Chapter 6, Problem 6.92P

To determine

(a)

The air flow rate in $m^{3}h$.

Answer to Problem 6.92P

$Q=1544.4m^3/h$

Explanation of Solution

Given information:

The room pressure is $10Pa$

The pressure drop is defined as,

$\Delta P=\rho g{h}_f=\rho g\left(f\frac{L}{D}\left(\frac{V^2}{2g}\right)\right)$

In above equation,

$\rho$ - Density of the fluid

$h_f$ - Head loss

$L$ - Length of pipe

$D$ - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

$Q=VA$

In above equation,

$V$ - Velocity of the flow

$A$ -Area

The hydraulic diameter is defined as,

$D_h=\frac{4A}{{\rm P}}$

$D_h$ - Hydraulic diameter of pipe

$A$ - Area of the pipe

${\rm P}$ - Perimeter

Calculation:

First of all, to find the hydraulic diameter,

$D_h=\frac{4A}{{\rm P}}=\frac{4\left(900cm\right)}{120cm}=30cm=0.3m$

To find volumetric flow rate,

Assume,

$f=0.02$

$\begin{array}{l}\Delta P=\rho \left(f\frac{L}{D}\left(\frac{V^2}{2}\right)\right)\\ 10N/m^2=\left(1.2kg/m^3\right)\left(\left(0.02\right)\frac{12m}{0.3m}\left(\frac{V^2}{2}\right)\right)\\ V=4.564m/s\end{array}$

To find the exact value of the velocity, further iterations needs to be done. For that we need to find the friction factor,

From the definitions, $b/a=1.0$,

Therefore the effective diameter will be equal to,

$D_{eff}=\left(\frac{64}{56.91}\right)D_h=\left(\frac{64}{56.91}\right)\left(0.3m\right)=0.337m$

To find the Reynolds’s number,

$R_e=\frac{V{D}_{eff}}{\nu }=\frac{\left(4.564m/s\right)\left(0.337m\right)}{1.5111\times {10}^{-5}{m}^{2}s}=101784.66$

To find the roughness ratio,

$\frac{\epsilon }{D_{eff}}=\frac{0.046\times {10}^{-3}m}{0.337m}=1.22\times {10}^{-4}$

To find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{101784.66}+{\left(\frac{\text{}1.22\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ \approx 0.0183\end{array}$

Therefore, to find the exact value of the velocity,

$\begin{array}{l}\Delta P=\rho \left(f\frac{L}{D}\left(\frac{V^2}{2}\right)\right)\\ 10N/m^2=\left(1.2kg/m^3\right)\left(\left(0.0183\right)\frac{12m}{0.3m}\left(\frac{V^2}{2}\right)\right)\\ V=4.772m/s\\ \end{array}$

To find the flow rate,

$Q=VA=\left(4.772m/s\right)\left(0.09m^2\right)=0.429m^3/s=1544.4m^3/h$

Conclusion:

The flow rate is equal to $1544.4m^3/h$

(b)

The room pressure if the flow rate is $1200m^3/h$.

$\Delta P=6.24N/m^2$

Given information:

Assume the flow rate as $1200m^3/h$

The pressure drop is defined as,

$\Delta P=\rho g{h}_f=\rho g\left(f\frac{L}{D}\left(\frac{V^2}{2g}\right)\right)$

In above equation,

$\rho$ - Density of the fluid

$h_f$ - Head loss

$L$ - Length of pipe

$D$ - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

$Q=VA$

In above equation,

$V$ - Velocity of the flow

$A$ -Area

The hydraulic diameter is defined as,

$D_h=\frac{4A}{{\rm P}}$

$D_h$ - Hydraulic diameter of pipe

$A$ - Area of the pipe

${\rm P}$ - Perimeter

Calculation:

Compute the mean velocity,

$V=\frac{Q}{A}=\frac{0.333m^3/s}{\left(0.09m^2\right)}=3.7m/s$

To find the Reynolds’s number,

$R_e=\frac{V{D}_{eff}}{\nu }=\frac{\left(3.7m/s\right)\left(0.337m\right)}{1.5111\times {10}^{-5}{m}^{2}s}=82516.05$

To find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{82516.05}+{\left(\frac{\text{}1.22\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ \approx 0.019\end{array}$

Therefore, according to the obtained results,

To find the room pressure,

$\begin{array}{l}\Delta P=\rho \left(f\frac{L}{D}\left(\frac{V^2}{2}\right)\right)\\ =\left(1.2kg/m^3\right)\left(\left(0.019\right)\frac{12m}{0.3m}\left(\frac{{\left(3.7m/s\right)}^2}{2}\right)\right)\\ =6.24N/m^2\end{array}$

Conclusion:

The room pressure is equal to $6.24N/m^2$.

To determine

(b)

The room pressure if the flow rate is $1200m^3/h$.

Answer to Problem 6.92P

$\Delta P=6.24N/m^2$

Explanation of Solution

Given information:

Assume the flow rate as $1200m^3/h$

The pressure drop is defined as,

$\Delta P=\rho g{h}_f=\rho g\left(f\frac{L}{D}\left(\frac{V^2}{2g}\right)\right)$

In above equation,

$\rho$ - Density of the fluid

$h_f$ - Head loss

$L$ - Length of pipe

$D$ - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

$Q=VA$

In above equation,

$V$ - Velocity of the flow

$A$ -Area

The hydraulic diameter is defined as,

$D_h=\frac{4A}{{\rm P}}$

$D_h$ - Hydraulic diameter of pipe

$A$ - Area of the pipe

${\rm P}$ - Perimeter

Calculation:

Compute the mean velocity,

$V=\frac{Q}{A}=\frac{0.333m^3/s}{\left(0.09m^2\right)}=3.7m/s$

To find the Reynolds’s number,

$R_e=\frac{V{D}_{eff}}{\nu }=\frac{\left(3.7m/s\right)\left(0.337m\right)}{1.5111\times {10}^{-5}{m}^{2}s}=82516.05$

To find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{82516.05}+{\left(\frac{\text{}1.22\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ \approx 0.019\end{array}$

Therefore, according to the obtained results,

To find the room pressure,

$\begin{array}{l}\Delta P=\rho \left(f\frac{L}{D}\left(\frac{V^2}{2}\right)\right)\\ =\left(1.2kg/m^3\right)\left(\left(0.019\right)\frac{12m}{0.3m}\left(\frac{{\left(3.7m/s\right)}^2}{2}\right)\right)\\ =6.24N/m^2\end{array}$

Conclusion:

The room pressure is equal to $6.24N/m^2$.