Unit Operations of Chemical Engineering
Unit Operations of Chemical Engineering
7th Edition
ISBN: 9780072848236
Author: Warren McCabe, Julian C. Smith, Peter Harriott
Publisher: McGraw-Hill Companies, The
Question
Book Icon
Chapter 9, Problem 9.1P
Interpretation Introduction

Interpretation:

The given motor power is to be verified to be adequate to drive the agitator at the given speed.

Concept Introduction:

The formula to calculate the Reynolds number for an agitator is:

  Re=nda2ρμ   ........ (1)

Here, n is the speed of the agitator in r/s, da is the diameter of the agitator, ρ is the density of the fluid, and

  μ is the viscosity of the fluid.

Now, power consumption by the agitator is calculated as:

  P=NPn3da5ρ   ........ (2)

Here, NP is the power number that depends on the Reynolds number of the fluid. Its value is taken from the standard graph specific for a type of an agitator.

Expert Solution & Answer
Check Mark

Answer to Problem 9.1P

The given motor with 8 kW power is not adequate to drive the agitator at the required speed as the power required is more than the available power of the motor.

Explanation of Solution

Given information:

Tank diameter, d=1.2 m

Height of the tank, h=2 m

Depth of latex filled in the tank, hf=1.2 m

Viscosity of latex, μ=10 P

Density of latex, ρ=800 kgm3

The tank is not baffled.

Diameter of the propeller, dp=360 mm

Height of the propeller in the tank, hp=360 mm

Pitch is 1:1

Agitator speed, n=800 r/min

Three-blade propeller is used.

Available power from the motor, Pavailable=8 kW

Convert the speed of the propeller from r/min to r/s as:

  n=800 rmin×min60 s=13.33 rs

Use equation (1) to calculate the Reynolds number for the propeller as:

  Re=ndp2ρμ=(13.33 rs)(360 mm×m1000 mm)2(800 kgm3)10 P×Pas10 P=1382

From the plot of NP versus Re for unbaffled propeller with three blades, the value of NP corresponding to the calculated value of Re is taken as:

  NP=0.73

Use equation (2) to calculate the required power for the given agitator as:

  Preq=NPn3dp5ρPreq=(0.73)(13.33 rs)3(360 mm×m1000 mm)5(800 kgm3)Preq.=8364 WPreq.=8.364 kW

Since the required power is more than the available power, the motor available to drive the agitator is not adequate.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 9 Solutions

Unit Operations of Chemical Engineering

Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The