Chapter 7, Problem 7.1P

Interpretation Introduction

Interpretation: The pressure drop through packed tubes is to be calculated for the different conditions of the pellets.

Concept introduction: The pressure drop through the packed tubes is calculated by the Ergun equation which is a combination of the Kozeny-Carman equation and Burke-Plummer equation.

The Kozeny-Carman equation is applicable for the particles in a bed whose Reynold’s number is less than 1. It is given as,

  $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{{\text{150<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}{\text{μ(1-ε)}}^{\text{2}}}{{\varphi }_{\text{s}}{}^{\text{2}}{\text{D}}_{\text{p}}{}^{\text{2}}{\text{ε}}^{\text{3}}}\mathrm{.......}(1)$

In the above formula, the notations used are,

  $\begin{array}{l}\text{ΔP}\text{=}\text{Pressure}\text{drop}\text{through}\text{packed}\text{tubes}\\ \text{L}\text{=}\text{Length}\text{of}\text{tube}\\ {\text{<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}\text{=}\text{Superficial}\text{velocity}\\ \text{μ}\text{=}\text{Viscosity}\text{of}\text{fluid}\\ \text{ε}\text{=}\text{Bed}\text{porosity}\\ {\text{D}}_{\text{p}}\text{=}\text{Pellet}\text{diameter}\\ {\varphi }_{\text{s}}\text{=}\text{Sphericity}\text{of}\text{particles}\end{array}$

The Burke Plummer equation is applicable for the particles in a bed whose Reynold’s number is more than 1000. It is given as,

  $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{1}{\text{.75<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}{}^{\text{2}}\text{ρ(1-ε)}}{{\varphi }_{\text{s}}{\text{D}}_{\text{p}}{\text{ε}}^{\text{3}}}\mathrm{.......}\text{(2)}$

  $\text{ρ=}\text{Density}\text{of}\text{air}$

The density of air can be given as,

  $\text{ρ}\text{=}\frac{\text{PM}}{\text{RT}}\mathrm{.......}(3)$

In the equation (3), notations used are,

P is the pressure at which air enters

M is the molecular weight of air

R is the Universal Gas constant

T is the temperature at which air enters

Combining equation (1) and equation (2) we get Ergun equation.

  $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{{\text{150<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}{\text{μ(1-ε)}}^{\text{2}}}{{\varphi }_{\text{s}}{}^{\text{2}}{\text{D}}_{\text{p}}{}^{\text{2}}{\text{ε}}^{\text{3}}}+\frac{\text{1}{\text{.75<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}{}^{\text{2}}\text{ρ(1-ε)}}{{\varphi }_{\text{s}}{\text{D}}_{\text{p}}{\text{ε}}^{\text{3}}}$ ……. (4)

Answer to Problem 7.1P

The pressure drop for diameter of pellets 0.003 m is 18465.6 Pa and when the diameter of pellet is increased to 0.004 m, the pressure drop is reduced by 5667.1 Pa or 30.7%.

Explanation of Solution

The data given for the air density calculation is,

  $\begin{array}{l}\text{P}\text{=}\text{2}\text{atm}\text{=}\text{2×101325}\text{Pa}\text{=}\text{202650}\text{Pa}\\ \text{M}\text{=}\text{30}\text{g/mol}\text{=}\text{0}\text{.030}\text{kg/mol}\\ \text{R}\text{=}\text{8}\text{.314}{\text{Pa-m}}^{\text{3}}\text{/mol}\text{.K}\\ \text{T}\text{=}\text{350°C}\text{=}\text{623K}\end{array}$

Substitute this data in equation (3),

  $\begin{array}{l}\text{ρ}\text{=}\frac{\text{(202650}\text{Pa)×(0}\text{.030}\text{kg/mol)}}{\text{(8}\text{.314}{\text{Pa-m}}^{\text{3}}\text{/mol}\text{.K)×}\left(\text{623K}\right)}\\ \text{ρ}\text{=}\text{1}\text{.174}{\text{kg/m}}^{\text{3}}\end{array}$

Now other data given to calculate pressure drop is,

  $\begin{array}{l}{\text{<mover accent="true"><mo>V</mo><mo>¯</mo></mover>}}_{\text{0}}\text{=}\text{1}\text{m/s}\\ \text{ε}\text{=}\text{0}\text{.4}\\ {\text{D}}_{\text{p}}\text{=}\text{3}\text{mm}\text{=}\text{0}\text{.003}\text{m}\\ {\varphi }_{\text{s}}\text{=}\text{1}\text{(For}\text{spherical}\text{particles)}\end{array}$

For air at the given temperature, the viscosity from appendix is, $\text{μ}\text{=}{\text{3×10}}^{\text{-5}}\text{kg/m}\text{.s}$ .

Substitute all the required data from above in equation (4),

  $\frac{\text{ΔP}}{\text{2}}\text{=}\frac{\text{150×(1}{\text{m/s)×(3×10}}^{\text{-5}}\text{kg/m}\text{.s)×(1-0}{\text{.4)}}^{\text{2}}}{{\text{(1)}}^{\text{2}}\text{×(0}\text{.003}{\text{m)}}^{\text{2}}\text{×(0}{\text{.4)}}^{\text{3}}}\text{+}\frac{\text{1}\text{.75×(1}\text{.174}{\text{kg/m}}^{\text{3}}\text{)×(1}{\text{m/s)}}^{\text{2}}\text{×(1-0}\text{.4)}}{\text{1×(0}\text{.003}\text{m)×(0}{\text{.4)}}^{\text{3}}}$

  $\begin{array}{l}\text{Solving}\text{above}\text{equation,}\\ \text{ΔP}\text{=}\text{(2}\text{m)×}\left[\text{(2812}\text{.5}{\text{kg/m}}^{\text{2}}{\text{.s}}^{\text{2}}\text{)}\text{+}\text{(6420}\text{.3}{\text{kg/m}}^{\text{2}}{\text{.s}}^{\text{2}}\text{)}\right]\\ \text{ΔP}\text{=}\text{18465}\text{.6}{\text{kg/ms}}^{\text{2}}\\ \text{ΔP}\text{=}\text{18465}\text{.6}\text{Pa}\end{array}$

When diameter is increased from 0.003 m to 0.004 m, the increase in diameter is 1.33 times the original so pressure drop will reduce by the same factor as all conditions are identical as previous conditions.

Thus,

  $\begin{array}{l}\text{ΔP}\text{=}\text{2×}\left[\left(\frac{\text{2812}\text{.5}\text{Pa}}{{\text{(1}\text{.333)}}^{\text{2}}}\right)\text{+}\left(\frac{\text{6420}\text{.3}\text{Pa}}{\text{1}\text{.333}}\right)\right]\\ \text{ΔP}\text{=}\text{2}\text{×}\left[\text{(1582}\text{.82}\text{Pa)}\text{+}\text{(4816}\text{.43}\text{Pa)}\right]\\ \text{ΔP}\text{=}\text{12798}\text{.5}\text{Pa}\mathrm{.......}\text{(6)}\end{array}$

The reduction in pressure is the difference between equation (5) and equation (6).

Reduction in pressure = 18465.6 Pa − 12798.5 Pa = 5667.1 Pa

  $\begin{array}{l}\text{\%}\text{Reduction}\text{=}\frac{\text{Reduction}\text{in}\text{Pressure}}{\text{Initial}\text{Pressure}}\text{×100}\\ \text{\%}\text{Reduction}\text{=}\frac{\text{5667}\text{.1}\text{Pa}}{\text{18465}\text{.6}\text{Pa}}\text{×100}\\ \text{\%}\text{Reduction}\text{=}\text{30}\text{.7}\text{\%}\end{array}$