5.16.4. Appendix 5.4: Air Blast Heat Exchanger Stack Momentum Equation

This appendix derives an expression for the air mass flowrate through the natural convection stack. The stack contains an opening at its base through which air is drawn in, the air passes over the finned tubes of the air blast heat exchanger and then rises to be exhausted at the top of the stack.

The one-dimensional steady-state momentum equation for flow in a channel of uniform cross section is

(5.16-46) $$\frac{\text{dP}}{\text{dz}} = \frac{- \text{d}}{\text{dz}} \left( \rho v^{2} \right) - \rho g \sin{\theta} - \tau \frac{P_{\text{w}}}{A}~ ,$$

where

$p$ =pressure

$\rho$ =density

$v$ =velocity

$\tau$ =wall shear stress

$P_{\text{w}}$ =wetted perimeter

$A$ =flow area

$\theta$ =channel inclination relative to horizontal

Integrating Eq. (5.16-46) gives the pressure change along the channel

(5.16-47) $$\Delta P = - \left( \frac{w}{A} \right)^{2} \left( \frac{1}{\rho_{\text{o}}} - \frac{1}{\rho_{\text{i}}} \right) - \rho_{\text{m}}g\mathcal{l} \sin{\theta} - \frac{K}{2\rho_{\text{m}}} \left( \frac{w}{a} \right)^{2}$$

where

$w$ =channel mass flowrate

$\mathcal{l}$ =channel length

$\rho_{\text{o}}$ =outlet density

$\rho_{\text{i}}$ =inlet density

$\rho_{\text{m}}$ =mean density

$K$ =flow loss coefficient

Using Eq. (5.16-47), the pressure change form stack inlet to above the heat exchanger is

(5.16-48) $$\Delta P = - \left( \frac{w}{A_{\text{R}}} \right)^{2} \left\lbrack \frac{K_{\text{SI}}}{2\rho_{\text{c}}}\left( \frac{A_{\text{R}}}{A_{\text{SI}}} \right)^{2} + \frac{K_{\text{HX}}}{2\rho_{\text{c}}}\left( \frac{A_{\text{R}}}{A_{\text{HX}}} \right)^{2} \right\rbrack$$

where

$A_{\text{SI}}$ = stack inlet cross-sectional area

$A_{\text{R}}$ = riser cross-sectional area

$A_{\text{HX}}$ = flow area at heat exchanger

$K_{\text{SI}}$ = stack inlet loss coefficient

$K_{\text{HX}}$ = heat exchanger loss coefficient

$\rho_{\text{c}}$ =inlet air density

The gravity and acceleration terms have been neglected.

Similarly, the pressure change from the start of the riser to the stack outlet is

(5.16-49) $$\Delta P = - \left( \frac{w}{A_{\text{R}}} \right)^{2} \frac{\left( K_{\text{SO}} + K_{\text{R}} \right)}{2\rho_{\text{h}}} - \rho_{\text{h}} g\mathcal{l}$$

where

$K_{\text{SO}}$ = stack outlet loss coefficient

$K_{\text{R}}$ = riser loss coefficient

$\rho_{\text{h}}$ = riser air density

$\mathcal{l}$ = riser length

The pressure change from the stack outlet through the outside air back to the stack inlet is approximately

(5.16-50) $$\Delta P = \rho_{\text{c}} g \mathcal{l}$$

The above three pressure changes, Eq. (5.16-48) through Eq. (5.16-50), must sum to zero since they are taken around a closed circuit. Solving for the air flowrate yields

(5.16-51) $$w^{2} = \frac{\left( \rho_{\text{c}} - \rho_{\text{h}} \right) g \mathcal{l} A_{\text{R}}^{2}}{\frac{K_{\text{SI}}}{2\rho_{\text{c}}} \left( \frac{A_{\text{R}}}{A_{\text{SI}}} \right)^{2} + \frac{K_{\text{HX}}}{2\rho_{\text{c}}} \left( \frac{A_{\text{R}}}{A_{\text{HX}}} \right)^{2} + \frac{\left( K_{\text{SO}} + K_{\text{R}} \right)}{2\rho_{\text{h}}}}$$