Lecture 3 SF

Lecture 3 SF 1

Part 2: Lecture 3

Santiago Faucher

Lecture 3 SF 2

Course Outline

1. Overview of Separation Processes (Course Notes + Chapter 1 Separation Processes by King)

2. Evaporation (Chapter 8 + PCEH 11-110)

3. Drying (Chapter 9)

4. Adsorption (Chapter 12)

5. Liquid-liquid extraction (Chapter 12)

6. Crystallization (Chapter 12)

7. Supercritical fluid extraction

8. Membrane processes (Chapter 13)

9. Filtration, Settling, Centrifugation (Chapter 14)

Lecture 3 SF 3

Part 2: Evaporation (Course Notes + Geankoplis Chapter 8 + Perry’s 11-110)

Design Criteria and Processing Factors

Evaporator Types and Applications

Single Effect Evaporator Calculations

Boiling Point Rise of Solutions

Enthalpy-Concentration Diagrams for Solutions

Multiple Effect Evaporator Calculations

Lecture 3 SF 4

Evaporator Design Criteria and Processing Factors

Heat Transfer: Most important factor in evaporator design as it affects heat transfer area which is the largest component of evaporator cost.

The best evaporator is the one that has the highest heat transfer cost coefficient: J/s K per $ of installed cost.

A circulation (pump) system is only attractive if the energy to run the pump is offset by a higher heat transfer cost coefficient.

Do you know how to calculate pumping cost?

Lecture 3 SF 5

Evaporator Design Criteria and Processing Factors

Vapour-liquid Separation: Must prevent entrainment due to product loss

Contamination of the vapour phase (polution)

Condensation of vapour onto surfaces (corrosion and fouling)

Overhead mist or spray may cause troublesome deposits

Vortices increase pump head requirements and therefore equipmentconfiguration

Short circuiting a big problem as it presents the problem of cavitation(there must be a net positive suction head)

Lecture 3 SF 6

Why are the feeds, tubing, and pump located where they are?

Have you learnt aboutNPSHR and NPSHA?

www.wintek-corp.com

Lecture 3 SF 7

NPSHR and NPSHANPSHR: Net Positive Suction Head Required

From the pump curve.

NPSHA: Net Positive Suction Head Available

Lecture 3 SF 8

NPSHR and NPSHA

NPSHA > NPSHR, otherwise we will flash.

Flash occurs when Pressure Inside the pump inlet side is less than the vapourpressure.

Pressure inside the pump is: Inlet Head (atm. + hydraulic head) – friction head loss – NPSHR

Therefore:

Inlet Head (atm. + hydraulic head) – friction head loss – NPSHR > Vapour pressure

Restating:

Inlet Head (atm. + hydraulic head) – friction head loss – Vapour Pressure > NPSHR

i.e. NPSHA > NPSHRNPSHA

Lecture 3 SF 9

Example 3.1: An evaporator is operating at atmospheric pressure using a forced circulation pump at a flowrate of 400 gpm. The feed to the recirculatingsystem is fed into the top of the evaporator at a rate of 40 gpm and at a temperature of 140 F. The head losses in the return piping system are 9 ft H2O. Calculate the height required between the pump and the top of the liquid level in the evaporator.

NPSHA = Inlet Head (atm. + hydraulic head) – friction head loss – VapourPressure

= 33.9 ft H2O + Height – 9 – 33.9 ft H2 = Height – 9

NPSHR from pump curve = 1 foot

Therefore: NPSHA = Height – 9 > 1 ft = NPSHR

So Height > 10 ft ! << You would want to add some fat to this so say 15 ft to be sure.

Lecture 3 SF 10

Example 3.2: An evaporator is operating at atmospheric pressure using a forced circulation pump at a flowrate of 400 gpm. The feed to the recirculatingsystem is fed before the pump at a rate of 40 gpm and at a temperature of 100 F. The head losses in the return piping system are 9 ft H2O. Calculate the height required between the pump and the top of the liquid level in the evaporator.

Note inlet temperature to pump has dropped from 212 F to approximately (40*Cp*(100-32)+360*Cp*(212-32))/400*Cp + 32 = 200 F. At this temperature from the steam tables the vapour pressure has been reduced from 14.7 psi to 11.52 psi or 33.9 ft to 26.6 ft H2O

NPSHA = Inlet Head (atm. + hydraulic head) – friction head loss – VapourPressure = 33.9 ft H2O + Height – 9 – 26.6 ft H2 = Height – 9 + 7.6 = Height –2.6

NPSHR from pump curve = 1 foot

Therefore: NPSHA = Height – 2.6 > 1 ft = NPSHR

So Height > 3.6 ft << You would want to add some fat to this so say 5 ft to be sure.

Lecture 3 SF 11

Other Design Criteria

Liquid concentration -> relates to viscosity and heat transfer

Temperature and Pressure Boiling temperature is inversely proportional to pressure.

Boiling points may increase as solution get concentrated (boiling point rise)

Foaming -> will determine the height of your freeboard in the design

Solubility of materials -> May be the limit to the concentration that you can achieve.

Scale deposits -> decrease your heat transfer coefficient.

Lecture 3 SF 12

Lecture 3 SF 13








Lecture 3 SF 14

Long Tube Vertical Natural Convection

-Cheapest per unit capacity

-High velocities and long tubes = one pass, short residence time

-Not ideal for crystallizations

Evaporator Types

Horizontal Tube

-Cheap

-Non-viscous mat’ls

having high U

-Liquids that do not deposit scale

Short Tube Vertical Natural Convection

-Cheap

-Non-viscous mat’ls

-Best at high ΔT = high U

-Large floor space

-Mild scaling

Lecture 3 SF 15

Falling Film

-Compact

-High heat transfer coeff.

-Low hold up and short residence time

-Not ideal for scale forming operations

Evaporator Types

Forced Convection

-Higher cost

-Suitable for viscous materials

-Highest U

-Compact

-Suitable for scaling or crystallization

Lecture 3 SF 16

Break time?

Planning a road trip? The Microjoule can make the journey between New York and Los Angeles over three times on just one gallon of gas.

Amazingly the Microjoule doesn't have any fancy tricks — it's just an aerodynamic, lightweight one-seater that runs off gasoline, but runs for a long time.

The Microjoule isn't necessarily new. The projected started in 1985 as an entry to the Shell Eco-Marathon, where cars compete to go the farthest on the least amount of gas, and it has won the competition every time since 1992 and broke the world record for the most miles-per-gallon six times.

A team of students from St. Joseph La Joliverie, a technical school in France, will take the MicroJoule to the Shell Eco-Marathon track at the French Nogaro Motor Circuit this year.

Lecture 3 SF 17








Lecture 3 SF 18

Single Effect Evaporators

q = UAΔT = UA(Ts-T1)

Lecture 3 SF 19

Approximations

Most evaporation problems deal with solutions that are fairly concentrated and as a result have differing:

Boiling points (higher)

Heat capacities

Enthalpies

… than pure water.

To resolve our heat balances we use the following approximations:The latent heat of evaporation of 1 kg mass of water from a solution can be approximated as that of pure water at the boiling point temperature of the solution (rather than use the latent heat at the equilibrium boiling temperature expected for pure water at the operating pressure).

If the heat capacities of the liquid and product stream are known we can use these to calculate the enthalpies of these streams.

Lecture 3 SF 20

Example 3.3 Heat Transfer Area in a Single-Effect Evaporator

A continuous single-effect evaporator concentrates 9072 kg/hr of a 1.0 wt% salt solution entering at 311K to a final concentration of 1.5 wt%. The vapour space of the evaporator is at 1 atm and the steam supplied is saturated at 143.3 kPa. The overall coefficient U = 1704 W/m2 K. Calculate the heat-transfer area required. Assume that, since it is dilute, the solution has the same boiling point as water.

Lecture 3 SF 21

Space for calculation

Lecture 3 SF 22


Lecture 3 SF 23


Lecture 3 SF 24

Effect of Process Variables

Effect of feed temperatureIn Example 3.3 if we had operated at a feed temperature of 350K (up from 311K), what reduction in evaporator size would have been achieved? (Area decreases from 150 m2 to 126 m2)

Effect of pressureAt 1 atm the ΔT was 10K, if we reduce the pressure to 0.41 atmwhat would the new surface area have been? (ΔT becomes 33K and area decreases from 150 m2 to 42 m2, however now require condenser and vacuum pump)

Effect of steam pressureUsing superheated steam will also reduce the size of the evaporator but this is an expensive option.

Lecture 3 SF 25

Example 3.4 Heat Transfer Coefficient in a Single-Effect Evaporator

A continuous single-effect evaporator concentrates 4535 kg/hr of a 2.0 wt% salt solution entering at 311K to a final concentration of 3.0 wt%. The vapour space of the evaporator is at 1 atm and the steam supplied is saturated at 383.2K. The area of the evaporator is 69.7m2. The heat capacity of the feed is Cp = 4.1 kJ/kg K. Calculate the heat-transfer coefficient required. Assume the solution to be dilute, since it is dilute, the solution has the same boiling point as water.

Lecture 3 SF 26


Lecture 3 SF 27


Lecture 3 SF 28


Lecture 3 SF 29

Break time?

There are 62,000 miles of blood vessels in the human body – laid end to end they would circle the earth 2.5 times

Lecture 3 SF 30








Lecture 3 SF 31

Boiling Point Rise

As mentioned previously most solutions being evaporated differ significantly from the behaviour of pure water in regard to:

Their boiling point at a specific pressure

Their heat capacities or enthalpies.

Dühring’s rule is a useful approach to predict the boiling point of solutions. Basically there is a linear relationship between the boiling point of a solution and that of pure water.

All that is needed therefore is to know the boiling point temperature of the solution at two pressures and one can then estimate any intermediary boiling points.

Lecture 3 SF 32

Dühring’s rule

Lecture 3 SF 33

Example 3.5 Use of Dühring Chart for Boiling Point Rise

The pressure in an evaporator is given as 1 atm and a solution of 50 wt% NaOH is being boiled. Determine the boiling point temperature and subsequently the boiling proint rise of this solution.

In this case the boiling point is292°F and so the BPR is 292°F – 212°F = 80°F.

The latent heat of the vapourwould be estimated as 1177 Btu/lbmfrom the saturated steam tables at 292°F or alternatively from thesuperheated steam tables at 292°FAnd 14.7 psia as 1189 Btu/lbm.

Lecture 3 SF 34

Other Boiling Point Rise Plots (PCEH)

Lecture 3 SF 35


When applied to example 3.5 where the BP was292°F we quickly get aBPR of 80°F.

Lecture 3 SF 36


The real useful aspect of this chart is determiningsolids contents fromboiling point data.

As we know that at 1 atmthe BP of water is 212°F and having measured a BP of the solution of 292°F, we can determine from this chart that the solids content is: 50 wt%.

Lecture 3 SF 37








Lecture 3 SF 38

Dale Carnegie Principle No. 3How do you make people want to do something?

“First, arouse in the other person an eager want. He who can do this has the whole world with him. He who cannot walks a lonely way”

What does that mean?

Find a way/mechanism/logic to make the other person want to do the task that you need them to accomplish. You cannot force someone to do something; the result will be poor. But if the individual sees value in it for themselves then they will eagerly take on the task.

Principle 3: Arouse in the other person an eager want.

Dale Carnegie, 1936

Lecture 3 SF 39

Dale Carnegie Summary of Fundamental Techniques for Handling People

PRINCIPLE 1Don’t criticize, condemn, or complain.

PRINCIPLE 2Give honest and sincere appreciation.

PRINCIPLE 3Arouse in the other person an eager want.

Dale Carnegie, 1936

Lecture 3 SF 40

End of Lecture 3 (Part 2)

Lecture 3 SF

Documents

Transcript of Lecture 3 SF