Chapter 7, Problem 118P

An electrostatic precipitator (ESP) is a device used in various applications to clean particle-laden air. First, the dusty air passes through the charging stage of the ESP. where dust particles are given a positive charge q_p(coulombs) by charged ionizer wires (Fig. P7-118). The dusty air then enters the collector stage of the device, where it flows between two oppositely charged plates. The applied electric field strength between the plates is E_f(voltage difference per unit distance). Shown in Fig. P7-118 is a charged dust particle of diameter D_p. It is attracted to the negatively charged plate and moves toward that plate at a speed called the drift velocity w. If the plates are long enough, the dust panicle impacts the negatively charged plate and adheres to it. Clean air exits the device. It turns out that for very small particle the drift velocity depends only on q_p. E_f,D_p, and air viscosity $\mu$ , (a) Generate a dimension less relationship between the drift velocity through the collector stage of the ESP and the given parameters. Show all your work. (b) If the electric field strength is doubled, all else being equal, by what factor will the drift velocity change? (c) For a given ESP, if the panicle diameter is doubled, all else being equal, by what factor will the drift velocity change?

To determine

(a)

The dimensionless relationship between the drift velocity through the collector stage of the ESP and the given parameters.

Answer to Problem 118P

The relationship between drift velocity and other parameter is

  $\frac{\omega \mu D_p}{q_p{E}_f}=\text{constant}$

Explanation of Solution

Given:

  $\begin{array}{l}Driftvelocity-\omega \\ Particlediameter-D_p\\ positivech\arg etoparticles-q_p\\ electricfieldstrength-E_f\\ vis\cos ity-\mu \end{array}$

Concept Used:

Buckinghams Pi Theorem

Calculation:

The electrostatic precipitalor is used to clean the air. Drift velocity of dust particle depends on qp, Ef, Dpand $\mu$.

There are 5 parameters, n=5

  $\begin{array}{l}Driftvelocity-\omega \\ Particlediameter-D_p\\ positivech\arg etoparticles-q_p\\ electricfieldstrength-E_f\\ vis\cos ity-\mu \end{array}$

W is a function of $D_p,q_p,E_f,\mu$

  $w=f(D_p,q_p,E_f,\mu )$

The primary dimension of each term

  $\begin{array}{l}\left\{w\right\}=\left\{\frac{dis\tan ce}{time}\right\}=\left\{\frac{L}{T}\right\}=\left\{L^1{T}^{-1}\right\}\\ \left\{D_p\right\}=\left\{L\right\}\\ \left\{q_p\right\}=\{current\times time\}=\{I\times T\}=\left\{I^1{T}^1\right\}\\ \left\{\mu \right\}=\left\{M^1{L}^{-1}{T}^{-1}\right\}\\ \left\{E_f\right\}=\left\{\frac{voltage}{dis\tan ce}\right\}=\left\{\frac{\frac{(\frac{work}{time})}{current}}{dis\tan ce}\right\}=\left\{\frac{M^1{L}^1}{T^3{I}^1}\right\}\\ \left\{E_f\right\}=\left\{M^{1}L{T}^{-3}{I}^{-1}\right\}\end{array}$

No. of primary dimensions are, j=4 (M, l, T, I)

No. of expected Pi = k= n-j =1

As j=4 we have to select four repeating parameters.

They are, $D_p,q_p,E_f,\mu$

Combining repeating parameter with remaining parameters.

Now, dependent $\pi$

  $({\pi }_1)=\omega q{p}^{a_1}E{f}^{b_1}{\mu }^{c_1}D{p}^{d_1}$

The primary dimensions of the above term are

  $({\pi }_1)=\left\{M^0{L}^0{T}^0{I}^0\right\}$

  $\left\{\omega q{p}^{a_1}E{f}^{b_1}{\mu }^{c_1}D{p}^{d_1}\right\}=\{(L^1{T}^{-1}){(I^1{T}^1)}^{a_1}{(M^1{L}^1{T}^{-3}{I}^{-1})}^{b1}{(M^1{L}^{-1}{T}^{-1})}^{c_1}{(L)}^{d_1}$

Equation becomes

  $\left\{M^0{L}^0{T}^0{I}^0\right\}=\{(L^1{T}^{-1}){(I^1{T}^1)}^{a_1}{(M^1{L}^1{T}^{-3}{I}^{-1})}^{b1}{(M^1{L}^{-1}{T}^{-1})}^{c_1}{(L)}^{d_1}$

Equating each primary dimension to solve $a_1{b}_1{c}_1{d}_1$

Current:

  $\begin{array}{l}\left\{I^0\right\}=\left\{I^{a_1}{I}^{b_1}\right\}\\ 0=a_1-b_1\\ b_1=a_1\end{array}$

Mass:

  $\begin{array}{l}\left\{M^0\right\}=\left\{M^{b_1}{M}^{c_1}\right\}\\ 0=b_1+c_1\\ b_1=-c_1\end{array}$

Time:

  $\begin{array}{l}\left\{T^0\right\}=\left\{T^{-1}{T}^{a_1}{T}^{-3b_1}{T}^{-c_1}\right\}\\ 0=-1+a_1-3b_1-c_1\\ 0=-1-2a_1-c_1\\ a_1=-1\\ b_1=-1\\ c_1=1\end{array}$

Length:

  $\begin{array}{l}\left\{L^0\right\}=\left\{L^1{L}^{b_1}{L}^{-c_1}{L}^{d_1}\right\}\\ 0=1+b_1-c_1+d_1\\ d_1=1\end{array}$

Putting values in $\pi$ the equation,

  $\begin{array}{l}({\pi }_1)=\omega q{p}^{-1}E{f}^{-1}{\mu }^{1}D{p}^1\\ {\pi }_1=\frac{\omega \mu Dp}{qpEf}\end{array}$

Conclusion:

Thus, by using Buckinghams Pi theorem, we can develop the dimensionless relationship between drift velocity and other parameters.

To determine

(b)

The factor by which the drift velocity will change if the electric field strength is doubled.

Answer to Problem 118P

Drift velocity will be double if electric field strength is doubled.

Explanation of Solution

Given:

Electric field strength =Ef

Concept Used:

  $\frac{\omega \mu Dp}{qpEf}=cons\tan t,$

Where,

  $\begin{array}{l}Driftvelocity-\omega \\ vis\cos ity-\mu \\ Particlediameter-D_p\\ ch\arg e-q_p\\ electricfieldstrength-E_f\end{array}$

Calculation:

We have, the relation between drift velocity and another parameter as follows

  $\frac{\omega \mu Dp}{qpEf}=cons\tan t,$

Where,

  $\begin{array}{l}Driftvelocity-\omega \\ vis\cos ity-\mu \\ Particlediameter-D_p\\ ch\arg e-q_p\\ electricfieldstrength-E_f\end{array}$

Hence,

  $\omega =cons\tan [\frac{IpEf}{\mu D_p}]$

If electric is strength is doubled, $Ef=2Ef$

  $\begin{array}{l}{w}^1=cons\tan t[\frac{qp(2Ef)}{\mu Dp}]\\ w^1=2[cons\tan \frac{q_p{E}_f}{\mu D_p}\\ w^1=2w\end{array}$

Conclusion:

Thus, drift velocity will be doubled if electric strength is doubled.

To determine

(c)

The factor by which the drift velocity will change for a given ESP if the particle diameter is doubled.

Answer to Problem 118P

Drift velocity will get half when particle diameter is doubled.

Explanation of Solution

Given:

Particle diameter −D

Concept Used:

  $\frac{\omega \mu Dp}{qpEf}=\text{constant},$

Where,

  $\begin{array}{l}Driftvelocity-\omega \\ vis\cos ity-\mu \\ Particlediameter-D_p\\ ch\arg e-q_p\\ electricfieldstrength-E_f\end{array}$

Calculation:

We have relation between drift velocity and other parameter as follows:

  $\frac{\omega \mu Dp}{qpEf}=cons\tan t,$

Where,

  $\begin{array}{l}Driftvelocity-\omega \\ vis\cos ity-\mu \\ Particlediameter-D_p\\ ch\arg e-q_p\\ electricfieldstrength-E_f\end{array}$

Hence,

  $\omega =cons\tan [\frac{IpEf}{\mu D_p}]$

If particle diameter is doubled, $D_p=2D_p$

  $\begin{array}{l}{w}^1=cons\tan t[\frac{qpEf}{\mu 2Dp}]\\ w^1=\frac{1}{2}[cons\tan \frac{q_p{E}_f}{\mu D_p}]\\ w^1=\frac{1}{2}w\end{array}$

Conclusion:

Thus particle diameter is inversely proportional to drift velocity. Then drift velocity will get half when particle diameter is doubled.