Raw Data
CHG 3122: Chemical Engineering Practice
Centrifugal Pumps Lab
This spreadsheet calculates the total dynamic head, the net positive suction head,
the hydraulic power, the efficiency, the motor power and the pressure, the
velocity head and the tota

CHG 3111
Unit Operations
Summer 2017
Tutorial Practice Problems Dryers
Problem 1
The hot air stream has temperature of 87C and the total pressure of 101.3 kPa abs; it contains water vapor
with a partial pressure pA = 8.02 kPa. For this air stream, calcula

CHG 3111
Unit Operations
Summer 2017
Tutorial Practice Problems Distillation
Problem 1
A mixture n-pentane (A) and n-heptane (B) is vaporized at atmospheric pressure under batch differential
conditions. Equilibrium data for this mixture is given below. If

CHG 3111
Unit Operations
Summer 2017
Tutorial Practice Problems Mass Transfer
Problem 1
Air at 25oC with a dew-point temperature of 0oC flows past the open end of vertical tube filled with liquid
water maintained at 25oC. The tube inside diameter is 2.1 c

Practice problems in LIQUID-LIQUID EXTRACTION selected solutions
1. Acetone Water Methyl isobutyl ketone
Q1. Acetone (A) Water (B) MIK (C)
F 1000kg S 500kg
x AF 0.235 x AS 0
xBF 0.765
xBS 0
xCF 0
xCS 1
M 1500kg
xAM
FxAF SxAS 1000 0.235 500 0
0.157
FS
1

Practice problems in ADSORPTION and ION EXCHANGE Solutions
Q1. Adsorption isotherms
We considered three forms of isotherms:
Linear:
q Kc
- gives straight line on plot of q vs. c [1]
Freundlich:
q Kc n
- gives straight line on log-log plot of q vs. c [2]
L

Practice problems in ABSORPTION
1. CO2 Air Water
V1
yA1
V2
yA2
P = 1.0 atm
T = 293 K
L0
xA0
Given:
L1
xA1
L0 300 kmol/h
x A0 0
V2 100 kmol/h
y A2 0.20
Find:
L1, V1, xA1 and yA1
x
y
x
y
L ' A0 V ' A2 L ' A1 V ' A1
1 y A2
1 xA1
1 y A1
1 x A0
(

Practice problems in LEACHING Solutions
Q1. Single-stage leaching
The known process variables are as follows:
Feed solvent: V2 = 80 kg
xA2 = 0.03 (feed solvent contains 3 wt % oil)
xC2 = 0.97
Feed slurry:
Total mass (including solids) is 100 kg, containin

Practice problems in LIQUID-LIQUID EXTRACTION
Notes:
- The following pages provide several recommended practice problems in liquid-liquid
extraction, including both single-stage and multi-stage equilibrium processes.
- Final answers will be provided for a

Practice problems in DRYING
1. Properties of humid air:
a.
H = 0.021 kg H2O / kg dry air, T = 32.2C, P = 101.3 kPa
Using Table A.2-9, PAS will be: PAS = 4.824 kPa
18.02 PAS
18.02
4.824
0.0311 kg H 2 O / kg dry air
28.97 P PAS 28.97 101 .3 4.824
H
100 % 0

Practice problems in ABSORPTION and HUMIDIFICATION
(See last page for additional equilibrium data.)
1. (Tutorial February 8th) A gas mixture at 1.0 atm pressure abs containing air and CO2 is
contacted in a single-stage mixer continuously with pure water a

Practice Problems in Distillation Part A Solutions
Q1.
First, the unknown flow rates must be determined.
=+
= +
100(0.45) = (0.95) + (100 )(0.1)
= 41.2
= 58.8
Since R is known and so is xD, the enriching line equation can be found.
+1 =
4
0.95
+
=

Practice problems in DISTILLATION
PART A: MCCABE-THIELE GRAPHICAL METHOD
Notes: - After the lecture on March 6th, we have covered the material needed to solve problems
1, 2(a) and (b), and 3(a), and to start setting up 3(d).
- Problem 4 is for the tutoria

Practice problems in LIQUID-LIQUID EXTRACTION final answers
1. Acetone Water Methyl isobutyl ketone
Problem 1
Water phase: R = 853
xA = 0.12
xB = 0.855
xC = 0.025
MIK phase:
E = 647
yA = 0.205
yB = 0.04
yC = 0.755
Problem 2
(a) Minimum solvent flow rate:

Practice problems in ADSORPTION and ION EXCHANGE
1. Equilibrium isotherm data for adsorption of glucose from an aqueous solution to activated
alumina are as follows:
c (g / cm3)
q (g solute / g alumina)
0.0040
0.026
0.0087
0.053
0.019
0.075
0.027
0.082
0.

Practice problems in LEACHING
1. In a single-stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of
soybeans containing 22 wt % oil is leached with 80 kg of solvent containing 3 wt % soybean
oil. The value of N for the slurry underflow

SOLUTIONS TO DRYERS
Solution to Problem 1
1
T2 = 44.74 oC
2
Solution to Problem 2
3
4
Solution to Problem 3
Note: one can assume any surface area for the calculations. The area will cancel out
5

CHG3111 UNIT OPERATIONS
Chapter 3: VapourLiquid Distillation
Andrew Sowinski, PhD
genie.uOttawa.ca | engineering.uOttawa.ca
CHG3111:UnitOperations BoguslawKruczek&AndrewSowinski,2017
Summer 2017
INTRODUCTION
Distillation
Distillation a process of separa

CHG3111 UNIT OPERATIONS
Chapter 2: Mass Transfer Review
Andrew Sowinski, PhD
genie.uOttawa.ca | engineering.uOttawa.ca
CHG3111:UnitOperations BoguslawKruczek&AndrewSowinski,2017
Summer 2017
INTRODUCTION
Separation by Phase Contact
Many chemical processe

Pump ED 101
Waterhammer - - Part 1 What & Why
Joe Evans, Ph.D
http:/www.pumped101.com
In this first of a two part column, we will define waterhammer and take a look at
the events that lead to its occurrence. We will also try to gain some perspective
on th

Pump ED 101
Waterhammer - - Part 2 Causes & Variables
Joe Evans, Ph.D
http:/www.pumped101.com
Last month, we used a hypothetical example to illustrate the onset and effects of
waterhammer. This month we will investigate its major causes and gain an even
b

Pump ED 101
How Lower Efficiency Can Reduce Overall Cost Vortex Action
Joe Evans, Ph.D
http:/www.pumped101.com
When we start the pump selection process for a particular application, one of our major
concerns is efficiency. If several different models, of

Pump ED 101
Specific Gravity & Viscosity Part 2
Joe Evans, Ph.D
http:/www.PumpEd101.com
Last month we learned that the power required by a centrifugal pump is directly
proportional to the specific gravity (SG) of the pumped liquid. We also learned
that, a

Pump ED 101
Specific Gravity & Viscosity Part 1
Joe Evans, Ph.D
http:/www.PumpEd101.com
For many of us, water is the only liquid that flows through our centrifugal pumps.
Because of this, specific gravity (SG) and viscosity are not factors when sizing
the

Pump ED 101
Positive Displacement Pumps
Part II Rotary Pumps
Joe Evans, Ph.D
http:/www.pumped101.com
Rotary pumps are positive displacement pumps that utilize rotary, rather
than reciprocating, motion in their pumping action. They can be designed to
pump

SELFPRIMINGPUMPS
Pump ED 101
The self priming centrifugal pump is designed to lift water from some level
belowthepumpsuctionwithouthavingtofillthesuctionpipingwithliquid.It
accomplishes this by creating a partial vacuum at the pump suction which
removesth

Pump ED 101
Constant Pressure Boosters
Joe Evans, Ph.D
[email protected]
http:/www.pumped101.com
Introduction
Although there are many different water pumping applications, most tend to fall
into three basic categories - constant pressure, constant flow,

Practice problems in ABSORPTION
1. CO2 Air Water
V1
yA1
V2
yA2
P = 1.0 atm
T = 293 K
L0
xA0
Given:
L1
xA1
L0 300 kmol/h
x A0 0
V2 100 kmol/h
y A2 0.20
Find:
L1, V1, xA1 and yA1
x
y
x
y
L ' A0 V ' A2 L ' A1 V ' A1
1 y A2
1 xA1
1 y A1
1 x A0
(

CHG3111 UNIT OPERATIONS
Chapter 1: Drying
Andrew Sowinski, PhD
genie.uOttawa.ca | engineering.uOttawa.ca
CHG3111:UnitOperations BoguslawKruczek&AndrewSowinski,2017
Summer 2017
INTRODUCTION
Drying: Removal of relatively small amount of water or other liqu