feed rate x Density of solution)

Answer 8

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Answer 9

Interpretation Introduction

(b)

INTERPRETATION:

The balance on sodium nitrate needs to be determined and converted into equation for dx/dt by providing initial condition.

Concept introduction:

Sodium Nitrate is denoted by NaNO_3.

By using derivation equation we will calculate the mass of fraction of NaNO_3.

So, Considered, x(t,ṁ) equal to mass of fraction of NaNO_3.

Answer 10

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 11

Interpretation Introduction

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 12

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 13

Interpretation Introduction

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 14

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Answer 15

Interpretation Introduction

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Answer 16

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 17

Interpretation Introduction

(f)

Interpretation:

The shapes of the plots of x versus t expected with the impeller on and off that shows the difference between the two curves at small and large t values needs to be sketched on the same chart.

CONCEPT INTRODUCTION:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

Answer 18

Expert Solution & Answer

Answer 19

Expert Solution & Answer

Answer 20

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Check out a sample textbook solution

See solution

Answer 21

Ch. 10 - A solution containing hydrogen peroxide with a...Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.6P Ch. 10 - Prob. 10.7P Ch. 10 - Water is added at varying rates to a 300-liter...Ch. 10 - Prob. 10.9P Ch. 10 - 10.10. A ventilation system has been designed for...

Solution Summary: The author explains how the mass balance on sodium nitrate needs to be determined and converted into equation for dx/dt.

Concept explainers

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)

Want to see the full answer?

Chapter 10 Solutions

Solution Summary: The author explains how the mass balance on sodium nitrate needs to be determined and converted into equation for dx/dt.

Concept explainers

(c) Interpretation: The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t. Concept introduction: The graph will be first drawn for x vs t. For x vs t graph dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(d) Interpretation: The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph. Concept introduction: The graph will be first drawn for x vs t. For x vs t graph dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(e) Interpretation: The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated. Concept introduction: The following equation will be taken to find the equation for time calculation: lnx0.45=−m200t The derivation will produce: t=−2ln(xf/0.45)

Want to see the full answer?

Chapter 10 Solutions

(c)

Interpretation:

The shapes of the plots gives mass rates needs to be sketched on a single graph of x versus t.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(d)

Interpretation:

The expression for x(t,m.) needs to be obtained and the plots of x versus t for given mass rates needs to be generated on a single graph.

Concept introduction:

The graph will be first drawn for x vs t.

For x vs t graph

dx/x = m/Vρdt−ln x=m/Vρ×ρ×t + C

(e)

Interpretation:

The time taken to flush out 90%, 99% and 99.9% of the sodium nitrate needs to be calculated.

Concept introduction:

The following equation will be taken to find the equation for time calculation:

lnx0.45=−m200t

The derivation will produce:

t=−2ln(xf/0.45)