Fundamentals of Momentum, Heat, and Mass Transfer
Fundamentals of Momentum, Heat, and Mass Transfer
6th Edition
ISBN: 9781118947463
Author: James Welty, Gregory L. Rorrer, David G. Foster
Publisher: WILEY
Question
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Chapter 11, Problem 11.1P
Interpretation Introduction

Interpretation: Dimensionless groups to determine the power output is to be discussed.

Concept Introduction: One of the ways to conduct dimension analysis is by using the Buckingham pi method. In this method, some core groups are determined. These core groups contain some fixed variables and some changing variables. However, the ultimate dimensions of these pi groups are zero which means they are supposedly dimensionless.

Expert Solution & Answer
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Answer to Problem 11.1P

Using the Buckingham pi method, the dimensionless expressions are π1=η , π2=H/D , π3=g/Dω2 , π4=Q/D3ω and π5=P/D2ω3

Explanation of Solution

Firstly, a table is constructed on the variables involved along with their dimensions as shown below.

    VariableSymbolDimension
    DiameterDM0Lt0
    Height HM0Lt0
    DensityρML3
    Gravitational accelerationgM0Lt-2
    Angular velocityωM0L0t-1
    DischargeQM0L3t-1
    PowerPML2t-3
    Efficiencyηdimensionless

With respect to the table above, the dimensional matrix formed is as below.

  [A]=000001111033002110123

Next, the rank r of the matrix shown above is to be determined. In the present case, rank is the number of rows i.e., 3

Therefore, the number of independent dimensionless groups i is determined by the following expression.

  i=nr=83=5

Here, n is the number of variables.

Say, π1 , π2 , π3 , π4 , and π5 are some groups. These groups are supposedly dimensionless. This means that the ultimate dimension of these groups is M0L0t0

Based on the data available so far, the following core groups can be proposed.

If it depends only on efficiency thereby making it automatically dimensionless that is π1=η

For the rest of the variables, there are some fixed variables and some changing variables. For example, let density, diameter, and angular velocity are common. Then, the rest of the variables are defined in the following manner.

Say,

  π2=ρaDbωcH

Using dimensional analysis, the same group may be re-expressed in the following manner.

  M0L0t0=[ML3t0]a[M0L1t0]b[M0L0t1]c[M0L1t0]

Upon comparing coefficients of individual dimensions, the following set of equations is obtained.

  a(0)+b(0)+c(0)+1=0a(3)+b(1)+c(0)+1=0a(0)+b(0)+c(1)+0=0

Hence, the following simplification is obtained.

  a=03a+b+1=0c=0

Simply put,

  a=0b=1c=0

This means that this particular group may have the following expression.

  π2=H/D

Say,

  π3=ρaDbωcg

Using dimensional analysis, the same group may be re-expressed in the following manner.

  M0L0t0=[ML3t0]a[M0L1t0]b[M0L0t1]c[M0L1t2]

Upon comparing coefficients of individual dimensions, the following set of equations is obtained.

  a(1)+b(0)+c(0)+0=0a(3)+b(1)+c(0)+1=0a(0)+b(0)+c(1)2=0

Hence, the following simplification is obtained.

  a=03a+b+1=0c=2

Simply put,

  a=0b=1c=2

This means that this particular group may have the following expression.

  π3=ρ0D1ω2g=g/Dω2

Say,

  π4=ρaDbωcQ

Using dimensional analysis, the same group may be re-expressed in the following manner.

  M0L0t0=[ML3t0]a[M0L1t0]b[M0L0t1]c[M0L3t1]

Upon comparing coefficients of individual dimensions, the following set of equations is obtained.

  a(1)+b(0)+c(0)+0=0a(3)+b(1)+c(0)+3=0a(0)+b(0)+c(1)1=0

Hence, the following simplification is obtained.

  a=03a+b+3=0c=1

Simply put,

  a=0b=3c=1

This means that this particular group may have the following expression.

  π4=ρ0D3ω1Q=Q/D3ω

Say,

  π5=ρaDbωcP

Using dimensional analysis, the same group may be re-expressed in the following manner.

  M0L0t0=[ML3t0]a[M0L1t0]b[M0L0t1]c[M1L2t3]

Upon comparing coefficients of individual dimensions, the following set of equations is obtained.

  a(1)+b(0)+c(0)+1=0a(3)+b(1)+c(0)+2=0a(0)+b(0)+c(1)3=0

Hence, the following simplification is obtained.

  a=03a+b+2=0c=3

Simply put,

  a=0b=2c=3

This means that this particular group may have the following expression.

  π5=ρ0D2ω3P=P/D2ω3

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Chapter 11 Solutions

Fundamentals of Momentum, Heat, and Mass Transfer

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