Chapter 7, Problem 7.2CP

To determine

(a)

Find the pressure drop without small plates?

Answer to Problem 7.2CP

$\Delta P=1.47Pa/m$

Explanation of Solution

Given information:

The fluid is air at $20°C$

Air flows at a speed of $5m/s$

Distance between the parallel plates is $10cm$

The length of interrupter plate is $1cm$

Plates are $1m$ wide into the paper

The pressure drop $\Delta P$ is defined as below,

$\Delta P=f\frac{L}{D_h}\left(\frac{\rho V^2}{2}\right)$

In above equation,

$f$ - Friction factor

$L$ - Length of duct

$D_h$ - Hydraulic diameter

The hydraulic diameter of parallel plates is defined as,

$D_h=2h$

$h$ - Distance between the plates

The Reynolds’s number is defined as,

${\mathrm{Re}}_{D_h}=\frac{\rho V{D}_h}{\mu }$

$\mu$ - Dynamic viscosity

The friction factor for smooth flow is defined as,

$\frac{1}{f^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2.0\log \left({\mathrm{Re}}_D{f}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-0.8$

Assume air at $20°C$ will have,

$\begin{array}{l}\rho =1.2kg/m^3\\ \mu =1.8\times {10}^{-5}kg/ms\end{array}$

Calculation:

Calculate the Hydraulic diameter,

$D_h=2h=0.2m$

Calculate the Reynolds’s number,

${\mathrm{Re}}_{D_h}=\frac{\rho V{D}_h}{\mu }=\frac{\left(1.2kg/m^3\right)\left(5m/s\right)\left(0.2m\right)}{1.8\times {10}^{-5}kg/ms}=66666.67$

Calculate the friction factor for smooth flow,

$\begin{array}{l}\frac{1}{f^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2.0\log \left({\mathrm{Re}}_D{f}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-0.8\\ =2.0\log \left(\left(66666.67\right)f^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-0.8\\ f=0.0196\end{array}$

Calculate the pressure drop,

$\Delta P=f\frac{L}{D_h}\left(\frac{\rho V^2}{2}\right)=\left(0.0196\right)\frac{1m}{0.2m}\left(\frac{\left(1.2kg/m^3\right){\left(5m/s\right)}^2}{2}\right)=1.47Pa/m$

Conclusion:

The pressure drop is equal to $1.47Pa/m$.

To determine

(b)

The number of small plates per meter of channel length that will cause the pressure drop to rise to $10Pa/m$ ?

Answer to Problem 7.2CP

$124$ Plates

Explanation of Solution

Given information:

The fluid is air at $20°C$

Air flows at a speed of $5m/s$

Distance between the parallel plates is $10cm$

The length of interrupter plate is $1cm$

Plates are $1m$ wide into the paper

The Reynolds’s number based on length is defined as,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }$

$\mu$ - Dynamic viscosity

The drag co-efficient for a laminar is defined as,

$C_D=\frac{1.328}{\sqrt{{\mathrm{Re}}_L}}$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

Where,

$C_D$ - Drag co-efficient

$\rho$ - Density of the fluid

$V$ - Velocity

$A$ - Characteristic area

Calculation:

Calculate the Reynolds’s number,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }=\frac{\left(1.2kg/m^3\right)\left(5m/s\right)\left(0.01m\right)}{1.8\times {10}^{-5}kg/ms}=3333.33$

Calculate the drag co-efficient,

$C_D=\frac{1.328}{\sqrt{{\mathrm{Re}}_L}}=\frac{1.328}{\sqrt{3333.33}}=0.023$

Calculate the drag force in a small plate,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A=\left(0.023\right)\frac{\left(1.2kg/m^3\right)}{2}{\left(5m/s\right)}^2\left(0.01m\right)\left(1m\right)=3.45\times {10}^{-3}N$

The drag of the plate for both sides will be equal to,

$\left(3.45\times {10}^{-3}\right)\left(2\right)=6.9\times {10}^{-3}N$

According to sub-part a,

The pressure drop without the small plates is equal to,

$\Delta P=1.47Pa/m$

Therefore, the pressure drop to rise to $10Pa/m$,

The small plates should provide,

$10Pa/m-1.47Pa/m=8.53Pa/m$

Therefore, number of plates $N$ required to provide this pressure drop will be equal to,

$N=\Delta p\frac{A}{F_{drag}}=\left(8.53Pa/m\right)\frac{0.1m^2}{6.9\times {10}^{-3}N}=124$

Conclusion:

The heat exchanger requires $124$ plates to provide a pressure drop of $10Pa/m$.